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For a cell, \mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \mathrm{M}\right| \mathrm{Ag}(\mathrm{s}) ) the cell potential is found to be 0.43 \mathrm{~V} at 298 \mathrm{~K }. The magnitude of standard electrode potential for \mathrm{Cu}^{2+} / \mathrm{Cu} is ______ \times 10^{-2} \mathrm{~V}.

\mathrm{\left[\text { Given : } \mathrm{E}_{\mathrm{Ag}^{\ominus} / \mathrm{Ag}}=0.80 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right]}

Option: 1

34


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

For the cell

\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s} )

    \mathrm{Anode: }\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2 +}+2 e^{-}
\mathrm{Cathode:} \frac{2 \mathrm{Ag}^{+}+2 e^{-} \longrightarrow 2 \mathrm{Ag}}{\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}}

\mathrm{E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{M} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{[\mathrm{Ag^{+}}]^{2}} }
\mathrm{ 0.43=E_{cell}^{0}-\frac{0.06}{2} \log \frac{10^{-3}}{\left(10^{-2}\right)^{2}}}
\mathrm{E_{\text {cell }}^{0}=0.43+0.03 \log 10 \\ =0.46 \mathrm{~V} }

\mathrm{E_{\text {cell }}^{0}=E^{0}_{Ag^{+}|Ag}-E_{Cu^{2+}|Cu} ^{0}}
\mathrm{0.46= 0.8-E^{0}_{Cu^{2+}|Cu}}

\mathrm{E^{0}_{Cu^{2+}|Cu}=0.34 \mathrm{~V}=34 \times 10^{-2} \mathrm{~V}}

Hence answer is 34.

Posted by

Shailly goel

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