Calculate the standard cell potential of the galvanic cell in which the following reaction takes place :
`2Cr(s) + 3Cd^(2+) (aq) to 2Cr^(3+) (aq) + 3Cd (s)`
Also calculate the `Delta_(r)G^(@)` value of the reaction.
[Given: `E_(Cr^(3+)//Cr)^(@) = -0.74 V: E_(Cd^(2+)//Cd)^(@) = -0.40 V` and `F = 96500 C mol^(-1)`]

Explanation

The electrode reactions may be represented as under :
`2Cr (s) to 2Cr^(3+)(aq) + 6e^(-)`
`3Cd^(2+)(aq) + 6e^(-) to 3Cd(s)`
Thus, n =6
Standard cell potential may be obtained as under :
`E_("cell")^(@) = E_(Cd^(2+)//Cd)^(@) -E_(Cr^(3+)//Cr)^(@) =-0.40 V -(-0.74 V)`
or `E_("cell")^(@) =0.34 V`
To calculate `Delta_(r)G^(@)` , applying the following relation and substituting the values, we get:
`Delta_(r)G^(@) =-nE^(@) F = -6 xx 0.34 V xx 96500 C`
or `Delta_(r)G^(@) = -196.86 kJ mol^(-1)`