Question

What is the standard electrode potential for the electrode, $Mn{O}_4^{-}|Mn{O}_2$ in an acid solution?


$E_{Mn{O}_4^{-}/M{n}^{2+}}^0=1.51V{E}_{Mn{O}_2/M{n}^{2+}}^0=1.23V$

• A.
  $2.45V$
• B. $3.35V$
• C.
  $0.87V$
• D.
  $1.70V$

Answer 1

The correct option is D

$1.70V$
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O;.....\left(1\right)$ $E^0=1.51V$
$\therefore \Delta G^0=-5\left(1.51\right)F=-7.55F$

$Mn{O}_2+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O;.......\left(2\right)$ $E^0=1.23V$
$\therefore \Delta G^0=-2\left(1.23\right)F=-2.46F$

On subtracting equation (2) from (1), we get,

$Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$

$\therefore \Delta G^0=-7.55F-(-2.46F)=-5.09F$

$\therefore$ $E_{Mn{O}_4^{-},Mn{O}_2}^o=\frac{\Delta G^O}{-nF}=\frac{-5.09F}{-3F}=1.70V$