In the cell Pt_{(s)} | H_2(g, 1 , bar) | HCl_ | vedclass

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(D) The cell reaction is: $\frac{1}{2} H_2(g) + AgCl(s) \rightarrow H^+(aq) + Cl^-(aq) + Ag(s)$.Applying the Nernst equation: $E_{cell} = E^o_{cell} -

Vedclass · Accepted Answer

$0.20$

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(D) The cell reaction is: $\frac{1}{2} H_2(g) + AgCl(s) ightarrow H^+(aq) + Cl^-(aq) + Ag(s)$.Applying the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.06}{n} \log Q$.Here,$n = 1$ and $Q = [H^+][Cl^-] / (P_{H_2})^{1/2}$.Given $[H^+] = 10^{-6} \, m$ and $[Cl^-] = 10^{-6} \, m$,so $Q = 10^{-6} 	imes 10^{-6} = 10^{-12}$.$0.92 = E^o_{cell} - 0.06 \log(10^{-12})$.$0.92 = E^o_{cell} - 0.06 	imes (-12)$.$0.92 = E^o_{cell} + 0.72$.$E^o_{cell} = 0.92 - 0.72 = 0.20 \, V$.Since $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{AgCl/Ag, Cl^-} - E^o_{H^+/H_2}$,and $E^o_{H^+/H_2} = 0 \, V$,we get $E^o_{AgCl/Ag, Cl^-} = 0.20 \, V$.

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