Question

The standard electrode potential in V for the electrode $Mn{O}_4^{-}/Mn{O}_2$ in solution is: (round off the answer to the nearest integer)
Given: $E_{Mn{O}_4^{-}/M{n}^{2+}}^o=1.51V$ and $E_{Mn{O}_2/M{n}^{2+}}^o=1.23V$

Answer 1

$Mn{O}_4^{-}+8H^{-}+5e\to M{n}^{2+}+4H_{2}O;E_1^o=1.51V$ ........... (i)

$\therefore -\Delta G_1^o=5\times 1.51\times F$

$=7.55F$

$Mn{O}_2+4H^{+}+2e\to M{n}^{2+}+2H_{2}O;E_2^o=1.23V$ ........... (ii)

$\therefore -\Delta G_2^o=2\times 1.23\times F$

$=2.46F$

Subtracting equation (ii) from (i)

$Mn{O}_4^{-}+4H^{+}+3e\to 2H_{2}O+Mn{O}_2;E_3^o=?$
or $-\Delta G_3^o=n_3{E}_3^{o}F$

$\therefore -\Delta G_3^o=-\Delta G_1^o-(-\Delta G_2^o)$

$3E_3^{o}F=7.55F-2.46F$

$\therefore E_3^o=\frac{5.09}{3}=1.70volt$