Question

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained 0.1M $Mn{O}_4^{-}$ and 1.72M $H^{+}$, and was treated with $F{e}^{2+}$, necessary to reduce 90% of $KMn{O}_4$ to $M{n}^{2+}$. $[E_{Mn{O}_4-/M{n}^{2+}}^0=1.51V]$

• A. 1.4V
• B. 1.5V
• C. 1.6V
• D. 1.3V

Answer 1

The correct option is B 1.5V

$Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}$
0.1 M 1.72 M 0 M
0.01 M 1.0 M 0.09 M

$E_{RP\left(Mn{O}_4^{2-}/M{n}^{2+}\right)}^0>E_{RP\left(H^{+}/H_2\right)}^0$

So, the permanganate electrode behaves as the cathode.

$E_{cell}^0=1.51V$

Using Nernst's equation : $(n=5)$
$E_{Mn{O}_{4^{2-}}/M{n}^{2+}}=1.51+\frac{0.059}{5}log\frac{\left[Mn{O}_{4^{-}}\right][H^{+}{]}^8}{\left[M{n}^{2+}\right]}$

$E_{cell}=1.51-0.011=1.499\approx 1.5V$