Question

$I_{2}\left(s\right)\left|\right.I^{-}\left(0 . 1 M\right)$ half cell is connected to a $H^{+}\left(\right.aq\left.\right)\left|\right.H_{2}\left(1 bar\right),Pt$ half cell and e.m.f. is found to be $0.7714V$ . If $E_{I_{2} / I^{-}}^{o}=0.535V$ , find the $pH$ of $H^{+}\left|\right.H_{2},Pt$ half cell.

Answer 1

The cell reaction is
$H_{2 \left(\right. g \left.\right)}+I_{2 \left(\right. s \left.\right)}\rightleftharpoons2H_{\left(\right. a q \left.\right)}^{+}+2I_{\left(\right. a q \left.\right)}^{-}$
Here $n=2$ (n = number of electons involved )
$E=E^\circ -\frac{0.0591}{n}log\frac{\left[\right. H^{+} \left]\right.^{2} \left[\right. I^{-} \left]\right.^{2}}{P_{H_{2}}}$
$0.7714=0.535-\frac{0.0591}{2}log\frac{\left[\right. H^{+} \left]\right.^{2} \left[\right. 0.1 \left]\right.^{2}}{1}$
On solving $\left(\right.H^{+}\left.\right)=10^{- 3}$
$pH=-log\left(\right.H^{+}\left.\right)=-log\left(\right.10^{- 3}\left.\right)$
So $pH=3$