Question

At what $pH$, given half cell $MnO _4^{-}(01 M ) \mid Mn ^{2+}(0001 M )$ will have electrode potential of $1282\, V$? (Nearest Integer)
(Given: $E _{ MnO _{4}^-| Mn ^{2+}}^{ o }=154 \,V , \frac{2303 RT }{ F }=0059\, V$)

Answer 1

Correct Answer: 3

$Mn{O}_4^{-}+8H^{+}+5e^{-}\rightleftharpoons M{n}^{2+}+4H_{2}O$
$E=E^{\circ }-\frac{0.059}{5}\log \frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]{\left[H^{+}\right]}^8}$
$1.282=1.54-\frac{0.059}{5}\log \frac{10^{-3}}{10^{-1}\times {\left[H^{+}\right]}^8}$
$\frac{0.258\times 5}{0.059}=\log \frac{10^{-2}}{{\left[H^{+}\right]}^8}$
$\Rightarrow 21.86=-2+8pH$
$\therefore pH=2.98$
$\simeq 3$

So, the correct answer is 3.