If for the half cell reaction `E^(@)` value are given
`Cu^(2+)+e^(-)rarrCu^(+),E^(@)=0.15V`
`Cu^(2+)+2e^(-)rarrCu^(+),E^(@)=0.34V`
calculate `E^(@)` of the half cell reaction
`Cu^(+)+e^(-)rarrCu`

Options

A. `+0.53 V`
B. `-0.53 V`
C. `-0.80 V`
D. `-0.28 V`

Explanation

`Cu^(2+)+e^(-)rarrCu^(+), E^(@)=0.15 V`
`(i) Cu^(+) rarr Cu^(2+)+e^(-), E_(1)^(@)=-0.15 V n_(1)=1`
`(ii) Cu^(2+)+2e^(-)rarrCu, E_(2)^(@)=0.34 V n_(2)=2`
From Eqs i and ii
`Cu^(+)+e^(-)rarr Cu , E_(3)^(@)= N_(3)=1`
`therefore triangle G_(3)^(@)=triangle G_(1)^(@) +triangle G_(2)^(@)`
`E_(3)^(@)=-n_(1)E_(1)^(@)-n_(2)(E_(2)^(2))/(n_(3))`
`E_(3)^(@)=(1xx(-0.15)+2xx(0.34))/(1)=+0.053 V`