Question

In the cell
$P{t}_{(s)}\mid H_2(g,1$ bar $)\left|HC{l}_{(aq)}\right|AgC{l}_{(s)}\left|A{g}_{(s)}\right|P{t}_{(s)}$ The cell potential is $0.92V$ when a $10^{-6}$ molal $HCl$ solution is used. The standard electrode potential of $\left(AgCl/Ag,C{l}^{-}\right)$electrode is______ $V$
$($ Given, $\frac{2.303RT}{F}=0.06V$ at $298K)$

Answer 1

Answer: 0.2

At anode: $H_{2(g)}⟶2H_{(aq)}^{+}+2e^{-}\times 1$
At cathode: $AgC{l}_{(s)}+e^{-}⟶A{g}_{(s)}+C{l}_{(aq)}^{-}\times 2$
Net reaction:
$H_{2(g)}+2AgC{l}_{(s)}\stackrel{(s)}{\to }2H_{(aq)}^{+}+2A{g}_{(s)}+2C{l}_{(aq)}^{-}$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.06}{2}{\log }_{10}{\left[H^{+}\right]}^2{\left[C{l}^{-}\right]}^2$
$0.92=\left(E_{H_2/H^{+}}^{\circ }+E_{AgCl/Ag,C{l}^{-}}^{\circ }\right)$
$-\frac{0.06}{2}{\log }_{10}\left[{\left(10^{-6}\right)}^2\times {\left(10^{-6}\right)}^2\right]$
$0.92=\left(0+E_{AgCl/Ag,C{l}^{-}}^{\circ }\right)-0.03{\log }_{10}{\left(10^{-6}\right)}^4$
$E_{AgCl/Ag,C{l}^{-}}^{\circ }=0.92+0.03\times (-24)=0.2V$