For the following cell,
`Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)||Cu(s)`
When the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG`
(in J `"mol"^(-1)`)
[F is Faraday constant, R is gas constant] T is temperaure, `E^(@)("cell")=1.1V`

Options

A. `2.303RT+1.1F`
B. `1.1F`
C. `2.303RT-2.2F`
D. `-2.2F`

Answer

C

Explanation

Cell reaction is
`Zn+Cu^(2+)toZn^(2+)+Cu`
`DeltaG=DeltaG^(@)+2.303RTlogQ`
`=DeltaG^(@)+2.303RTlog([Zn^(2+)])/([Cu^(2+)])`
`DeltaG^(@)=-nFE_(cell)^(@)=-2F(1.1)`
`thereforeDeltaG=-2F(1.1)+2.303RTlog10`
`=2.303RT-2.2F`