To determine the standard cell potential (E^o) for the reaction 3Mn²⁺ → Mn + 2Mn³⁺ and predict whether it will occur, we need to combine the given half-cell reactions appropriately and use the relationship between Gibbs free energy and cell potential.

Step 1: Write the Target Reaction

The target reaction is: $3{\mathrm{Mn}}^{2+}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+}$

Step 2: Analyze the Given Half-Reactions

Given half-reactions:

1. Reduction: ${\mathrm{Mn}}^{2+}+2e^{-}\to \mathrm{Mn}$; $E^{\circ }=-1.18\text{V}$

2. Reduction: ${\mathrm{Mn}}^{3+}+e^{-}\to {\mathrm{Mn}}^{2+}$; $E^{\circ }=+1.51\text{V}$ (Note: The given is for 2 such reactions, but E^o is per mole of electrons, so it remains +1.51 V for one.)

Step 3: Reverse Reactions to Match the Target

The target reaction involves oxidation of Mn²⁺ to Mn and reduction of Mn²⁺ to Mn³⁺? Wait, let's see: In 3Mn²⁺ → Mn + 2Mn³⁺, one Mn²⁺ is reduced to Mn, and two Mn²⁺ are oxidized to Mn³⁺. So, we need:

- The reduction half-reaction: ${\mathrm{Mn}}^{2+}+2e^{-}\to \mathrm{Mn}$ (as given, E^o_red = -1.18 V)

- The oxidation half-reaction: ${\mathrm{Mn}}^{2+}\to {\mathrm{Mn}}^{3+}+e^{-}$ (reverse of the second given half-reaction). For oxidation, E^o_ox = -E^o_red = -1.51 V. Since two such oxidations are needed, the total E^o_ox remains -1.51 V (as E^o is intensive).

Step 4: Combine the Half-Reactions

To get the net reaction, add the reduction and oxidation half-reactions after multiplying to balance electrons:

Reduction (cathode): ${\mathrm{Mn}}^{2+}+2e^{-}\to \mathrm{Mn}$ (E^o_red = -1.18 V)

Oxidation (anode): $2{\mathrm{Mn}}^{2+}\to 2{\mathrm{Mn}}^{3+}+2e^{-}$ (E^o_ox = -1.51 V)

Net cell reaction: ${\mathrm{Mn}}^{2+}+2{\mathrm{Mn}}^{2+}+2e^{-}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+}+2e^{-}$, which simplifies to $3{\mathrm{Mn}}^{2+}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+}$.

Step 5: Calculate E^o_cell

The standard cell potential is E^o_cell = E^o_cathode - E^o_anode = E^o_red - E^o_red(anode) = (-1.18) - (1.51) = -2.69 V.

Alternatively, using oxidation potentials: E^o_cell = E^o_ox(anode) + E^o_red(cathode) = (-1.51) + (-1.18) = -2.69 V.

Step 6: Predict Spontaneity

Since E^o_cell is negative (-2.69 V), the reaction is non-spontaneous under standard conditions. Thus, it will not occur.

Final Answer

The E^o for $3{\mathrm{Mn}}^{2+}\to \mathrm{Mn}+2{\mathrm{Mn}}^{3+}$ is -2.69 V, and the reaction will not occur. So, the correct option is: –2.69 V; the reaction will not occur.

Related Topics and Formulae

Nernst Equation: Used to calculate cell potential under non-standard conditions: $E=E^{\circ }-\frac{RT}{nF}\ln Q$, where Q is the reaction quotient.

Gibbs Free Energy and Cell Potential: ΔG° = -nFE°; if E° < 0, ΔG° > 0, reaction is non-spontaneous.

Electrochemical Series: The tendency of a species to gain electrons (reduction potential) determines the direction of spontaneous reaction.