Question

In the cell $P t(s) \mid H_{2}(g, 1 b a r|H C l(a q)| A g(s) \mid P t(s)$ the cell potential is $0.92$ when a $10^{-6}$ molal $HCl$ solution is used. THe standard electrode potential of $\left( AgCl / Ag , Cl ^{-}\right)$ electrode is:
$\left\{\right.$ given,$\frac{2.303 R T}{F}=0.06 V$ at $\left.298 K \right\}$

• A. 0.20 V
• B. 0.76 V
• C. 0.40 V
• D. 0.94 V

Answer 1

Answer: (A)

$Pt ( s ) H _{2}( g , 1 bar ) HCl ( aq ) AgCl ( s ) Ag ( s ) \mid Pt ( s )$
Anode: $H _{2} \xrightarrow{10^{-6} m }2 H ^{+}+2 e \times 1$
Cathode : $e ^{-}+ AgCl ( s ) \longrightarrow Ag ( s )+ Cl ^{-}( aq )$
$H _{2}( g ) l + AgCl ( s ) \longrightarrow 2 H ^{+}+2 Ag ( s )+2 Cl ^{-}( aq )$
$E _{\text {cell }}= E _{\text {cell }}^{0}-\frac{0 \cdot 06}{2} \log _{10}\left(\left( H ^{+}\right)^{2} \cdot\left( Cl ^{-}\right)^{2}\right)$
$.925=\left( E _{ H _{2} / H ^{+}}^{0}+E_{A g C l / A g, C l^{-}}^{0}\right)-\frac{0.06}{2} \log _{10}\left(\left(10^{-6}\right)^{2}\left(10^{-6}\right)^{2}\right)$
$.92=0+E_{A g C l / A g, C l^{-}}^{0}-0.03 \log _{10}\left(10^{-6}\right)^{4}$
$E _{ AgCl }^{0} / Ag , Cl ^{-}=\cdot 9 \cdot 2+\cdot 03 \times-24=0 \cdot 2 V$