Question

In the cell
$Pt _{( s )} \mid H _{2}( g , 1$ bar $)\left| HCl _{( aq )}\right| AgCl _{( s )}\left| Ag _{( s )}\right| Pt _{( s )}$ The cell potential is $0.92\, V$ when a $10^{-6}$ molal $HCl$ solution is used. The standard electrode potential of $\left( AgCl / Ag , Cl ^{-}\right)$electrode is______ $V$
$\left(\right.$ Given, $\frac{2.303 RT }{ F }=0.06\, V$ at $\left.298\, K \right)$

Answer 1

At anode: $H _{2( g )} \longrightarrow 2 H _{( aq )}^{+}+2 e ^{-} \times 1$
At cathode: $AgCl _{( s )}+ e ^{-} \longrightarrow Ag _{( s )}+ Cl _{( aq )}^{-} \times 2$
Net reaction:
$H _{2( g )}+2 AgCl _{( s )} \xrightarrow{( s )} 2 H _{( aq )}^{+}+2 Ag _{( s )}+2 Cl _{( aq )}^{-}$
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.06}{2} \log _{10}\left[ H ^{+}\right]^{2}\left[ Cl ^{-}\right]^{2}$
$0.92=\left( E _{ H _{2} / H ^{+}}^{\circ}+ E _{ Ag Cl / Ag , Cl ^{-}}^{\circ}\right)$
$-\frac{0.06}{2} \log _{10}\left[\left(10^{-6}\right)^{2} \times\left(10^{-6}\right)^{2}\right]$
$0.92=\left(0+ E _{ AgCl / Ag , Cl ^{-}}^{\circ}\right)-0.03 \log _{10}\left(10^{-6}\right)^{4}$
$E _{ AgCl / Ag , Cl ^{-}}^{\circ}=0.92+0.03 \times(-24)=0.2 V$