1. Write the given half-reactions with their standard reduction potentials

   $Mn^{2+} + 2e^- \to Mn \qquad E^\circ = -1.18\,V$

   $Mn^{3+} + e^- \to Mn^{2+} \qquad E^\circ = +1.51\,V$

   (The factor 2 written in the question does not change the electrode potential.)

2. Target reaction

   We need:

   $3Mn^{2+} \to Mn + 2Mn^{3+}$

   Split it into oxidation and reduction parts:

   • Reduction: $Mn^{2+} + 2e^- \to Mn \qquad E^\circ_{\text{red}} = -1.18\,V$

   • Oxidation: $Mn^{2+} \to Mn^{3+} + e^-$

     Since $Mn^{3+} + e^- \to Mn^{2+} \qquad E^\circ = +1.51\,V$ reversing it gives $Mn^{2+} \to Mn^{3+} + e^- \qquad E^\circ_{\text{ox}} = -1.51\,V$

3. Balance electrons

   Multiply the oxidation half-reaction by 2:

   $2Mn^{2+} \to 2Mn^{3+} + 2e^-$

   Reduction half-reaction remains:

   $Mn^{2+} + 2e^- \to Mn$

   Adding:

   $3Mn^{2+} \to Mn + 2Mn^{3+}$

4. Calculate standard cell potential

   $E^\circ_{\text{cell}} = E^\circ_{\text{red(cathode)}} + E^\circ_{\text{ox(anode)}}$

   $E^\circ_{\text{cell}} = (-1.18) + (-1.51) = -2.69\,V$

5. Interpretation

   Since $E^\circ_{\text{cell}} < 0$ the reaction as written is non-spontaneous under standard conditions, so it will not occur spontaneously.

6. Check options

   • A: $-0.33\,V$; will not occur ❌
   • B: $-0.33\,V$; will occur ❌
   • C: $-2.69\,V$; will not occur ✅
   • D: $-2.69\,V$; will occur ❌

Therefore, the correct option is C.