In the cell $\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}, 1$ bar $) / \mathrm{HCl}(\mathrm{aq}) \| \mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mid \mathrm{Pt}(\mathrm{s})$, the cell potential is $0.92 \mathrm{~V}$ when a $10^{-6}$ molal $\mathrm{HCl}$ solution is used. The standard electrode potential of $\left(\mathrm{AgCl} / \mathrm{Ag}, \mathrm{Cl}^{-}\right)$ electrode is:
$\left\{\right.$ Given $: \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$ at $\left.298 \mathrm{~K}\right\}$
Correct Option: , 4
Given that:
$\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}, 1$ bar $) / \mathrm{HCl}(\mathrm{aq}) \| \mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mid \mathrm{Pt}(\mathrm{s})$
$\mathrm{E}_{\text {cell }}=0.92 \mathrm{~V}$
Now, $\mathrm{E}_{\text {cell }}=E_{\mathrm{H}_{2}(\mathrm{~g}) / \mathrm{H}^{+}(\mathrm{aq})}^{\circ}+E_{\mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mathrm{Cr}^{-}}^{\circ}-\frac{0.06}{n} \log Q$
Cell reaction: $\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-}$
$\mathrm{AgCl}(\mathrm{s})+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq})$
Net cell reaction:
$\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{Cl}^{-}(\mathrm{aq})$
$\therefore \quad Q=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Cl}^{-}\right]}{\left(\mathrm{P}_{\mathrm{H}_{2}}\right)^{1 / 2}}$
Here, $10^{-6}$ molal $\mathrm{HCl}$ solution is used
So $Q=\frac{10^{-6} \times 10^{-6}}{1}=10^{-12}$
(assuming molality = molarity)
Now, $0.92=E_{\mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}), \mathrm{Cl}^{-}}^{\circ}-\frac{0.06}{1} \log 10^{-12}$
$E_{\mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mathrm{Cl}^{\circ}}^{\circ}=0.92+[0.06 \times(-12)]$
$=0.92-0.72=0.20 \mathrm{~V}$