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Question

Given below are the half cell reactions:
Mn2+(aq)+2eMn(s); E0=1.18 V
2(Mn+3(aq)+eMn2+(aq)); E0=+1.51 V

The E0 for 3Mn2+(aq)Mn(s)+2Mn3+(aq) cell reaction will be:

A
0.33 V
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B
+2.69 V
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C
+0.33 V
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D
2.69 V
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Solution

The correct option is D 2.69 V
The standard reduction potential of the half cells are given.
E0Mn2+(aq)/Mn(s)=1.18 V
E0Mn3+(aq)/Mn2+(aq)=+1.51 V

ΔG0=nFE0

Mn2+(aq)+2eMn(s)......(1); ΔG01=2F×1.182Mn3+(aq)+2e2Mn2+(aq)...(2); ΔG02=2F×1.51

Reverse equation (2)
2Mn2+(aq)2Mn3+(aq)+2e....(3); ΔG03=+2F×1.51

Adding equation (1) and (3), we get,
3Mn2+(aq)Mn(s)+2Mn3+(aq)......(4); ΔG04=2F×E0

ΔG04=ΔG01+ΔG03
2F×E0=2F×1.18++2F×1.51
E0=2.69 V
E0=2.69 V



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