Question
Calculate the electrode potential of copper electrode dipped in 0.1M $CuSO_4$ solution at 298K. Given that $\text{E}^{\circ}_{\text{cu}^{2}+/\text{cu}}=0.34\text{V}.$
✓
Answer
$\text{Cu}^{2+}+2\text{e}^-\xrightarrow{ \ \ \ \ \ \ }\text{Cu};\text{n}=2$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=\text{E}^{\circ}_{\text{Cu}^{2+}/\text{Cu}}-\frac{0.059}{\text{n}}\log\frac{1}{[\text{Cu}^{2+}]}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=0.34-\frac{0.059}{2}\log\frac{1}{0.1}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=0.34-0.0295=0.3105\text{V}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=\text{E}^{\circ}_{\text{Cu}^{2+}/\text{Cu}}-\frac{0.059}{\text{n}}\log\frac{1}{[\text{Cu}^{2+}]}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=0.34-\frac{0.059}{2}\log\frac{1}{0.1}$
$\text{E}_{\text{Cu}^{2+}/\text{Cu}}=0.34-0.0295=0.3105\text{V}$
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