Given below are the half-cell reactions:
\(\text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \quad (E^0 = -1.18\text{ V})\)
\(2(\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}) \quad (E^0 = +1.51\text{ V})\)
The \(E^0_{cell}\) for \(3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+}\) will be _____________.

The given target reaction is \(3\text{Mn}^{2+} \rightarrow \text{Mn} + 2\text{Mn}^{3+}\)

This reaction can be split into two half-reactions:
Cathode (Reduction): \(\text{Mn}^{2+} + 2e^- \rightarrow \text{Mn}\)
Anode (Oxidation): \(2\text{Mn}^{2+} \rightarrow 2\text{Mn}^{3+} + 2e^-\)

The standard reduction potentials are:
\(E^0_{\text{cathode}} = E^0_{\text{Mn}^{2+}/\text{Mn}} = -1.18\text{ V}\)
\(E^0_{\text{anode}} = E^0_{\text{Mn}^{3+}/\text{Mn}^{2+}} = +1.51\text{ V}\)

The standard cell potential is given by:
\(E^0_{cell} = E^0_{\text{cathode}} - E^0_{\text{anode}}\)

\(E^0_{cell} = -1.18\text{ V} - 1.51\text{ V} = -2.69\text{ V}\)

Since \(E^0_{cell}\) is negative, the standard Gibbs free energy change \(\Delta G^0 = -nFE^0_{cell}\) is positive. Therefore, the reaction is non-spontaneous and will not occur.

Answer: \(-2.69\) V, the reaction will not occur (Non-Spontaneous)