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Chemistry Question on Electrochemistry

Given below are the half-cell reactions Mn2++2e−→Mn;E∘=−1.18 VMn ^{2+}+2 e ^{-} \rightarrow Mn ; E ^{\circ}=-1.18 \,V (Mn3++e−→Mn2+);E∘=+1.51 V\left( Mn ^{3+}+ e ^{-} \rightarrow Mn ^{2+}\right) ; E ^{\circ}=+1.51 \,V The E∘E ^{\circ} for 3 Mn2+→Mn+2Mn3+3 \,Mn ^{2+} \rightarrow Mn +2 Mn ^{3+} will be

A

−2.69 V-2.69\, V; the reaction will not occur

B

−2.69 V- 2.69\, V; the reaction will occur

C

−2.69 V-2.69 \,V; the reaction will occur

D

−0.33 V- 0.33\, V; the reaction will occur

Answer

−2.69 V-2.69\, V; the reaction will not occur

Explanation

Solution

(1) Mn2++2e→Mn;E∘=−1.18VMn ^{2+}+2 e \rightarrow Mn ; E ^{\circ}=-1.18 V;

ΔG1∘=−2F(−1.18)=2.36F\Delta G _{1}^{\circ}=-2 F (-1.18)=2.36 F

(2) Mn3++e→Mn2+;E∘=+1.51 V Mn ^{3+}+ e \rightarrow Mn ^{2+} ; E ^{\circ}=+1.51 \,V ;

ΔG2∘=−F(1.51)=−1.51 F\Delta G _{2}^{\circ}=- F (1.51)=-1.51 \,F
(1)−2×(2)(1)-2 \times (2)
3Mn2+→Mn+2Mn3+3 Mn ^{2+} \rightarrow Mn +2 Mn ^{3+} ;

ΔG3∘=ΔG1∘−2ΔG2∘\Delta G _{3}^{\circ} =\Delta G _{1}^{\circ}-2 \Delta G _{2}^{\circ}
=[2.36−2(−1.51)]F=[2.36-2(-1.51)] F
=(2.36+3.02)F=(2.36+3.02) F
=5.38 F=5.38 \,F

But ΔG3∘=−2FE∘ \Delta G_{3}^{\circ}=-2 FE ^{\circ}
⇒5.38F=−2FE∘\Rightarrow 5.38 F =-2 FE ^{\circ}
⇒E∘=−2.69 V\Rightarrow E ^{\circ}=-2.69 \,V

As E∘E ^{\circ} value is negative reaction is non spontaneous