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

Options

A. increases by 30 mV
B. decreases by 30 m V
C. increases by 59 m V
D. decreases by 59 m V

Answer

B

Explanation

`(E_(Cu^(2+)|Cu))_(1) = E_(Cu^(2+)|Cu)^(0) + (0.059)/(2) [Cu^(2+)] log [Cu^(2+)]_(1)`
`[Cu^(2+) (aq) + 2e^(-) to Cu (s)` and `(E_(Cu^(2+) |Cu))_(2) = E_(Cu^(2+) |Cu)^(0) + (0.059)/(2) "log" [Cu^(2+)]_(2)`
`implies Delta E_(Cu^(2+)|Cu) = - (0.059)/(2) "log" ((C_(2))/(C_(1)))`
`= - (0.059)/(2) "log" 10 = - 0.030 V = -30 m V`