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)` is [F is Faraday constant , R is gas constant , T is temperature , `E_(cell)^(@) = 1.1V`]

Options

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

Answer

A

Explanation

`Zn (s) + Cu^(2+) (aq) to Zn^(2+) (aq) + Cu (s)`
`Delta G = Delta G^(@) + RT ln Q`
`Delta G= Delta G^(@) + 2.303 RT log Q`
`Q = ([Zn^(2+)])/([Cu^(2+)]) = (10)/(1)`
`Delta G^(@) = - n F E_(cell)^(@) = - 2 xx F xx 1.1`
`therefore Delta G = - 2 xx F xx 1.1 + 2.303 RT log (10)/(1)`
`=2.303 RT - 2.2 F`