Question
The $\text{E}^\ominus(\text{M}^{2+}/\text{M})$ value for copper is positive $(+0.34V).$ What is possibly the reason for this? $($Hint: consider its high $\Delta_\text{a}\text{H}^\ominus$ and low $\Delta_\text{hyd}\text{H}^\ominus).$

Answer

$\text{E}^\ominus(\text{M}^{2+}/\text{M})$ for any metal is related to the sum of enthelpy changes taking place in following steps$:$
$\text{M(s)}+\Delta_\text{a}\text{H}\rightarrow\text{M(g)}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
$Cu$ has a high enthelpy of atomisation $(\Delta_\text{a}\text{H})$ and a low enthelpy of hydration $(\Delta_\text{hyd}\text{H})$. The high energy required to transform $Cu(s)$ to $Cu^{2+}(aq)$ is not balenced by its hydration enthelpy.
Hence $\text{E}^\ominus(\text{Cu}^{2+}/\text{Cu})$ is positive.

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