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in Oxidation-Reduction Reactions by (49.4k points)

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)

1 Answer

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(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.

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