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

Options

A. increases by 29.5 m V
B. decreases by 29.5 m V
C. increases by 529 m V
D. decreases by 592 m V

Answer

D

Explanation

A/C to Nernst equation
`E = E^(@) + (0.0592)/(n) "log " [M^(n+)]`
`[M^(n+)]` decreases by 100 times
`therefore E = E^(@) + (0.0592)/(2) "log" (1)/(100) = E^(@) + (0.0592)/(2) "log" 10^(-2)`
`E = E^(@) - (0.0592)/(2) xx 2 = E^(@) = -0.0592 V `
`E = E^(@) - 592` m V
E decreases by 529 m V