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 Cu2+(aq) is not balenced by its hydration enthelpy. Hence $\text{E}^\ominus(\text{Cu}^{2+}/\text{Cu})$ is positive.
$\text{M(s)}+\Delta_\text{a}\text{H}\rightarrow\text{M(g)}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
$\text{M(g)}+\Delta_\text{i}\text{H}\rightarrow\text{M}^{2+}{\text{(g)}}$
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 Cu2+(aq) is not balenced by its hydration enthelpy. Hence $\text{E}^\ominus(\text{Cu}^{2+}/\text{Cu})$ is positive.
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