Blue colour of copper sulphate is slowly discharged when an iron rod is dipped into it. Explain this by calculating `DeltaG^(@)` with the help of the following data :
`[E_(Cu^(2+)//Cu)^(@) = 0.34 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V` and 1 Faraday = 96500 C `mol^(-1)`]

Explanation

`E_("cell")^(@) = E_("Cu^(2+)//Cu)^(@) - E_(Fe^(2+)//Fe)^(@) = +0.34 V - (-0.44 V) = +0.78 V`
We have the equation
`DeltaG^(@) =-nFE_("cell")^(@) = -2 xx 96500 C mol^(-1) xx 0.78 V = -19301 J mol^(-1) = -19.3 kJ mol^(-1)`
As the value of `DeltaG^(@)` is negative, the reaction will take place. That means blue colour of copper sulphate is slowly discharged.