In the electrochemical cell: `Znabs(ZnSO_(4)(0.01M))abs(CuSO_(4)(1.0M))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 followings, 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 uarrE_1`
D. `E_1 = E_2`

Answer

B

Explanation

`E_("cell")=E_("cell")^@-(0.0591)/2log.([Zn^(2+)])/([Cu^(2+)])`
`E_1=E_("cell")^@-(0.0591)/2log.(10^(-2))/1" "Zn(s)rarrZn^(2+)Zn^(2+)(aq)+2e^(-)`
`E_1=E_("cell")^@+0.0591........(1) " "Cu^(2+)(aq)+2e^(-)rarrCu(s)`
`E_2=E_("ceell")^@-(0.0591)/2log.(1)/10^(-2)" "Zn(s)+Cu^(2+)(aq)rarrZn^(2+)(aq)+Cu(s)`
`E_2=E_("cell")^@-0.0591.......(2)`
`:.E_1 gt E_2`