Find $E ^{0}$ for the following reaction : $MnO _{4}^{-}+4 H ^{+}+3 e ^{-} \longrightarrow MnO _{2}+2 H _{2} O$

Question

Given :

(i) $MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \longrightarrow Mn ^{2+}+4 H _{2} O \quad E ^{0}=x_{1} V$

(ii) $MnO _{2}+4 H ^{+}+2 e ^{-} \longrightarrow Mn ^{2+}+2 H _{2} O \quad E ^{0}= x _{2} V$

Find $E ^{0}$ for the following reaction :

$MnO _{4}^{-}+4 H ^{+}+3 e ^{-} \longrightarrow MnO _{2}+2 H _{2} O$

(A) $x_{2}-x_{1}$

(B) $x_{1}-x_{2}$

(C) $\frac{5 x_{1}-2 x_{2}}{3}$

(D) $\frac{2 x_{1}-5 x_{2}}{3}$

Answer 1

The E⁰ for the above reaction will be $\frac{5 x_{1}-2 x_{2}}{3}$. So, The correct option of this question is (C).