Redox and Equivalent ConceptHard

Question

Consider the following reaction:
x MnO4- + y C2O42- + z H+ → x Mn2+ + (2y) CO2 + (z/2) H2O
The values of x, y and z in the reaction are respectively :-

Options

A.5,2 and 16
B.2,5 and 8
C.2, 5 and 16
D.5,2 and 8

Solution

(i) MnO4- + 8H+ + 5e- → Mn+2 + 4H2O
(ii) C2O4-2 → 2CO2 + 2e- 
⇒ (i)×2 + (ii)× 5

  2MnO4- + 16H+ + 5C2O4-2 → 2Mn+2 + 8H2O + 10CO2

Given equation :

On comparing with balanced equation
we get ; x = 2, y = 5, z = 16.

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