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  • When   molal <img alt="\mathrm{HCL}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BH

When 10^{-4}  molal \mathrm{HCL}  solution is used the cell potential for the reaction \mathrm{\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_2(\mathrm{~g}, \mid \mathrm{bar}) /\mathrm{HCl}( aq ) 11 \mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mid \mathrm{Pt}(\mathrm{s})}  What will be the standard electrode potential of \mathrm{(AgCl (\mathrm{Ag ,Cl}),}   given that \mathrm{\frac{2.303 R T}{F}=0.6 \mathrm{~V}}  at  \mathrm{298 \mathrm{~K}} 

Option: 1

0.84V


Option: 2

0.32V


Option: 3

0.62V


Option: 4

0.44V


Answers (1)

best_answer

\mathrm{Ecell=E_{H_2}(g) / H^{+}(a q)+E_{A g C l}(s) / A g(s), Cl^{-}-\frac{0.06}{n} \log Q}

\mathrm{\mathrm\mathrm{Cell}\ \mathrm{rxm} \ \frac{1}{2} \mathrm{H}_2(g) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+e^{-}}

\mathrm{\mathrm{AgCl}(\mathrm{s})+e^{-} \rightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}^{-} \text { (aq) }}

\mathrm{ \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\mathrm{AgCl}(s) \rightarrow \mathrm{Ag}(\mathrm{s})+H^{+}+\dot{Q}^{-}(\mathrm{aq}) }

\mathrm{ Q=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Cl}^{-}\right]}{\left(\mathrm{PH}_{\mathrm{H}_2}\right)^{1 / 2}} }

Here,  \mathrm{ 10^{-8} }   molal \mathrm{ HCL } sol is used 

\mathrm{ Q=\frac{10^{-4} \times 10^{-4}}{1}=10^{-86} }

assuming molality = molarity
\mathrm{ 0.92=E_{A g C l(s) / A g(s)/ Cl^-}-\frac{0.06}{1} \log 10^{-8} }


\mathrm{ \text { E{AgCl}} / \mathrm{Ag (s)}, Cl^{-}=0.92+[0.06 \times(-8)] }
\mathrm{ =0.92-0.48 \\ }

\mathrm{ =0.44 \mathrm{~V} \\ }.

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Gunjita

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