Question

The following two redox reactions are given:
i) $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O;E^o=X_{1}V$
ii) $Mn{O}_2+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O;E^o=X_{2}V$
Find $E^o$ for the following reaction.
$Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$

• A. $X_2-X_1$
• B. $X_1-X_2$
• C. $\frac{5X_1-2X_2}{3}$
• D. $\frac{2X_1-5X_2}{3}$

Answer 1

The correct option is B $X_1-X_2$

i) $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O;E^o=X_{1}V$......(1)

ii) $Mn{O}_2+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O;E^o=X_{2}V$......(2)

Reverse equation (2)
iii) $M{n}^{2+}+2H_{2}O\to Mn{O}_2+4H^{+}+2e^{-},E^o=-X_{2}V$......(3)

Add equation (3) to equation (1),
$Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$

$E^o$ for the above reaction is $X_1-X_2$