1. Identify the half-cells

The cell is:

$\text{Pt}/M/M^{3+}(0.001\,\text{M})/Ag^+(0.01\,\text{M})/Ag$

Given:

• Cell emf: $E_{\text{cell}} = 0.421\,\text{V}$
• $E^\circ_{Ag^+/Ag} = 0.80\,\text{V}$
• We need $E^\circ_{M^{3+}/M}$

The silver electrode has higher reduction potential, so it acts as cathode:

$Ag^+ + e^- \to Ag$

The metal electrode acts as anode:

$M \to M^{3+} + 3e^-$

So overall reaction is:

$M + 3Ag^+ \to M^{3+} + 3Ag$

2. Write Nernst equation for the cell

For the overall reaction,

$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$

Here,

• $n=3$
• Reaction quotient:

$Q = \frac{[M^{3+}]}{[Ag^+]^3}$

Substitute concentrations:

$Q = \frac{10^{-3}}{(10^{-2})^3} = \frac{10^{-3}}{10^{-6}} = 10^3$

Thus,

$\log Q = 3$

So,

$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{3}(3)$

$E_{\text{cell}} = E^\circ_{\text{cell}} - 0.0591$

Given $E_{\text{cell}}=0.421$,

$0.421 = E^\circ_{\text{cell}} - 0.0591$

$E^\circ_{\text{cell}} = 0.421 + 0.0591 = 0.4801\,\text{V}$

3. Relate standard cell potential to electrode potentials

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

So,

$0.4801 = 0.80 - E^\circ_{M^{3+}/M}$

Hence,

$E^\circ_{M^{3+}/M} = 0.80 - 0.4801 = 0.3199\,\text{V}$

$E^\circ_{M^{3+}/M} \approx 0.32\,\text{V}$

4. Check options

• A: 0.38 V ❌
• B: 0.32 V ✅
• C: 1.28 V ❌
• D: 0.66 V ❌

Therefore, the correct answer is:

$\boxed{0.32\,\text{V}}$

So, Option B is correct.