Read the following passage for the evaluation of `E^(@)` when different number of electrons are involved. Consider addition of the following half reactions
(1) `Fe^(3+)(aq)+3e^(-)rarr Fe(s) E_(1)^(@)=0.45V`
(2) `Fe(s)rarr Fe^(2+)(aq)+2e^(-) E_(2)^(@)=-0.04V`
(3) `Fe^(3+)(aq)+e^(-)rarr Fe^(2+)(aq) E_(3)^(@)= ?`
Because half-reactions (1) and (2) contains a different number of electrons, the net reaction (3) is another half-reaction and `E_(3)^(@)` can't be obtained simply by adding `E_(1)^(@)` and `E_(3)^(@)`. The free - energy changes however, are additive because G is a state function : `Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(2)^(@)`
For the reactions
(1) `MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2)+4H_(2)O`
`E^(@)=1.51 V`
(2) `MnO_(2)+4H^(+)+2e^(-)rarr Mn^(2+)2H_(2)O`
`E^(@)=1.32` then for the reaction
(3) `MnO_(4)^(-)+4H^(+)+3e^(-)rarrMnO_(2)+2H_(2)O, E^(@)` is

Options

A. `1.70 V`
B. `5.09 V`
C. `0.28 V`
D. `0.842`

Answer

A

Explanation

`Eq(3)=Eq(1)-Eq(2)`
Thus
`n_(1)=5, n_(2)=2, n_(3)=3, G_(1)^(@)=1.5V, E_(2)^(@)=-1.23V`
`therefore E_(3)^(@)=(1.51xx5-1.23xx2)/(2)=1.70V`