The electrode pptenticals for
` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) `
and ` Cu^+ (aq) + e^- rarr Cu (s)`
are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.

Options

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

Answer

B

Explanation

`Cu^(2+) + e^(-) to Cu^+ E_(1)^(@) = 0.15 V " " … (i)`
`therefore " " Delta G_(1)^(@) = - n FE_(1)^(@) = - 1 xx F xx 0.15`
`Cu^(+) + e^- to Cu E_(2)^(@) = 0.50 V " " … (ii)`
`Delta G_(2)^(@) = - n F E_(2)^(@) = - 1 xx F xx 0.50`
Adding eqn. (i) and (ii)
`Cu^(2+) + 2e^(-) to Cu`
`Delta G^(@) = - n FE^(@) = - 2 xx FE^(@)`
`- 2 xx F xx E^(@) = - 1 xx F xx 0.15 + (-1 xx F xx 0.50)`
`E^(@) = (0.15 +0.50)/(2) = 0.325 V`