Calculate `DeltaG and E_(cell)` for the cell:
`AI//AI^(3+)(0.01M)||Fe^(2+)(0.02M)//Fe` given that `E^(@)(Al^(3+)//Al)=-1.66V and E^(@)(Fe^(2+)//Fe)=-0.44V`

Explanation

`Al|Al^(3+)||Fe^(2+)|Fe`
Cell reactions are
`Al to Al^(3+)+3e^(-)]xx2`
`Fe^(2+)+2e^(-) to Fe]xx3`
`2Al+2e^(-) to Fe]xx3`
`2Al+3Fe^(2+)to3Fe+2Al^(3+)`
Here, `n=6 and Q_(C)=([Al^(3+)]^(2))/([Fe^(2+)]^(3))`
`EMF^(@)=E_(R)^(@)-E_(L)^(@)`
`=[-0.44-(-1.66)]V`
`=(1.66-0.44)V=1.22V`