The electrode potential for `Cu^(2+)(aq)+e rarr Cu^(+)(aq) and Cu^(+)(aq)+e rarr Cu(s)` are 0.15V 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+)(aq)+erarr Cu^(+)(aq), DeltaG_(1), E_(1)^(@)=+0.15V`
`Cu^(+)(aq)+e rarr Cu(s), DeltaG_(2), E_(2)=+0.50V`
`Cu^(2+)(aq)+2e rarr Cu(s), DeltaG_(3), E_(3)^(@)=?`
`DeltaG_(3)=DeltaG_(1)+DeltaG_(2)`
or, `-2FE_(3)^(@)=-1F xx0.15+(-1Fxx0.50)`
`E_(3)^(@)=(0.65)/(2)=0.325V`