Question

At what $pH$, given half cell $Mn{O}_4^{-}(0.1M)\mid M{n}^{2+}(0.001M)$ will have electrode potential of $1.282V$ ?______ (Nearest Integer)
Given $E_{Mn{O}_4^{-}|M{n}^{2+}}^o=1.54V,\frac{2.303RT}{F}=0.059V$

Answer 1

Answer: 3

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