Find the equivalent weight of Na2C2O4 andKMnO4 in the following redox reaction.Na2C2O4+KMnO4⟶2CO2+Mn2+ (acidic medium) (at.wt., Na=23,C=12,O=16,K=39,Mn=55 ).

Solution:

${Na}_{2} C_{2} O_{4} + {KMnO}_{4} ⟶ 2 {CO}_{2} + {Mn}^{2 +}$

(i) $C_{2}^{+ 3} ⟶ 2 C^{+ 4} + 2 e^{-}$

$\begin{aligned} ∴ Eq. wt. of {Na}_{2} C_{2} O_{4} = \frac{Mol. wt. of {Na}_{2} C_{2} O_{4}}{no. of electrons lost by} \\ one molecule of {Na}_{2} C_{2} O_{4} \\ = \frac{(2 \times 23) + (2 \times 12) + (4 \times 16)}{2} \end{aligned}$

$\begin{aligned} = \frac{46 + 24 + 64}{2} \\ = \frac{134}{2} = 67 Ans. \\ (ii) {Mn}^{+ 7} + 5 e^{-} ⟶ {Mn}^{+ 2} \\ ∴ Eq. wt. of {KMnO}_{4} = \frac{Mol. wt. of {KMnO}_{4}}{no. of electrons gained by} \\ one molecule of {KMnO}_{4} \\ = \frac{39 + 55 + (4 \times 16)}{5} = \frac{158}{5} \\ = 31.6 Ans. \end{aligned}$