## Balancing Of Chemical Equations Involving Redox Reactions

Redox reaction can be balanced with the help of two methods. These are the

- Ion-electron method
- 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.
- 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, the 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] 2Cr^{3+}

**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, H_{2}O or H^{+} is used. First, oxygen atoms are balanced by adding H_{2}O molecules to the side that needs O-atoms.

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) of the reaction occurs in an alkaline medium, for balancing the H and O -atoms, H_{2}O 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:**

1. In the presence of H_{2}SO_{4}, potassium dichromate (K_{2}Cr_{2}O_{2}) and ferrous sulphate (FeSO_{4}) react together to produce ferric sulphate [Fe_{2}(SO_{4})_{3}] and chromic sulphate [Cr_{2}(SO_{4})_{3}].

**Reaction:**

K_{2}Cr_{2}O_{7} + FeSO_{4} + H_{2}SO_{4}→ K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + Fe_{2}(SO_{4})_{3} + H_{2}O

**The reaction can be expressed in ionic form as:**

Cr^{3+} + Fe^{3+} + H_{2}O

** Oxidation half-reaction:** Fe^{2+}→Fe^{3+}+ e ……………………(1)

** Reduction half-reaction:** Cr_{2}O_{7}^{2-}+ Cr^{3+ }……………………(2)

1. Balancing the Cr -atom: Cr_{2}O_{7}^{2-}– Cr^{3}+ 7H_{2}O

2.To equalise the number of O -atoms on both sides, 7 water molecules are to be added to the right side. 2Cr^{3+} + 7H_{2}O

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

∴ One water molecule is required for each O-atom.

3. 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.

4. For equalising the charge on both sides, 6 electrons are to be added to the left side.

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

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.

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

6FeSO_{4} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} → 3Fe_{2}(SO_{4})_{3} + Cr_{2}(SO_{4})_{3}+ K_{2}SO_{4}+ 7H_{2}O

∴ For 2H^{+} ions, one H_{2}SO_{4 }molecule is required

2. In presence of H_{2}SO_{4}, KMnO_{4} and FeSO_{4} react together to produce MnSO_{4} and Fe_{2}(SO_{4})_{3}.

** Reaction:** KMnO_{4} + FeSO_{4} + H_{2}SO_{4} → K_{2}SO_{4} + MnSO_{4}+ Fe_{2}(SO_{4})3 + H_{2}O

The equation can be expressed in ionic form as:

MnO_{4} + Fe^{2+} + H^{+}→-Mn^{2+} + Fe^{3+} + H_{2}O

**Oxidation half-reaction:** Fe^{2+} — Fe^{3+} + e ………………………..(1)

**Reduction half-reaction:** MnO_{4}^{–} + 8H^{+} + 5e → Mn^{2+} + 4H_{2}O………………………..(2)

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.

As one Fe_{2}(SO_{4})_{3} molecule contains two Fe -atoms, the equation is multiplied by 2

10Fe^{2+} + MnO_{4}^{–} + 16H^{+} →10Fe^{3+} + 2Mn^{2+} + 8H_{2}O

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

10FeSO_{4} + 2KMnO_{4}+ 8H_{2}SO_{4} → 5Fe_{2}(SO_{4})_{3} + 2MnSO_{4} + 8H_{2}O

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

10FeSO_{4}+ 2KMnO_{4} + 8H_{2}SO_{4} → 5Fe_{2}(SO_{4})_{3} + 2MnSO_{4} + K_{2}SO_{4} + 8H_{2}O

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

3. In the reaction between K_{2}Cr_{2}O_{7} acidified with dilute H_{2}SO_{4} and KI, Cr_{2}(SO_{4})_{3} and I_{2} are formed.

K_{2}Cr_{2}O_{7}+KI + H_{2}SO_{4} → K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + I_{2} + H_{2}O

The equation can be expressed in ionic form as— Cr_{2}O_{7}^{2-}+I^{–} + H^{+} — Cr^{3+} + I_{2} + H_{2}O

**Oxidation half-reaction:** 2I^{–}→ I_{2} + 2e……………..(1)

**Reduction half-reaction:** Cr_{2}O_{7} ^{2-}+ I^{–}+14H^{+} + 6e — 2Cr^{3+} + 7H_{2}O ……………………….(2)

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

This is the balanced equation of the reaction in ionic form. The above ionic reaction can be expressed in molecular form as follows

6KI + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4}→3I_{2} + Cr_{2}(SO_{4})_{3} + 7H_{2}O

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

6KI + K_{2}Cr_{2}O_{7}+ 7H_{2}SO_{4} → 3I_{2} + Cr_{2}(SO_{4})_{3} + 4K_{2}SO_{4} + 7H_{2}O

4. In the reaction between KMnO_{4}, acidified with dilute H_{2}SO_{4} and oxalic acid (H_{2}C_{2}O_{4}), MnSO_{4} and CO_{2} were produced.

Reaction: KMnO_{4} + H_{2}C_{2}O_{4} + H_{2}SO_{4} → K_{2}SO_{4} + MnSO_{4} + CO_{2} + H_{2}O

**Oxidation half-reaction:** C_{2}O_{4}→ 2CO_{4}^{2-}+ 2e ……………………..(1)

**Reduction half-reaction:** MnO_{4}^{–} +8H^{+}+ 5e → Mn^{2+}+ 4H_{2}O……………………..(2)

Now, multiplying equation (1) by 5 and equation (2) by 2 and then adding them, we get,

5C_{2}O_{4} ^{2-}+ 2MnO_{4} + 16H^{+}+10CO_{2} + 2Mn^{2+} + 8H_{2}O

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

5H_{2}C_{2}O_{4} + 2KMnO_{4}+ 3H_{2}SO_{4}→ 10CO_{2} + Mn^{2+} +8H_{2}O

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

5H_{2}C_{2}O_{4} + 2KMnO_{4} + 3H_{2}SO_{4} → 10CO_{2} + 2MnSO_{4} + K_{2}SO_{4} + 8H_{2}O

5. In NaOH solution, Zn reacts with NaNO_{3} to yield Na_{2}ZnO_{2}, NH_{3} and H_{2}O.

**Reaction:**

Zn + NaNO_{3} + NaOH — Na_{2}ZnO_{2} + NH_{3} + H_{2}O The equation can be expressed in ionic form as—

Zn + NO^{–} + OH^{–}→ ZnO_{2–} →+ NH_{3} +H_{2}O

**Oxidation half-reaction:** Zn + 4OH^{–}→ ZnO_{2}^{–} + 2H_{2}O + 2C ……………………..(1)

**Reduction half-reaction:** NO_{3} + 6H_{2}O + 8C — NH_{3} + 9OH^{– }……………………..(2)

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

4Zn + 16OH- + NO, + 6H_{2}O→ 4ZnO_{2} + NH_{3} + 90H- + 8H_{2}O

Or, 4Zn + 7OH^{–}+ NO^{–}_{3}→ 4ZnO_{2}^{–} + NH_{3} + 2H_{2}O

It is the balanced equation of the reaction in ionic form. Expressing the above equation in molecular form

4Zn + 7NaOH + NaNO_{3}→ 4Na_{2}ZnO_{2} + NH_{3} + 2H_{2}O

It is the molecular form of the balanced equation of the reaction.

6. In the presence of HNO_{3}, sodium bismuthatic (NaHO_{3}) reacts with Mn(NO_{3})_{2} to produce coloured sodium permanganate (NaMnO_{4}) and itself gets reduced to bismuth nitrate.

**Reaction:** NaBIO_{3} + Mn(NO_{3}) + UNO_{2} →NaMnO_{4} + Bi(NO_{3}) + H_{2}O The equation can be expressed in ionic form as— BIO_{2} + Mn^{2+} → Bl^{3+} + MnO_{4} + H_{2}O

**Oxidation half-reaction:** Mn^{2+} + 4H_{2}O → MnO_{4}^{–}+8H^{+}→+5e ……………………(1)

**Reduction half-reaction:** BiO_{3} + 6H^{+} + 2e — Bi^{3+} + 3H_{2}O…………………….(2)

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

2Mn^{2+} + 8H_{2}O + 5BiO_{3} ^{–} +30H^{+}→ 2MnO_{4}^{–} + 16H^{+} + 5Bi^{3+} +15H_{2}O

2Mn^{2+} + 5BiO_{3} + 14H^{+}→ 5Bi^{3+}+ + 2MnO_{4}^{–} +7HO

This is the balanced ionic equation of the reaction. The equation in the molecular form stands as—

2Mn(NO_{3})_{2} + 5NaBiO_{3}+ 14HNO_{3} → 5Bi(NO_{3})_{3} + 2NaMnO_{4} + 7H_{2}O

**Ionic reaction:** IO_{3}^{–} + I^{–}+ H^{+}→ I_{2} + H_{2}O

**Oxidation half-reaction:** 2I^{–} → I_{2} + 2e ……………………….(1)

** Reduction half-reaction:** 2 IO_{3}^{–} + 12H^{+}+ 10 e → I_{2} + 6H_{2}O ……………………….(2)

Multiplying equation (1) by 5 and then adding to equation (2) we get,

10I^{–} + 2IO_{3}^{–}+ 12H^{+} → 6I_{2}+ 6H_{2}O or, 5I^{–} + 1O_{3}^{–}+ 6H^{+} — 3I_{2} + 3H_{2}O

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.
- 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 FI_{2}O molecules are added to the other side.

**Example **

1. Copper dissolves in concentrated HNO_{3} to form Cu(NO_{3})_{2}, NO_{2} and H_{2}O

**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 Cu(NO_{3})_{2} _{ }molecules in the reaction should be 1:2. So the equation may be written as

Cu + 2HNO_{3} → Cu(NO_{3})_{2} + 2NO_{2} + H_{2}O

Now, to produce one molecule of Cu(NO_{3})_{2} two NO_{2} radicals i.e. two molecules of UNO_{2} are required. Hence in the reaction further addition of two molecules of HNO_{3} is necessary. So the balanced equation is expressed as

Cu + 4HNO_{3}→ Cu(NO_{3})_{2} + 2NO_{2} + 2H_{2}O

Now, to produce one molecule of Cu(NO_{3})_{2}, two NO_{3} radicals i.e. two molecules of UNO_{2} are required.

**Hence in the reaction further addition of two molecules of HNO _{2} is necessary. So the balanced equation is expressed as**

Cu+4HNO_{3} Cu(NO_{3})^{→ 2+}2NO^{2+}2H_{2}O

**When H _{2}S gas is passed through chlorine water H_{2}SO_{4} 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 Cl_{2} -2 unit.

To neutralise the effect of Increase and decrease In oxidation number in the given equation, the number of molecules of H_{2}S and Cl_{2} should be in the ratio of 2: i.e., 1: <1.

**Therefore, the equation becomes**

H_{2}S+4CI_{2}+HCl+H_{2}SO_{4}

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

3. NH_{3} gas when passed over heated Cut) produces Cu, N_{2} and H_{2}O.

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 NH_{3} in the reaction should be in the ratio of 3:2. Hence, the balanced equation will be—

3CuO+2NH_{3}→3Cu+N_{2}+3H_{2}O

4. In the reaction between KMnO_{4} and H_{2}O_{2}, the products obtained were K_{2}SO_{2} MnSO_{2}, H_{2}O And O_{2}.

**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 H_{2}O_{2}

To balance the decrease and increase in oxidation numbers, the ratio of the number of KMnO_{2} and H_{2}O, molecules in the equation for the reaction will be 2:5.

Again from 2 molecules of KMnO_{4} and 5 molecules of H_{2}O_{2}, 2 molecules of MnSO_{2} and 5 molecules of O_{2} are produced respectively. Thus the equation becomes—

2KMnO_{4} + 5H_{2}O_{2}+ H_{2}SO_{4}→ K_{2}SO_{4}+ 2MnSO_{4} + 5O_{2} + H_{2}O

Again, for the formation of 1 molecule of K_{2}SO_{4} and 2 molecules of MnSO_{4}, three SO_{4}– radicals are required and hence three H_{2}SO_{4} molecules are necessary on the left-hand side. Besides this, the total number of H-atoms in 5 molecules of H_{2}O_{2} and 3 molecules of H_{2}SO_{4} = 16.

These H-atoms produce water molecules. Therefore, 8 molecules of H_{2}O are to be placed on the right-hand side. So the balanced equation will be—

2KMnO_{4} + 5H_{2}O_{2} + 3H_{2}SO_{4} → K_{2}SO_{4} + 2MnSO_{4} + 5O_{2}+ 8H_{2}O

**5. White phosphorus and concentrated NaOH react together to yield NaH _{2}PO_{2} and PH_{3}. Reaction**

The increase in oxidation number of P [P to NaH_{2}PO_{2} ] = +1- 0 = 1 unit (oxidation). The decrease in oxidation number of P [P to PH_{3}] = 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.

Now, in the oxidation of P, NaH_{2}PO_{2} and its reduction, PH_{3} are formed. So the oxidation of three P atoms forms 3 molecules of NaH_{2}PO_{9} and for this, three NaOH molecules are required. Again 1 atom of P reduction produces 1 molecule of PH_{3}. So the equation will be

P_{4}+3NaOH + H_{2}O→ 3NaH_{2}PO_{2} + PH_{3}

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 H_{2}O will be necessary. Hence, the balanced equation will be

P_{4}+ 3NaOH + H_{2}O → 3NaH_{2}PO_{2} + PH_{3}

6. In NaOH solution, Zn reacts with NaNO_{3} to yield Na_{2}ZnO_{2}, NH_{3} and H_{2}O.

**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 NaNO_{3} should be in the ratio of 4:1. Now, 1 molecule of NaNO_{3} and 4 atoms of Zn produce 1 molecule of NH3 and 4 molecules of Na_{2}ZnO_{2} respectively. Therefore the reaction is—

4Zn + NaNO_{3} + NaOH →4Na_{2}ZnO_{2} + NH_{3} + H_{2}O

Again formation of 4 molecules of Na_{2}ZnO_{2} requires 8 Na -atoms, out of which 1 atom is supplied by 1 molecule of NaNO_{3}. Additional 7 Na -atoms come from NaOH on the left side. To balance H -atoms on both sides, 1 H_{2}O molecule is to be placed on the right side. Thus the balanced equation will be —

4Zn + NaNO_{3} + 7NaOH →4Na_{2}ZnO_{2} + NH_{3} + 2H_{2}O

**7. In the reaction between Cr _{2}O_{3} and Na_{2}O_{2}, Na_{2}CrO_{4 }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 Cr_{2}O_{3} and Na_{2}O_{2} should be =1:3. Now 2 molecules of Na_{2}CrO_{4} are produced from 1 molecule of Cr_{2}O_{3}. Hence the equation will be as follows—

Cr_{2}O_{3}+3Na_{2}O_{2}+H_{2}O→ 2Na_{2}CrO_{4}+NaOH

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

Cr_{2}O3+ 3Na_{2}O_{2} + H_{2}O→ 2Na_{2}CrO_{4} + 2NaOH

8. White phosphorus reacts with copper sulphate solution to produce Cu, H_{3}PO_{4} and H_{2}SO_{4}.

**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 CuSO_{4} molecules should be in the ratio of 2: 5. Again 2 molecules of H_{3}PO_{4} and five Cu -atoms will be produced respectively from two P atoms and five CuSO_{4} molecules. As a result, the equation becomes—

2P + 5CuSO_{4} + H_{2}O→ 5Cu + 2H_{3}PO_{4} + H_{2}SO_{4}

To balance the number of SO²‾_{4} radicals on both sides of the equation, 5 molecules of H_{2}SO_{4} are to be added to the right-hand side ofthe equation.

2P + 5CuSO_{4} + H_{2}O→5Cu + 2H_{3}PO_{4} + 5H_{2}SO_{4}

Now, the total number of H-atoms present in 2 molecules of H_{3}PO_{4} and 5 molecules of H_{2}SO_{4} =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 + 5CuSO_{4} + 8H_{2}O→5Cu + 2H_{3}PO_{4} + 5H_{2}SO_{4}

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

**Reaction:**

Aluminium is oxidised in this reaction to produce sodium aluminate. On the other hand, the H -atoms of NaOH and H_{2}O are reduced to produce H_{2}. 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 H_{2}O =3 unit.

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:**

If in 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, (P_{4}→PH_{3}), 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 P_{4}.

⇒ \(\frac{M}{4}+\frac{M}{12}=\frac{4 \times 31}{4}+\frac{4 \times 31}{12}=31+10.33 \text {= } 41.33\)

∴ The 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**

**1. Permanganometry titration**

A titration in which KMnO_{4} 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 (Fe^{2+}) solution can be estimated by titrating the solution with a standard solution of KMnO_{4}.

** Reaction:**

MnO_{4} + 5Fe^{2+} + 8H+ → Mn^{2+} + 5Fe3+ + 4H_{2}O …………………….(1)

**Oxidation reaction:** Fe^{2+}→ Fe3+ + e

**Reduction reaction:** MnO_{4} ^{–}+ 8H^{+} + 5e → Mn^{2+} + 4H_{2}O

So, in this reaction, the equivalent mass of Fe^{2+} — an atomic mass of Fe and the equivalent mass of KMnO_{4} \(=\frac{1}{5} \times\) Molecular or formula mass of KMnO_{4}

According to the reaction (1), 1 mol of KMnO_{4} = 5 mol of Fe^{2+} ions or, 1000 mLof 1 mol of KMnO_{4} solution = 5 × 55.85g of Fe^{2+} ions or, 1 mLof l(M) KMnO_{4} solution = 0.2792g of Fe^{2+} ions mL of 5(N) KMnO_{4} solution = 0.2792g of Fe^{2+} ions. [In the given reaction, the normality of KMnO_{4} solution is five times its molarity.]

lmL of (N) KMnO_{4} solution = 0.05585g of Fe^{2+} ions

**2. Dichromatometry titration**

A ptration in which a standard solution of potassium dichromate (K_{2}Cr_{2}O_{7}) 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 (Fe^{2+}) solution can be calculated by titrating the solution with a standard solution of K_{2}Cr_{2}O_{7}.

**Reaction: **

Cr_{2}O_{7}+ 14H+ + 6Fe^{2+}→ 2Cr^{3+}+ 6Fe^{2+} + H_{2}O …………….(1)

In reaction (1), the equivalent mass of Fe^{2+} is equal to the atomic mass of Fe, & the equivalent mass of K_{2}Cr_{2}O_{7} is equal to one-sixth of its molecular or formula mass.

According to the reaction (1), 1 mol of K_{2}Cr_{2}O_{7} H 6mol of Fe^{2+} ions

Or, 1000mL of 1M K_{2}Cr_{2}O_{7} = 6 × 55.85 g of Fe^{2+} ions

Or, lmLof1M K_{2}Cr_{2}O_{7} solution s 0.3351g of Fe^{2+} ions

Or, lmL of 6N K_{2}Cr_{2}O_{7} solution s 0.3351g of Fe^{2+} ions

[In the given reaction, the normality of K_{2}Cr_{2}O_{7} solution is six times its molarity.]

lmL of IN K_{2}Cr_{2}O_{7} solution = 0.05585g of Fe^{2+} ions

**3. 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 I_{2}. The liberated I_{2} is then titrated with a standard Na_{2}S_{2}O_{3} solution using starch as an indicator.

The amount of liberated iodine is calculated from the volume of standard Na_{2}S_{2}O_{3} 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 Cu^{2+} ions. The addition of excess KI to a neutral or an acidic solution of Cu^{2+} ions results in oxidation of 1 to I2 and reduction of Cu^{2+} 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 I _{2 }with Na_{2}S_{2}O_{3} 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:** I_{2} + 2e → 2I^{–}

Therefore, the equivalent mass of Na_{2}S_{2}O_{3}

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

= molecular or formula mass of Na_{2}S_{2}O_{3}

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

According to the reactions (1) and (2), 2mol of Cu^{2+} = 1 mol of I2 and 2 mol of Na_{2}S_{2}O_{3} s 1 mol of I_{2 }

Therefore, 2mol of Na_{2}S_{2}O_{3}= 2 mol of Cu^{2+} or, 1 mol of Na_{2}S_{2}O_{3} = l mol of Cu^{2+} = 63.5g of Cu^{2+} or, l mol of lM Na_{2}S_{2}O_{3} solution = 63.5 g of Cu^{2+} or, 1 mol of IN Na_{2}S_{2}O_{3} solution = 63.5gof Cu^{2+} [As in the reaction of Na_{2}S_{2}O_{3} with I_{2}, the equivalent mass of Na_{2}S_{2}O_{3} is equal to its molecular mass].