Derive expression to calculate emf (reduction) of the following half cells at `25^@C`
`Cu^(2+)|Cu(s)`

Explanation

To derive the expression for the electromotive force (emf) of the half-cell reaction \( Cu^{2+} | Cu(s) \) at \( 25^\circ C \), we will use the Nernst equation. Here are the steps to arrive at the expression:

### Step 1: Write the half-reaction
The reduction half-reaction for copper is:
\[ Cu^{2+} + 2e^- \rightarrow Cu(s) \]
This indicates that cupric ions (\( Cu^{2+} \)) gain two electrons to form solid copper (\( Cu(s) \)).
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