Calculate the electrode potential of copper electrode dipped in 0.1M CuSO4 solution at 298K. Given that E^ _ cu^ 2 +/cu =0.34V.

Cu^ 2+ +2e^- Cu;n=2 E_ Cu^ 2+ /Cu =E^ _ Cu^ 2+ /Cu -0.059 n 1 [Cu^ 2+ ] E_ Cu^ 2+ /Cu =0.34-0.059 2 1 0.1 E_ Cu^ 2+ /Cu =0.34-0.0295=0.3105V

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Question

Calculate the electrode potential of copper electrode dipped in 0.1M CuSO₄ solution at 298K. Given that $\text{E}^{\circ}_{\text{cu}^{2}+/\text{cu}}=0.34\text{V}.$

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Answer

$\text{Cu}^{2+}+2\text{e}^-\xrightarrow{ \ \ \ \ \ \ }\text{Cu};\text{n}=2$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=\text{E}^{\circ}_{\text{Cu}^{2+}/\text{Cu}}-\frac{0.059}{\text{n}}\log\frac{1}{[\text{Cu}^{2+}]}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=0.34-\frac{0.059}{2}\log\frac{1}{0.1}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=0.34-0.0295=0.3105\text{V}$

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