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Question

The standard electrode potential in V for the electrode MnO4/MnO2 in solution is: (round off the answer to the nearest integer)
Given: EoMnO4/Mn2+=1.51V and EoMnO2/Mn2+=1.23V

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Solution

MnO4+8H+5eMn2++4H2O;Eo1=1.51V ........... (i)
ΔGo1=5×1.51×F
=7.55F

MnO2+4H++2eMn2++2H2O;Eo2=1.23V ........... (ii)
ΔGo2=2×1.23×F
=2.46F
Subtracting equation (ii) from (i)
MnO4+4H++3e2H2O+MnO2;Eo3=?
or ΔGo3=n3Eo3F

ΔGo3=ΔGo1(ΔGo2)

3Eo3F=7.55F2.46F

Eo3=5.093=1.70 volt

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