(i) `Q=([Hg_(2)^(2+)])/([Ag^(+)]^(2))=10^(-1)/([10^(-4)]^(2))=10^(7)`
`E^(@)=E_(Ag^(+)//Ag)^(@)-E_(Hg_(2)^(2+)//Hg)^(@)`
`=0.80-0.79=0.01 V`
`E=E^(@)-0.059/n log Q`
`=0.01-0.059/2 log 10^(7)`
`=-0.1965 V`
Negative value shows that the reaction will proceed from right to left, i.e., in backward direction.
(ii) `Q=([Hg_(2)^(2+)])/([Ag^(+)]^(2))=10^(-4)/([10^(-1)]^(2))=10^(-2)`
`n=2`
`E^(@)=0.01" volt"`
`E=E^(@)-0.059/n log_(10) Q`
`=0.01-0.059/2 log_(10) 10^(-2)`
`=0.01 + 0.059 V`
`=0.069 V`
Since, the value of cell potential is positive, the reaction will proceed spontaneously in forward direction.