The solution of `CuSO_(4)` in which copper rod is immersed is diluted to times. The reduction electrode potential

Options

A. Increase by 0.030 V
B. Decreasse by 0.030V
C. Increase by 0.059 V
D. Decrease by 0.059 V

Answer

B

Explanation

`Cu^(2+) +2e to Cu`
`E_(1)=E_(1)^(9) -(0.0591)/(2) log (1)/(Cu^(2+))_(1)`
`E_(2)=E_(1)^(@)-(0.0591)/(2) log (1)/(Cu^(2+))_(2)`
`E_(2)=E_(1)^(@)-(0.0591)/(2) log (10)/((Cu_(1)^(2+))`
`=E_(1)^(@)-(0.0591)/(2) log (1)/((Cu_(1)^(2+))-(0.0591)/(2) log 10`
`=E_(1)-0.030V`