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

Answer

A

Explanation

(a) `Cu^(2+)+e^(-) to Cu^(+), E_(1)^(@)=0.1 5 V, DeltaG_(1)^(@)=-n_(1)E_(1)^(@)F`
`Cu^(+)+e^(-) to Cu, E_(2)^(@)=0.50 V, DeltaG_(2)^(@)=-n_(2)E_(2)^(@)F`
`Cu^(2+)+2e^(-) to Cu, E^(@)=?, DeltaG^(@)=-nE^(@)F`
` DeltaG^(@) =DeltaG_(1)^(@)+DeltaG_(2)^(@)`
`-nE^(@)F=-n_(1)E_(1)^(@)F-n_(2)E_(2)^(@)F`
`or -2E^(@)F=-Fxx0.15+(-1Fxx0.50)`
`or -2E^(@)F=-0.15F-0.50F`
` or -2FE^(@)=-F(0.15+0.50)`
`:. E^(@)=(0.65)/(2)=0.325V`