1. Identify the half-cells

The cell is

Left electrode is the standard hydrogen electrode, so $E_{\text{left}} = E^\circ_{H^+/H_2} = 0\,\text{V}$

Right electrode involves the couple: $M^{3+} + 2e^- \rightarrow M^+$ with $E^\circ_{M^{3+}/M^+} = 0.2\,\text{V}$

Since the hydrogen electrode is on the left and the metal ion electrode is on the right, $E_{\text{cell}} = E_{\text{right}} - E_{\text{left}} = E_{\text{right}}$ Thus, $E_{\text{right}} = 0.1115\,\text{V}$

2. Apply Nernst equation to the right electrode

For the reduction reaction $M^{3+} + 2e^- \rightarrow M^+$

The reaction quotient is $Q = \frac{[M^+]}{[M^{3+}]}$

So the Nernst equation is

Given $\frac{[M^+]}{[M^{3+}]} = 10^a$ so

Hence,

But given $E = 0.1115\,\text{V}$ Therefore,

3. Solve for $a$

4. Final answer

5. Comparison with stored correct answer

Stored correct answer = 3

Our derived answer = 3

So the answer matches.