The standard potential is given below, of the following at 25∘C. MnO2→Mn3+E∘=0.95V Mn+3→Mn2+E∘=1.51V
The standard potential of MnO2→Mn2+ is :
A
−0.56V
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B
−2.46V
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C
−1.23V
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D
1.23V
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Solution
The correct option is D1.23V 2e−+MnO2→Mn2+,E∘=? ΔG=−2FE∘ (i)e−+MnO2→Mn3+ΔG01=−FE0=−0.95V (ii)e−+Mn3+→Mn2+ΔG02=−FE0=−1.51V
Equation (i) + (ii) give the desired equations ∴−2FE0=(−0.95−1.51)F E0=1.23V