Question

What is the standard electrode potential for the electrode $Mn{O}_4^{-}/Mn{O}_2$ in solution?
(Given: $E_{Mn{O}_4^{-}/M{n}^{2+}}^{\circ }=1.51volt,E_{Mn{O}_2/M{n}^{2+}}^{\circ }=1.23volt$).

Answer 1

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O;E^{\circ }=1.51volt$
$Mn{O}_2+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O;E^{\circ }=1.23volt$
Subtracting
$\overline{Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O;E^{\circ }=?}$
$E^{\circ }=\frac{5\times 1.51-2\times 1.23}{3}=\frac{7.55-2.46}{3}=\frac{5.09}{3}=1.70volt$.