Calculate `DeltaG^(@)` for the reaction : `Cu^(2+)(aq) +Fe(s) hArr Fe^(2+)(aq) +Cu(s)`. Given that `E_(Cu^(2+)//Cu)^(@) = 0.34 V`,
`E_(Fe^(+2)//Fe)^(@) =- 0.44 V`

Explanation

The cell reactions are:
`Fe(s) rarr Fe^(2+) (aq) +2e^(-)` .... at (anode)
`Cu^(2+) (aq) +2e^(-) rarr Cu(s)` ....at (cathode)
We know that: `DeltaG^(@) =- nF E_(cell), n = 2 ` and `F = 96500 C`
`E_(cell)^(@) = [E_((Cu^(2+)//Cu))^(@) -E_((Fe^(2+)//Fe))^(@)] = (+0.34V)-(-0.44V)=+0.78V`
`:. DeltaG^(@) =- nF E_(cell)^(@)`
`=- (2) xx (96500C) xx (+0.78 V)`
`=- 150540 CV ` ( 1CV = 1J)