WBCHSE Class 11 Chemistry Redox Reactions Questions And Answers

Redox Reactions Long Answer Type Questions

Question 1. In the following redox reactions, identify the oxidation half-reactions and reduction half-reactions along with the oxidants and reductants

⇒ \(\begin{aligned}
& \mathrm{Cl}_2(g)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_2(g) \\
& \mathrm{Sn}^{2+}(a q)+2 \mathrm{Fe}^{3+}(a q) \rightarrow \mathrm{Sn}^{4+}(a q)+2 \mathrm{Fe}^{2+}(a q) \\
& 2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3(a q)+\mathrm{I}_2(s) \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6(a q)+2 \mathrm{NaI}(a q)
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{Fe}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{H}_2(g) \\
& \mathrm{H}_2 \mathrm{~S}(a q)+\mathrm{Cl}_2(g) \rightarrow \mathrm{S}(s)+2 \mathrm{HCl}(a q) \\
& 2 \mathrm{FeCl}_2(a q)+\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{FeCl}_3(a q) \\
& 2 \mathrm{Hg}^{2+}(a q)+\mathrm{Sn}^{2+}(a q) \rightarrow \mathrm{Hg}_2^{2+}(a q)+\mathrm{Sn}^{4+}(a q)
\end{aligned}\)
Answer:

Oxidation half-reaction: 2l-(aq)→I₂(s) + 2e
Reduction half-reaction: Cl₂(g) + 2e→2Cl-(aq)
Oxidant: Cl₂(g); Reductant: l-(aq)

Oxidation half-reaction: Sn²⁺(aq)→Sn⁴⁺(aq) + 2e
Reduction half-reaction: Fe³⁺(aq) + e→Fe²⁺(aq)
Oxidant: Fe³⁺(aq); Reductant: Sn2+(aq)

Redox Reactions Questions

Oxidation half-reaction:
\(\mathrm{S}_2 \mathrm{O}_3^{2-}(a q) \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}(a q)+2 e\)
Reduction half-reaction: I2(s) + 2e→2l-(aq)
Oxidant: I₂(s); Reductant: Na₂S₂O₃(aq)

Oxidation half-reaction: Fe(s)→Fe²⁺(aq) + 2e
Reduction half-reaction: 2H+(aq) + 2e→H2(g)
Oxidant: H+(aq); Reductant: Fe(s)

Oxidation half-reaction: S2-(aq)→S(s) + 2e
Reduction half-reaction: Cl₂(g) + 2e→CI-(aq)
Oxidant: Cl₂(g); Reductant: H2S(aq)

Oxidation half-reaction: Fe²⁺(aq)→Fe³⁺(aq)
Reduction half-reaction: Cl₂(g) + 2e→2Cl-(aq)
Oxidant: Cl₂; Reductant: FeCl₂

Oxidation half-reaction: Sn²⁺(aq)→Sn⁴⁺(aq) + 2e
Reduction half-reaction: \(2 \mathrm{Hg}^{2+}(a q)+2 e \rightarrow \mathrm{Hg}_2^{2+}(a q)\)
Oxidant: Hg²⁺(aq); Reductant: Sn²⁺(aq)

Question 2. Give an example of an oxygen-containing compound for each of the following oxidation states of oxygen: \(+1,-\frac{1}{2},-1\)
Answer: The oxidation states of O in O₂F₂, KO₂  And H₂O₂ are \(+1,-\frac{1}{2} \text { and }-1\) respectively

Question 3. A compound is composed of three elements A, B, and C. The oxidation numbers of A, B, and C in the compound are 1 1, +5, and -2, respectively. Which one of the following formulas represents the molecular formula of the compound? A₂BC₄ or A₂(BC₃).
Answer: The algebraic sum of the oxidation numbers of all atoms present in a molecule is equal to zero.

In an A₂BC₄ molecule, the algebraic sum ofthe oxidation numbers of all the constituent atoms

= 2 X (+1) +1 X (+5) + 4 X (-2) = -1

In A₂(BC₃)₂ molecule, the algebraic sum of the oxidation numbers of all the atoms

Redox Reactions Questions

= 2 X (+ 1) + 2[5 + 3 x (-2)] = 0

Therefore, A₂(BC₃)₂ represents the molecular formula ofthe compound.

Question 4.  Give two examples of nitrogen-containing compounds, in one of which, the oxidation state of N-atom is +1, while in the other compound, N-atoms exist in two different oxidation states.
Answer:

1. The oxidation number of N in N₂O is +1.
2. The oxidation numbers of two N-atoms in NH₄NO₃ are -3 and +5 respectively

Question 5. A compound is composed of three elements A, B, and C. The oxidation numbers of A, B, and C in the compound are 11, +5, and -2, respectively. Which one of the following formulas represents the molecular formula of the compound? A₂BC, or A₂(BC₃)
Answer: The algebraic sum of the oxidation numbers of all atoms present in a molecule is equal to zero.

In an A₂BC₄ molecule, the algebraic sum ofthe oxidation numbers of all the constituent atoms. = 2 X (+1) +1 X (+5) + 4 X (-2) = -1

In A₂(BC₃)₂ molecule, the algebraic sum of the oxidation numbers of all the atoms = 2 X (+ 1) + 2[5 + 3 x (-2)] = 0

Redox Reactions Questions

Therefore, A₂(BC₃)₂ represents the molecular formula ofthe compound.

Question 6. Give two examples of nitrogen-containing compounds, in one of which, the oxidation state N-atom is +1, while in the other compound, N-atoms exist in two different oxidation states.
Answer: The Oxidation number of N in N₂O is +1. The Oxidation Numbers Of Two N-atoms in NH₄NO₃ are -3 and +5 respectively

Question 7. Among the reactions given below, identify the redox reactions and also mention the oxidant and the reductant in each case

⇒ \(\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g})\)

Answer: \(\stackrel{+3}{\mathrm{Fe}_2} \mathrm{O}_3(s)+3 \stackrel{+2}{\mathrm{C}} \mathrm{O}(g) \rightarrow 2 \stackrel{0}{\mathrm{Fe}}(s)+\stackrel{+4}{3} \mathrm{CO}_2(g)\)

The Oxdination number of Fe decreases from +3 to 0 while that of C increases from +2 to +4. so in this reaction, Fe₂O₃ undergoes reduction and Co undergoes Oxdination. Hence, it’s a redox reaction in which Fe₂O₃ acts as an Oxdiant and Co as Reductant.

2. \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3(a q)+\mathrm{I}_2(s) \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6(a q)+2 \mathrm{NaI}(a q)\)

⇒ \(2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}} \mathrm{O}_3(a q)+\stackrel{0}{\mathrm{I}}(s) \rightarrow \mathrm{Na}_2 \stackrel{+2.5}{\mathrm{~S}_4} \mathrm{O}_6(a q)+2 \mathrm{NaI}^{-1}(a q)\)

The oxidation number of S increases from +2 to +2.5 while that of decreases from 0 to -l. So, in this reaction, Na₂S₂O₃ undergoes oxidation and I₂ undergoes reduction. Hence, it is a redox reaction in which Na2S203 acts as a reductant and I₂ as an oxidant.

In this reaction, there occurs no change in oxidation number for any element. So, it is not a redox reaction.

Redox Reactions Questions

⇒ \(\stackrel{+1 .}{\mathrm{C}} \mathrm{u}_2 \mathrm{~S}(\mathrm{~s})+\stackrel{0}{\mathrm{O}}(\mathrm{g}) \rightarrow \stackrel{0}{\mathrm{C} \mathrm{Cu}}(\mathrm{s})+\stackrel{+4-2}{\mathrm{SO}_2}(\mathrm{~g})\)

The oxidation number of Cu decreases from +1 to 0 while that of S increases from -2 to +4. Also, the oxidation number of 0 decreases from O to -2. So, in this reaction, Cu₂S undergoes oxidation as well as reduction and O₂ undergoes reduction. Hence, it is a redox reaction in which Cu₂S serves as both oxidant and reductant and O2 acts as an oxidant.

⇒ \(\stackrel{+1}{\mathrm{C}} \mathrm{u}_2 \stackrel{-2}{\mathrm{~S}}(s)+\stackrel{0}{\mathrm{O}}(g) \rightarrow 2 \stackrel{0}{\mathrm{C}}(s)+\stackrel{+4-2}{\mathrm{SO}_2}(g)\)

The oxidation number of Cu decreases from +1 to 0 while that of S increases from -2 to +4. Also, the oxidation number of 0 decreases from 0 to -2. So, in this reaction, Cu₂S undergoes oxidation as well as reduction and O₂ undergoes reduction. Hence, it is a redox reaction in which Cu₂S serves as both oxidant and reductant, and O₂ acts as an oxidant.

Redox Reactions Questions

In this reaction, no change in oxidation number for any element takes place. So, it is not a redoxreaction.

The oxidation number of S increases from -2 to +6 while that of N decreases from +5 to +4. So, in this reaction, H₂S undergoes oxidation and HNO₃ undergoes reduction. Hence, it is a redox reaction in which H₂S acts as a reductant and HNO₃ acts as an oxidant

Hence, it is a redox reaction in which H₂S acts as a reductant and HNO₃ acts as an oxidant.

Question 8. Identify the following half-reactions as oxidation half¬ reactions and reduction half-reactions:

1. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 2 \mathrm{Cr}^{3+}\)
2. \(\mathrm{Cr}(\mathrm{OH})_4^{-}(a q) \rightarrow \mathrm{CrO}_4^{2-}(a q)\)
3. \(\mathrm{IO}_3^{-}(a q) \rightarrow \mathrm{IO}_4^{-}(a q)\)
4. \(\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{Cl}^{-}(a q)\)
5. \(\mathrm{MnO}_4^{-}(a q) \rightarrow \mathrm{MnO}_2(s)\)
6. \(\stackrel{+6}{\mathrm{C}}_2 \mathrm{O}_7 \rightarrow 2^{+3} \mathrm{Cr}\)

The oxidation number of Cr changes from +6 to +3. This indicates that the reaction involves the reduction of Cr₂O²¯. Hence, it is a reduction half-reaction.

⇒ \(\mathrm{Cr}^{+3}(\mathrm{OH})_4^{-} \rightarrow{ }^{+6} \mathrm{CrO}_4^{2-}(a q)\) The reaction involves the oxidation of Cr(OH)₄ because the oxidation number of Cr increases from +3 to +6. So, this reaction represents an oxidation half-reaction.

⇒ \(\stackrel{+5}{\mathrm{I}}_3^{-}(a q) \rightarrow \stackrel{+7}{\mathrm{IO}_4^{-}}(a q)\) This reaction represents an oxidation half-reaction since the oxidation number of I increases from +5 to +7 in the reaction.

⇒ \(\stackrel{+1}{\mathrm{ClO}^{-}}(a q) \rightarrow \mathrm{Cl}^{-1}(a q)\)

This reaction represents a reduction half-reaction because the oxidation number of Cl decreases from +1 to -1 in the reaction.

Redox Reactions Questions

⇒ \(\stackrel{+7}{\mathrm{MnO}_4^{-}}(a q) \rightarrow \stackrel{+4}{\mathrm{MnO}_2}(s)\)

⇒ \(\mathrm{Cr}^{+3}(\mathrm{OH})_4^{-} \rightarrow{ }^{66} \mathrm{CrO}_4^{2-}(a q)\) The reaction involves the oxidation of Cr(OH)4 because the oxidation number of Cr increases from +3 to +6. So, this reaction represents an oxidation half-reaction.

⇒ \(\stackrel{+5}{\mathrm{IO}_3^{-}}(a q) \rightarrow \stackrel{+7}{\mathrm{IO}_4^{-}}(a q)\) This reaction represents an oxidation half-reaction since the oxidation number of I increases from +5 to +7 in the reaction.

⇒ \(\stackrel{+1}{\mathrm{Cl}} \mathrm{O}^{-}(a q) \rightarrow \stackrel{-1}{\mathrm{Cl}}^{-}(a q)\) This reaction represents a reduction half-reaction because the oxidation number of Cl decreases from +1 to -1 in the reaction.

⇒ \(\stackrel{+7}{\mathrm{MnO}_4^{-}}(a q) \rightarrow \stackrel{+4}{\mathrm{MnO}_2}(s) .\) This is a reduction half¬ reaction because the oxidation number of Mn decreases from +7 to +4.

Question 9. Give an example of a disproportionation reaction. Calculate the volume of /0.225(M) KMnO₄ solution that can completely react with 45mL of a 0.125(M) I:eS0₄ solution in an acid medium.
Answer: Second part: \(\left[\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+e\right] \times 5\)

Reduction reaction: \(\begin{aligned}
& \mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e \longrightarrow \\
& \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l) \times 1
\end{aligned}\)

Net reaction: \(\begin{aligned}
& 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q) \rightharpoondown \\
& 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

5 mol of FeSO₄ = 1 mol of MnO₄

or, 1000mL of5 M FeSO₄ = 1000 ml limit of 1M KMnO₄

or, 1mL of 5M FeSO₄ s lmL of lMKMnO₄

or, 1mL of 1M FeSO₄ \(\equiv \frac{1}{5} \mathrm{~mL}\) of 1M KMno4

or, 45mL of 0.125M FeSO₄ = 45 X 0.125 ,\(\times \frac{1}{5} \mathrm{~mL}\) of 1M KMnO₄ \(\equiv \frac{9 \times 0.125}{0.225} \mathrm{~mL}\) = 5mL of0.225M KMnO₄

= The volume of KMnO₄ required = 5mL

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Question 10. For an element to undergo a disproportionation reaction, at least how many oxidation states should the clement exhibit?
Answer: When an element undergoes disproportionation reaction, oxidation state the element changes in the following way

Intermediate- Higher oxidation + Lower oxidation

Example: The reaction of Cl₂ with cold and dilute NaOH is a disproportionation reaction.

So. an element will be able to undergo disproportionation reactionist exhibits at least three oxidation states

Question 11. A clement has three oxidation numbers, +6, +7, and +4. If it exhibits a +7 oxidation number in a compound, will the compound be able to participate In a disproportionation reaction?
Answer: The compound cannot undergo a disproportionation reaction. This is because the element in the compound exists in its highest oxidation state.

The compound would have been able to undergo a disproportionation reaction if the element existed in the +6 oxidation state in the compound as this oxidation state lies between the oxidation states +7 and +4.

Redox Reactions Questions

Question 12. An element can show 0, 1, and +5 oxidation states. The oxidation numbers of the element In two compounds are -1 and +5. Is a comproportionation reaction Involving these two compounds possible?
Answer: In a comproportionation reaction, two reactants in which a particular element exists in different oxidation states, react to form a substance in which that element exists in an intermediate oxidation state.

The given oxidation states of die elementin two compounds are -1 and +5. So, if these two compounds together undergo a compro-portionation reaction they will form a substance in which the element will exist in a zero oxidation state. This oxidation state lies between -1 and +5. Hence, the two compounds together can undergo a comproportionation reaction.

Question 13. Identify the following reactions as disproportionation and comproportionation reactions—

⇒ \(\mathrm{Ag}^{2+}(a q)+\mathrm{Ag}(s) \rightarrow 2 \mathrm{Ag}^{+}(a q)\)

⇒ \(\begin{aligned}
& 2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(\mathrm{~g}) \\
& 4 \mathrm{KClO}_3(\mathrm{~s}) \rightarrow \mathrm{KCl}(\mathrm{s})+3 \mathrm{KClO}_4(\mathrm{~s})
\end{aligned}\)

⇒ \(\begin{aligned}
& 2 \mathrm{MnO}_4^{2-}(a q)+ 2 \mathrm{H}_2 \mathrm{O}(l) \rightharpoondown \\
& 2 \mathrm{MnO}_4^{-}(a q)+\mathrm{MnO}_2(s)+4 \mathrm{OH}^{-}(a q)
\end{aligned}\)

⇒ \(\begin{aligned}
& 2 \mathrm{NH}_4 \mathrm{NO}_3(s) \rightarrow \mathrm{N}_2(g)+4 \mathrm{H}_2 \mathrm{O}(g)+\mathrm{O}_2(g) \\
& 1 \mathrm{O}_3^{-}(a q)+5 \mathrm{I}^{-}(a q)+6 \mathrm{H}^{+}(a q) \rightarrow 3 \mathrm{I}_2(s)+3 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

Answer: \({\stackrel{+2}{\mathrm{Ag}^2}}^{2+}(a q)+\stackrel{0}{\mathrm{Ag}}(s) \rightarrow \stackrel{+1}{\mathrm{Ag}^{+}}(a q)\)

In this reaction, the resulting species, Ag+, exists in an oxidation state (+1) that lies between the oxidation states of Ag2+(+2) and Ag(0). Hence, it is a comproportio¬ nation reaction.

⇒ \(2 \mathrm{H}_2 \mathrm{O}_2^{-1}(l) \rightarrow 2 \mathrm{H}_2 \stackrel{-2}{\mathrm{O}}(l)+\stackrel{0}{\mathrm{O}}_2(\mathrm{~g})\) In this reaction, H₂O₂ (oxidation number of 0 is -1 ) decomposes to form H₂O (oxidation number of O is -2) and O₂ (oxidation number of O is zero). The oxidation state -1 lies between 0 and -2. So, this reaction represents a disproportionation reaction.

Redox Reactions Questions

⇒ \(4 \mathrm{~K}^{+5} \mathrm{IO}_3(s) \rightarrow \mathrm{KC} \stackrel{-1}{\mathrm{I}}+3 \mathrm{~K}^{+7} \mathrm{ClO}_4(s)\) In this reaction, the oxidation number of Cl decreases (+5 to-1 ) as well as increases (+5 to +7). This means KC1O₃ undergoes both oxidation and reduction in the reaction. Hence, this reaction is a disproportionation reaction.

⇒ \(\begin{aligned}
2 \mathrm{MnO}_4^{2-}(a q) & +2 \mathrm{H}_2 \mathrm{O}(l) \\
& \stackrel{+7}{2} \mathrm{MnO}_4^{-}(a q)+\stackrel{+4}{\mathrm{MnO}_2}(s)+4 \mathrm{OH}^{-}(a q)
\end{aligned}\)

In this reaction, the oxidation number of Mn increases (+6 to +7) and decreases (+6 to +4) as well. ‘Hus’ means MnO- undergoes both oxidation and reduction in the reaction. Therefore, this reaction is a disproportionation reaction.

⇒ \(2 \stackrel{-3}{\mathrm{NH}_4} \stackrel{+5}{\mathrm{NO}_3}(\mathrm{~s}) \rightarrow \stackrel{0}{\mathrm{~N}} \mathrm{~N}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g})\)

NH₄NO₃ molecule has two N-atoms, out of which one has an oxidation number of -3 and the other has an oxidation number of +5—the decomposition of NH₄NO₃ results in N₂(g). So, in this reaction, the oxidation number of one N-atom increases from -3 to 0 while the oxidation number of other N-atom decreases from +5 to 0. Since the oxidation number 0 lies between the oxidation numbers -3 and +5, the reaction represents a comproportionation reaction.

⇒ \(\stackrel{+5}{\mathrm{IO}_3^{-}}(a q)+\stackrel{-1}{5}^{-1}(a q)+6 \mathrm{H}^{+}(a q) \rightarrow 3 \stackrel{0}{\mathrm{I}}(s)+3 \mathrm{H}_2 \mathrm{O}(l)\)

In this reaction, 10J (oxidation number of I = +5) reacts with I- (oxidation number of = -1 ) to form I₂ (oxidation number of = 0 ). Since the oxidation number 0 lies between -1 and +5, the reaction represents a comproportionation reaction.

Question 14. Determine the equivalent masses of the following underlined compounds by both oxidation number and electronic methods

SO₂ + 2H₂O→H₂SO₄

HNO₃→NO₂ + H₂O

HNO₃ + 3H⁺→ NO + 2H₂O

MnO₂ + 4H⁺→ Mn²⁺→ + 2H₂O

⇒ \(\xrightarrow{\mathrm{KMnO}_4}+\frac{\mathrm{FeSO}_4}{\mathrm{~K}_2} \mathrm{SO}_4+\mathrm{HnSO}_4+\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}\)

Answer: Oxidation number method: \(\stackrel{+4}{\mathrm{SO}_2}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4\)

In this reaction, the oxidation number of S increases from +4 to +6. The change in oxidation number per molecule of S02 = 6-4 = 2 units.

∴ Equivalent mass of SO₂

⇒ \(=\frac{\text { Molecular mass of } \mathrm{SO}_2}{2}=\frac{64}{2}=32\)

Electronic method: SO, + 2H₂O→4H+→+ SO2- + 2e

Number of electrons lost by a molecule of SO₂ for its oxidation = 2.

Equivalent mass of SO₂ \(=\frac{64}{2}=32\)

Oxidation number method: \(\stackrel{+5}{\mathrm{HN}} \mathrm{O}_3 \rightarrow \stackrel{+4}{\mathrm{~N}} \mathrm{O}_2+\mathrm{H}_2 \mathrm{O}\)

The change in oxidation number per molecule of HNO₃ = 5-4 = unit:

∴ Equivalent mass of HNO₃

⇒ \(=\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{1}=63\)

Electronic method: NO₃ + 2H+ → NO₂+ HaO.

The number of electrons involved in the die reduction of NO₃ is 1.

Redox Reactions Questions

∴ Equivalent mass of HNO₃

⇒ c\(=\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{1}=63\)

Oxidation number method:

⇒ \(\mathrm{H}^{+5} \mathrm{~S}_3+3 \mathrm{H}^{+} \rightarrow \stackrel{+2}{\mathrm{~N} O}+2 \mathrm{H}_2 \mathrm{O}\)

In this reaction, the oxidation number of N decreases from +5 to +2. So, the change in oxidation number per molecule of I-INOg = 5-2 = 3 units.

∴ Equivalent mass of HNO₃

⇒ \(=\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{3}=21\)

Electronic method: \(\mathrm{NO}_3^{-}+4 \mathrm{H}^{+}+3 e \rightarrow \mathrm{NO}+2 \mathrm{H}_2 \mathrm{O}\)

The number of electrons involved in the reduction of 1 molecule of HNOg = 3

∴ Equivalent mass of HNO₃

⇒ \(=\frac{\text { Molecular mass of } \mathrm{HNO}_3}{1}=\frac{63}{3}=21\)

Oxidation number method:

⇒ \(\stackrel{+4}{\mathrm{MnO}_2}+4 \mathrm{H}^{+} \rightarrow \stackrel{+2}{\mathrm{M}}{ }^{2+}+2 \mathrm{H}_2 \mathrm{O}\)

In this reaction, the oxidation number decreases from +4 to +2. So, the change in oxidation number per molecule of MnO₂ = 4-2 = 2 units.

Equivalent mass of MnO₂

⇒ \(=\frac{\text { Molecular mass of } \mathrm{MNO}_2}{1}=\frac{87}{2}=43.5\)

Electronic method:

⇒ \(\mathrm{MnO}_2+4 \mathrm{H}^{+}+2 e \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O}\)

The number of electrons involved in the reduction of a molecule of MnO₂ =-2

⇒ \(=\frac{\text { Molecular mass of } \mathrm{MnO}_2}{1}=\frac{87}{2}=43.5\)

Oxidation number method:

⇒ \(\begin{aligned}
& \stackrel{+7}{\mathrm{~K}} \mathrm{nO}_4+\stackrel{+2}{\mathrm{~F}} \mathrm{eSO}_4+\mathrm{H}_2 \mathrm{SO}_4 \\
& \mathrm{~K}_2 \mathrm{SO}_4+\stackrel{+2}{\mathrm{MnSO}_4+\stackrel{+3}{\mathrm{Fe}} \mathrm{e}_2}\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}
\end{aligned}\)

In this reaction, KMnO₄ undergoes reduction because the oxidation number of Mn decreases from +7 to +2. So, the change in oxidation number of Mn= 7-2=5 units.

Equivalent mass of KMnO₄

⇒ \(\begin{aligned}
& =\frac{\text { Molecular mass of } \mathrm{KMnO}_4}{\begin{array}{c}
\text { Change in oxidation number per molecule } \\
\text { of } \mathrm{KMnO}_4 \text { because of its reduction }
\end{array}} \\
& =\frac{158}{5}=31.6
\end{aligned}\)

In the reaction, FeSO₄ undergoes oxidation because the oxidation number of Fe increases from +2 to +3. So, the change in oxidation number of Fe = 3-2 = 1 unit.

Equivalent mass of FeSO₄

⇒ \(\begin{aligned}
& =\frac{\text { Molecular mass of } \mathrm{FeSO}_4}{\begin{array}{c}
\text { Change in oxidation number per molecule } \\
\text { of } \mathrm{FeSO}_4 \text { because of its oxidation }
\end{array}} \\
& =\frac{151.85}{1}=151.85
\end{aligned}\)

Electronic method:

MnO₄ + 8H+→+ 5e→ Mn²⁺  + 4H₂O

Equivalent mass of KMnO₄

⇒ \(\begin{aligned}
& =\frac{\text { Molecular mass of } \mathrm{KMnO}_4}{\begin{array}{c}
\text { Number of electrons gained in reduction of } \\
\text { one molecule of } \mathrm{KMnO}_4
\end{array}} \\
& =\frac{158}{5}=31.6
\end{aligned}\)

Fe²⁺ →Fe³⁺ + e

Equivalent mass of FeSO₄

⇒ \(\begin{aligned}
& =\frac{\text { Molecular mass of } \mathrm{FeSO}_4}{\begin{array}{c}
\text { Number of electrons lost in oxidation of } \\
\text { one molecule of } \mathrm{FeSO}_4
\end{array}} \\
& =\frac{151.85}{1}=151.85
\end{aligned}\)

Question 15. Determine the equivalent mass of Br₂(Z) [Molecular mass =159.82/in the given reaction:

⇒ \(\begin{aligned}
& 2 \mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+\mathrm{Br}_2(l) \rightharpoondown \\
& 2 \mathrm{Mn}^{2+}(a q)+2 \mathrm{BrO}_3^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

In the given reaction, the oxidation half-reaction is —

⇒ \(\mathrm{Br}_2(l)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{BrO}_3^{-}(a q)+12 \mathrm{H}^{+}(a q)+10 e\)

The number of electrons involved in the oxidation of one molecule of Br₂ = 10 Equivalent mass of Br₂.

⇒ \(\begin{aligned}
& =\frac{\text { Molecular mass of } \mathrm{Br}_2}{\begin{array}{c}
\text { Number of electrons involved per } \\
\text { molecule of } \mathrm{Br}_2 \text { in its oxidation }
\end{array}} \\
& =\frac{159.82}{10}=15.982
\end{aligned}\)

Question 16. \(\mathrm{MnO}_4^{2-}\) undergoes a disproportionation reaction in an acidic medium but MnO₄ docs do not. Give reason.
Answer: The oxidation number of Mn in MnO₄ is +7, which is the highest oxidation number that Mn can possess. So, it does not undergo the disproportionation reaction.

Again, in the case of \(\mathrm{MnO}_4^{2-}\) the oxidation number of Mn is +6. Therefore, Mn in \(\mathrm{MnO}_4^{2-}\) can increase its oxidation number to +7 or decrease it to some lower value. So, \(\mathrm{MnO}_4^{2-}\) undergoes a disproportionation reaction as given below—

In the above reaction, the oxidation number of Mn increases from +6 in \(\mathrm{MnO}_4^{2-}\) to +7 in MnO₄ and decreases to +4 in MnO₄.

Question 17. What amount of K₂Cr₂O₇? (in mmol) is required to oxidize 24 mL 0.5 M Mohr’s salt?
Answer: The number of mmol of Mohr’s salt in 24 mL 0.5MMohr’s salt solution =24 x 0.5 = 12.

So, the balanced redox reaction is

⇒ \(\begin{aligned}
& \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+6\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot \mathrm{FeSO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}+7 \mathrm{H}_2 \mathrm{SO}_4 \\
& \mathrm{~K}_2 \mathrm{SO}_4+6\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+3 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+43 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

So, from the balanced equation, we see that 6mmol Mohr’s salt gets oxidized by 1mmol K₂Cr₂O₇.

12 mmol Mohr’s salt gets oxidized by \(\frac{1}{6} \times 12\) = 2 mmol K₂Cr₂O₇.

Question 18. Explain with reaction mechanism why the reaction between 03 and H2O2 is written as— \(\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})\)
Answer: The reaction between O₃ and H₂O₂ isasfollowsFirst step: O₃(g)→O₂(g) + O(g)’

Second step: H₂O₂(g) + O(g)→H₂O(g) +O₂(g)

In the first step, ozone produces nascent oxygen which is H₂O₂ in the second step. So, the overall reaction is as follows H₂O₂ + O₃→H₂O + O₂ + O₂

So, in the overall reaction, O₂ Is written twice because a total of two molecules of O₂ are produced during the reaction.

Question 19. 12.53 cm3 0.051M SeO₂ reacts completely with 25.5 cm3 0.1 M CrSO₄ to produce Cr₂(SO₄)₃. What is the change in the oxidation number of Se in this redox reaction?
Answer: Let the oxidation number of Se in the newly produced compound be x.

The redox reaction is as follows—

⇒ \(\begin{gathered}
{\left[\mathrm{Se}^{4+}+x e \longrightarrow \mathrm{Se}^{4-x}\right] \times 1} \\
{\left[\mathrm{Cr}^{2+} \longrightarrow \mathrm{Cr}^{3+}+e\right] \times x}
\end{gathered}\)

⇒ \(\mathrm{Se}^{4+}+x \mathrm{Cr}^{2+} \longrightarrow \mathrm{Se}^{4-x}+x \mathrm{Cr}^{3+}\)

Now, 12.53 cm3 0.051M SeO₂ = 12.53 x 0.051 = 0.64 mmol SeO₂

25.5 cm3 0.1 M CrSO₄=25.5 x 0.1 = 2.55 mmol CrSO₄ However according to the balanced equation, 1 mol SeO₂ gets reduced by x mol CrSO₄.

2.55 mmol CrSO₄ is reduced by \(\frac{2.55}{x}\) mmol SeO₂

Redox Reactions Questions

But 0.64 mmol SeO₂ gets reduced

⇒ \(\text { So, } \frac{2.55}{x}=0.64 \quad \text { or, } x=4\)

The change in oxidation number of Se -atom = 4- (4- x) = x = 4.

Question 20. 30 ml 0.05 M KMnb4 is required for the complete oxidation of 0.5 g oxalate in an acidic medium. Calculate ) tl,e percent amount of oxalate in that salt sample.
Answer: \(2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

According to the equation, 2 mol MnO-4 = 5 mol C₂O₄

⇒ \(1 \mathrm{~mol} \mathrm{MnO}_4^{-} \equiv \frac{5}{2} \mathrm{~mol} \mathrm{C}_2 \mathrm{O}_4^{2 \mathrm{ris}}\)

Again, 1000 mL 0.05(M)KMnO4 =→ 0.05 mol KMnO₄

30 mL 0.05(M)MnO4 \(\Rightarrow \frac{0.05 \times 30}{1000}\)

= 1.5×10-3 mol KMnO₄

Now, 1 mol \(\mathrm{MnO}_4=\frac{5}{2} \mathrm{~mol} \mathrm{C}_2 \mathrm{O}_4\)

⇒ \(1.5 \times 10^{-3} \mathrm{~mol} \mathrm{MnO}_4=\frac{5}{2} \times 1.5 \times 10^{-3} \mathrm{~mol} \mathrm{C}_2 \mathrm{O}_4^2\)

⇒ \(=\frac{5}{2} \times 1.5 \times 10^{-3} \times 38 \mathrm{~g} \mathrm{C}_2 \mathrm{O}_4^{2-}=0.33 \mathrm{~g} \mathrm{C}_2 \mathrm{O}_4^{2-}\)

⇒ \(\text { Percentage of } \mathrm{C}_2 \mathrm{O}_4^{2-} \text { in the sample }=\frac{0.33 \times 100}{0.5}=66 \%\)

Question 21. What will be the nature ofthe suit formed when 2 mol Nil Is added to the pigeon’s solution of mol pyrophosphoric? Clive equation?

Answer: From the structure of pyrophosphoric acid, it Is clear that it contains four replaceable hydrogen atoms.

So, the reaction between and 2 mol NaOH will be as follows— \(-\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7+2 \mathrm{H}_2 \mathrm{O}\) 2HaO There are two replaceable 11-atoms in Na., H₂PO₇ fonned due to the above reaction. So, it is an acidic salt.

Redox Reactions Questions

Question 22. Find the oxidation state of C-l and C-2 In CH₃CH₂OH.
Answer: The oxidation number of eadi of three H -atoms attached to C-2 = +1. Therefore, the total oxidation number of three Hatorns = +3. For the C — C bond, the oxidation number of the C-2 atom does not change.

So die oxidation number of the C-2 atom =-3. Now, die total oxidation number of two H -atoms attached to the C-l atom =+2. Again the oxidation number of the —OH group attached to the C-l atom =-l.

Hence the total oxidation number of two H -atoms and one linked to the C-l atom =+→2 + (-1) = +1. Thus, the oxidation number of C-l Hence, die oxidation staties’&f C-l and C-2 in CH₃CH, OH are -1 and -3 respectively.

Question 23. 1 mol N₂H₄ loses 10 mol of electrons with, the formation of 1 mol of a new compound Y. If the new compound contains the same number of N-atoms then what will be the oxidation number of nitrogen in the new compound? (Assume that the oxidation number of the H -atom does not change.
Answer: The oxidation number of each N -atom in N₂H₂ = -2 As given, 1 mol N₂H₄→1 mol Y + 10 mol e.

Suppose, oxidation million of N in its molecule of Y – x.

Total oxidation uninbor of two N -Moms In Y molecule -Oxidation number of two N atoms In N₂H₄ molecule = 2 x (-2) + IO or, x = + 3

The oxidation number of each N-atom In compound Y = +3

Question 24. Oxidation million of the elements A, It mid-C are 12,1 to mid -2 respectively. Which one will lie the formula of the compound containing these three elements? \(\mathrm{A}_2\left(\mathrm{BC}_2\right)_2, \mathrm{~A}_3\left(\mathrm{~B}_2 \mathrm{C}\right)_2, \mathrm{~A}_3\left(\mathrm{BC}_4\right)_2\)
Answer: The total oxidation number of all the elements in a compound should be zero (0).

In A₂(BC₂) molecule, the sum of oxidation numbers of all the atoms =2x (+ 2) + 2 X (+ 5) + 4 X (-2) = + 6

In the A₂(B₂C)₂ molecule, the sum of oxidation numbers of all the atoms = 3 X (+ 2) + 4 X (+ 5) + 2 X (-2) = + 22

In A₃(BC₄)₂ molecule, the sum of oxidation numbers of all the atoms = 3 X (+ 2) + 2 x (+ 5) + 8 x (-2) = 0.

∴ The correct formula of the compound will be A₃(BC₄)₂

Question 25. In an acidic medium, for the reduction of each NO₃ ion in the given reaction, how many electrons will be required? NO₃ NH₂OH
Answer: NO₃— NH₂OH; For equalizing the number of O -atoms on both sides, two H₂O molecules are added to the right side (having a lesser number of O -atoms) and two H+ ions are added to the left side for each molecule of H₂O added.

⇒ \(\mathrm{NO}_3^{-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{NH}_2 \mathrm{OH}+2 \mathrm{H}_2 \mathrm{O}\)

For equalizing the number of H -atoms on both sides, three additional H+ ions are required on the left side

Redox Reactions Questions

so we get, \(\mathrm{NO}_3^{-}+7 \mathrm{H}^{+} \longrightarrow \mathrm{NH}_2 \mathrm{OH}+2 \mathrm{H}_2 \mathrm{O}\)

To balance the charge on both sides, 6 electrons are added to the left side ofthe equation.

⇒ \(\mathrm{NO}_3^{-}+7 \mathrm{H}^{+}+6 e \longrightarrow \mathrm{NH}_2 \mathrm{OH}+2 \mathrm{H}_2 \mathrm{O}\)

Hence, for the reduction of each NO₃ ion into an NH₂OH molecule, 6 electrons are required.

Question 26. CO₃O₄ is an oxide of CO₃ and CO₂ If its formula is Cox(2)Co,(m)O₄, then what is the value of x and y?
Answer: The sum of the oxidation number of the elements in a compound is equal to zero. So, for Cox(2)COy(3)O₄, 2x+3y-4 x2 = 0 or, 2x+ 3y = 8 The only solution for this equation is x = 1 and y = 2.

Question 27. How many electrons should A₂H₃ liberate so that In the new compound, A shows an oxidation number of \(-\frac{1}{2}\)?
Answer: Let, A₂H₃ will liberate x electrons.

Therefore \(2 \times\left(-\frac{1}{2}\right)+3 \times(+1)=+x\)

or, -1+3=+x or, x=2.

Very Short Answer Type Questions

Question 1. Identify the redox reactions among the following:

• 2CuSO₄ + 4KI→2CuI + I₂ + 2K₂SO₄
• BaCl₂ + Na₂SO₄→BaSO₄ + 2NaCl
• 2NaBr + Cl₂→2NaCl + Br₂
• NH4NO₂→+N₂ + 2H₂O
• CUSO₄ + 4NH₃→[CU(NH₃)₄]SO₄
• 3I₂ + 6NaOH→NaIO₃ + 5NaI + 3H₂O

Answer: In this reaction, the oxidation number of Cu decreases (+ 2 → +1) and the oxidation number of I₄ increases (-1→ 0) i.e. reduction of CuSO₄ and oxidation of KI take place. Thus, it is a redoxreaction.

⇒ \(\stackrel{+2}{\mathrm{CuSO}_4}+\stackrel{-1}{\mathrm{KI}} \rightarrow \stackrel{+1}{\mathrm{CuI}}+\stackrel{0}{\mathrm{I}} 2+\mathrm{K}_2 \mathrm{SO}_4\)

This reaction does not involve any change in the oxidation number of any element i.e., in this reaction, oxidation or reduction does not take place. Hence, it is not a redoxreaction

Redox Reactions Questions

⇒ \(\stackrel{+2}{\mathrm{BaCl}_2}+\stackrel{+1}{\mathrm{Na}_2} \mathrm{SO}_4 \rightarrow \stackrel{+2}{\mathrm{BaSO}}{ }_4+2 \mathrm{NaCl}^{-1}\)

In this reaction, the oxidation number of bromine increases from -1 to 0 and the oxidation number of chlorine decreases from 0 to -1. In this case, NaBr gets oxidized whereas Cl₂ gets reduced. Hence, this reaction is a redox reaction.

⇒ \(2 \mathrm{NaBr}+\stackrel{0}{\mathrm{C}} \mathrm{l}_2 \rightarrow 2 \mathrm{Na}{ }^{-1} \mathrm{Cl}+\stackrel{0}{\mathrm{Br}}{ }_2\)

In NH₄NO₂, the oxidation number of N in NH₄ increases from -3 to 0, while the oxidation number of N in NO₂ decreases from +3 to 0, i.e., in this reaction, simultaneous oxidation of NH+ and reduction of NO₂ in the compound occur. So, this reaction is a redoxreaction.

⇒ \(\stackrel{-3}{\mathrm{NH}_4}{\stackrel{+3}{\mathrm{~N}} \mathrm{O}_2 \rightarrow \stackrel{0}{\mathrm{~N}}}_2+2 \mathrm{H}_2 \mathrm{O}\)

This reaction does not involve any change in the oxidation number of any element. So, it is not a redox reaction.

⇒ \(\left(\stackrel{+2}{\mathrm{CuSO}_4}+4 \mathrm{NH}_3\right) \rightarrow\left[\stackrel{+2}{\mathrm{Cu}}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4\)

In the given reaction, the oxidation number of iodine increases from 0 to +5 and decreases from 0 to -1 i.e., both oxidation and reduction occur in this reaction. Thus, it is a redox reaction.

⇒ \(3 \stackrel{0}{\mathrm{I}}_2+6 \mathrm{NaOH} \rightarrow \mathrm{NaIO}_3^{+5}+5 \mathrm{NaI}^{-1}+3 \mathrm{H}_2 \mathrm{O}\)

Question 2. Which of the following reactions are disproportionation reactions and comproportionation reactions?

• 4KC1O₃→KC1 + 3KC1O₄
• 3K₂MnO₄ + 2H₂O→2KMnO₄ + MnO₂ + 4KOH
• KIO₃ + 5KI + 6HCI→3I₂ + 6KCI + 3H₂O
• 2C₆H₅CHO + NaOH C₆H₅COONa + C₆H₅CH₂OH
• Ag²⁺ + Ag→2Ag⁺

Answer: \(\stackrel{45}{4 \mathrm{KClO}_3} \rightarrow \stackrel{-1}{\mathrm{KCl}}+3 \stackrel{+7}{\mathrm{KClO}_4}\)

In KC1O₃, the oxidation number of Cl = + 5. The oxidation numbers of Cl in KC1 and KC1O₄ are -1 and + 7 respectively. So in this reaction, KC1O₃ undergoes simultaneous oxidation and reduction producing KC1O₄ and KC1 respectively. Hence, it is a disproportionation reaction.

Redox Reactions Questions

⇒ \(3 \mathrm{~K}_2 \stackrel{+6}{\mathrm{MnO}}_4+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KMnO}_4+\stackrel{+7}{\mathrm{MnO}_2}+4 \mathrm{KOH}\)

The oxidation number of Mn in K₂MnO₄ = + 6. On the other hand, the oxidation numbers of Mn in the products KMnO₄ and MnOa are + 7 and + 4 respectively. Therefore, in the reaction, K₂MnO₄ is oxidized and reduced at the same time forming KMn04 and MnOa. Thus, it is a disproportionation reaction.

⇒ \(\stackrel{+55}{\mathrm{KIO}_3}+5 \stackrel{-1}{\mathrm{KI}}+6 \mathrm{HCl} \rightarrow 3 \stackrel{0}{\mathrm{I}}_2+6 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}\)

The oxidation numbers of iodine in K103 and KI are + 5 and -1 respectively and the oxidation number of iodine in 12 is zero. This oxidation number lies between + 5 and -1, which is an intermediate oxidation state. So, it is a comproportionation reaction.

⇒ \(2 \mathrm{C}_6 \mathrm{H}_5^{+1} \mathrm{C} \mathrm{HO}+\mathrm{NaOH} \underset{\mathrm{C}_6 \mathrm{H}_5+3}{+3} \mathrm{COONa}+\mathrm{C}_6 \mathrm{H}_5^{-1} \mathrm{CH}_2 \mathrm{OH}\)

In this reaction, the oxidation number of carbon in the benzene ring does not change. However, the oxidation number of carbon atoms in the —CHO group (the oxidation number of carbon in —CHO is +1 ) changes to a higher oxidation number of +3 in the —COONa group and a lower oxidation number of -1 in the —CH₂OH group. Here, —the CHO group gets simultaneously oxidized & reduced. Hence, it is a disproportionation reaction.

Redox Reactions Questions

⇒ \(\stackrel{+2}{\mathrm{Ag}^{2+}}+\stackrel{0}{\mathrm{Ag}} \rightarrow 2 \mathrm{Ag}^{+1}\)

The oxidation numbers of the reactants Ag²⁺ and Ag are + 2 and O respectively and the oxidation number of the product is +1. This oxidation number is intermediate between the oxidation numbers +2 and O . so it is a comproportionation reaction.

Question 3. Identify the redox reaction(s) and also the oxidants B and the reductants from the following reaction(s).

⇒ \(2 \mathrm{MnO}_4^{-}+5 \mathrm{SO}_2+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{SO}_4^{2-}+2 \mathrm{Mn}^{2+}+4 \mathrm{H}_3 \mathrm{O}^{+}\)

⇒ \(\begin{aligned}
& \mathrm{NH}_4^{+}+\mathrm{PO}_4^{3-} \longrightarrow \mathrm{NH}_3+\mathrm{HPO}_4^{2-} \\
& \mathrm{HClO}+\mathrm{H}_2 \mathrm{~S} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{S}
\end{aligned}\)
Answer: In this reaction, the oxidation number of Mn decreases from (+7→+2) and the oxidation number of S increases from (+4→+6). So this reaction causes a reduction of Mn04 and oxidation of SO₂. Hence, the reaction is a redox reaction. Here Mnt)ÿ acts as an oxidant and SO₂ as a reductant.

This reaction does not involve any change in the oxidation number of any element. So it is, not a redoxreaction.

This reaction involves an increase in the oxidation number of S from -2 to 0, and a decrease in the oxidation number of, Cl from +1 to -1. So it is a redox reaction. Here HCIO is an oxidising agent and H₂S is a reducing agent.

Question 4. Determine the equivalent masses of Na₂S₂O₃.5H₂0 +2 and KBrO₃ In the following reactions.

⇒ \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

⇒ \(\mathrm{BrO}_3^{-}+6 \mathrm{H}^{+}+6 e \longrightarrow \mathrm{Br}^{-}+3 \mathrm{H}_2 \mathrm{O}\)
Answer: In this reaction, two \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) ions are oxidised by losing 2 electrons. So, for the oxidation of one \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) ion, one electron is given up

Equivalent mass of Na₂S₂O3-5H₂O

⇒ \(=\frac{\text { Molecular mass of } \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \cdot 5 \mathrm{H}_2 \mathrm{O}}{1}=248\)

This reaction produces Br- from BrO₃. In the reduction of each molecule of KBr03, 6 electrons are accepted.

Redox Reactions Questions

In the given reaction, equivalent mass of KBrO₃

⇒ \(=\frac{\text { Molecular mass of } \mathrm{KBrO}_3}{6}=\frac{167}{6}=27.8\)

Question 5. Determine the equivalent weights of the underlined compounds in the following two reactions:

⇒ \(\mathrm{FeSO}_4+\frac{\mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightharpoondown}{\mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}}\)

⇒ \(\begin{aligned}
& \mathrm{MnO}_2+\mathrm{HCl} \longrightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \\
& {[K=39, M n=55,0=16]} \\
&
\end{aligned}\)

Answer: In this reaction, the decrease in oxidation number of Mn =(7-2) = 5 units. Equivalent weight of Mn.

⇒ \(=\frac{\text { Molecular weight of } \mathrm{KMnO}_4}{\text { Change in oxidation number }}=\frac{158}{5}=31.6\)

⇒ \(\stackrel{+4}{\mathrm{MnO}}{ }_2 \longrightarrow \stackrel{+2}{\mathrm{MnCl}}{ }_2\) Here the oxidation number ofMn decreases by (4-2) = 2 unit.

∴ Equivalent weight of MnO₂

⇒ \(=\frac{\text { Molecular weight of } \mathrm{MnO}_2}{\text { Change in oxidation number }}=\frac{87}{2}=43.5\)

Question 6. Balance the following equation with the help of the oxidation number method.

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{CO} \rightarrow \mathrm{FeO}+\mathrm{CO}_2\)

Answer: In Fe3O₄, the oxidation number of each Fe -atom =+2.67. In the reaction, the oxidation number of each Fe -atom decreases from + 2.67 to 2. So, decrease in oxidation number. of eachFe -atom = +0.67

Therefore for three Fe -atoms, the total decrease in oxidation number 3 x (+ 0.67) =+ 2 unit.

On the other hand, the oxidation number of C increases from +2 to +4. Hence, an increase in the oxidation number of C= 2 units.

Therefore, in the reaction, Fe₃O4 and CO will react with each other in the molar ratio of 2: 2 or 1: 1 . Again 1 molecule of Fe₃O₄ will produce 3 molecules of F₂O. Hence, the balance equation will be—Fe₃O₄ + CO →3FeO + CO₂.

Question 7. Balance by ion-electron method: MnO₂ + HCl→+Mn²⁺ + Cl₂ + H₂O
Answer: Oxidation half-reaction: 2C1¯→ Cl₂+ 2e

• Reduction half-reaction: MnO₂ + 4H⁺ + 2e→Mn²⁺ + 2H₂O
• Net reaction: MnO₂ + 4H+ →+ 2Cl →Mn₂+ Cl₂ + 2H₂O or, MnO₂ + 4H⁺ +4C1→ Mn²⁺+ 2C1^– + C1₂ + 2H₂O

Redox Reactions Questions

Therefore the balanced chemical equation is—

MnO₂ + 4HC1 MnCl₂ + Cl₂ + 2H₂O

Question 8. In a basic medium, balance the half-reactions below:

⇒ \(\mathrm{Cr}(\mathrm{OH})_3 \rightarrow \mathrm{CrO}_4^{2-} \text { (2) } \mathrm{Cl}_2 \mathrm{O}_7 \rightarrow 2 \mathrm{ClO}_2^{-}\)

Answer: To equalize the number of O -atoms, one H₂O molecule is added to the right side (because it has an excess O -atom) and for this 1 molecule of H₂O, two OH Now, we get Cr(OH)₂ + 2OH^– ions are added to the left side. CrO^2- +H₂O. In this equation, five H -atoms are on the left side, and two H atoms are present on the right side. To equalize the number of H -atoms on both sides, three OH-(aq) and three H₂O molecules are added to the left side and right side respectively. Finally, we get—

⇒ \(\begin{aligned}
& \mathrm{Cr}(\mathrm{OH})_3+2 \mathrm{OH}^{-}+3 \mathrm{OH}^{-} \rightarrow \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}+3 \mathrm{H}_2 \mathrm{O} \\
& \text { or, } \mathrm{Cr}(\mathrm{OH})_3+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

For balancing the charge, 3 electrons are added to the right side. Therefore, the final half-reaction becomes,

⇒ \(\mathrm{Cr}(\mathrm{OH})_3+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}+3 e\)

To balance the number of 0 -atoms on both sides, 3 molecules of H₂O(Z) are added to the left side (because there are excess O -atoms on this side) and for these three H₂O(l) molecules, six OH- (aq) ions are added to the right side. Then we get—

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}\)

To balance the charge, electrons are added to the left side. Hence, the balanced equation is

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+3 \mathrm{H}_2 \mathrm{O}+8 e \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}\)

Question 9. Balance the following reaction in acidic and alkaline
medium: \(\mathrm{SO}_3^{2-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)\) 
Answer: Addlemedium: \(\mathrm{SO}_3^{2-}(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)\)

In Tills reaction, this left side of the equation is deficient in one () -atom. So the number of O -atoms on both sides is equalized by adding one H2O(aq) molecule to the left side.

⇒ \(\mathrm{SO}_3^{2-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{SO}_4^{2-}(a q)\)

Answer: To equalize the number of O -atoms, one H₂O molecule is added to the right side (because it has an excess 0 -atom) and for this 1 molecule of H₂O, two OH- ions are added to the left side. Now, we get \(\mathrm{Cr}(\mathrm{OH})_3+2 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}\)

In this equation, five H -atoms are on the left side and two H atoms are present on the right side. To equalize the number of H -atoms on both sides, three OH-(aq) and three H20 molecules are added to the left side and right side respectively. Finally, we get-

Redox Reactions Questions

⇒\(\begin{aligned}
& \mathrm{Cr}(\mathrm{OH})_3+2 \mathrm{OH}^{-}+3 \mathrm{OH}^{-} \rightarrow \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}+3 \mathrm{H}_2 \mathrm{O} \\
& \text { or, } \mathrm{Cr}(\mathrm{OH})_3+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

For balancing the charge, 3 electrons are added to the right side. Therefore, the final half-reaction becomes,

⇒ \(\mathrm{Cr}(\mathrm{OH})_3+5 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_4^{2-}+4 \mathrm{H}_2 \mathrm{O}+3 e\)

To balance the number of 0 -atoms on both sides, 3 molecules of H₂O(Z) are added to the left side (because there is excess O -atoms on this side) and for these three H₂O(l) molecules, six OH-(aq) ions are added to the right side. Then we get—

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}\)

For balancing the charge, 8 electrons are added to the left side. Hence, the balanced equation is,

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+3 \mathrm{H}_2 \mathrm{O}+8 e \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}\)

Question 10. Determine the values of x and y in the following balanced equation:
\(5 \mathrm{H}_2 \mathrm{O}_2+x \mathrm{ClO}_2+2 \mathrm{OH}^{-} \longrightarrow x \mathrm{Cl}^{-}+y \mathrm{O}_2+6 \mathrm{H}_2 \mathrm{O}\) 
Answer: Oxidation half-reaction: \(\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+2 e\)

Reduction half-reaction:

⇒\(\mathrm{ClO}_2+2 \mathrm{H}_2 \mathrm{O}+5 e \longrightarrow \mathrm{Cl}^{-}+4 \mathrm{OH}^{-}\)

Multiplying equations (1) and (2) by 5 and 2 respectively and adding the equations, we get—

⇒ \(\begin{aligned}
5 \mathrm{H}_2 \mathrm{O}_2+10 \mathrm{OH}^{-}+2 \mathrm{ClO}_2+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \\
2 \mathrm{Cl}^{-}+5 \mathrm{O}_2+10 \mathrm{H}_2 \mathrm{O}+8 \mathrm{OH}^{-}
\end{aligned}\)

5H₂O₂ + 2C1O₂ + 2OH→2Cr + 5O₂ + 6H₂O ….[3]

Comparing equation (3) with the given equation, we get x = 2 and y – 5.

Question 11. In the given reaction determine the equivalent weight of AS₂S₃: [Assume that M.W. of As2S3 =AM]

⇒ \(\begin{aligned}
\mathrm{As}_2 \mathrm{~S}_3+7 \mathrm{ClO}_3^{-}+ & 7 \mathrm{OH}^{-}-7 \\
& 2 \mathrm{AsO}_4^{3-}+7 \mathrm{ClO}^{-}+3 \mathrm{SO}_4^{2-}+6 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

The increase in oxidation number of each As -atom =5-3 = 2 units. So, the total increase in oxidation number of two As -atoms =2×2 = 4 units.

Tlw Increase In oxidation number of each S-atom =+6-(-2) Therefore, the total Increase in oxidation number of three S -atoms = 3 x B = 24 unit.

Hence, the total Increase In oxidation number for each As2S,t molecule In Its oxidation Is a (24 + 4) = 20 unit.

Equivalent weight of AS₂S₃ In the given reaction

⇒ \(=\frac{\text { Molecular weight of } \mathrm{As}_2 \mathrm{~S}_3}{\text { Increase in oxidation number }}=\frac{M}{28}\)

Question 12. Determine the equivalent mass of Fe.,04 in the given Mn = 6-4 = 2 units. reaction: FeO₄ + KMn0_4↓Fc₂O₃ + MnO₂ (Assume that the molecular mass of FC₃O₄ =M
Answer: \(\stackrel{+\mathrm{F} / 3}{\mathrm{Fe}} \mathrm{e}_3 \mathrm{O}_4 \rightarrow \stackrel{+3}{\mathrm{~F}} \mathrm{e}_2 \mathrm{O}_3\stackrel{+\mathrm{F} / 3}{\mathrm{Fe}} \mathrm{e}_3 \mathrm{O}_4 \rightarrow \stackrel{+3}{\mathrm{~F}} \mathrm{e}_2 \mathrm{O}_3\)

In this reaction, the increase in oxidation number for each Pc -atom \(=3-\frac{8}{3}=\frac{1}{3} \text {. }\)

So, the total increase in oxidation number for 3 Fe -atoms

⇒ \(=3 \times \frac{1}{3}=1 \text { unit. }\)

Therefore, the equivalent mass of Fe3O4

⇒ \(=\frac{\text { Molecular mass of } \mathrm{Fe}_3 \mathrm{O}_4}{\text { Increase in oxidation number }}=\frac{M}{1}=M\)

Question 13. An oxidizing agent KH(IO₃)₂ in the presence of 4.0 (N) HCI gives IC1 as a product. Determine the equivalent weight of KH(IO₃)₂. [K = 39,1 = 127]
Answer: According to the question, \(\stackrel{+5}{\mathrm{KH}}\left(\mathrm{IO}_3\right)_2 \xrightarrow{+1} \mathrm{ICl} ;\) In the reaction, the change in oxidation number of each I -atom = + 5- (+1) = 4 unit. So, for two -atoms present in 1 molecule of KH(IO₃)₂, the total change in oxidation number =2×4 = 8 units. So, in acid medium, the equivalent weight of KH(IO₃)₂

⇒ \(=\frac{\text { Molecular weight of } \mathrm{KH}\left(\mathrm{IO}_3\right)_2}{\text { Total change in oxidation number }}=\frac{390}{8}=48.75\)

Question 14. Find the oxidation number of carbon in methanal and methanoic acid.
Answer: Methanal: The oxidation number of C in the HCHO molecule = 0.

Methanol acid: Suppose the oxidation number of C in methanoic acid.

2 X (+1) + x + 2 x (-2) = 0 (for two H-atoms)(for two O-atoms) Hence, oxidation number of C in HCOOH molecule = +2.

Redox Reactions Questions

Question 15. What will be the change in the oxidation number of Mn when MnO₂ is melted with solid KNO₃ & NaOH?
Answer: Potassium manganate is produced when MnO₂ is melted in the presence of solid KNO₃ and NaOH.

⇒ \(\stackrel{+4}{\mathrm{MnO}_2}+2 \mathrm{KOH}+\mathrm{KNO}_3 \longrightarrow \mathrm{K}_2 \stackrel{+6}{\mathrm{MnO}}{ }_4+\mathrm{KNO}_2+\mathrm{H}_2 \mathrm{O}\)

Soin this reaction, the oxidation number of Mn increases from +4 to +6 i.e., a change in oxidation number of Mm= 6-4=2 unit

Question 16. What is the ratio of equivalent weights of MnO₄ in acidic, basic & neutral mediums?
Answer: The reaction that the MnO₄ ion undergoes in acidic, basic, and neutral medium are as follows. Acidic Medium:

⇒ \(\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e \rightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_2 \mathrm{O}(l)\)

• Equivalent weight \(E_1=\frac{M}{5}\) where M = KMn04.
• Basic Medium: MnO₄ (aq) + e→MnO₄ (aq)
• Equivalent weight \(E_2=\frac{M}{1}\)

Neutral Medium:

⇒ \(\mathrm{MnO}_4^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)+3 e \rightarrow \mathrm{MnO}_2(s)+4 \mathrm{OH}^{-}(a q)\)

Therefore \(E_1: E_2: E_2=\frac{1}{5}: 1: \frac{1}{3}=3: 15: 5\)

Question 17. MnO₄ reacts with Ax+ to form AO-, Mn2+, and O2. One mole of MnO- oxidizes 1.25 moles of Ax+ to AO₃. What is the value of x?
Answer: \(\mathrm{MnO}_4^{-}+\mathrm{A}^{x+} \rightarrow \mathrm{AO}_3^{-}+\mathrm{Mn}^{2+}+\mathrm{O}_2\)

The change in oxidation number of = 5 unit (+7→+2) and that of A = (5- x) unit [+x→+5 in AO_3^-3]

1 mole of MnO₃ reacts completely with 1.25 mole of Ax+

Therefore, 1×5 = 1.25(5 -x)

Redox Reactions Questions

Solving for x gives x = +1

Thus, the value of x = +1

Question 18. 20 mL solution of 0.1 (M) FeSO₂ as completely oxidized using a suitable oxidizing agent. What is the number of electrons exchanged?
Answer: 20 mL of 0.1(M) FeSO₄ \(\begin{aligned}
& \equiv \frac{0.1}{1000} \times 20 \\
& \equiv 2 \times 10^{-3} \mathrm{~mol} \text { of } \mathrm{Fe}^{2+}
\end{aligned}\)

Fe²⁺ is oxidized to Fe³⁺, leaving 1 electron.

Hence, the number of electrons exchanged by 2 x 10-3 mol of Fe²⁺ is— 2 X 10-3 x 6.022 x 10₂₃ = 1.2044 x 10₂₁ electrons

Question 19. Give an example of a compound In which The mime element exists In two different oxidation slates.
Answer: NH₄NO₃

Question 20. Given an example of a compound In which the oxidation number of N=+1.
Answer: N₂O

Question 21. Give an example of a compound In which the oxidation number of O r₂.
Answer: OF₂

Redox Reactions Questions

Question 22. What Is the oxidation number of Fe In Po(CO)s?
Answer: Zero(0)

Question 23. What is the oxidation number of sodium In sodium amalgam?
Answer: Zero(0)

Question 24. Give an example of a compound In which the oxidation number and valency of an element In the compound are the same.
Answer: CCl₄

Question 25. What are the oxidation numbers of three C -atoms In C3°2?
Answer: \(\mathrm{O}=\stackrel{+2}{\mathrm{C}}=\stackrel{0}{\mathrm{C}}=\stackrel{+2}{\mathrm{C}}=\mathrm{O},\)

Question 26. What Is the oxidation number of carbon In CUCOCH₃?
Answer: -4/3

Question 27. What type of redox reaction does the given reaction belong to? \(2 \mathrm{H}_2 \mathrm{~S}+\mathrm{SO}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{~S}\)
Answer: Comproportlonation

Question 28. Which of the following Is unable to participate in disproportionation reaction? CIO-, C1O₂, C1O₃, ClO₄
Answer: CIO¯₄

Question 29. In which of the following ions does Fe exist in the same oxidation state? \(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-},\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}^{2-}\right.\)
Answer: [Fe(CN₆)]_4‾ and [Fc(CN)gNO]^2-

Question 30. What is the equivalent mass of CuSO₄ in the given reaction? 2CuSO₄ + \(2 \mathrm{CuSO}_4+4 \mathrm{KI} \rightarrow \mathrm{Cu}_2 \mathrm{I}_2+\mathrm{I}_2+\mathrm{K}_2 \mathrm{SO}_4\)
Answer: M

Question 31. What is the average oxidation number of Fe in \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)?
Answer: \(+\frac{18}{7}\)

Redox Reactions Questions

Question 32. Which one is the oxidizing agent in the given reaction: Which element In oxidized and which element Is reduced In the reaction, \(\mathrm{AsO}_2^{-}+\mathrm{Sn}^{2+} \rightarrow \mathrm{As}+\mathrm{Sn}^{4+}+\mathrm{H}_2 \mathrm{O}\)
Answer: AsO‾²

Question 33. Which element In oxidized and which element Is reduced In the reaction, 4KCIO₂(g) → 3KClO₄(g) + KCi(g)?
Answer: Cl is reduced and oxidized simultaneously

Short Answer Type Questions

Question 1. What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

1. Kl₃
2. CH₃COOH

Answer: The chemical structure of KI₃ is K+[1-1→1]¯

In Kl₃-1 ion forms a coordinate Bond with I₂. The Osditaion Number Of Each 1 atom in 12 molecules is zero, and the oxidation number of K+ id +1, therefore from the oxidation number of one I- will be -1

2. The C-1 atom is linked to one O-atom and one OH group, For O -atom and OH -group, the oxidation numbers are 2 and I respectively, As C-l and C-2 are the atoms of the name element, the covalent linkage between them makes no change In oxidation number of either atom.

So, the oxidation number of C- 1 would be +3 since the total oxidation number of one O-atom and one Oil -group is -3. The total oxidation number of three 11 -atoms linked to the C-2 atom Is +3. So, the oxidation number of C-2 would be -3.

Question 2. Justify the following reactions are redox reactions:

⇒ \(\begin{aligned}
& \mathrm{CuO}(s)+\mathrm{H}_2(g) \rightarrow \mathrm{Cu}(s)+\mathrm{H}_2 \mathrm{O}(g) \\
& \mathrm{Fe}_2 \mathrm{O}_3(s)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_2(g)
\end{aligned}\)

Redox Reactions Questions

⇒ \(\begin{aligned}
& 4 \mathrm{BCl}_3(g)+3 \mathrm{LiAlH}_4(s)- \\
& 2 \mathrm{~B}_2 \mathrm{H}_6(g)+3 \mathrm{LiCl}(s)+3 \mathrm{AlCl}_3(s)
\end{aligned}\)

⇒ \(\begin{aligned}
& 2 \mathrm{~K}(s)+\mathrm{P}_2(g) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(s) \\
& 4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_2 \mathrm{O}(g)
\end{aligned}\)
Answer: In the reaction, the oxidation number of Cu decreases (+2→0) indicating the reduction of CuO, and the oxidation number of 2 increases (0 →+1), indicating the oxidation of H₂. Hence, it is a redox reaction.

In the reaction, the oxidation number of Fe decreases (+3→+O) and that of C increases
→+4). Therefore, the reaction involves the reduction of Fe₂O₃ and the oxidation of CO. Hence, it is a redox reaction.

The reaction involves the reduction of BC1₃ because the oxidation number of B decreases from +3 to -3 and the oxidation of LiAH₄ as the oxidation number increases from -1 to +1. So, it is a redox reaction.

⇒ \(2 \stackrel{0}{\mathrm{~K}}(s)+\stackrel{0}{\mathrm{~F}}_2(g) \rightarrow \stackrel{+1}{2} \stackrel{-1}{\mathrm{~K}}^{-} \mathrm{F}^{-}(s)\)

In the reaction, the increase in the oxidation number of K (0→+1) and the decrease in the oxidation number of F (0 to -1) indicate that the former undergoes oxidation and the latter reduces. Hence, the given reaction is a redox reaction.

⇒ \(4 \stackrel{-3}{\mathrm{~N}}_3(\mathrm{~g})+5 \stackrel{0}{\mathrm{O}}_2(\mathrm{~g}) \rightarrow 4 \stackrel{+2}{\mathrm{~N}} \mathrm{O}^{-2}(\mathrm{~g})+6 \stackrel{+1}{\mathrm{H}}_2^{-2} \mathrm{O}(\mathrm{g})\)

In the reaction, NH₃ undergoes oxidation because the oxidation number of N-atom increases (-3 to +2), while O₂ undergoes reduction, as is evident from the decrease in the oxidation number of O (0→-2). So, it is a redox reaction.

Question 3. Fluorine reacts with ice and results in the change: H₂O(s) + F₂(g)-+HF(g) + HOF(g); Justify that this reaction is a redox reaction.
Answer: \(\stackrel{+1}{\mathrm{H}}_2 \stackrel{-2}{\mathrm{O}}^{-2}(s)+\stackrel{0}{\mathrm{~F}}{ }_2(g) \rightarrow \stackrel{+1}{\mathrm{H}} \stackrel{-1}{\mathrm{~F}}^{}(g)+\stackrel{+1}{\mathrm{HOF}}^{0-1}(g)\)

Redox Reactions Questions

In the reaction, H₂O undergoes oxidation because the oxidation number,0 increases (-2 to 0 and F₂ undergoes reduction as the oxidation number of F decreases (0 to -1). Hence, it is a redox reaction.

Question 4. Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) & NO-₃. Suggest the structure of these compounds. Count for the
Answer: \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\): If the oxidation number of Cr in \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) be x, then \(\begin{gathered}
2(x)+7(-2)=-2 \\
(\text { For } 0)
\end{gathered}\)

∴ x=6

Therefore, there is no fallacy.

NO-₃: According to a conventional method, the oxidation number of N in NO₃: x+ 3(-2) = -1 or, x = +5 (where x = +3 -1. oxidation number of N in NO₃)

However, according to the chemical bonding method, the structure of NO-3 is if the oxidation number of N is -0=0 the above structure is x, then, x+ (-1) + (-2) + (-2) = 0 (for O-)(for= O)(for-O)

or, x = +5

So, in both conventional and chemical bonding methods. the oxidation number of N in NO-₃ is +5. Therefore, there is no fallacy.

Question 5. Write formulas for the following compounds:

1. Mercury chloride
2. Nickel sulphate
3. Tin 4
4. Oxide
5. Thallium sulphate Iron(3)
6. Sulphate Chromium(3) oxide

Answer: HgCl₂

1. NiSO₄
2. SnO₂
3. TI₂SO₄
4. Fe₂(SO₄)₃
5. Cr₂O₃.

Question 6. Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5
Answer: \(\begin{aligned}
& \text { C: } \stackrel{-4}{\mathrm{C}} \mathrm{H}_4, \stackrel{-3}{\mathrm{C}} \mathrm{H}_3-\stackrel{-3}{\mathrm{C}} \mathrm{H}_3, \stackrel{-2}{\mathrm{C}} \mathrm{H}_3 \mathrm{Cl}, \stackrel{-1}{\mathrm{C}} \mathrm{H} \equiv \stackrel{-1}{\mathrm{C}} \mathrm{H}, \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2 \mathrm{Cl}_2 \text {, } \\
& \stackrel{+1}{\mathrm{C}} 2_2 \mathrm{Cl}_2 \text { or } \stackrel{+1}{\mathrm{C}}_6 \mathrm{Cl}_6, \stackrel{+3}{\mathrm{C}} \mathrm{Cl}_6, \stackrel{+4}{\mathrm{C}} \mathrm{Cl}_4 \\
&
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{N}: \stackrel{-3}{\mathrm{~N}} \mathrm{H}_3, \quad \stackrel{-2}{\mathrm{~N}} \mathrm{H}_2-\stackrel{-2}{\mathrm{~N}} \mathrm{H}_2, \quad \stackrel{-1}{\mathrm{~N}} \mathrm{H}_2 \mathrm{Cl}, \stackrel{\ominus}{\mathrm{N}}, \stackrel{-1}{\mathrm{~N}} \mathrm{O} \mathrm{O}, \stackrel{+2}{\mathrm{NO}} \text {, } \\
& \stackrel{+3}{N}_2 \mathrm{O}_3, \stackrel{4}{N}_2 \mathrm{O}_4, \stackrel{+5}{\mathrm{~N}_2} \mathrm{O}_5 \\
&
\end{aligned}\)

Question 7. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. why?
Answer: A species can act as both oxidant and reductant am one of its constituent atoms has an intermediate value of oxidation number. So, in a reaction the atom can increase or decrease its oxidation number i.e., it can act as an oxidant as well as a; reductant.

In SO₂, the oxidation number of S is +4. The highest and lowest oxidation numbers of S are +6 and -2 respectively. Therefore, the S-atom in SO₂ can increase its oxidation number in a reaction in which SO₂ acts as a reductant and decrease its oxidation number in a reaction in which SO₂ plays the role of an oxidant. Hence, SO₂ can act as an oxidant as well as a reductant.

In H₂O₂, the oxidation number of 0 is —1 1 The highest and lowest oxidation numbers of oxygen are -2 and O respectively. Therefore, the oxygen atom in H₂O₂ is capable of increasing or decreasing its oxidation number. In the reaction in which H₂O₂ acts as an oxidant, the oxidation number of oxygen decreases from -1 to -2 and in the reaction in which it acts as a reductant, the oxidation number of oxygen increases from -1 to 0. Hence, H₂O₂ can act both as an oxidant and a reductant.

In O₃, the oxidation number of oxygen is zero. Oxygen can show two oxidation numbers, -1 and -2. So, the oxidation number of oxygen O₃ can reduce to -1 or -2, but it can never increase. Hence, O₃ can act only as an oxidant.

Redox Reactions Questions

In HNO₃, the oxidation number of nitrogen is +5. It is the maximum oxidation number that nitrogen can exhibit. So, the only opportunity for nitrogen in HNO₃ is to decrease its oxidation number. Hence, HNO₃ can act only as an oxidant.

Question 8. Consider the reactions:

⇒ \(\begin{aligned}
& 6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(a q)+6 \mathrm{O}_2(\mathrm{~g}) \\
& \mathrm{O}_3(g)+\mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+2 \mathrm{O}_2(\mathrm{~g})
\end{aligned}\)

Why it is more appropriate to write these reactions as:

⇒ \(\begin{aligned}
& 6 \mathrm{CO}_2(g)+12 \mathrm{H}_2 \mathrm{O}(l) \\
& \quad \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(a q)+6 \mathrm{H}_2 \mathrm{O}(l)+6 \mathrm{O}_2(g)
\end{aligned}\)

⇒ \(\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(g)+\mathrm{O}_2(g)\)Also, suggest a technique to investigate the path of the above and redox reactions.
Answer: The reaction shown by the equation takes place in the photosynthesis process. From the equation, it may seem that the reaction involves only the consumption of H₂O. However, if we look at the steps that are supposed to be involved in the photosynthesis reaction, it becomes evident that consumption as well as formation of H₂O takes place in the photosynthesis reaction. The proposed steps of the photosynthesis reaction are

Decomposition of H₂O into H₂ and O₂

12H₂O(Z)→12H₂(g) + 6O₂(g)

Formation of C6H1206 and H₂O due to reduction of CO₂(g) by H₂(g) produced in step.

Redox Reactions Questions

6CO₂(g) + 12H₂(g)→C6HI₂O₆(s) + 6H₂O(l) -[2]

Combining equations (1) and (2) gives the complete reaction for the photosynthesis process.

⇒ \(\begin{aligned}
& 6 \mathrm{CO}_2(g)+12 \mathrm{H}_2 \mathrm{O}(l) \rightharpoondown \downarrow \\
& \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l) \cdots[3]
\end{aligned}\)

Thus, equation (3) will be more appropriate for representing the photosynthesis reaction because it gives the actual stoichiometry of the reactants and the products involved in the given reaction.

From the equation, the source of O₂ formed in the reaction is not obvious. One may think O₂ is formed from O₃ or H₂O₂ or both O₃ and H₂O₂. The detailed steps of this reaction as shown below reveal that O₂ is formed from both O₃ and H₂O₂.

⇒\(\begin{aligned}
& \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{O}_2(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \\
& \frac{\mathrm{H}_2 \mathrm{O}_2(l)+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(\mathrm{~g})}{\mathrm{H}_2 \mathrm{O}_2(l)+\mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})} \\
&
\end{aligned}\)

Therefore, it is appropriate to represent the reaction by equation (1).

To investigate the paths of the reaction 1 and 2, we adopt the tracer technique method. In this method, we use H₂OI₈ instead of H₂O for reaction 1 and \(\mathrm{H}_2 \mathrm{O}_2^{18}\) instead of H₂O₂ for reaction 2.

Question 9. The compound AgF₂ is unstable. However, if formed the compound acts as a very strong oxidising agent. Why?
Answer: AgF₂ can be prepared although it is not a stable compound. This is because the oxidation state of Ag in AgF₂ is +2, which is not the stable oxidation state of Ag. The stable oxidation state of Ag is +1. As a result, Ag2+ in AgF₂ quickly reduces to Ag+ by gaining an electron (Ag2+ e→Ag+). This brings about the instability of AgF₂ and makes it a very strong oxidising agent.

Redox Reactions Questions

Question 10. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of a lower oxidation state is formed if the reducing agent is in excess and a compound of a higher oxidation state is formed if the oxidising agent is in excess. Justify this statement by giving three illustrations.
Answer: The given statement can be justified from the examples of j reactions mentioned below.

The reaction of C (reductant) with O₂ (oxidant) may result in CO or CO₂ or a mixture of CO and CO₂. However, if this reaction is initiated with the excess amount of C, the only product that forms is CO. On the other hand, if O₂ is taken in excess in the reaction, only CO₂ is formed. In CO, the oxidation state of C is +2, and in CO₂, it is +4.

Thus, we see that taking an excess amount of reductant leads to the formation of a compound lower oxidation state. Conversely, a compound of a higher oxidation state is formed when the oxidant is taken in excess.

The reaction of P₄ (reductant) with Cl₂ (oxidant) results in PCl₃ when P₄ is taken in excess, while it results in PC1₅ when Cl₂ is taken in excess.

The oxidation state of PC1₃ is +3 and that in PC1₅ is +5. Thus, an excess amount of reductant produces an oil compound lower oxidation state and an excess amount of oxidant produces a compound with a higher oxidation state.

Redox Reactions Questions

The same thing happens when Na (reductant) is reacted with O₂ (oxidant).In the presence of excess Na, the resulting compound is Na₂O, in which the oxidation state of oxygen is -2 in the presence of excess O₂, the resulting compound is Na202, in which the oxidation state of oxygen is -1.

[In this reaction, the mass of Na is 46g and that of oxygen is 64 g]

Question 11. How do you count for the following observations? Although alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet In the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent-smelling gas HC1, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer: The reaction in acid medium:

Redox Reactions Questions

Multiplying equation (1) by 6 and equation (2) by 5 and then adding them together, we have

The reaction in alkaline or neutral medium:

Multiply equation (3) by 2 and then adding it to equation (4), we have

Even though toluene oxidises to benzoic acid in the presence of acidic or alkaline KMnO₄, the manufacture of benzoic acid from toluene is usually carried out by using alcoholic KMnO₄ as an oxidant.

Redox Reactions Questions

This is because ofthe following advantages:

The use of alcoholic KMnO₄ is cost-effective because carrying out the reaction in the presence of it does not require adding either acid or alkali in the reaction medium. In a neutral medium, OH- ions are produced during the reaction.

Both KMnO₄ and toluene are soluble in alcohol and they form a homogeneous mixture. This facilitates the reaction and contributes towards speeding up the reaction.

When concentrated H₂SO₄ is added to an inorganic mixture containing chloride, HC₁, which has a pungent smell, is produced.

Question 13. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions

\(2 \mathrm{AgBr}(s)+\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_2(a q) \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{HBr}(a q)+\mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_2(a q)\) \(\mathrm{HCHO}(l)+2\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_0\right]^{+}(a q)+3 \mathrm{OH}^{-}(a q) \rightarrow 2 \mathbf{A g}(s)+\mathrm{HCOO}^{-}(a q)+4 \mathrm{NH}_3(a q)+2 \mathrm{H}_2 \mathrm{O}(l)\) \(\mathrm{HCHO}(l)+2 \mathrm{Cu}^{2+}(a q)+5 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{Cu}_2 \mathrm{O}(s)+\mathrm{HCOO}^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)\) \(\mathrm{N}_2 \mathrm{H}_4(l)+2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{N}_2(g)+4 \mathrm{H}_2 \mathrm{O}(l)\) \(\mathrm{Pb}(s)+\mathrm{PbO}_2(s)+2 \mathrm{H}_2 \mathrm{SO}_4(a q) \rightarrow 2 \mathrm{PbSO}_4(s)+2 \mathrm{H}_2 \mathrm{O}(l)\)

Question 14. Consider the reaction:

⇒ \(\begin{aligned}
& 2 \mathrm{~S}_2 \mathrm{O}_3^{2-}(a q)+\mathrm{I}_2(s) \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}(a q)+2 \mathrm{I}^{-}(a q) \\
& \mathrm{S}_2 \mathrm{O}_3^{2-}(a q)+2 \mathrm{Br}_2(l)+5 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \downarrow \\
& 2 \mathrm{SO}_4^{2-}(a q)+4 \mathrm{Br}^{-}(a q)+10 \mathrm{H}^{+}(a q)
\end{aligned}\)
Answer: The standard reduction potential for Br2/2Br- system is greater than that for I2/2I- system \(\left(E_{\mathrm{Br}_2 / 2 \mathrm{Br}}^0=1.09 \mathrm{~V}\right.\) and \(E_{1_2 / 21^{-}}^0=0.54 \mathrm{~V}\) indicating Br2 is a stronger oxidising agent than I2. The average oxidation number of S in \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) is +2. and that in \(\mathrm{S}_4 \mathrm{O}_6^{2-}\) is 2.5, while the oxidation number of S in SO₂– is +6. The oxidation number per S-atom changes by 0.5 unit in the reaction \(\mathrm{S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}\) while in the reaction \(\mathrm{S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}\) this change occurs by 4 unit. Being a stronger oxidising agent, Br₂ is capable of increasing the oxidation number of S in \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) to the maximum oxidation number of6, thereby leading to the formation of SO₂– ion. On the other hand, I2, being a weaker oxidising agent, increases the oxidation number of Sin S20|~ to an oxidation number of 2.5 and results in the formation of \(\mathrm{S}_4 \mathrm{O}_6^{2-}\) ion.

Question 15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant
Answer: The standard redox potentials (or standard reduction potentials) ofthe redox couples formed by halogens are—

Are doxcouple consists of a reduced form and an oxidised form. The larger the value of E° for a redox couple, the greater the tendency of its oxidised form to get reduced and the smaller the tendency of its reduced form to get oxidised. The reverse is true when the value of E° for a redox couple is small.

Combining this idea with standard electrode potentials of the redox couples given, we can infer that the tendency of oxidised forms (i.e., F₂, Cl₂, Br₂ and I₂) to get reduced or the strength of oxidising power of the oxidised forms follows the order: F₂ > Cl₂ > Br₂ > I₂, and the tendency of reduced forms {i.e., F-, Cl-, Br- and I-) to get oxidised or the strength of reducing power ofthe reduced forms follows the order:

Redox Reactions Questions

I¯> Br¯> Cl¯ > F¯

As the oxidising power of F₂ is highest among the halogens, it is capable of oxidising other halides to the corresponding halogens. No other halogen except F₂, has this ability

⇒ \(\begin{aligned}
& \mathrm{F}_2(g)+2 \mathrm{Cl}^{-}(a q) \rightarrow 2 \mathrm{~F}^{-}(a q)+\mathrm{Cl}_2(g) \\
& \mathrm{F}_2(g)+2 \mathrm{Br}^{-}(a q) \rightarrow 2 \mathrm{~F}^{-}(a q)+\mathrm{Br}_2(l) \\
& \mathrm{F}_2(g)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{~F}^{-}(a q)+\mathrm{I}_2(s)
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{Cl}_2(g)+2 \mathrm{Br}^{-}(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{Br}_2(l) \\
& \mathrm{Cl}_2(g)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_2(s) \\
& \mathrm{Br}_2(l)+2 \mathrm{I}^{-}(a q) \rightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_2(s)
\end{aligned}\)

Therefore, F₂ has the strongest oxidising power among the halogens. The oxidation of a hydrohalic acid produces its halogen. The tendency of a hydrohalic acid to get oxidised or the reducing power of a hydrohalic acid is high when the halide ion of the hydrohalic acid exhibits a greater tendency to get oxidised. As the tendency ofhalide ions to get oxidised follows the order I¯ > Br¯ > Cl¯ > F-, the reducing power of hydrohalic acids will follow the order HI>HBr>HCl>HF.

This is confirmed by the following reactions: HI or HBr can reduce H₂SO₂ to SO₂, but HC1 or HF cannot reduce it.

⇒ \(\begin{aligned}
& 2 \mathrm{HI}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{I}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \\
& 2 \mathrm{HBr}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Br}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

I- can reduce Cu2+ to Cu+, but Br- cannot.

⇒ \(\begin{aligned}
& 2 \mathrm{Cu}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightarrow \mathrm{Cu}_2 \mathrm{I}_2(s)+\mathrm{I}_2(s) \\
& \mathrm{Cu}^{2+}(a q)+2 \mathrm{Br}^{-}(a q) \rightarrow \text { No reaction }
\end{aligned}\)

Therefore, we can conclude that HI is the strongest reducing agent among the hydrohalic acids.

Question 16. Why does the following reaction occur?

⇒ \(\begin{aligned}
& \mathrm{XeO}_6^{4-}(a q)+2 \mathrm{~F}^{-}(a q)+ 6 \mathrm{H}^{+}(a q) \longrightarrow \\
& \mathrm{XeO}_3(g)+\mathrm{F}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

What conclusion about the compound Na4Xe06 (of which XeO3 is a part) can be drawn from the reaction?
Answer: \(\begin{aligned}
& \stackrel{+8}{\mathrm{XeO}} \mathrm{O}_6^{4-}(a q)+2 \mathrm{~F}^{-1}-(a q)+6 \mathrm{H}^{+}(a q) \\
& \stackrel{+6}{\mathrm{X}} \mathrm{OO}_3(g)+\stackrel{0}{\mathrm{~F}}{ }_2(g)+3 \mathrm{H}_2 \mathrm{O}(l) \\
&
\end{aligned}\)

The oxidation numbers of Xe in XeO4-6 and XeO3 are +8 and +6 respectively. Thus, in the reaction the oxidation number of Xe decreases and hence XeOg- undergoes reduction and acts as an oxidising agent. On the other hand, the oxidation number of F increases from -1 to 0.

Redox Reactions Questions

Therefore, in the reaction fluorine undergoes oxidation and hence it acts as a reductant. As the reaction is spontaneous and XeO- oxidises F-, it can be concluded that Na₄XeOg has stronger oxidising power than F₂.

Question 17. Consider the reactions:

⇒ \(\begin{aligned}
\mathrm{H}_3 \mathrm{PO}_2(a q)+ & 4 \mathrm{AgNO}_3(a q)+2 \mathrm{H}_2 \mathrm{O}(l)- \\
& \mathrm{H}_3 \mathrm{PO}_4(a q)+4 \mathrm{Ag}(s)+4 \mathrm{HNO}_3(a q)
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{H}_3 \mathrm{PO}_2(a q)+2 \mathrm{CuSO}_4(a q)+2 \mathrm{H}_2 \mathrm{O}(l)- \\
& \mathrm{H}_3 \mathrm{PO}_4(a q)+2 \mathrm{Cu}(s)+\mathrm{H}_2 \mathrm{SO}_4(a q)
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}(l)+2\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}(a q)+3 \mathrm{OH}^{-}(a q)- \\
& \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}(a q)+2 \mathrm{Ag}(s)+4 \mathrm{NH}_3(a q)+2 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Answer: The reaction involves the reduction of the Ag⁺ ion to Ag and the oxidation of H₃PO₂ to H₃PO₄. Thus, in this reaction, the Ag+ ion behaves as an oxidant. It oxidises H₃PO₂ to H₃PO₄.

The reaction involves the reduction of Cu²⁺ ion to Cu and the oxidation of H₃PO₂ to H₃PO₄. Thus, in this reaction, the Cu²⁺ ion behaves as an oxidant. It oxidises H₃PO₂ to H3PO₄.

The reaction involves the oxidation of C₆H₅CHO to C₆H₅COOH and the reduction of [Ag(NH₂₃)2]+ to Ag. Thus, in this reaction, the Ag+ ion acts as an oxidant. It oxidizes C₆H₅CHO to C₆H₅COOH.

This reaction indicates that the Cu²⁺ ion is not capable of oxidizing C₆H₅CHO. This explains why the Cu²⁺ ion is a weaker oxidising agent than the Ag+ ion.

Question 20. What sorts of information can you draw from the following reaction?
Answer: Let us first analyze whether (CN)₂ in the reaction gets oxidised or reduced or simultaneously oxidized as well as reduced. To know this the knowledge of the oxidation states of C in (CN)₂, CN and CNO- are required.

The oxidation state of C in (CN)₂ is \(+3\left[(\stackrel{+3-3}{(\mathrm{CN}})_2\right]\) is +2 \(\mathrm{CN}^{-}\left[\mathrm{CN}^{-}\right] \text {and }+4 \text { in } \mathrm{CNO}^{-}\left[\mathrm{C}^{+3-32} \mathrm{CN}^{-}\right]\)

Now let us consider the given equation \((\stackrel{+3}{\mathrm{C}} \mathrm{N})_2(g)+2 \mathrm{OH}^{-}(a q) \xrightarrow[+2]{\mathrm{C}} \mathrm{N}^{-}(a q)+\stackrel{+4}{\mathrm{C}} \mathrm{NO}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

Redox Reactions Questions

From this equation, we see that (CN)₂ reduced to the oxidation number of C reduces (+3→+2) in this change, and It gets oxidised to CNO- because the oxidation number of C Increases (+3→+4) In this change. Thus, from this reaction, we have the following information: CD In an alkaline medium cyanogen gas dissociates into cyanide Ion (ON-) and cyanate Ion (CNO-).

It Is a redox reaction. More particularly, it is a disproportionation reaction because (CN)₂ undergoes oxidation and reduction simultaneously. Cyanogen is a pseudohalogen. It acts as a halogen on reacting with alkalis.

Question 21. The Mn3+ 1 ion Is unstable In solution and undergoes disproportionation to give Mn2+, MnO₂, and H+ ions. Write a balanced ionic equation for the reaction.
Answer: Mn³⁺ is unstable in an aqueous medium and undergoes a disproportionation reaction, forming Mn³⁺, MnO₂ and H⁺. So, the reaction is—

Here the oxidation reaction is—

Redox Reactions Questions

⇒ \(\mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{MnO}_2(s)+4 \mathrm{H}^{+}(a q)+e \quad \cdots[1]\)

Adding equation (1) and equation (2), we have

⇒ \(2 \mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{MnO}_2(s)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}^{+}(a q)\)

This is the balanced equation for the disproportionation reaction that the Mn³⁺ ion undergoes in an aqueous medium.

Question 22. Consider the elements: Cs, Ne, I and F

• Identify the element that exhibits only a negative oxidation state.
• Identify the element that exhibits only a positive oxidation state.
• Identify the element that exhibits both positive and negative oxidation states.
• Identify the element which exhibits neither the negative nor the positive oxidation state

Answer: Being an element of the highest electronegativity, F always shows a negative oxidation state. It exhibits only a -1 oxidation state

Cs is an alkali metal and shows a strong electropositive character. As a result, it always shows a positive oxidation state. It exhibits only a +1 oxidation state.

Like other halogens, I also show a -1 oxidation state. In addition, it shows positive oxidation states, +1, +3, +5 and +7 when it forms compounds with more electronegative elements.

Ne is an inert element and does not tend to gain or lose electrons. As a result, it shows neither a positive oxidation state nor a negative oxidation state.

Question 23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating it with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer: The reaction that occurs when S02 is used to remove excess Cl2 in drinking water is—

⇒ \(\mathrm{Cl}_2(a q)+\mathrm{SO}_2(a q) \rightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{SO}_4^{2-}(a q)\)

Oxidation reaction: \(\stackrel{+4}{\mathrm{SO}_2}(a q) \xrightarrow{+6} \mathrm{SO}_4^{2-}(a q)\)

To balance O-atoms, we add 2H₂O to the left-hand side and 4H+ to the right-hand side.

⇒ \(\mathrm{SO}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{SO}_4^{2-}(a q)+4 \mathrm{H}^{+}(a q)\)

Balancing charges on both sides, we have

Redox Reactions Questions

Reduction reaction: Cl₂(aq) + 2e→2Cl-(aq)….2

Adding equation (1) to equation (2), we have

⇒ \(\begin{aligned}
\mathrm{Cl}_2(a q)+\mathrm{SO}_2(a q)+ & 2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \\
& \mathrm{SO}_4^{2-}(a q)+2 \mathrm{Cl}^{-}(a q)+4 \mathrm{H}^{+}(a q)
\end{aligned}\)

This is the balanced equation for the reaction.

Question 24. Refer to the periodic table given in your book and now answer the following questions:

Select the possible non-metals that can show a disproportionation reaction. Select three metals that can show a disproportionation reaction.
Answer: Non-metals such as P4(s), Cl₂(g) and Br₂(Z) undergo disproportionation reaction.

⇒ \(\begin{gathered}
\mathrm{P}_4(s)+3 \mathrm{NaOH}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \downarrow \\
\mathrm{PH}_3(g)+3 \mathrm{NaH}_2 \mathrm{PO}_2(a q)
\end{gathered}\)

⇒ \(\begin{gathered}
\mathrm{Cl}_2(g)+6 \mathrm{NaOH}(a q)(\mathrm{Hot}) \longrightarrow \\
5 \mathrm{NaCl}(a q)+\mathrm{NaClO}_3(a q)+3 \mathrm{H}_2 \mathrm{O}(l)
\end{gathered}\)

⇒ \(\begin{aligned}
\mathrm{Br}_2(l)+6 \mathrm{NaOH}(a q)(\mathrm{Hot}) \\
5 \mathrm{NaBr}(a q)+\mathrm{NaBrO}_3(a q)+3 \mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

Three metals that can show disproportionation reactions are copper, gallium and manganese.

⇒ \(\begin{aligned}
2 \stackrel{+1}{\mathrm{C}} u^{+}(a q) & \rightarrow \stackrel{0}{\mathrm{Cu}}(s)+\stackrel{+2}{\mathrm{Cu}^{2+}}(a q) \\
3 \mathrm{Ha}^{+}(a q) & \rightarrow \stackrel{+3}{\mathrm{Ga}}^{3+}(a q)+2 \stackrel{\ominus}{\mathrm{Ga}}(s)
\end{aligned}\)

Redox Reactions Questions

⇒ \(\begin{aligned}
& 2 \mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \\
& \mathrm{MnO}_2(s)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}^{+}(a q)
\end{aligned}\)

Question 25. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen?
Answer: Reaction: \(\begin{gathered}
4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
4 \times 17 \mathrm{~g}=68 \mathrm{~g} \quad 5 \times 32 \mathrm{~g}=160 \mathrm{~g} 4 \times 30 \mathrm{~g}=120 \mathrm{~g}
\end{gathered}\)

Therefore, 160 g O₂ is required to oxidise 68 g NH₃.

20g O₂ is required to oxidise \(\frac{68}{160} \times 20 \mathrm{~g}\)

= 8.5 g of NH₃

So, here O₂ is the limiting reagent. The amount of nitric oxide produced depends upon the amount of oxygen taken and not on the amount of NH3 taken

According to the above equation,

160 g O₂ produces 120 g NO

⇒ \(20 \mathrm{~g} \mathrm{O}_2 \text { produces } \frac{120}{160} \times 20 \mathrm{~g} \mathrm{NO}=15 \mathrm{~g} \mathrm{NO}\)

Thus the reaction between 10 g NH₃ and 20gO₂ produces a maximum amount of 15 g NO.

Question 26. Identify whether the following reactions are redox reactions or not:

\(\begin{array}{r}
\mathrm{Pb}\left(\mathrm{NO}_3\right)_2(a q)+\mathrm{K}_2 \mathrm{CrO}_4(a q) \longrightarrow \\
\mathrm{PbCrO}_4(s)+2 \mathrm{KNO}_3(a q)
\end{array}\)

NH₄NO₂(s)→N₂(g) + 2H₂O(g)

Redox Reactions Questions

\(\begin{aligned}
& \mathrm{CaCO}_3(s)+2 \mathrm{HCl}(a q) \longrightarrow \\
& \mathrm{CaCl}_2(a q)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

2HgO(s)→2Hg(l) + O₂(g)

2HNO₃(aq) + P₂O₅(s)→N₂O₅(g) + 2HPO₃(a<7)

3CuSO₄(aq) + 2PH₃(g)→Cu₃P₂(s) + 3H2S04(aq)

Answer: Redox Reaction

Question 27. Identify the oxidant and reductant in the following reactions:

N₂H₄ (l)+ ClO₂(aq)→NO(g) + Cl-(aq)

\(\begin{aligned}
\mathrm{Cl}_2(g)+2 \mathrm{NaOH}(a q) \\
\mathrm{NaCl}(a q)+\mathrm{NaOCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

Cu₂S(s) + 2Cu₂O(s)→6Cu(s) + SO₂(g)

2HgCl₂(aq) + SnCl2(aq)→Hg₂Cl₂(aq) + SnCl₄(aq)

HOCl(aq) + H₂S(aq)→S(s) + H₃O+(a<7) + Cl-(aq)

Redox Reactions Questions

Answer: Oxidant

1. CIO₃
2. Cl₂
3. Cu₂O
4. CI₂
5. HOCl
6. K₂MnO₄
7. KIO₃

Reductant:

1. N₂H₄
2. Cl₂
3. Cu₂S
4. SnCl₂
5. H₂S
6. K₂MnO₄
7. K₁

Question 28. Which are oxidized in the following reactions? Give reasons.

• 2Na + H₂→2NaH
• H₂O₂ + Q₃ H₂O + O₂

Answer: Na H₂O₂

Question 29. In which of the following two reactions does HNPO₃ not act as an oxidizing agent? Explain 
Answer: 2

Question 30. Which one of the following two reactions is a redox reaction?
Answer: 2

Redox Reactions Questions

Question 31. Arrange the following compounds in increasing order of the oxidation number of S. Na₂S₄O₆, H₂S₂O₇, H₂SO₃, Na₂S₂O₃
Answer: Na₂S₂O₃ < Na₂S₄Og < H₂SO₃ < H₂S₂O₇

Question 32. Arrange the following compounds in increasing order of the oxidation number of N. Mg₃N₂, NH₂OH, (N₂H₅)₂SO₄, [CO(NH₃)5CI]CI₂
Answer: \(\mathrm{Mg}_3 \mathrm{~N}_2<\left(\mathrm{N}_2 \mathrm{H}_5\right)_2 \mathrm{SO}_4<\mathrm{NH}_2 \mathrm{OH}<\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 ;\) 

Question 33. What are the values of a and b in the given redox reaction?
\(a \mathrm{KMnO}_4+\mathrm{NH}_3 \rightarrow b \mathrm{KNO}_3+\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{H}_2 \mathrm{O}\)
Answer: a=8,b=3

Question 34. Write the half-reactions ofthe given redox reaction \(\mathrm{UO}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{UO}_2^{2+}+\mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}\)
Answer: \(\mathrm{UO}^{2+}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{UO}_2^{2+}+2 \mathrm{H}^{+}+2 e\) \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

Question 35. How many moles of electrons will be required to reduce 1 mol N03 ion to hydrazine in the acidic medium?
Answer: 7 mol

Question 36. In an alkaline medium, C1O₂ oxidizes H₂O₂ to O₂ and itself reduces to Cl- ion. How many moles of H₂O₂ will be oxidized by 1 mol of C₁O₂?
Answer: 2.5 mol

Redox Reactions Questions

Question 37. In the basic medium, KNO₂ is oxidized by KMnO₄, forming KNO₃. How many moles of KMnO₄ are required to oxidize 1 mol of KNO₂?
Answer: 2/3 mol

Question 38. Calculate— the number of mol of KMnO₄ required to oxidize 1 mol of ferrous oxalate in an acidic medium, The number of mol of K₂Cr₂O₇ required to oxidize 1 mol of ferrous oxalate in an acidic medium
Answer: 0.6 mol

Question 39. Calculate the equivalent mass of the underlined Compounds.
Answer: 183.75 16.5

Question 40. Iodine reacts with sodium sulfate a neutral medium. Write the balanced equation of this reaction. Calculate the equivalent mass of sodium thiosulfate in this reaction. (Assume molecular mass of sodium thiosulfate =M).
Answer: \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\) Equivalent mass = M.

Redox Reactions Questions

Fill In The Blanks

Question 1. Oxidation number of N in Mg₃N₂ is_________________.
Answer: -3

Question 2. An element can exist in +1, 0, +5 oxidation states. In a compound, if the element exists in a state, then the compound can participate in a disproportionation reaction.
Answer: +1

Question 3. The oxidation number of S in (CH₃)₂SO is_________________.
Answer: +4

Question 4. Number of electron(s) involved in the reaction, IO¯₃ →I‾ in basic medium is_________________.
Answer: 6

Question 5. The equivalent mass of SO₂ in the reaction, SO₂ →\(\mathrm{SO}_4^{2-}\) in acidic medium is_________________.
Answer: 32

Question 6. \(\mathrm{CrO}_4^{2-}+x \mathrm{H}_2 \mathrm{O}+y e^{-} \rightarrow\left[\mathrm{Cr}(\mathrm{OH})_x\right]^{-}+x \mathrm{OH}^{-}\), where y =_________________.
Answer: 3

Question 7. The oxidation number of Mo in [Mo₂O₄-(C₂H₄)2(H₂O₂)]²⁺ is_________________.
Answer: +5

Redox Reactions Multiple Choice Questions

Question 1. Oxidation numbers of S in peroxonosulphric and peroxonosulphric acids respectively are-

1. +3 and +3
2. +4 and +6
3. +6 and +6
4. +8 and +7

Answer: 2. +4 and +6

Question 2. The oxidation number of pyrophosphoriqaeid is-

1. +1
2. +3
3. +4
4. +5

Answer: 4. +5

Redox Reactions Questions

Question 3. When SO₂ gas is passed through an acidic solution of K₂Gr₂O₇, the oxidation number of S changes by

1. 2 unit
2. 3 unit
3. 4 unit
4. 6 unit

Answer: 1. 2 unit

Question 4. When manganous salt is fused with KNO₃ and solid NaOH, the oxidation number changes from—

1. +2 to +3
2. +2 to +4
3. +2 to +6
4. +2 to +7

Answer: 3. +2 to +6

Question 5. Which ofthe following reactions is not a redox reaction—

1. 2CuSO₄ + 4KI→ CU₂I₂ + 2K₂SO₄ + I₂
2. SO₂ + H₂O→H₂SO₃
3. CUSO₄ + 4NH₃→ [CU(NH₃)4]SO₄
4. 4KC1O₃ →3KC1O₄ + KC1

Answer: 3. CUSO₄ + 4NH₃→ [CU(NH₃)4]SO₄

Question 6. Which of the following reactions does the reaction, Ag²⁺ + Ag→Ag⁺, belong to—

1. Reduction
2. Oxidation
3. Comproportionation
4. Disproportionation

Answer: 3. Comproportionation

Redox Reactions Questions

Question 7. In the following reaction, the oxidation half-reaction gives—

⇒ \(\begin{aligned}
2 \mathrm{KMnO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \\
\mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

1. MnSO₄
2. CO₂
3. K₂SO₄
4. H₂O

Answer: 2. CO₂

Question 8. The amount of electrons required to reduce 1 mol of nitrate ions to hydrazine is-

1. 7 mol
2. 6 mol
3. 5 mol
4. 4 mol

Answer: 1. 7 mol

Question 9. The reaction of C1O₂ with H₂O₂ in an alkaline medium results in O2 and Cl- ions. In this reaction, C1O₂ acts as an oxidant. The number of mol of H₂O₂ oxidised by1 mol of C1O₂ is-

1. 1.0
2. 1.2
3. 2.5
4. 2.8

Answer: 3. 2.5

Question 10. In the balanced equation of the reaction, \(\mathrm{Zn}+\mathrm{NO}_3^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{ZnO}_2^{2-}+\mathrm{NH}_3\), the coefficients of Zn, NO3 and OH- respectively are—

1. 1,4 and 8
2. 8,3 and 2
3. 4,1 and 7
4. 5,2 and 8

Answer: 3. 4,1 and 7

Redox Reactions Questions

Question 11. The amount of iodine that liberates in the reaction of 0.1 mol of K₂Cr₂O? with an excess of K₃ in an acidic solution is

1. 0.1 mol
2. 0.2 mol
3. 0.3 mol
4. 0.4 mol

Answer: 3. 0.3 mol

Question 12. In a strong alkaline solution, the equivalent mass of KMnO₄ (molecular mass =M) is—

1. M/5
2. M/2
3. M/2
4. M

Answer: 4. M

Question 13. In the balanced equation for the reaction,
\(a \mathrm{KMnO}_4+b \mathrm{NH}_3 \rightarrow \mathrm{KNO}_3+\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{H}_2 \mathrm{O},\) the
values of a and b respectively are-

1. 3 and 7
2. 8 and 3
3. 5 and 2
4. 6 and 8

Answer: 2. 8 and 3

Redox Reactions Questions

Question 14. In the reaction of KMnO₄ with ferrous ions in an acidic medium, KMnO₄ oxidizes ferrous ions to ferric ions and itself gets reduced to manganous salt. The number of ferrous ions oxidized by 100 mL of 0.2(N) KMnO₄ solution is—

1. 1.117 g
2. 1.562 g
3. 2.173 g
4. 1.934 g

Answer: 1. 1.117 g

Question 15. The oxidation number of B in NaBH₄ is —

1. -3
2. +3
3. +2
4. -4

Answer: 2. +3

Question 16. The equivalent mass of the oxidant in the reaction, 3C1₂ + 6NaOH- 5NaCl + NaC1O₃ + 3H₂O is

1. 71
2. 14.2
3. 7.1
4. 35.5

Answer: 3. 7.1

Redox Reactions Questions

Question 17. \(\mathrm{Cr}(\mathrm{OH})_3+\mathrm{IO}_3^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{CrO}_4^{2-}+\mathrm{H}_2 \mathrm{O}+\mathrm{I}_2\) In the
balanced equation of this reaction, the coefficient of H₂O is-

1. 2
2. 3
3. 4
4. 5

Answer: 4. 5

Question 18. In the balanced equation for the reaction-

⇒ \(\begin{aligned}
& \mathrm{As}_2 \mathrm{~S}_3+a \mathrm{ClO}_3^{-}+b \mathrm{OH}^{-} \\
& x \mathrm{AsO}_4^{3-}+y \mathrm{ClO}^{-}+z \mathrm{SO}_4^{2-}+6 \mathrm{H}_2 \mathrm{O} \\
&
\end{aligned}\)

1. x+y+z=a
2. a+x+z=b
3. a-x-z=y
4. b-a=y-z

Answer: 2. a+x+z=b

Question 19. In the reaction, Fe3O₄ + KMnO₄ → Fe₂O₃ +MnO₂, the equivalent mass of Fe₃O₄ is—

1. 116
2. 232
3. 773
4. 154.6

Answer: 2. 232

Redox Reactions Questions

Question 20. Which of the following requires an oxidant—

1. Cu²⁺→ Cu
2. Cu₃P₂→2PH₃
3. \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}\)
4. \(\mathrm{SO}_3 \rightarrow \mathrm{SO}_4^{2-}\)

Answer: 3. \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}\)

Question 21. In the presence of HCl(aq), K₂Cr₂O₇ oxidizes tin (Sn) into Sn⁴⁺ ions. The amount of that will be oxidisedby1 mol K₂Cr₂O₇ is —

1. 1.0 mol
2. 1.5 mol
3. 2.0 mol
4. 2.5 mol

Answer: 2. 1.5 mol

Question 22. The amount of Na₂S₂O₃ to be required for reducing iodine produced by the reaction of mol ofKI with H₂O₂ in an acid medium is—

1. 0.5 mol
2. 1 mol
3. 2 mol
4. 2.5 mol

Answer: 3. 2 mol

Redox Reactions Questions

Question 23. The ratio of equivalent masses of KMnO₄ in acidic, strong alkaline, and neutral solutions is—

1. 3:5:15
2. 3:15:5
3. 5:5:3
4. 3:3:5

Answer: 2. 3:15:5

Question 24. The amount of H₂O₂ required for decolorizing 1 mol of KMnO₄ in an acid solution is—

1. 1.5 mol
2. 2.0 mol
3. 2.5 mol
4. 3.0 mol

Answer: 3. 2.5 mol

Redox Reactions Questions

Question 25. Fe has the lowest oxidation state in—

1. FeSO₄(NH₄)2S04-6₂O
2. K4[Fe(CN)g]
3. Fe(CO)₅
4. Fe0.9₄O

Answer: 3. Fe(CO)₅

Question 26. A compound of Xe and F is found to have 53.5% Xe. What is the oxidation number of Xein in this compound—

1. -4
2. 0
3. +4
4. +6

Answer: 4. +6

Redox Reactions Questions

Question 27. Disproportionation reaction is not possible for—

1. ASH₃
2. SF₄
3. H₅10₆
4. PC1₃

Answer: 3. PC1₃

Question 28. When lmol of KC.103 accepts 4 t nol of electrons, the expected product is—

1. C10₂
2. or
3. C10₄
4. cr

Answer: 3. cr

Redox Reactions Questions

Question 29. \(\mathrm{M}^{x+}+\mathrm{MnO}_4^{-} \rightarrow \mathrm{MO}_3^{-}+\mathrm{Mn}^{2+}+\frac{1}{2} \mathrm{O}_2\) If 1 mol of Mn04 oxidises 2.5 mol of M-v+, then the value of X is —

1. 5
2. 3
3. 4
4. 1

Answer: 4. 1

Question 30. During the reaction between KC1O₃ and (COOH)₂ in an acidic medium, the tire element that undergoes a maximum change in the oxidation number is—

1. K
2. O
3. Cl
4. C

Answer: 3. Cl

Redox Reactions Questions

Question 31. If the oxidation numbers of Cr in CrOg, K₂CrO₄  K,Cr₂O- and [Cr(NH₃)4Cl₂]Cl are +a, +b, +c and +d respectively, then—

1. a>c>b>d
2. a=x>b>d
3. a=b>c>d
4. a=b=c>d

Answer: 2. a=x>b>d

Question 32. The oxidation number of Pt in [Pt(C₂H₄)Cl₃]- is—

1. +3
2. +4
3. +2
4. O

Answer: 3. +2

Question 34. For the reaction: \(\mathrm{H}_2 \mathrm{O}_2+x \mathrm{ClO}_2 \rightarrow x \mathrm{Cl}^{-}+y \mathrm{O}_2+\mathrm{H}_2 \mathrm{O}\) the value of y/x is—

1. 2.0
2. 2.5
3. 1.0
4. 1.5

Answer: 2. 2.5

Redox Reactions Questions

Question 35. The oxidation number of Ba(H₂PO₂)₂ is—

1. +3
2. +2
3. +1
4. -1

Answer: 3. +1

Question 36. \(a \mathrm{Mn}^{2+}+b \mathrm{BiO}_3^{-}+c \mathrm{H}^{+}=\mathrm{I}^{-} \mathrm{MnO}_4^{-}+b \mathrm{Bi}^{3+}+d \mathrm{H}_2 \mathrm{O}\)

1. a=3
2. b=5
3. c=10
4. d=6

Answer: 2. b=5

Redox Reactions Questions

Question 37. The mixture of NaOH solution and white P on heating produces PH₃ gas and Na₂H₂PO₂. The above reaction is an example of

1. Oxidation reaction
2. Reduction reaction
3. Comproportionation reaction
4. Disproportionation reaction

Answer: 4. Disproportionation reaction

Question 38. For the reaction:

⇒ \(\begin{aligned}
& \mathrm{Zn}(\mathrm{s})+\mathrm{HNO}_3(a q) \longrightarrow \\
& \quad \mathrm{Zn}\left(\mathrm{NO}_3\right)_2(a q)+\mathrm{NH}_4 \mathrm{NO}_3(a q)+\mathrm{H}_2 \mathrm{O}(l)
\end{aligned}\)

Redox Reactions Questions

The change in oxidation number per mole HNO₃ is—

1. Increases by 6 unit
2. Decreases by 4 unit
3. Decreases by 8 unit
4. Decreases by 6 unit

Answer: 3. Decreases by 8 unit

Question 39. To balance the chemical equation \(\mathrm{Cl}_2 \mathrm{O}_7(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(l)+x e \rightarrow \mathrm{ClO}_2^{-}(a q)+\mathrm{OH}^{-}(a q)\) the value of x should be-

1. 8
2. 6
3. 5
4. 4

Answer: 2. 6

Redox Reactions Questions

Question 40. Find the equivalent mass of Na₂S₂O₃ for the reaction,

⇒ \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3(a q)+\mathrm{I}_2(\mathrm{~s}) \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6(a q)+\mathrm{NaI}(a q)\)

[Assume that the molecular mass of Na₂S₂O₃ is M]—

1. M/8
2. M
3. M/2
4. M/4

Answer: 2. M

Question 41. Which of die following substances undergo disproportionation reactions in basic medium—

1. F₂
2. P₄
3. S₂
4. Br₂

Answer: 2. P₄

Redox Reactions Questions

Question 42. In which of the following compounds, the oxidation number of oxygen is fractional—

1. B₄O₁₀
2. B₂H₆
3. CSO₂
4. KO₃

Answer: 1. B₄O₁₀

Question 43. Wlien Cl₂ is passed through NaOH in the cold, the oxidation number of changes from—

1. 0 to -1
2. 0 to +2
3. 0 to -2
4. 0 to -1

Answer: 1. 0 to -1

Redox Reactions Questions

Question 44. In which of the following cases equivalent mass of a reductant is equal to its molecular mass—

1. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\)
3. \(2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{NaI}\)
4. \(\mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}\)

Answer: 2. \(\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\)

Question 45. Identify the redox reactions—

1. 2CuSO₄ + 4KI→2CuI + 12 + 2K₂SO₄
2. BaCl₂ + Na₂SO₄→BaSO₄ + 2NaGl
3. 3I₂ + 6NaOH→NaIO3 + 5Nal + 3H₂O
4. CuSO₄ + 4NH₃→[Cu(NH₃)₄]SO₄

Answer: 1. 2CuSO₄ + 4KI→2CuI + 12 + 2K₂SO₄

Redox Reactions Questions

Question 46. When ammonium nitrate is heated, the oxidation numbers of the N-atoms present it change from—

1. -3 to +1
2. -3 to 0
3. -2 to +4
4. +5 to 0

Answer: 2. -3 to 0

Question 47. For the reaction, \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}-\)

1. \(\mathrm{S}_2 \mathrm{O}_3^{2-} \text { gets oxidised to } \mathrm{S}_4 \mathrm{O}_6^{2-}\)
2. \(\mathrm{S}_2 \mathrm{O}_3^{2-} \text { gets reduced to } \mathrm{S}_4 \mathrm{O}_6^{2-}\)
3. I₂ gets oxidized to I¯
4. I₂ gets reduced to I¯

Answer: 1. \(\mathrm{S}_2 \mathrm{O}_3^{2-} \text { gets oxidised to } \mathrm{S}_4 \mathrm{O}_6^{2-}\)

Redox Reactions Questions

Question 48. Which of the following statements about the following reaction is wrong—

1. 2Cu₂O + cu₂s→6cu + SO₂
2. Both cu₂O and cu₂s are reduced
3. Only cu₂s is reduced
4. Cu₂s is the oxidation
5. Only cu₂O is reduced

Answer: 2. Both cu₂O and cu₂s are reduced

Question 49. Which of the following orders represents the correct descending order of oxidation numbers—

1. HNO₂ > NO > NH₄C1 > N₂
2. HNO₃ > NO > N₂ > NH₂C1
3. H₂S₂O₇ > Na₂S₂O₃ > Na₂S4O₆ > S₈
4. H₂SO₅ > H₂SO₃ > SC1₂ > H₂S

Answer: 2. HNO₃ > NO > N₂ > NH₄C1

Redox Reactions Questions

Question 50. Which ofthe following reactions are not reactions—

1. SO₂(g) +H₂O(f) H₂SO₃(aq)
2. Ca(s) + H₂(g) → CaH₂(s)
3. 2H₂S(aq) + SO₂(g)→2H20(l) + 3S(s)
4. \(\begin{aligned}
   2 \mathrm{PCl}_5(g)+ & \mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow \\
   & 2 \mathrm{POCl}_3(a q)+2 \mathrm{HCl}(a q)+\mathrm{SO}_2 \mathrm{Cl}_2(g)
   \end{aligned}\)

Answer: 1. SO₂(g) +H₂O(f) H₂SO₃(aq)

Question 51. In which compounds does Cr exist +6 oxidation state

1. CrO₂Cl₂
2. Na₂[Cr(CN)₆]
3. CrO₅
4. K₂Cr₂O₇

Answer: 1. CrO₂Cl₂

Redox Reactions Questions

Question 52. When ammonium nitrite (NH₂NO₂) is heated—

1. Oxidation of nitrogen takes place
2. Reduction of nitrogen takes place
3. The overall reaction is a disproportionation reaction
4. The overall reaction is a double decomposition reaction

Answer: 2. Reduction of nitrogen takes place

Question 53. In which compounds does an atom exist in two different oxidation states

1. H₂SO₅
2. NH₄NO₃
3. Fe₂O₃
4. H₂O₂

Answer: 1. H₂SO₅

Redox Reactions Questions

Question 54. In the balanced equation for the reaction—

⇒ \(\mathrm{H}_2 \mathrm{SO}_4+x \mathrm{HI} \rightarrow \mathrm{H}_2 \mathrm{~S}+y \mathrm{I}_2+z \mathrm{H}_2 \mathrm{O}\)

1. x=y-z
2. y=z
3. x=2y
4. z=2x

Answer: 2. y=z

Question 55. In the reaction,

⇒ \(\mathrm{KMnO}_4+\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{MnO}_2+\mathrm{SO}_4^{2-}+\mathrm{OH}^{-}\)

(Assume formula masses of KMnO₂ and Na₂S₂O₃ M₁ and M₂ respectively)—

1. The equivalent mass of KMnO₂ = M₁/3
2. The equivalent mass of Na₂S₂O₃ = M₂
3. The equivalent mass of KMnO₄ = M₁/5
4. The equivalent mass of Na₂S₂O₂ = M₂/8

Answer: 1. The equivalent mass of KMnO₄ = M1/3

Redox Reactions Questions

Question 56. In the balanced equation for the reaction,

⇒ \(\mathrm{UO}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{UO}_2^{2+}+\mathrm{Cr}^{3+}+\mathrm{H}_2 \mathrm{O}\) the coefficient of-

1. UO²⁺
2. \(\mathrm{UO}_2^{2+} \text { is } 3\)
3. \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} \text { is } 1\)
4. H₂O is 7

Answer: 2. \(\mathrm{UO}_2^{2+} \text { is } 3\)

Redox Reactions Questions

Question 57. The disproportionation of 1 mol of \(\mathrm{MnO}_4^{2-}\) ions in a neutral aqueous solution results in—

1. 1/3 mol of MnO¯₄
2. 2/3 mol of MnO₂
3. 2/3 mol of MnO₄
4. 1/3 mol of MnO₂

Answer: 3. 2/3 mol of MnO₄

Question 58. In the reaction the oxidation number of marked with (*)-

1. Increases by 2 units
2. Increases by 1 unit
3. Decreases by 2 units
4. Decreases by 3 units

Answer: 1. Increases by 2 units

Question 59. For the reaction: SO₂ + 2H₂S→3S + 2H₂O —

1. The equivalent mass of the oxidant is 64
2. Equivalent mass ofoxidantis 16
3. The number of electrons accepted oxidant is 4
4. The number of electrons lost by reductant is 6

Answer: 2. Equivalent mass of oxidant is 16

Redox Reactions Questions

Question 60. The species that cannot be reducing agents are—

1. SO₃
2. \(\mathrm{SO}_3^{2-}\)
3. H₂SO₄
4. S^2-

Answer: 1. SO₃

Question 61. Which are conserved all redox reactions—

1. Charge
2. Mass
3. Either charger or Mass
4. Neither charge nor mass

Answer: 1. Charge

Question 62. The equivalent weight of K₂Cr₂O₇ in an acidic medium is expressed in terms of its molecular weight (M) as—

1. \(\frac{M}{3}\)
2. \(\frac{M}{4}\)
3. \(\frac{M}{6}\)
4. \(\frac{M}{7}\)

Answer: 3. \(\frac{M}{6}\)

In an acidic medium, K₂Cr₂O₇ undergoes reduction, forming a Cr³⁺ ion.

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

In this reaction, the equivalent weight of K₂Cr₂O₇

⇒ \(=\frac{\text { Molecular weight of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7}{\begin{array}{c}
\text { No of electrons gained by a molecule of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \\
\text { in its reduction }
\end{array}}=\frac{M}{6}\)

Redox Reactions Questions

Question 63. If Cl₂ is passed through hot aqueous NaOH, the products formed have Cl in different oxidation states. These are indicated as

1. -1 and +1
2. -1 and +5
3. -1 and +5
4. -1 and +3

Answer: 2. -1 and +5

Reaction: Cl₂ + 6NaOH→NaCl + 5NaC1O₃ + 3H₂O The oxidation number of Cl in NaCl is -1 and that in NaCIO₃ is +5.

Question 64. In an aqueous alkaline solution, two-electron reductions of HO‾₂ give—

1. HO‾
2. H₂O
3. O₂
4. O‾₂

Answer: 1. HO‾

Redox Reactions Questions

Question 65. Consider the following reactions

⇒ \(x \mathrm{MnO}_4^{-}+y \mathrm{C}_2 \mathrm{O}_4^{2-}+z \mathrm{H}^{+} \rightarrow x \mathrm{Mn}^{2+}+2 y \mathrm{CO}_2+\frac{z}{2} \mathrm{H}_2 \mathrm{O}\)

The values of x, y, and z in the reaction are respectively

1. 5,2 And 8
2. 5,2 and 16
3. 2,5 and 8
4. 2,5 and 16

Answer: 4. 2,5 and 16

⇒ \(\begin{aligned}
& {\left[\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2} \\
& \frac{\left[\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2+2 e\right] \times 5}{2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}}
\end{aligned}\)

‍∴ x=2, y=5 and z=16

Question 66. In which of the following reactions, H₂O₂ acts as a reducing agent-

1. H₂O₂ + 2H+ + 2e→2H₂O
2. H₂O₂-2e→ O₂ + 2H⁺
3. H₂O₂ + 2e→2OH¯
4. H₂O₂ + 2OH¯-2e→O₂ + 2H₂O

Choose the correct option 

1. 2,4
2. 1,2
3. 3,4
4. 1,3

Answer: 1. 2,4

In the reaction, H₂O₂ → O₂ + 2H+ + 2e, electrons are lost by H₂O₂ and hence H₂O₂ acts as a reductant. In the reaction, H₂O₂+2OH¯→ O₂ + 2H₂O + 2e, electrons are lost by H₂O₂ and hence H₂O₂ acts as a reductant.

Redox Reactions Questions

Question 67. The Pair in which phosphorus atoms have a formal oxidation state of +3 is-

1. Orthophosphoric and Pyrophosphoric acid
2. Pyrophosphorus and Hypophosphoric acid
3. Orthophosphoric and Hypophosphoric acid
4. Pyrophosphorus and Pyrophosphoric acid

Answer: 1. Orthophosphorus and pyrophosphoric acid

Orthophosphoric acid: \(\begin{aligned}
& +1+3-2 \\
& \mathrm{H}_3 \mathrm{PO}_3
\end{aligned}\)

Pyrophosphorus acid:

Let the oxidation number of P in pyrophosphoric acid be x.

So, 4(+1) + 2x + 5(-2) = 0

or, 2x = 6 or, x = +3

Question 68. Which of the following reactions is an example of a redox creation:

1. XeFG + H₂O→XeOF4 + 2HF
2. XeF₆ + 2H₂O→XeO₂F₂ + 4HF
3. XeF₄ +O₂F₄→XeF₆ + O₂
4. XeF₂ + PF₅→[XeF]+[PF₆]-

Answer: 3. XeF₄ + O₂F₄→XeF₆ + O₂

⇒ \(\stackrel{+6-1}{\mathrm{XeF}_6}+\mathrm{H}_2 \stackrel{-2}{\mathrm{O}} \rightarrow \mathrm{XeOF}_4^{-2-1}+2 \mathrm{HF}\)

⇒ \(\stackrel{+6}{\mathrm{XeF}_6-1}+2 \mathrm{H}_2 \stackrel{-2}{\mathrm{O}} \xrightarrow[\rightarrow]{+6} \mathrm{XeO}_2^{-2-1} \mathrm{~F}_2+4 \mathrm{HF}\)

⇒ \(\stackrel{+2}{\mathrm{XeF}_2^{-1}}+\stackrel{+5-1}{\mathrm{PF}_2^{-1}} \rightarrow\left[\stackrel{+2}{\mathrm{XeF}} \mathrm{F}^{-1}\right]\left[\mathrm{PF}_6^{-1}\right]\)

For these reactions, there is no change in the oxidation number of the respective elements. So these reactions are not redox reactions.

⇒ \(\stackrel{+4}{\mathrm{XeF}_4}+\stackrel{+4}{\mathrm{O}_2} \mathrm{~F}_4^{-1} \rightarrow \stackrel{+6}{\mathrm{X}} \mathrm{XeF}_6-\stackrel{0}{\mathrm{O}}_2\)

Redox Reactions Questions

Question 69. The oxidation states of Cr in [Cr(H₂O) ]C1₃ [Cr(C₆HG)₂] and K2[Cr(CN₂)(0)₂(O₂)(NH₃)] respectively are—

1. +3,+4 and +6
2. +3,+2 and +4
3. +3,0 and +6
4. +3,0 and +4

Answer: 3. +3,0 and +6

⇒ \(\left[\stackrel{+3}{\mathrm{Cr}}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3,\left[\stackrel{0}{\mathrm{Cr}}\left(\mathrm{C}_6 \mathrm{H}_6\right)_2\right]\)

In K₂[Cr(CN)₂(O₂)(0)₂(NH₃)] compound, let, the oxidation number of Cr be x.

or, 2 + X-2-4-2 + 0

or, x = +6

Question 70. A mixture of potassium, Oxalic acid, and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number-

1. S
2. H
3. Cl
4. C

Answer: 3. Cl

Question 71. In which of the following compounds, nitrogen exhibits the highest oxidation state—

1. N₂H₄
2. NH₃
3. N₃H
4. NH₂OH

Answer: 3. N₃H

⇒ \(\stackrel{-2}{\mathrm{~N}_2} \mathrm{H}_4, \stackrel{-3}{\mathrm{~N}} \mathrm{H}_3, \stackrel{-1 / 3}{\mathrm{~N}} \mathrm{H}_3 \mathrm{H}, \stackrel{-1}{\mathrm{~N}} \mathrm{H}_2 \mathrm{OH}\)

Redox Reactions Questions

Question 72. In acidic medium, H₂O₂ changes \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) to CrO₅ which has two (—0—0— ) bonds. The oxidation state of Cr in CrO₅ is-

1. +5
2. +3
3. +6
4. -10

Answer: 3. +6 let the oxidation number of Cr in CrO₅ be x

Question 73.

1. H₂O₂ + O₃ →H₂O + 2O₂
2. H₂O₂ + Ag₂O→ 2Ag + H₂O + O₂

The role of hydrogen peroxide in the above reactions is respectively—

1. Oxidisingin (1) and reducing (2)
2. Reducing (1) and oxidising (2)
3. Reducing (1) and (12)
4. Oxidisingin (1) and (2)

Answer: 3. Reducing (1) and (12)

⇒ \(\mathrm{H}_2 \stackrel{-1}{\mathrm{O}}_2+\stackrel{0}{\mathrm{O}}_3 \rightarrow \mathrm{H}_2 \mathrm{O}+2 \stackrel{0}{\mathrm{O}}_2\)

In this reaction, H₂O₂ undergoes oxidation and forms O₂. Hence, it acts as a reductant

Question 74. Assuming complete ionization, the same moles of which of the following compounds will require the least amount of acidified KMnO₄ for complete oxidation—

1. FeSO₄
2. FeSO₃
3. FeC₂O₄
4. Fe(NO₂)₂

Answer: 1. FeSO₄ will require the least amount of acidified KMnO₄ for complete oxidation.

Redox Reactions Questions

Question 75. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior-

1. Cu + 2H₂SO₄→CuSO₄ + SO₂ + 2H₂O
2. S + 2H₂SO₄→3SO₂ + 2H₂O
3. C + 2H₂SO₄→CO₂ + 2SO₂ + 2H₂O
4. CaF₂ + H₂SO₄→CaSO₄ + 2HF

Answer: 4. CaF₂ + H₂SO₄→CaSO₄ + 2HF

In this reaction, there is no change in the oxidation number of any elements, present. Thus, it is not a redox reaction.

Question 76. For the redoxreaction—

⇒ \(\mathrm{MnO}_4^{-}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

The correct coefficients of the reactants for the balanced equation are:

1. 16,5,2
2. 2,5,16
3. 2,16,5
4. 5,16,2

Answer: 2. 2,5,16

⇒ \(\begin{aligned}
& {\left[\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2} \\
& {\left[\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2+2 e\right] \times 5} \\
& 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O} \\
&
\end{aligned}\)

Redox Reactions Questions

Question 77. When KMnO₄ reacts with KBr in an alkaline medium and gives a bromate ion, the oxidation state of Mn changes from +7 to—

1. +6
2. +4
3. +3
4. +2

Answer: 2. +4

⇒ \(2 \mathrm{MnO}_4^{-}+\mathrm{Br}^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \stackrel{+4}{2} \mathrm{MnO}_2+\mathrm{BrO}_3^{-}+2 \mathrm{OH}^{-}\)

Question 78. K₂Cr₂O₇ in an acidic medium converts into

1. Cr²⁺
2. Cr²⁺
3. Cr⁴⁺
4. Cr⁵⁺

Answer: 2. Cr²⁺

Question 79. The oxidation state of iron in hemoglobin is—

1. 0
2. +2
3. -2
4. +3

Answer: 2. +2

Question 80. What is the oxidation number of Br in KBrO₂

1. +6
2. +7
3. +5
4. +8

Answer: 2. +7

Redox Reactions Questions

Question 81. Substances that are oxidized and reduced in the following reaction are respectively—

⇒ \(\mathrm{N}_2 \mathrm{H}_4(l)+2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow \mathrm{N}_2(g)+4 \mathrm{H}_2 \mathrm{O}(l)\)

1. N₂H₄H₂O
2. N₂H₄H₂O₂
3. N₂H₂O₂
4. H₂O₂N₂

Answer: 2. N₂H₄H₂O₂