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Question: Given below are the half-cell reactions: \( M{n^{2 + }} + 2{e^ - } \to Mn;{E^\circ } = - 1.18V \)...

Given below are the half-cell reactions:
Mn2++2e−→Mn;E∘=−1.18VM{n^{2 + }} + 2{e^ - } \to Mn;{E^\circ } = - 1.18V
2(Mn3++e−→Mn2+);E∘=+1.51V2(M{n^{3 + }} + {e^ - } \to M{n^{2 + }});{E^\circ } = + 1.51V
The E∘{E^\circ } for 3Mn2+→Mn+2Mn3+3M{n^{2 + }} \to Mn + 2M{n^{3 + }} will be:
(a)−0.33V(a) - 0.33V ; The reaction will not occur.
(b)−0.33V(b) - 0.33V ; The reaction will occur.
(c)−2.69V(c) - 2.69V ; The reaction will not occur.
(d)−2.69V(d) - 2.69V ; The reaction will occur. (d)−2.69V(d) - 2.69V

Explanation

Solution

Before calculating the cell potential, we should review a few definitions. The anode half reaction, which is defined by the half-reaction in which oxidation occurs, and the cathode half-reaction, defined as the half-reaction in which reduction takes place.

Complete answer:
The standard electrode potential is denoted by E∘{E^\circ } . Either standard reduction potential or standard oxidation potential can be calculated for an electrode using standard hydrogen electrodes. Standard cell potential is the difference between standard reduction potential of the two half-cells or half reactions. The equation will be:
E∘cell=Ered∘−Eoxd∘{E^\circ }_{cell} = {E_{red}}^\circ - {E_{oxd}}^\circ
The electrode potential of 2(Mn3++e−→Mn2+)2(M{n^{3 + }} + {e^ - } \to M{n^{2 + }}) Is +1.51V+ 1.51V
To get the main equation we have to reverse the second equation and add both the equations. The reversed equation will be:
2Mn3+→2Mn2++2e−2M{n^{3 + }} \to 2M{n^{2 + }} + 2{e^ - } , the electrode potential will become, E∘=−1.51V{E^\circ } = - 1.51V
Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction. After doing that, the overall equation is obtained.
3Mn2+→Mn+2Mn3+3M{n^{2 + }} \to Mn + 2M{n^{3 + }}
The cell potential will be equal to, E∘cell=Ered∘−Eoxd∘{E^\circ }_{cell} = {E_{red}}^\circ - {E_{oxd}}^\circ
E∘cell=−1.51+(−1.18){E^\circ }_{cell} = - 1.51 + ( - 1.18)
E∘cell=−2.69V{E^\circ }_{cell} = - 2.69V
The reaction is not possible as the ΔG\Delta G will come positive for this case and that indicates the reaction is nonspontaneous.
The correct option is (C)−2.69V(C) - 2.69V ; The reaction will not occur.

Note:
The half-cell with the higher reduction potential will undergo reduction within the cell. The half-cell with the lower reduction potential will undergo oxidation. Using the standard cell notation, the condition such as the concentration of the ions in the solution can be represented.