Correct Answer - B
The relationship between `E_("cell")^(@)` for a galvanic cell and `Delta_(r)G^(@)`, the standard Gibbs energy change, for the chemical reaction of the cell is
`Delta_(r)G^(@) = -nFE_("cell")^(@)`
For the cell reaction
`2Ag^(+)(aq.)+Cu(s) rarr Cu^(2+)(aq.)+2Ag(s)`
We have
`n = 2 mol`
`E_("cell")^(@) = +0.46 V` Therefore
`Delta_(r)G^(@) = (-2 mol)(96,500 C mol^(-1))(+0.46 V)`
`= -88780 J = -88.780 kJ`
`= -89.0 kJ`