Question

At what $pH$, given half cell $MnO _4^{-}(0.1 M ) \mid Mn ^{2+}(0.001 M )$ will have electrode potential of $1.282\, V$ ?______ (Nearest Integer)
Given $E _{ MnO _{4}^-| Mn ^{2+}}^{ o }=1.54 \,V , \frac{2.303 RT }{ F }=0.059\, V$

Answer 1

$MnO _4^{-}+8 H ^{+}+5 e ^{-} \rightleftharpoons Mn ^{2+}+4 H _2 O$
$ E = E ^{\circ}-\frac{0.059}{5} \log \frac{\left[ Mn ^{2+}\right]}{\left[ MnO _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+8 pH $
$ \therefore pH =2.98 $
$ \simeq 3 $