Question

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

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

Answer 1

Answer: (A)

$Pt(s)H_2(g,1bar)HCl(aq)AgCl(s)Ag(s)\mid Pt(s)$
Anode: $H_2\stackrel{10^{-6}m}{\to }2H^{+}+2e\times 1$
Cathode : $e^{-}+AgCl(s)⟶Ag(s)+C{l}^{-}(aq)$
$H_2(g)l+AgCl(s)⟶2H^{+}+2Ag(s)+2C{l}^{-}(aq)$
$E_{\text{cell }}=E_{\text{cell }}^0-\frac{0\cdot 06}{2}{\log }_{10}\left({\left(H^{+}\right)}^2\cdot {\left(C{l}^{-}\right)}^2\right)$
$.925=\left(E_{H_2/H^{+}}^0+E_{AgCl/Ag,C{l}^{-}}^0\right)-\frac{0.06}{2}{\log }_{10}\left({\left(10^{-6}\right)}^2{\left(10^{-6}\right)}^2\right)$
$.92=0+E_{AgCl/Ag,C{l}^{-}}^0-0.03{\log }_{10}{\left(10^{-6}\right)}^4$
$E_{AgCl}^0/Ag,C{l}^{-}=\cdot 9\cdot 2+\cdot 03\times -24=0\cdot 2V$