Given below are the half-cell reactions:
$$Mn^{2+} + 2e^- \to Mn$$; $$E^\circ = -1.18$$ V
$$2(Mn^{3+} + e^- \to Mn^{2+})$$; $$E^\circ = +1.51$$ V
The $$E^\circ$$ for $$3Mn^{2+} \to Mn + 2Mn^{3+}$$ will be:

We have to find the standard electrode potential for the overall reaction

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

For this purpose the two relevant half-cell reactions, together with their standard reduction potentials, are supplied:

$$Mn^{2+} + 2e^- \rightarrow Mn \qquad E^\circ_{\text{red,1}} = -1.18\text{ V}$$

$$Mn^{3+} + e^- \rightarrow Mn^{2+} \qquad E^\circ_{\text{red,2}} = +1.51\text{ V}$$

In the desired overall reaction one portion of $$Mn^{2+}$$ is being reduced to $$Mn$$ and another portion is being oxidised to $$Mn^{3+}$$. Therefore we treat the first equation as a reduction (cathodic) half-reaction and the second equation in the reverse direction as an oxidation (anodic) half-reaction.

Writing the oxidation half-reaction explicitly, we reverse the second equation and change the sign of its potential:

$$Mn^{2+} \rightarrow Mn^{3+} + e^- \qquad E^\circ_{\text{ox}} = -\,E^\circ_{\text{red,2}} = -(+1.51\text{ V}) = -1.51\text{ V}$$

The reduction half-reaction remains as given:

$$Mn^{2+} + 2e^- \rightarrow Mn \qquad E^\circ_{\text{red}} = -1.18\text{ V}$$

Before adding the half-reactions we must equalise the number of electrons. The oxidation half-reaction contains one electron, whereas the reduction half-reaction contains two. Multiplying the oxidation half-reaction by 2 yields

$$2\big(Mn^{2+} \rightarrow Mn^{3+} + e^-\big)$$

or

$$2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^- \qquad E^\circ_{\text{ox}} = -1.51\text{ V}$$

Now we can add the two balanced half-reactions:

$$\big(2Mn^{2+} \rightarrow 2Mn^{3+} + 2e^-\big) + \big(Mn^{2+} + 2e^- \rightarrow Mn\big)$$

Canceling the $$2e^-$$ on both sides gives the overall balanced reaction

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

Exactly the reaction required. The standard cell potential for the overall process is obtained by adding the potentials of the individual half-reactions, because one is written as a reduction and the other as an oxidation:

$$E^\circ_{\text{cell}} \;=\; E^\circ_{\text{red}} \;+\; E^\circ_{\text{ox}}$$

Substituting the values, we get

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

$$E^\circ_{\text{cell}} \;=\; -2.69\text{ V}$$

Because the standard potential is negative, the reaction as written is non-spontaneous under standard conditions; in simple words, it will not occur on its own.

Hence, the correct answer is Option A.