WBCHSE Class 11 Chemistry Redox Reactions Notes

Redox Reactions Introduction

Redox Reactions Notes: Chemical reactions, according to their nature, are divided into different classes, such as combination reactions, decomposition reactions, elimination reactions, polymerisation reactions and many other types.

Oxidation-reduction or redox-reaction is a class of chemical reactions in which oxidation and reduction occur simultaneously. A large number of chemical and biological reactions belong to this class. Some processes that are associated with oxidation-reduction reactions are the rusting of iron, the production of caustic soda and chlorine by electrochemical method, the production of glucose by photosynthesis, the generation of electric power by battery, fuel cell, etc.

Redox Concept Reactions According To Electronic

Oxidation reaction: A chemical reaction in which an atom, an ion or a molecule loses one or more electrons is called an oxidation reaction. An atom, ion or molecule is oxidised by loss of electron (s).

Examples: Oxidation reactions involving—

Loss of electron(s) by an atom: Generally atoms of metallic elements such as Na, K, Ca, etc., undergo oxidation by losing electron(s), thereby producing positive ions.

⇒ \(\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+e ; \mathrm{K} \longrightarrow \mathrm{K}^{+}+e ; \mathrm{Ca} \longrightarrow \mathrm{Ca}^{2+}+2 e\)

Loss of electron(s) by a cation: Some cations such as Fe²⁺, Sn²⁺, Cu⁺, etc., undergo oxidation by losing one or more electrons, thereby forming cations with higher charges.

⇒ \(\begin{gathered}
\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e ; \mathrm{Sn}^{2+} \longrightarrow \mathrm{Sn}^{4+}+2 e ; \\
\mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+e
\end{gathered}\)

Loss of electron(s) by an anion: Anions such as I- and Br- ions oxidise to neutral atoms or molecules by losing electron(s).

⇒ \(2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 e ; 2 \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_2+2 e\)

Loss of electron(s) by a molecule: Neutral molecules such as H₂, H₂O₂ and H₂O oxidise to cations by losing one or more electrons

⇒ \(\begin{gathered}
\mathrm{H}_2 \longrightarrow 2 \mathrm{H}^{+}+2 e ; \mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{O}_2+2 \mathrm{H}^{+}+2 e \\
\mathrm{H}_2 \mathrm{O} \longrightarrow \frac{1}{2} \mathrm{O}_2+2 \mathrm{H}^{+}+2 e
\end{gathered}\)

Reduction reaction: A chemical reaction in which an atom, an ion or a molecule gains one or more electrons is called a reduction reaction. An atom, ion or molecule is reduced by the gain of electron(s).

Examples: Reduction reactions involving—

Gain of electron(s) by an atom: Atoms of different elements in particular, such as, atoms of chlorine, bromine, oxygen and other non-metals are reduced to anions by gaining electron(s).

⇒ \(\mathrm{Cl}+e \longrightarrow \mathrm{Cl}^{-} ; \mathrm{Br}+e \longrightarrow \mathrm{Br}^{-} ; \mathrm{O}+2 e \longrightarrow \mathrm{O}^{2-}\)

Gain of electron(s) by a cation: Cations such as H⁺, Fe²⁺, Fe³⁺, Cu²⁺ etc., are reduced to neutral atoms or cations with lower charges by accepting electron(s)

⇒ \(\begin{gathered}
\mathrm{H}^{+}+e \longrightarrow \mathrm{H} ; \mathrm{Fe}^{2+}+2 e \longrightarrow \mathrm{Fe} \\
\mathrm{Fe}^{3+}+e \longrightarrow \mathrm{Fe}^{2+} ; \mathrm{Fe}^{3+}+3 e \longrightarrow \mathrm{Fe} \\
\mathrm{Cu}^{2+}+2 e \longrightarrow \mathrm{Cu} ; \mathrm{Cu}^{2+}+e \longrightarrow \mathrm{Cu}^{+}
\end{gathered}\)

Gain of electron(s) by a molecule: Neutral molecules such as Cl₂, O₂, H₂O₂ etc., are reduced by gaining one or more electrons.

⇒ \(\begin{gathered}
\mathrm{Cl}_2+2 e \longrightarrow 2 \mathrm{Cl}^{-} ; \mathrm{O}_2+4 \mathrm{H}^{+}+4 e \longrightarrow 2 \mathrm{H}_2 \mathrm{O} \\
\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+}+2 e \longrightarrow 2 \mathrm{H}_2 \mathrm{O}
\end{gathered}\)

Oxidant and reductant in light of electronic concept

Oxidant: In a redox reaction, the species that itself gets reduced by accepting electron(s) but oxidises other substances is called an oxidant or oxidising agent. So, an oxidising agent is an electron acceptor. The greater the tendency of a substance to accept electrons, the greater the oxidising power it possesses.

Examples: Oxygen (O₂), hydrogen peroxide (H₂O₂), halogens (F₂, Cl₂, Br₂, 1₂), nitric acid (HNO₃), potassium permanganate (KMn04), potassium dichromate (K₂Cr₂O₇), sulphuric acid (H₂SO₄), etc.

Reductant: In a redox reaction, the species that itself gets oxidised by losing electron(s) but reduces other substances is called a reductant or a reducing agent. So, a reducing agent is an electron donor.

The substance having a high tendency to lose electrons acts as a strong reducing agent. Alkali metals (Na, K, Rb, Cs, etc.,) of group-IA of the periodic table show a strong tendency to lose electrons and hence behave as powerful reducing agents.

Example; Hydrogen (H₂), hydrogen sulphide (H₂S), carbon (C), some metals (Na. K, Fe. Al, etc.), stannous chloride (SnCl₂), sulphur dioxide (SO₂), oxalic add (H₂C₂O₄), sodium thiosulphate (Na₂S₂O₂), etc.

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According to an electronic concept;

1. Oxidation involves the loss of one or more electrons. Reduction involves the gain of one or more electrons.
2. Oxidants are electron acceptors. Reductants are electron donors.

Identification of oxidants and reductants with the help of electronic concept

Reaction: 2KI(aq) + Br₂(l)→2KBr(aq) + l₂(s)

The reaction can be represented in ionic form as—

⇒ \(\begin{aligned}
& 2 \mathrm{~K}^{+}(a q)+2 \mathrm{I}^{-}(a q)+\mathrm{Br}_2(l) \longrightarrow \\
& 2 \mathrm{~K}^{+}(a q)+2 \mathrm{Br}^{-}(a q)+\mathrm{I}_2(s)
\end{aligned}\)

This equation shows that in the reaction, the K4 ion does not undergo any change. The only function that it does In the reaction is to balance the charge. So, the K⁺ ion only acts as a spectator ion in the reaction.

Hence, the net ionic equation of the reaction is-

⇒ \(2 \mathrm{I}^{-}(a q)+\mathrm{Br}_2(l) \rightarrow \mathrm{I}_2(s)+2 \mathrm{Br}^{-}(a q)\)

Equation(2) shows that the 1^– ion produces I₂ by losing electrons, while Br₂ forms Br^– ions by gaining electrons. 1 lenco, In this reaction, the conversion of I- into [2l^–(aq) → l₂(s) + 2e] is an oxidation reaction. On the other hand, the conversion of Br₂ into Br^– →2Br (l)] is a reduction reaction. Thus, in this reaction, ion, i.e., Kl is the reductant and Br₂ is the oxidant.

Oxidation-reduction occur simultaneously

Neither an oxidation reaction nor a reduction reaction can occur alone. Oxidation and reduction reactions are complementary to each other. In a reaction system, if a reactant loses an electron, then there must be another reactant In the system that will gain the lost electron. Thus, oxidation and reduction reactions must occur together.

In a redox reaction, the reducing agent gets oxidised by losing electron(s), while the oxidising agent gets reduced by accepting the lost electron(s). For example, Metallic zinc reacts with copper sulphate in a solution to form metallic copper and zinc sulphate.

⇒ \(\mathrm{Zn}(s)+\mathrm{CuSO}_4(a q) \rightarrow \mathrm{ZnSO}_4(a q)+\mathrm{Cu}(s) \downarrow\)

In an aqueous solution, CuSO₄ and ZnSO₄ exist almost completely In dissociated state. So, the reaction can be expressed in the form of an ionic equation as

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

In this reaction, the Zn atom loses two electrons, forming the Zn^2- ion (oxidation). On the other hand, the Cu²⁺ ion accepts these two lost electrons to produce Cu^– atom (reduction)

In this reaction, the Zn-atom can lose electrons only because the Cu²⁺ ion present in the reaction system accepts those lost electrons. Alternatively, the Cu²⁺ ion can get reduced by accepting electrons only because the Zn-atom present in the reaction system lose electrons.

So, in a reaction, when a substance undergoes oxidation, another substance present in the reaction system undergoes reduction. Tints it can be said that oxidation and reduction occur simultaneously.

Half-reaction

Any redox reaction consists of two half-reactions, one is for an oxidation reaction and the other is for a reduction reaction Both of these reactions are called a half-reaction.

In a redox reaction, the half-reaction Involving oxidation Is called oxidation half-reaction, and the half-reaction involving reduction is called reduction half-reaction.

Redox reaction = Oxidation half-reaction Reduction half-reaction

Reaction: Zn (s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s)

Oxidation Half Reaction: Zn(s)→Zn²⁺(Aq)+2e

Reduction Half- reaction: Cu²⁺(aq)+2e→Cu(s)

Oxidation State And Oxidation Number

According to electronic theory, redox reactions can be explained in terms of electron transfer. The electronic theory can be applied in the case of ionic compounds because, in the formation of ionic compounds, one reactant gives up the electron(s) and another reactant accepts those electron(s).

However, this theory cannot be applied in the case of a redox reaction involving covalent compounds due to the absence of cations or anions in such compounds. Thus this theory is unable to Identify the oxidant and reductant in such types of reactions.

To overcome this problem, the concept of oxidation number has been introduced. All types of redox reactions can be explained based on oxidation number. Each constituent element of any compound has a definite valency. Similarly, it can be assumed that each atom of any element has a definite oxidation number.

Oxidation state

An atom of an element converts into an ion when it loses or gains an electron. The loss of one or more electrons by an atom results in the formation of a cation, while the gain of one or more electrons produces an anion.

Cation is the oxidised state and anion is the reduced state of an element For example, the Na+ ion is the oxidised state of the Na atom and the Cl- ion represents the reduced state of the Cl -atom. The state of oxidation or reduction of an element presenting a compound is called the oxidation state of that element.

Oxidation state Definition: The state of oxidation or reduction in which an atom of an element exists in a compound is called the oxidation state of the element in that compound.

From the element charge of the compound ion of an element present in an ionic compound, the oxidation state of that element can be easily determined. If the element exists as a cation in the compound, the element is said to be in the oxidised state. On the other hand, if the element exists as an anion in the compound, then the element is said to be in a reduced state.

Example: In the formation of the compound ZnCl₂, the Zn atom loses two electrons to produce a Zn²⁺ ion and two CIatoms accept two electrons, one electron each to yield two Cl- ions. Zn²⁺ ion combines with two Cl- ions to form a ZnCl₂ molecule.

Thus, in ZnCl₂, the Zn -atom exists in the oxidised state, while the Cl -atom exists in the reduced state. Cations or anions derived from the same element may exist in different oxidation states in different compounds. For example, in CuCl₂, Cu exists as Cu²⁺, while in CuCl, it exists as Cu+ ion. Thus, the oxidation state of Cu in CuCl₂ is higher than that in CuCl.

Oxidation number

Oxidation number Definition: The oxidation number of an element in a compound is a definite number, which indicates the extent of oxidation or reduction to convert an atom of the element from its free state to its bonded state in the compound.

If oxidation is necessary to effect such a change, then the oxidation number will be positive. The oxidation number will be negative when such a change requires reduction. The oxidation number of an atom or molecule in its free state is considered to be zero (0).

The number expressing the die oxidation state of the atom of an element in a compound denotes the oxidation number of the elements in the compound.

The oxidation number of elements in electrovalent compounds:

The charge that an atom of an element In the molecule of an electrovalent compound carries, is equal to the oxidation number of the element In a compound, According to the nature (positive or negative) of the charge, the oxidation number may be positive or negative. The number of electrons (s) lost or gained by an atom during the formation of an Ionic compound determines the oxidation number of the clement in that compound.

Examples: ln NaCl, sodium and chlorine exist ns Na + and (11- ions, respectively. So the oxidation numbers of sodium and chlorine are +1 and – 1, respectively. In PeCI₂> Iron and chlorine are present as PO²⁺ and Cl- Ions, so the oxidation numbers of iron and chlorine are +2 and -1, respectively.

The oxidation number of elements in covalent compounds: The formation of a covalent compound does not involve the direct transfer of electron(s) between the participating atoms; instead, a covalent bond Is formed by the sharing of electrons.

When two atoms of different electronegativities form a covalent bond(s) through the sharing of one or more electron pairs, they do not get an equal share of the electrons.

The more electronegative atom acquires a greater possession of the shared electron pair (s) than the less electronegative atom.

As a result, the more electronegative atom acquires a partial negative charge. It Is assumed that the atom of the more electronegative element has gained electron(s) i.e. it has been reduced and the less electronegative atom has lost electron(s) i.e., It has been oxidised.

The oxidation number of an atom In a covalent compound is considered to be equal to the number of electron pair(s) the atom shares with one or more atoms of different electronegativities in the compound, If the atom concerned is of higher electronegativity, then Its oxidation number is taken as negative, and if It is of lower electronegativity, its oxidation number is taken as positive.

In the case of a molecule of an element, such as H₂, N₂, O₂, Cl₂ etc., the two atoms of the same electronegativity are covalently linked by sharing electron pair (s) between them. Hence, these two atoms use the electron pair(s) equally and none of the atoms acquire a positive or negative charge. Therefore, the oxidation number of the atoms In the molecule of these elements is regarded as zero.

Example: In hydrogen chloride molecule (HC1) one electron pair exists between H and Cl-atoms, As the electronegativity of chlorine is more than that of hydrogen, the oxidation number of H – +1 and that of Cl =-J,

In water (H₂O) molecule, the O -atom shares two electron palms, one each with two separate H -atoms. As oxygen is more electronegative than hydrogen, the oxidation number of H -atom -+l and that of O -atom =-2.

In a carbon dioxide molecule (CO₂), a carbon atom shares four electron pairs, to each with two separate O atoms. As oxygen has higher electronegativity than carbon, in CO₂ molecule, the oxidation number of C = + 4 and that of O =-2

In the ethylene (C₂H₄) molecule, two equivalent carbon atoms share two C=C electron pairs between themselves, Since these two electron pairs are equally shared by two C -atoms, they have no role in determining the oxidation number of C.

Again, each carbon atom shares two electron pairs with two separate H -atoms. As carbon is more electronegative than hydrogen, the oxidation number of each H -atom = + 1 and the oxidation number of each C -atom =-2.

Rules For Calculating the Oxidation Number Of An Element

The following rules are to be followed in determining the oxidation number of an element in a compound.

The oxidation number of an element in its free or elementary state is taken as zero (0).

Example: \(\stackrel{\oplus}{\mathrm{Z}} \mathrm{n}, \stackrel{0}{\mathrm{Cu}}, \stackrel{0}{\mathrm{C}} \mathrm{L}_2, \stackrel{\ominus}{\mathrm{P}}_4, \stackrel{0}{\mathrm{~S}}_8 \text {, etc. }\) Etc.,

The oxidation number of a monoatomic ion in an ionic compound is equal to its charge.

Example: In FeCl₂, iron and chlorine exist as Fe²⁺ and Cl-. So, in i.e., Cl₂, the oxidation number of Fe and Cl are +2 and -1 respectively.

In the case of a polyatomic ion, the sum of the oxidation numbers of all the atoms present in it is equal to the charge of the Ion.

Examples: The sum of oxidation numbers of all the atoms presenting [Fe(CN)₈]4- =-1.0 The sum of the oxidation numbers of the atoms Cr₂O^2-₇  ion =-2.

Determination of the oxidation numbers of the atoms in covalent compounds has been discussed in

The algebraic sum ofthe oxidation numbers of all atoms in a neutral molecule is zero(O).

Example: In the FCC1₃ molecule, the oxidation number of Fe is +3 and the oxidation number of Cl is -1. So, the total oxidation number of the atoms in the FeCl₃ molecule =(+3) + 3 X (- 1) = 0.

The oxidation number of hydrogen:

In metallic hydride, it is always \(\stackrel{+1}{\mathrm{NaH}}, \stackrel{+2}{\mathrm{C}} \mathrm{CaH}_2\)

In all hydrogen-containing compounds except for metallic hydrides, it is +1.

Example; \(\stackrel{+1}{\mathrm{NH}}_3, \stackrel{+1}{\mathrm{H}} \mathrm{O}, \stackrel{+1}{\mathrm{H}_2} \mathrm{SO}_4, \mathrm{NaHCO}_3 \text {, etc. }\)

The oxidation number of oxygen in its compounds: The oxidation number of oxygen in most compounds=-2

Example:

1. In peroxide compounds (H₂O₂ Na₂O₂), the oxidation number of oxygen =-1.
2. In superoxides (Example; KO₂), the oxidation number of oxygen =-1/2.
3. Since fluorine is more electronegative than oxygen, the oxidation number of oxygen in F₂O = +2.
4. In all fluorine-containing compounds, the oxidation number of fluorine =- 1.
5. There are some elements which always show, fixed oxidation numbers in their compounds.

Example: Alkali metals (Li, Na, K, Rb, Cs) always show a +1 oxidation state in their compounds. The oxidation number of alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra ) is +2 in their compounds. The oxidation number of Zn in its compounds is +2 and the oxidation number of A1 in its compounds is +3.

The maxim oxidation1 number of an element cannot exceed its group numbering in the periodic table.

The following always have definite oxidation numbers in their compound

Calculation of oxidation number in some compounds

The oxidation number of an element in a compound can be calculated if the oxidation numbers of other elements in the compound are known.

Oxidation number of S in H₂SO₄: Suppose, the oxidation number of S in H₂SO₄ = x.

Total oxidation number of two H-atoms in H₂S0₄ molecule = 2 X (+1) = +2 Total oxidation number of four O -atoms in H₂SO₄ molecule = 4 X (-2) =-8 .% Total oxidation number of all atoms in a H₂SO₄ molecule = +2 + x + (-8) = x- 6.

Now, the sum of the oxidation numbers of all atoms in a molecule = 0.

Therefore, x-6 = 0 or, x = +6

∴ The oxidation number of S in H₂SO₄ = +6

Redox Reactions Notes

The oxidation number of Cl in KC1O₄: If the oxidation number of Cl = x, then the total oxidation number of all the atoms in the KC1O₄ molecule =+1+x+ 4x (-2) = x- 7 Sum of oxidation numbers) all the atoms present in a molecule = 0.

∴ x-7 = 0 or, x = +7

Hence, the oxidation number of Cl in KC1O₄ =

The oxidation number of N in NH₄NO₃: NH₄NO₃ is an ionic compound in which NH₄ and NO-₃ are the cation and anion, respectively.

Let the oxidation no. of in NH⁺₄ ion be x. Then the total oxidation no. of the atoms present in this ion = x + 4.

Redox Reactions Notes

For a polyatomic ion, the oxidation no. of the ion is equal to its charge i.e., the oxidation number of NH₄ ion = +1. x + 4 = +1 or, x = -3 Again, if the oxidation number of N in NO₃ be y, then y + 3 X (-2) = -1 or, y = +5.

Hence, in NH₄NO₃ the oxidation number of one Natom is -3 and that of another N -atom is +5.

The oxidation number of Cl in Ca(OCl)Cl: In this compound. The cl -atom in OCl- is linked with the O-atom, and another Cl atom exists as the Cl- ion. The oxidation number of the Cl -atom that exists as Cl- ion =-1.

Let the oxidation number of Cl atom in OCP be x.

∴ – 2+ x =-1 or, x =+l

So, in Ca(.OCDCl, the oxidation number of one Cl atom is 1 and that of another Cl -atom is +1.

Redox Reactions Notes

The oxidation number of Mn in KMnO₄: Suppose, the oxidation number ofMn.in KMn04 = x.

Total oxidation number ofthe atoms in KMnO4 =+1 + x+ 4 x (-2) = x-7

∴ x-7 = 0 or, x = +7

Hence, the oxidation number of in KMnO₄ =+7

The oxidation number of P in H₄P₂O₇: Let the oxidation number of P in H₄P₂O₇ be x. Hence, the total oxidation number of P-atom in H₄P₂O₇ molecule = 4 x (+1) -+2 x x+ 7(-2) = 4-+2x-14 =2x— 10 2x- 10 = 0 or, x = 5

Therefore, the oxidation number of P in H₄P₉O2 = +5

The oxidation number of Fe in Fe(CO)₅: CO is a neutral ligand (molecule) and its oxidation number = 0. So, the oxidation number of Fein Fe(CO)- is zero (0).

Redox Reactions Notes

The oxidation number of Fe in K₄[Fe(CN)6]: Suppose, the oxidation number of Fe in K4[Fe(CN)6] =x. Hence, 4 x (+1) + x+ 6 x (-1) = 0

The oxidation number of an element in 3 compounds may be zero(0); In compounds like C₆H₁₂O₆, HCHO, CH₂C₁₂ etc., the oxidation number of carbon is zero (0). Suppose, the oxidation number ofCin C₆H₁₂O₆ = x. So, 6x+ 12x(+1) + 6x(-2) = 0

∴ x=0

Some exceptions regarding the determination of oxidation number

The anomaly fractional oxidation state:

Since electrons can not be transferred fractionally, the fractional oxidation state of an element seems to be a hypothetical case. But in compounds like Fe₃O_{4, and} NaS₄O6 the oxidation states of Fe, and S are \(+\frac{8}{3}\) and +2.5, respectively.

Redox Reactions Notes

The fractional oxidation state is only the average oxidation state of an element when two or more of its atoms with different oxidation states are present in a compound. For such compounds, the actual oxidation state can be determined by knowing the structure of the compound.

The oxidation number of Cr in CrO₅: According to the usual method, the oxidation number of Cr in the CrO₅ molecule would be +10. TAT -2 [CrO₅, or x = +10]. However, the oxidation number of Cr can never exceed 6 because the total number of electrons in its 3d and 4s orbitals is 6. From the chemical structure of CrO₅, it can be shown that the oxidation No number of Cr in CrO₅ is, in fact, +6.

Let the oxidation number of Cr in the Cr05 molecule be x. x + lx (-2) + 4 X (-1) = 0 (for O) (for O-atoms linked O — O bond) -V x = +6. Hence, the oxidation number of Cr in Cr05 = +6.

The oxidation number of S in H₂SO₆: According to the tyre .usual method, the oxidation number of sulphur in the H₂SO₆ molecule is +8

Redox Reactions Notes

⇒ \(\left[\stackrel{+1}{\mathrm{H}_2} \stackrel{x-2}{\mathrm{SO}}_5^{-2}, x+2-10=0 \quad \text { or, } x=+8\right] \text {. }\)

The oxidation number of sulphur can never exceed + 6. The chemical structure of H₉SO₆ shows that the oxidation number of sulphur in H₂SO₅ is Suppose, the oxidation number of S in H₂SO₅=x

Hence, the oxidation number of the S -atom in H₂SO₅ = +6

The oxidation number of S in Na₂S₂O₃: According to the usual method, the average value of oxidation numbers of S in Na₂S₂O₃ molecule = +2 However the reaction of Na₂S₂O₃ with dilute H₂SO₄, one of the two S -atoms in Na₂S₂O₃ is precipitated as sulphur and the other is oxidised to SO₂.

Redox Reactions Notes

Therefore, the two S -atoms in the Na₂S₂O₃ molecule are not identical. Consequently, their oxidation numbers cannot be the same. The chemical structure of the Na₂S₂O₃ molecule indicates that the two sulphur atoms in it are linked by a coordinate bond.

The oxidation number of the S -atom accepting the electron pair in the coordinate bond is considered to have an oxidation number of -2. If the oxidation number ofthe other S -atom is taken as .v, then.

⇒ \(\begin{aligned}
& 2 \times(+1)+3 \times(-2)+x \times 1+1 \times(-2)=0 \\
& \text { (For Na-atoms) (For } \mathrm{O} \text {-atoms) (For } \mathrm{S} \text {-atom held } \\
& \text { by coordinate bond) } \\
&
\end{aligned}\)

∴ x=+6

Redox Reactions Notes

Hence, the oxidation number of S -atoms H₂SO₅= +6

The oxidation number of S in Na₂S₂O₃: According to the usual method, the average value of oxidation numbers of S in Na₂S₂O₃ molecule = +2

⇒ \(\left[\stackrel{+1}{\mathrm{Na}_2} \stackrel{x}{\mathrm{~S}}_2-2 \mathrm{O}_3, 2 \times(+1)+2 x+3 \times(-2)=0 \text { or, } x=+2\right] \text {. }\)

However, in the reaction of Na₂S₂O₃ with dilute H₂SO₄, one of the two S -atoms in Na₂S₂O₃ is precipitated as sulphur and the other is oxidised to SO₂. Therefore, the two S -atoms in the Na₂S₂O₃ molecule are not identical.

Redox Reactions Notes

Consequently, their oxidation numbers cannot be the same. The chemical structure of the Na₂S₂O₃ molecule indicates that the two sulphur atoms in it are linked by a coordinate bond. The oxidation number of the S -atom accepting the electron pair in the coordinate bond is considered to have an oxidation number of -2. If the oxidation number ofthe other S -atom is taken as .v, then

⇒ \(\begin{aligned}
& 2 \times(+1)+3 \times(-2)+x \times 1+1 \times(-2)=0 \\
& \text { (For } \mathrm{Na} \text {-atoms) (For } \mathrm{O} \text {-atoms) (For } \mathrm{S} \text {-atom held } \\
& \text { by coordinate bond) } \\
&
\end{aligned}\)

Therefore, in the Na₂S₂O₃ molecule, the oxidation number of one S -atom is -2 and that ofthe others +6.

Redox Reactions Notes

The oxidation number of S in Na₂S₄O₆: According to the usual method, the average value of oxidation numbers of S in Na₂S₄O₆ molecule would be + 2.5.

In atoms that molecule, are covalently the oxidation linked number is zero. If the two oxidation sulphur numbers of each of the remaining two s- atoms is x then

⇒ \(\begin{aligned}
& x \times 2+2 \times 0+6 \times(-2) \times 2 \times(+1)=0 \\
& \text { (For } \mathrm{S}) \text { (For } \mathrm{S}-\mathrm{S}) \text { (For } \mathrm{F} \text { ) } \\
&
\end{aligned}\)

Therefore, the oxidation number of each of the two remaining S -atoms in N₂S₄O (. is +5.

The oxidation number of Fe in Fe₂O, According to the usual method, the oxidation number of Fe in Fe,0, would be +[ 3 x .v + 4 x (-2) – 0]. This value of the oxidation number of Fe in Fe₃O expresses the average oxidation number of Fe.

Redox Reactions Notes

Fe, O₄ is a mixed oxide having the composition FeO>Fe₂O₃. Therefore, in FeO, the oxidation number of Fe is + 2, and in Fe₄O₂ the oxidation number of Fe is +3.

Explanation Of Oxidation-Reduction In Terms Of Oxidation Number

According to the concept of oxidation number, oxidation is a chemical reaction in which the oxidation number of atoms increases and reduction is a chemical reaction in which the oxidation number of an atom decreases.

Redox Reactions Notes

So, oxidation means an increase in oxidation number, whereas reduction means a decrease in oxidation number. Examples of oxidation and reduction are as follows.

Explanation of oxidation-reduction reaction

Consider the reaction of HNO₃ with H₂S, forming nitric oxide (NO) and sulphur. In this reaction, the oxidation number of N decreases from +5 in HNO₂ to +2 in NO and the oxidation number of S increases from -2 in H2S to 0 in S. So, the reaction brings about a reduction of HN03 and oxidation of H2S.

In a redox reaction, all the atoms of a participating reactant do not change their oxidation number. Only one atom of the reactant changes oxidation number. This element is called an effective or reactive element.

Redox Reactions Notes

In the previous example, only the N -atom of HNO₃ changes oxidation number. The oxidation numbers of hydrogen or oxygen remain the same before and after the reaction.

The reaction of FeSO₄ with KMnO4 acidified with dilute H₂SO₄ results in K₂SO₄, MnSO₄, FeSO₄ and H₂O.

In this reaction, the oxidation number of Mn decreases (+ 7→ + 2) and the oxidation number of Fe increases (+2 → + 3). Therefore, in the reaction, KMnO₄ is reduced and FeSO₄ is oxidised

Identification of oxidant and reductant based on oxidation number

In redox reactions, the substance that gets oxidised is the reducing agent and the substance that gets reduced is the oxidising agent. Based on oxidation number, it can be stated that in a redox reaction, the substance in which the oxidation number of an atom increases is the producing agent and the substance in which the oxidation number of an atom decreases is the oxidising agent.

Redox Reactions Notes

Example: In presence of H₂SO₄, K₂Cr₂O₇ reacts with KI to form I₂ and chromic sulphate [Cr₂(SO₄)₃]

Here, the oxidation number of Cr decreases from +6 to +3 and the oxidation number of 1- increases from -1 to 0. Therefore, K₂Cr₂O₇ acts as an oxidising agent and KI acts as a reducing agent.

How a redox reaction is identified: At first oxidation number of each of the constituent elements of the participating substances is assigned. If the oxidation numbers of the elements change, then the reaction is identified as a redox reaction. If none of the elements shows any change in oxidation number, then the reaction is not a redox reaction.

Example: Identify whether the given two reactions are redox reactions or not

In this reaction, the oxidation numbers of all the atoms of the participating substances remain the same. Hence, it is not a redox reaction.

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

Redox Reactions Notes

In this reaction, the oxidation number of N increases 0) and the oxidation number of oxygen decreases (0 →2). Hence, it is a redox reaction.

Auto Oxidation-Reduction Reactions

There are some redox reactions in which the same substance gets partially oxidised and reduced. This type of reaction is termed an auto oxidation-reduction reaction.

Example: Potassium chlorate (KC1O₃) on heating decomposes to produce KC₁ and O₂ gas:

In this reaction, the oxidation number of Cl decreases from +5 to -1 and the oxidation number of oxygen increases from- 2 to 0. So, in this reaction of KC1O₃, one atom (O) is oxidised and the other (Cl) is reduced.

Redox Reactions Notes

Lead nitrate undergoes thermal decomposition to produce PbO, N02 gas and 02 gas:

In this reaction, the oxidation number of the N -atom reduces from +5 to +4 and the oxidation number of the O -atom increases from -2 to zero (0 ). Hence, in the reaction, the N atom is reduced and the 0 -atom is oxidised. So, Pb(NO₃)₃ in its thermal decomposition undergoes oxidation and reduction simultaneously.

Ammonium nitrate (NH₄NO₃) on heating decomposes to produce water vapour, N₂ gas and O₂ gas

NFI₄NO₃ is an ionic compound, consisting of ammonium cation (NH+4) and nitrate anion (NO₃). The oxidation numbers of the N -atom in NH+ and NO₂ ions are -3 and 5 respectively. In this reaction, the oxidation number of N in the NH₄ ion increases (-3→ 0) and the oxidation number of N in NO₃ decreases (+ 5 →+ 0). Therefore, NH₄NO₃ undergoes oxidation and t In this reaction, reduction at die same time.

Redox Reactions Notes

Disproportionation And Comproportionation Reactions

Disproportionation reaction in which an element of a reactant undergoes oxidation and reduction simultaneously, resulting in two substances in which one of the elements exists in a higher oxidation state and the other exists in a lower oxidation state

Examples: In the reaction of chlorine with cold and dilute NaOH solution, sodium hypochlorite (NaOCl) and sodium chloride are formed. In this reaction oxidation number of chlorine decreases (0→ -1) and increases (0→ +1) at the same time. thus, chlorine is simultaneously oxidised and reduced in the reaction.

Therefore, this reaction is an example of a disproportionation reaction.

When white phosphorus is heated with a caustic soda solution, phosphine (PH₄) and sodium hypophosphite (NaH₂PO₂) and produced.

Here P. undergoes oxidation and reduction simultaneously. In one of the products (PH₄), P exists in a lower oxidation state (-3) and in the other product (NaH₄PO₂), it exists in a higher oxidation state (+1). Hence, tills reaction is an example of a disproportionation reaction.

Redox Reactions Notes

Comproportionation Reaction: It is a reaction in which two reactants containing a particular element but in two different oxidation states react with each other to produce a substance in which the said element exists in an intermediate oxidation state.

Therefore, a comproportionation reaction is the opposite of a disproportionation reaction. Example: KBrO₃ reacts with KBr in an acidic medium to produce Br₂.

In this reaction, the oxidation number of one Br-torn decreases (from +5 to. 0) and that of another Br-atom increases (from -1 to 0 ). Br₂ is formed by the oxidation of KBrO₃ and the reduction of KBr. The oxidation number of Br₂ is zero, which is intermediate between the oxidation numbers of Br atoms in KBrO, (+5 ) and KBr (-1 ). Therefore, it is a comproportionation reaction.

Equivalent Mass Of Oxidant And Reductant

The equivalent mass of oxidants and reductants is calculated by following two different methods. In one method, the calculation is done in terms of several electron (s) gained by an oxidant or the number of electrons (s) lost by a reductant. The other method takes into account the change in the oxidation number of an element present in the oxidising and reducing agents.

Redox Reactions Notes

Oxidation number method: Equivalent mass of an oxidant or reductant denotes the number obtained by dividing the molecular mass of the oxidant or reductant with the change in oxidation number of an element in the oxidant or reductant in their respective reduction or oxidation reaction.

⇒ \(\begin{gathered}
\text { Equivalent mass } \\
\text { of the oxidant }
\end{gathered} \frac{\text { Molecular or formula mass of oxidant }}{\begin{array}{c}
\text { Total change in oxidation number } \\
\text { of an element present in a molecule } \\
\text { of the oxidant during its reduction }
\end{array}}\)

Redox Reactions Notes

.Redox Reactions Notes

Balancing Of Chemical Equations Involving Redox Reactions

Redox reaction can be balanced with the help of two methods. These are the ion—electron method and the Oxidation number method.

Ion-electron method

Jade and Lamer in 1927 introduced this method. In the ion-electron method, only the molecules and ions which participate in the chemical reaction are shown.

In balancing redox reactions by this method the following steps are followed: 

The reaction is written in ionic form.

The reaction is divided into two half-reactions with the help of ions and electrons. One half-reaction is for oxidation reaction and the other half-reaction is for reduction reaction.

Redox Reactions Notes

While writing the oxidation reaction, the reducing agent and the oxidised substance are written respectively on the left and right of an arrow signing are written respectively on the left and right of the arrow sign

To denote the loss of electrons in oxidation half-reaction, the number of electrons (s) is written on the right of the arrow sign (→). While writing the reduction half-reaction, the number of electrons (s) gained is written on the left arrow sign ( →).

Thus, oxidation half-reaction is: Reducing agent – Oxidised substance +ne [where n = no. of electron (s) lost in oxidation reaction] Thus reduction half-reaction is Oxidising agent + ne Reduced substance [where n = no. of electron(s) gained reduction reaction] 2Cr³⁺

Then each half-reaction is balanced according to the following steps:

In each of the half-reactions, the number of atoms other than H and O -atoms on both sides ofthe arrow sign is balanced.

If a reaction takes place in an acidic medium, for balancing the number of H and O-atoms on both sides of the arrow sign, H₂O or H+ is used. First, oxygen atoms are balanced by adding H₂O molecules to the side that needs O-atoms.

Redox Reactions Notes

Then to balance the number of H-atoms, two H+ ions (2H+) for each molecule of water are added to the opposite side (i.e., the side deficient in hydrogen atoms). 0If the reaction occurs in an alkaline medium, for balancing the H and O -atoms, H₂O or OH- ion is used. Each excess oxygen atom on one side of the arrow sign is balanced by adding one water molecule to the same side and two ions to the other side.

If the hydrogen atom is still not balanced, it is then balanced by adding one OH- for every excess hydrogen atom on the side of the hydrogen atoms and one water molecule on the other side of the arrow sign in a half-reaction, both H+ and OH- ions cannot participate.

The charge on both sides of each half-reaction is balanced. This is done by adding an electron to that side which is a deficient negative charge.

To equalise the number of electrons of the two half-reactions, any one of the reactions or both reactions should be multiplied by suitable integers.

Now, the two half-reactions thus obtained are added. Cancelling the common term(s) on both sides, the balanced equation is obtained.

Examples: In the presence of H₂SO₄, potassium dichromate (K₂Cr₂O₂) and ferrous sulphate (FeSO₄) react together to produce ferric sulphate [Fe₂(SO₄)₃] and chromic sulphate [Cr₂(SO₄)₃].

Reaction: K₂Cr2O₇ + FeSO₄ + H₂SO₄→ K₂SO₄ + Cr₂(SO₄)₃ + Fe₂(SO₄)₃ + H₂O The reaction can be expressedin ionic form as: Cr3+ + Fe3+ + H₂O Oxidation half-reaction: Fe²⁺→Fe³⁺+ e Reduction half-reaction: Cr₂O^2-

Redox Reactions Notes

Balancing the Cr -atom: Cr₂O^2-–

To equalise the number of O -atoms on both sides, 7 water molecules are to be added to the right side. 2Cr³⁺ + 7H₂O

⇒ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

[One water molecule is required for each O-atom.]

To balance H-atoms on both sides, 14H+ ions are to be added to the left side.

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

2H+ ions are required for each water molecule] Jd] For equalising the charge on both sides, 6 electrons are to be added to the left side.

Now, for balancing the number of electrons in oxidation and reduction half-reactions, the balanced oxidation half-reaction is multiplied by 6 and the balanced reduction half-reaction by 1. Then these two equations are added.

⇒ \(\begin{aligned}
6 \mathrm{Fe}^{2+} & \longrightarrow 6 \mathrm{Fe}^{3++} 6 e \\
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e & \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\
6 \mathrm{Fe}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+} & \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Redox Reactions Notes

This balanced equation has been expressed in ionic form. This equation can be represented in molecular form as—

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

For 2H+ ions, one H₂SO₄ molecule is required]

In presence of H₂SO₄, KMnO₄ and FeSO₄ react together to produce MnSO₄ and Fe₂(SO₄)₃.

⇒ \(\begin{aligned}
& \text { Reaction: } \mathrm{KMnO}_4+\mathrm{FeSO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightharpoondown \\
& \mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Redox Reactions Notes

To balance the number of electrons lost in the oxidation half-reaction, the oxidation half-reaction is multiplied by 5 and then the two reactions are added.

⇒ \(\begin{aligned}
5 \mathrm{Fe}^{2+} & \longrightarrow 5 \mathrm{Fe}^{3+}+5 e \\
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e & \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\
\hline 5 \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+} & \longrightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

As one Fe₂(SO₄)₃ molecule contains two Fe -atoms, the equation is multiplied by 2

10Fe²⁺ + 2MnO-4 + 16H+→10Fe³⁺ + 2Mn²⁺ + 8H₂O

This is the balanced equation in ionic form. This equation when expressed in molecular form becomes—

⇒ \(\begin{aligned}
10 \mathrm{FeSO}_4+2 \mathrm{KMnO}_4+ & 8 \mathrm{H}_2 \mathrm{SO}_4 \rightharpoondown \\
& 5 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{MnSO}_4+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Redox Reactions Notes

Equalising the number of atoms of different elements and the sulphate radicals we get,

⇒ \(\begin{aligned}
& 10 \mathrm{FeSO}_4+2 \mathrm{KMnO}_4+8 \mathrm{H}_2 \mathrm{SO}_4 \rightharpoondown \\
& 5 \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3+2 \mathrm{MnSO}_4+\mathrm{K}_2 \mathrm{SO}_4+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

This is a balanced equation of the given reaction in molecular form.

In H₂SO₄ the reaction and KI, between Cr₂(SO₄) 3K₂Cr₂O₇ and I₂ are acidified formed.

The equation can be expressed in ionic form as— Cr₂O₂-7 +1- + H + — Cr3+ + 12 + H₂O

Oxidation half-reaction: 21→ I₂ + 2e Reduction half-reaction:

Cr₂O2→+ 14H+ + 6e — 2Cr3+ + 7H₂O

To balance the electrons, equation (1) is multiplied by 3 and added to equation (2). Thus the equation stands as—

⇒ \(\begin{gathered}
6 \mathrm{I}^{-} \longrightarrow 3 \mathrm{I}_2+6 e \\
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\
\hline 6 \mathrm{I}^{-}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_2+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
\end{gathered}\)

This is the balanced equation ofthe reaction in ionic form. The above ionic reaction can be expressed in molecular form as follows— 6KI + K₂Cr₂O₇ + 7H₂SO₄→3I₂ + Cr₂(SO₄)₃ + 7H₂O

Equalising the number of atoms of potassium and sulphate radical on the left and right sides, we have,

⇒ \(\begin{aligned}
& 6 \mathrm{KI}+ \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \\
& 3 \mathrm{I}_2+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+4 \mathrm{~K}_2 \mathrm{SO}_4+7 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

In the reaction between KMnO₄, acidified with dilute H₂SO₄ and oxalic acid (H₂C₂O₄), MnSO₄ and CO₂ were produced.

Redox Reactions Notes

⇒ \(\begin{aligned}
& \text { Reaction: } \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \\
& \mathrm{K}_2 \mathrm{SO}_4+\mathrm{MnSO}_4+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\
&
\end{aligned}\)

⇒ \(\begin{aligned}
& 5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \text { noils } \\
& 10 \mathrm{CO}_2+2 \mathrm{Mh} S \mathrm{O}_4+8 \mathrm{H}_2 \mathrm{O} \\
&
\end{aligned}\)

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

Equalising the number of atoms of potassium and sulphate radical we get,

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

This Is the balanced equation of the given reaction in molecular form.

In NaOH solution, Zn reacts with NaNO₃ to yield Na₂ZnO₂, NH₃ and H₂O.

Reaction: Zn + NaNO₃ + NaOH — Na₂ZnO₂ + NH₃ + H₂O The equation can be expressed in ionic form as—

Zn + NO^– + OH^–→ ZnO_2^– →+ NH₃ +H₂O

Oxidation half-reaction: Zn + 40H^–→ ZnO2^– + 2H₂O + 2C

Redox Reactions Notes

Reduction half-reaction: NO₃ + 6H₂O + 8c — NH3 + 90H^–

Now multiplying equation (1) by 4 and then adding to equation (2), we get,

⇒ \(\begin{aligned}
& 4 \mathrm{Zn}+16 \mathrm{OH}^{-}+\mathrm{NO}_3^{-}+ 6 \mathrm{H}_2 \mathrm{O} \longrightarrow \\
& 4 \mathrm{ZnO}_2^{-}+\mathrm{NH}_3+9 \mathrm{OH}^{-}+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
\)

or, 4Zn + 70H→+ NO₃→4ZnO-2 + NH₃ + 2H₂O It is the balanced equation of the reaction in ionic form. Expressing the above equation in molecular form— 4Zn + 7NaOH + NaNO₃→ 4Na₂ZnO₂ + NH₃ + 2H₂O It is the molecular form of the balanced equation of the reaction.

In the presence of HNO₃, sodium bismuthatic (NaHO₃) reacts with Mn(NO₃)₂ to produce coloured sodium permanganate (NaMnO₄) and itself gets reduced to bismuth nitrate.

Reaction: NaBIO₃ + Mn(NO₃) + UNO₂ →NaMnO₄ + Bi(NO₃) + H₂O The equation can be expressed in ionic form as— BIO₂ + Mn²⁺ → Bl³⁺ + MnO₄ + H₂O

Oxidation half-reaction: Mn²⁺ + 4H₂O → Mn04+8H+→+5c

Reduction half-reaction: BIO₂ + 6H+ + 2e — Bi3+ + 3H₂O.

Multiplying equation (1) by 2 and equation (2) by 5 and then adding them we get—

\(\begin{array}{r}
2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{BiO}_3^{-}+30 \mathrm{H}^{+} \longrightarrow \\
2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+5 \mathrm{Bi}^{3+}+15 \mathrm{H}_2 \mathrm{O} \\
\text { or, } 2 \mathrm{Mn}^{2+}+5 \mathrm{BiO}_3^{-}+14 \mathrm{H}^{+} \rightarrow 5 \mathrm{Bi}^{3+}+2 \mathrm{MnO}_4^{-}+7 \mathrm{H}_2 \mathrm{O}
\end{array}\)

This is the balanced ionic equation of the reaction. The equation in the molecular form stands as— \(\begin{aligned}
& 2 \mathrm{Mn}\left(\mathrm{NO}_3\right)_2+5 \mathrm{NaBiO}_3+14 \mathrm{HNO}_3 \\
& 5 \mathrm{Bi}\left(\mathrm{NO}_3\right)_3+2 \mathrm{NaMnO}_4+7 \mathrm{H}_2 \mathrm{O}
\end{aligned}\) Balancing the number of Na -atoms and the nitrate radicals, we get \(\begin{aligned}
& 2 \mathrm{Mn}\left(\mathrm{NO}_3\right)_2+5 \mathrm{NaBiO}_3+14 \mathrm{HNO}_3- \\
& 5 \mathrm{Bi}\left(\mathrm{NO}_3\right)_3+2 \mathrm{NaMnO}_4+3 \mathrm{NaNO}_3+7 \mathrm{H}_2 \mathrm{O}
\end{aligned}\) In an acidic medium, iodate ( IO-₃) oxidises iodide (I-) to iodine and itself gets reduced to iodine.

Ionic reaction: IO-₂ + I- + H+→ I₂ + H₂O

Oxidation half-reaction: 21- → I₂ + 2e

Reduction half-reaction: 2IO-3 + 12H++ 10 e →6I₂ + 6H₂O Multiplying equation (1) by 5 and then adding to equation (2) we get, 10I- + 21O-3+ 12H+ → 6I2 + 6H₂O or, 5I- + 1O-3 + 6H+ — 3I₂ + 3H₂O This is the balanced ionic equation of the reaction.

Oxidation number method

In any redox reaction, the increase in the oxidation number of some of the atoms is balanced by the decrease in the oxidation number of some other atoms.

The steps which are to be followed while balancing the oxidation-reduction equation by this method are given below— After identifying the oxidant and reductant, the skeleton equation for the reaction is written.

The elements of the reactants and the products changing oxidation number are identified and the oxidation number of the concerned atoms is mentioned.

The reactant in which the element undergoes a decrease in oxidation number is the oxidant, while the reactant in which the element undergoes an increase in oxidation number is the reductant.

Redox Reactions Notes

As oxidation and reduction are complementary to each other, die increase and decrease in oxidation numbers should necessarily be equal, For this reason, the respective formulae of the oxidants and reductants are multiplied by a possible suitable integer so that the changes in oxidation numbers arc equalised.

For balancing the equation, it may sometimes be necessary to multiply the formula of other substances participating in the reaction by a suitable integer.

If the reactions are carried out in an acidic medium, then, to balance the number of O -atoms, one molecule of O- is added for each O -atom to the side of the equation deficient in oxygen. To balance the number of -atoms, H+ ions are added to the side deficient in hydrogen.

In case of a reaction occurring in an alkaline medium, to balance the number of O -atoms on both sides of the equations, for each O -atom one molecule of water is added to the side deficient in O -atoms and to the opposite side two OH- ions for each water molecule are added.

Again for balancing the number of FI -atoms on both sides of the equation, for each 2 -atom one OH- ion is added to the side which contains excess 2 -atoms and the same number of FI₂O molecules are added to the other side.

Example 1. Copper dissolves in concentrated HNO₃ to form CU(NO₃)₂, NO₂ and H₂O

Reaction:

In the given reaction, the increase in oxidation number of Cu -atom =(+2)-0 = 2 unit (oxidation) and the decrease in oxidation number of N -atom =(+5)-(+4) = 1 unit (reduction).

To nullify the effect of increase and decrease in the oxidation numbers, the ratio of the number of Cu -atoms and UNO₂ molecules in the reaction should be 1:2. So the equation may be written as—

Cu + 2HNO₃ → Cu(NO₃)₂ + 2NO₂ + H₂O

Now, to produce one molecule of Cu(NO₃)₂ two NO₂ radicals i.e. two molecules of UNO₂ are required.

Hence in the reaction further addition of two molecules of HNO₃ is necessary. So the balanced equation is expressed as—

Cu + 4HNO₃→ Cu(NO₃)₂ + 2NO₂ + 2H₂O

Now, to produce one molecule of Cu(NO₃)₂, two NO₃ radicals i.e. two molecules of UNO₂ are required.

Hence in the reaction further addition of two molecules of HNO₂ is necessary. So the balanced equation is expressed as—

Cu+4HNO₃ Cu(NO₃)²⁺2NO²⁺2H₂O

When H₂S gas is passed through chlorine water H₂SO₄ is produced.

Reaction:

In this reaction, an increase in the oxidation number of S = (+6) — (— 2) = 2 units (oxidation) and a decrease In the oxidation number of Cl = 0 – (— 1 ) I unit (reduction). So decrease In oxidation number for two (‘,1 -atoms or I molecule of Cl2 -2 unit.

Redox Reactions Notes

To neutralise the effect of Increase and decrease In oxidation number in the given equation, the number of molecules of H₂S and Cl₂ should be in the ratio of 2: i.e., 1: <1.

Therefore, the equation becomes—

H₂S+4CI₂+HCl+H₂SO₄

Balancing the number of 11 and O -atoms on both sides gives the balanced equation —

H₂S+4Cl₂+4H₂O-8HCl+H₂SO₄

NH₃ gas when passed over heated Cut) produces Cu, N₂ and H₂O.

In this reaction, increase In oxidation number of N=0-(-3) = 3 unit (oxidation) and decrease In oxidation number of Cu =(+2)-0 = 2 unit (reduction). As, in a redox reaction, the total increase in oxidation number is equal to the total decrease In oxidation number, the number of molecules of CuO and Nil₃ in the reaction should be in the ratio of 3:2. Hence, the balanced equation will be—

3CuO+2NH₃→3Cu+N₂+3H₂O

In the reaction between KMnO₄ and H₂O₂, the products obtained were K₂SO₂ MnSO₂, H₂O And O₂.

Reaction:

In this reaction increases in oxdination number of O=0-(-1)=1 (oxidaxtion) and dexrease in oxidation number of MN= (+7)-(+20)=5 unit (reduction).

The total increase in the oxidation number of two 0 -atoms presents one molecule of H₂O₂To balance the decrease and increase in oxidation numbers, the ratio of the number of KMnO₂ and H₂O, molecules in the equation for the reaction will be 2:5. Again from 2 molecules of KMnO₄ and 5 molecules of H₂O₂, 2 molecules of MnSO₂ and 5 molecules of O₂ are produced respectively. Thus the equation becomes—

⇒ \(\begin{aligned}
& 2 \mathrm{KMnO}_4+5 \mathrm{H}_2 \mathrm{O}_2+ \mathrm{H}_2 \mathrm{SO}_4 \rightharpoondown \\
& \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+5 \mathrm{O}_2+\mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Redox Reactions Notes

Again, for the formation of 1 molecule of K₂SO₄ and 2 molecules of MnSO₄, three SO4- radicals are required and hence three H₂SO₄ molecules are necessary on the left-hand side. Besides this, the total number of H-atoms in 5 molecules of H₂O₂ and 3 molecules of H₂SO₄ = 16. These H-atoms produce water molecules. Therefore, 8 molecules of H₂O are to be placed on the right-hand side. So the balanced equation will be—

⇒ \(\begin{aligned}
& 2 \mathrm{KMnO}_4+5 \mathrm{H}_2 \mathrm{O}_2+3 \mathrm{H}_2 \mathrm{SO}_4-7 \\
& \mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+5 \mathrm{O}_2+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

White phosphorus and concentrated NaOH react together to yield NaH₂PO₂ and PH₃. Reaction

The increase in oxidation number of P [P to NaH₂PO₉ ] = +1- 0 = 1 unit (oxidation). The decrease in oxidation number of P [P to PH₃] = 0-(-3) = 3 unit (reduction). To balance the increase and decrease in oxidation number, three P atoms for oxidation and one P -atom for reduction are required. Thus four P -atoms are necessary.

Redox Reactions Notes

Now, in the oxidation of P, NaH₂PO₂ and its reduction, PH₃ are formed. So the oxidation of three P atoms forms 3 molecules of NaH₂PO₉ and for this, three NaOH molecules are required. Again 1 atom of P reduction produces 1 molecule of PH₃. So the equation will be—

P₄+3NaOH + H₂O→ 3NaH₂PO₂ + PH₃

On the right side of the equation, there are 6 oxygen atoms, out of which 3 atoms will come from 3 molecules of NaOH and for the rest three atoms, 3 molecules of H₂O will be necessary. Hence, the balanced equation will be

P₄+ 3NaOH + H₂O → 3NaH₂PO₂ + PH₃

In NaOH solution, Zn reacts with NaNO₃ to yield Na₂ZnO₂, NH₃ and H₂O.

Reaction:

In the reaction, an increase in the oxidation number of Zn =(+2) -0 = 2 unit (oxidation) and a decrease in the oxidation number of N =(+5)-(-3) = 8 unit (reduction). As the increase and decrease in oxidation number in the reaction must be equal, the number of Zn -atoms and the number of molecules of NaNO₃ should be in the ratio of 4:1. Now, 1 molecule of NaNO₃ and 4 atoms of Zn produce 1 molecule of NH3 and 4 molecules of Na₂ZnO₂ respectively. Therefore the reaction is—

4Zn + NaNO₃ + NaOH →4Na₂ZnO₂ + NH₃ + H₂O

Redox Reactions Notes

Again formation of 4 molecules of Na₂ZnO₂ requires 8 Na -atoms, out of which 1 atom is supplied by 1 molecule of NaNO₃. Additional 7 Na -atoms come from NaOH on the left side. To balance H -atoms on both sides, 1 H₂O molecule is to be placed on the right side. Thus the balanced equation will be —

4Zn + NaNO₃ + 7NaOH →4Na₂ZnO₂ + NH₃ + 2H₂O

In the reaction between Cr₂O₃ and Na₂O₂, Na₂CrO₄ and NaOH are produced.

Reaction:

In this reaction, an increase in the oxidation number of Cr = (+6)- (+3) = 3 unit (oxidation) and a decrease in the oxidation number of O =(- 1 )-(- 2) = 1 unit (reduction). Thus a total increase in the oxidation number of two Cr -atoms =3×2 = 6 units and the total decrease in the oxidation number of two O -atoms = 1×2 = 2 units.

To balance the increase and decrease in oxidation number, the ratio of Cr₂O₃ and Na₂O₂ should be =1:3.

Now 2 molecules of Na₂CrO₄ are produced from 1 molecule of Cr₂O₃. Hence the equation will be as follows—

Cr₂O₃+3Na₂O₂+H₂O-2Na₂CrO₂+NaOH

If Na, H and O- atoms are balanced on both sides, the balanced equation will stand as —

Cr₂O3+ 3Na₂O₂ + H₂O→ 2Na₂CrO₄ + 2NaOH

Redox Reactions Notes

White phosphorus reacts with copper sulphate solution to produce Cu, H₃PO₄ and H₂SO₄.

Reaction:

In this reaction, an increase in the oxidation number of P = (+5)-0 = 5 unit (oxidation) and a decrease in the oxidation number of Cu =(+2)-0 = 2 unit (reduction).

Since the increase and decrease in oxidation number must be equal, in the given reaction, the ratio of the number of atoms of P and the number of CuSO₄ molecules should be in the ratio of 2: 5. Again 2 molecules of H₃PO₄ and five Cu -atoms will be produced respectively from two P atoms and five CuSO₄ molecules. As a result, the equation becomes—

2P + 5CuSO₄ + H₂O→ 5Cu + 2H₃PO₄ + H₂SO₄

To balance the number of SO²‾₄ radicals on both sides of the equation, 5 molecules of H₂SO₄ are to be added to the right-hand side ofthe equation.

2P + 5CuSO₄ + H₂O→5Cu + 2H₃PO₄ + 5H₂SO₄

Now, the total number of H-atoms present in 2 molecules of H₃PO₄ and 5 molecules of H₂SO₄ =16. So, for balancing the number of H-atoms, 8 water molecules are to be placed on the left-hand side. So, the balanced equation will be

2P + 5CuSO₄ + 8H₂O→5Cu + 2H₃PO₄ + 5H₂SO₄

Aluminium powder when boiled with caustic soda solution yields sodium aluminate and hydrogen gas.

Reaction:

⇒ \(\stackrel{-0}{\mathrm{~A}} \mathrm{l}+\mathrm{NaOH}+\stackrel{+1}{\mathrm{H}} \mathrm{H}_2 \mathrm{O} \longrightarrow \stackrel{+3}{\mathrm{NaAlO}_2}+\stackrel{0}{\mathrm{H}_2}\)

Aluminium is oxidised in this reaction to produce sodium aluminate. On the other hand, the H -atoms of NaOH and H₂O are reduced to produce H₂. Therefore, the change in oxidation number in the reaction may be shown as follows—

The increase in oxidation number of A1 = (+3) -0 = 3 unit (oxidation), the decrease in oxidation number of 1 H atom of NaOH molecule \(=(+1)-\left(\frac{1}{2} \times 0\right)\) (reduction) and decrease in oxidation number of 2 H -atoms
of water molecule = 2 x (+1) -2×0 = 2 unit(reduction).

Hence, the total decrease in oxidation number for the Hatoms in 1 molecule of NaOH and 1 molecule of H₂O =3 unit.

Redox Reactions Notes

Since, in a chemical reaction, the increase and decrease in oxidation number are the same, the ratio of the number of A1 atoms, NaOH molecule and water molecules in the given reaction should be =1: 1: 1.

⇒ \(\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NaAlO}_2+\frac{1}{2} \mathrm{H}_2+\mathrm{H}_2\)

Hence, the given reaction may be represented as:

Now, to express the number of molecules of reactants and products in terms of whole numbers, both sides of the equation should be multiplied by 2.

So, the balanced equation will be as follows:

⇒ \(\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NaAlO}_2+\frac{1}{2} \mathrm{H}_2+\mathrm{H}_2\)

Determination of equivalent mass of an element or compound in disproportionation reaction: Ifin oxidation and reduction reaction, the change in oxidation number of any element or an element of any compound participating in a disproportionation reaction be n1 and n2 respectively and M be the molecular mass of that element or compound, then the equivalent mass of that element or compound \(=\frac{M}{n_1}+\frac{M}{n_2}\)

In oxidation reaction (P4— change in oxidation number of each P -atom = 1 unit. So the total change in oxidation number of four P-atoms =4×1 =4 units. In the reduction reaction, (P4->PH3), the change in oxidation number of each P-atom is 3 units. So the total change in oxidation number of four P-atoms =4×3 = 12 units. Thus in this reaction, the equivalent mass of P₄.

\(=\frac{M}{4}+\frac{M}{12}=\frac{4 \times 31}{4}+\frac{4 \times 31}{12}=31+10.33 \text {= } 41.33\) [since atomic mass of p = 31]

Redox Titration

A process by which a standard solution of an oxidant (or a standard solution of a reductant) is completely reacted with a solution of an unknown concentration of a reductant (or with a solution of an unknown concentration of an oxidant) in the presence of a suitable indicator is called redox titration.

In a redox titration, an oxidant (or a reductant) reacts completely with an equivalent amount of a reductant (or an oxidant). Therefore, in a redox titration, the number of grams equivalent of oxidant = number of grams equivalent of reductant.

Types of redox titrations

Permanganometry titration: A titration in which KMnO₄ solution is used as the standard solution. In this titration, no indicators are needed.

Example: The amount of iron present in an acidic ferrous ion (Fe²⁺) solution can be estimated by titrating the solution with a standard solution of KMnO₄.

⇒ \(\begin{aligned}
& \mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+ 8 \mathrm{H}^{+} \longrightarrow \\
& \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Oxidation reaction: Fe²⁺→ Fe3+ + e

Redox Reactions Notes

Reduction reaction: MnO-₄ + 8H+ + 5e→Mn²⁺ + 4H₂O

So, in this reaction, the equivalent mass of Fe²⁺ — an atomic mass of Fe and the equivalent mass of KMnO₄ \(=\frac{1}{5} \times\) Molecular or formula mass of KMnO₄ According to the reaction (1), 1 mol of KMnO4 = 5 mol of Fe²⁺ ions or, 1000 mLof1 mol of KMnO₄ solution = 5 x 55.85g of Fe²⁺ ions or, 1 mLof l(M) KMnO₄ solution = 0.2792g of Fe²⁺ ions mL of 5(N) KMnO₄ solution = 0.2792g of Fe²⁺ ions. [In the given reaction, the normality of KMnO₄ solution is five times its molarity.]

lmL of (N) KMnO₄ solution = 0.05585g of Fe²⁺ ions

Dichromatometry titration: A ptration in which a standard solution of potassium dichromate (K₂Cr₂O₇) is used.

In this titration, sodium or barium diphenylamine sulphonate or diphenylamine is used as an indicator.

Example: The amount of iron present in an acidic ferrous ion (Fe²⁺) solution can be calculated by titrating the solution with a standard solution of K₂Cr₂O₇.

⇒ \(\begin{aligned}
& \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \\
& 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{2+}+\mathrm{H}_2 \mathrm{O} \\
&
\end{aligned}\)

In reaction (1), the equivalent mass of Fe²⁺ is equal to the atomic mass of Fe, & the equivalent mass of K₂Cr2O₇ is equal to one-sixth of its molecular or formula mass. According to the reaction(l), 1 mol of K₂Cr₂O₇ H 6mol of Fe²⁺ ions or, 1000mL of 1M K₂Cr₂O₇ = 6 x 55.85 g of Fe²⁺ ions or, lmLof1M K₂Cr₂O₇ solution s 0.3351g of Fe²⁺ ions or, lmL of 6N K₂Cr₂O₇ solution s 0.3351g of Fe²⁺ ions [In the given reaction, normality of K₂Cr₂O₇ solution is six times its molarity.]

lmL of IN K₂Cr₂O₇ solution = 0.05585g of Fe²⁺ ions

Redox Reactions Notes

iodometry titration: In this titration, KI in excess is added to a neutral or an acidic solution of an oxidant. Consequently, the oxidant quantitatively oxidises I- ions (reductant), to form I₂. The liberated I₂ is then titrated with a standard Na₂S₂O₃ solution using starch as an indicator.

The amount of liberated iodine is calculated from the volume of standard Na₂S₂O₃ solution consumed in one titration. After the amount of liberated iodine is known, one can calculate the amount of oxidant by using the balanced chemical equation for the reaction of oxidant with iodine.

Example: Iodometric titration is often used for quantitative estimation of Cu²⁺ ions. The addition of excess KI to a neutral or an acidic solution of Cu²⁺ ions results in oxidation of 1 to I2 and reduction of Cu²⁺ to Cu⁺.

So, in the reaction(l), the equivalent mass of \(\mathrm{Cu}^{2+}=\frac{2 \times \text { atomic mass of } \mathrm{Cu}}{2}=\text { atomic mass of } \mathrm{Cu}\)

The reaction of I₂ with Na₂S₂O₃ is: \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

In this reaction, Oxidation reaction: \(2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 e\)

Reduction reaction: I₂ + 2e→2I-

Therefore, the equivalent mass of Na₂S₂O₃

⇒ \(=\frac{2 \times \text { molecular or formula mass of } \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}{2}\)

= molecular or formula mass of Na₂S₂O₃

Equivalent mass of I₂ \(=\frac{\text { molecular mass of } \mathrm{I}_2}{2}=\text { atomic mass of } \mathrm{I}\)

According to the reactions (1) and (2), 2mol of Cu²⁺ = 1 mol of I2 and 2 mol of Na₂S₂O₃ s 1 mol of I₂ Therefore, 2mol of Na₂S₂O₃= 2 mol of Cu²⁺ or, 1 mol of Na₂S₂O₃ = l mol of Cu²⁺ = 63.5g of Cu²⁺ or, l mol of lM Na₂S₂O₃ solution = 63.5 g of Cu²⁺ or, 1 mol ofIN Na₂S₂O₃ solution = 63.5gof Cu²⁺ [As in the reaction of Na2S203 with I2, the equivalent mass of Na₂S₂O₃ is equal to its molecular mass].