Question

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contains 0.1 M $Mn{O}_4^{-}$ and $0.8M{H}^{+}$ and which was treated with $F{e}^{2+}$ necessary to reduce $90\%$ of $Mn{O}_4^{-}$ to $M{n}^{2+}$. $E_{Mn{O}_4^{-}/M{n}^{2+}}^o=1.51V$. Give answer in $V$ in nearest integer.

Answer 1

The net cell reaction is
$Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$
The reduction half reaction is
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

	$Mn{O}_4^{-}$	$H^{+}$	$M{n}^{2+}$
Initial concentration (M)	0.1	0.8	0
Final concentration (M)	$0.1-x$	$0.8-8x$	$x$
Final concentration (M)	0.01	0.08	0.09

The expression for the cell potential is
$E{=}^0-\frac{0.0592}{n}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right][H^{+}{]}^8}$
Substitute values in the above expression.
$E=1.51-\frac{0.0592}{5}log\frac{0.09}{0.01\times (0.08{)}^8}$
$E=1.51-0.099=1.411V$