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  • Assume a cell with the following reaction <img alt="\begin{aligned} &\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}^{+}\left(1 \times 10^{-3} \mathrm{M}\right) \rightarrow \mathrm{Cu}^{2+}(0.250 \mathrm{M})+2 \mathrm{Ag}_{(\mathrm{s})} \\ &\mathrm{E}

Assume a cell with the following reaction \begin{aligned} &\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}^{+}\left(1 \times 10^{-3} \mathrm{M}\right) \rightarrow \mathrm{Cu}^{2+}(0.250 \mathrm{M})+2 \mathrm{Ag}_{(\mathrm{s})} \\ &\mathrm{E}_{\text {cell }}^{\oplus}=2.97 \mathrm{~V} \end{aligned} E_{\text {cell }} for the above reaction is_________ \mathrm{V} . (Nearest integer) [Given : \log 2.5=0.3979, \mathrm{~T}=298 \mathrm{~K}]
 

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best_answer

The reaction in the electrochemical cell is

\begin{aligned} &\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 10^{-3} \mathrm{M}\; \;\ \ \; \quad 0.25 \mathrm{M} \end{aligned}

Nernst equation gives the \mathrm{E_{cell}} as

\begin{aligned} \mathrm { E_{cell} } &=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left\{\frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}\right\} \\ &=2.97-0.03 \log \left(\frac{0.25}{10^{-6}}\right) \\ &=2.97-0.03 \log \left(2.5 \times 10^{5}\right) \end{aligned}

          \begin{aligned} &=2.97-0.03(5+\log 2.5) \\ &=2.97-0.03(5.3979) \\ &=2.8 \mathrm{~V} \end{aligned}

Hence, the answer is 3.

Posted by

sudhir.kumar

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