Question

In the cell
${\left(Pt\right)}_{(s)}|H_2(g,1bar)\left|{\left(HCl\right)}_{(aq)}\right|{\left(AgCl\right)}_{(s)}\left|{\left(Ag\right)}_{(s)}\right|{\left(Pt\right)}_{fs}$
The cell potential is $0.92V$ when a $10^{-6}$ molal $HCl$ solution is used. The standard electrode potential of $\left(AgCl/Ag,{\left(Cl\right)}^{-}\right)$ electrode is $bV.$ The value of $10b$ is..(Nearest integer)
(Given, $\frac{2.303RT}{F}=0.06V$ at $298K$ )

Answer 1

Answer: 2

At anode: $H_{2(g)}\to 2H_{(a)}^{+}+2e^{-}\times 1$
At Cathode: $AgC{l}_{(s)}+e^{-}\to A{g}_{(s)}+C{l}_{(a)}^{-}\times 2$
Net reaction: $H_{2(g)}+2AgC{l}_{(s)}\stackrel{(o)}{\to }2H_{(a)}^{+}+2A{g}_{(s)}+2C{l}_{(\text{ial })}^{-}$
$E_{\text{cell }}=E_{\text{cell }}^s-\frac{0.06}{2}{\log }_{10}{H}^{+2}[Cl{]}^2$
$0.92-E_{H_{2}M}^a+E_{A_{B}CiAg.Cr}^{\circ }-\frac{0.06}{2}$
${\log }_{10}10^{-6^2}\times 10^{-6^2}$
$0.92-01E_{\text{AgCiAe. C}}^{\circ }0.03{\log }_{10}10^{-6^4}$
$E_{AgCl/Ag,C{l}^{-}}^{\circ }=0.92+0.03\times (-24)=0.2V=bV10b=2$