Solve the following
Question:

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 $\left.298 \mathrm{~K}\right\}$

1. $0.94 \mathrm{~V}$

2. $0.76 \mathrm{~V}$

3. $0.40 \mathrm{~V}$

4. $0.20 \mathrm{~V}$

Correct Option: , 4

Solution:

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, $0.92=E_{\mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}), \mathrm{Cl}^{-}}^{\circ}-\frac{0.06}{1} \log 10^{-12}$

$E_{\mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mathrm{Cl}^{\circ}}^{\circ}=0.92+[0.06 \times(-12)]$

$=0.92-0.72=0.20 \mathrm{~V}$