The electrode potentials for
`Cu^(2+)+e^(-)toCu^(+)` and
`Cu^(+)e^(-)toCu_(s)`
are `+0.15`V and `+0.50V` respectively the value of `E_((Cu^(2+))/(Cu))^(@)` will be?

Options

A. 0.150V
B. 0.500V
C. 0.325V
D. 0.650V

Answer

C

Explanation

`Cu^(2+)+erarrCu^(+)`
`DeltaG_(1)=-nFE^(@)=-1xxFxx0.15`
`Cu^(+)+erarrCu,DeltaG_(2)=-1xxFxx0.50`
`Cu^(2+)+2erarrCu`
`DeltaG_(3)=-1xxFxxE_(Cu^(2+)//Cu)^(@)`
`DeltaG_(3)=DeltaG_(1)+DeltaG_(2)`
`-2FE_(Cu^(2+)//Cu)^(@)= -0.15F+(-0.50E)`
or `-2E_(Cu^(2+)//Cu)^(@)=-0.15-0.50=-0.65`
or `E_(Cu^(2+)//Cu)^(@)=(0.65)/(2)=0.325V`.