Question

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

Answer 1

Correct Answer - B
`Cu_((aq))^(2+) + e^(-) to Cu_((aq))^(+) , E_(1)^(@) = 0.15 V`
`Cu_((aq))^(+) + e^(-) to Cu_((s)) , E_(2)^(@) = 0.50 V`
`Cu^(2+) + 2e^(-) to Cu`
Now , `Delta G^(@) = Delta G_(1)^(@) + Delta G_(2)^(@)`
or , `- n FE^(@) =-n_(1) FE_(1)^(@) - n_(2) FE_(2)^(@)`
or , `E^(@) = (n_(1) E_(1)^(@) + n_(2) E_(2)^(@))/(n) = (1 xx 0.15+ 1 xx 0.50)/(2) = 0.325` V