Question

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained 0.1 M $Mn{O}_4^{-}$ and 0.8 M $H^{+}$ and which was treated with $F{e}^{2+}$ necessary to reduce 90% of the $Mn{O}_4^{-}$ to $M{n}^{2+}$.
$Mn{O}_4^{-}+8H^{+}+5e\to M{n}^{2+}+H_{2}O,E^{\circ }=1.51V$

• A. 1.39 V
• B. 2.39 V
• C. 1.59 V
• D. None of these

Answer 1

The correct option is A 1.39 V

Solving using Nerst equation:

$E=E^0-\frac{0.059}{n}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right].[H^{+}{]}^8}$

Aftre 90% reaction completion:

$E=1.51-\frac{0.059}{5}log\frac{0.09}{0.01\times {0.08}^8}$

$E=1.51-\frac{0.059}{5}(log9-8log0.08)$

$E=1.51-\frac{0.059}{5}\left(9.73\right)$

$E=(1.51-0.115)=1.395V$