Question

Balance the following reaction by oxidation number method and also identify oxidising agent.
(i) H₂S + MnO^-₄ ➝ S + Mn²⁺ + H₂O (Acidic medium)
(ii) Cl₂O₇ + H₂O₂ ➝ ClO^-₂ + O₂₍gas) (Basic medium)

Answer 1

(i)

(a) Write unbalanced equation,

(b) Oxidation number of Mn decreases from + 7 to +2. So, MnO^-₄ is reduced to Mn⁺² and oxidation number of S increase from -2 to 0 so, H₂S is oxidised to S. Hence, writing partial equation as,

Oxidation number decreases 5 (Reduction)

Oxidation number increases by 2 (Oxidation)
Multiplying equation 2 by 2 and equation 3 by 5, we get
2MnO^-₄ ➝ 2Mn⁺² … (4)
5H₂S ➝ 5S …(5)
Adding equations 4 and 5, we get
2MnO₄^– + 5H₂S ➝ 2Mn²⁺ + 5S … (6)
Change (-1) (0)  (+4) (0)

(c) Addition of water molecules in the direction of deficiency of oxygen.
2MnO₄ + 5H₂ S ➝ 2Mn²⁺ + 5S + 2Mn²⁺+5S + 8H₂O

(d) To balance hydrogen atom, H⁺ions are added to the left hand side of the reaction.
5H₂S + 2MnO^-₄ + 6H⁺ ➝ 5S + 2Mn²⁺ + 8H₂O
This is balanced equation.

(ii) 

(a) Write unbalanced equation,

(b) Oxidation number of Cl decrases from +7 to +3, so Cl₂O₇ is reduced to ClO^-₂ and oxidation number of O increases from -1 to 0. So, H₂O₂ is oxidised to O₂. Hence, writing partial equations,

Oxidation number decreases by 4 (Reduction)

Multiplying equation 2 by 1 and equation 3 by 4 and also balancing chlorine in ag (2)
Cl₂O₇ ➝ 2ClO^-₂   …(4)
4H₂O₂ ➝ 4O₂    ...(5)
Adding equation (4) and (5), we get
C1₂O₇ + 4H₂O₂ ➝ 2ClO₂ + 4O₂
Change (0) (0)  (-1) 0

(c) Addition of water molecular in the direction of deficiency of oxygen and addion of OH^– ions to left hand side
Cl₂O₇ + 4H₂O₂ + 2OH^– ➝ 2ClO₂^– + 4O₂ + 5H₂O
This is balanced equation.