Question

In the cell
$\left(Pt\right)_{\left(\right. s \left.\right)}\left|\right.H_{2}\left(\right.g,1bar\left.\right)\left|\left(HCl\right)_{\left(\right. aq \left.\right)}\right|\left(AgCl\right)_{\left(\right. s \left.\right)}\left|\left(Ag\right)_{\left(\right. s \left.\right)}\right|\left(Pt\right)_{fs}$
The cell potential is $0.92\,V$ 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, $\frac{2 . 303 RT}{F}=0.06\,V$ at $298\,K$ )

Answer 1

At anode: $H_{2(g)} \rightarrow 2 H_{(a)}^{+}+2 e^{-} \times 1$
At Cathode: $AgCl _{(s)}+e^{-} \rightarrow Ag _{(s)}+ Cl _{(a)}^{-} \times 2$
Net reaction: $H _{2(g)}+2 AgCl _{(s)} \xrightarrow{(o)} 2 H _{(a)}^{+}+2 Ag _{(s)}+2 Cl _{(\text {ial })}^{-}$
$E _{\text {cell }}= E _{\text {cell }}^s-\frac{0.06}{2} \log _{10} H ^{+2}[ Cl ]^2$
$0.92-E_{H_2 M}^a+E_{A_B CiAg . C r}^{\circ} -\frac{0.06}{2}$
$\log _{10} 10^{-6^2} \times 10^{-6^2}$
$0.92-01 E^{\circ}_{\text {AgCiAe. C}} 0.03 \log _{10} 10^{-6^4}$
$E_{A g C l / A g, Cl ^{-}}^{\circ}=0.92+0.03 \times(-24)=0.2 V=b V 10 b=2$