In the electrochemical cell : `Zn|ZnSO_4 (0.01M)|CuSO_4 (1.0 M)|Cu`, the emf of this Daniel cell is `E_1`. When the concentration of `ZnSO_4`, is changed to 1.0 M and that `CuSO_4` changed to 0.01M, the emf changes to `E_2`. From the following , which one is the relationship between `E_1 and E_2` ?

Options

A. `E_1 lt E_2`
B. `E_1 gt E_2`
C. `E_2 = 0 uarr E_1`
D. `E_1 = E_2`

Answer

B

Explanation

`E_("cell") = E_("cell")^(@) - (0.0591)/(2) "log" ([zn^(2+)])/([Cu^(2+)])`
`E_1 = E_("cell")^(@) - (0.0591)/(2) "log" (10^(-2))/(1) " " Zn (S) to Zn^(2+)(aq) + 2e^(-)`
`E_1 = E_("cell")^(@) + 0.0591 ………..(1) " " Cu^(2+)(aq) + 2e^(-) to Cu(s)`
`E_2 = E_("cell")^(@) - (0.0591)/(2) log 1/(10^(-2)) " " Zn(s) + Cu^(2+)(aq) to Zn^(2+)(aq) +Cu(s)`
`E_2 = E_("cell")^(@) - 0.0591 .............(2)`
`:. E_1 > E_2`.