For the following cell, calculate the emf:
`Al//Al^(3+) (0.01 M)"||" Fe^(2+) (0.02M)"|"Fe`
Given: `E_(Al^(3+)//Al)^(Theta) = -1.66V, E_(Fe^(2+)//Fe)^(Theta)= -0.44V`

Answer

`E_("cell")=1.2308V; DeltaG= -712.633kJ`

Explanation

Cell reaction is :
`{:(" "Al^(3+)+3e^(-) to Al"]"xx2),(" "Fe^(2+)+2e^(-) to Fe"]"xx3),(bar(2Al^(3+)+3Fe^(2+) to 2Al +3Fe^(2+),n=6)):}`
`E_("cell")^(Theta)=E_("cathode")^(Theta)-E_("anode")^(Theta)= -0.44 V- (-1.66V)=1.22V`
According to Nernst equation, `E_("cell")`
`=E_("cell")^(Theta) -(0.0591)/(n)"log"([Fe^(2+)]^(3))/([Al^(3+)]^(2))`
or `E_("cell")=1.22 -(0.0591)/(6)"log"((0.02)^(3))/((0.01)^(2))= 1.22 -(0.0591)/(6)log 0.08`
`=1.22 -(0.0591)/(6)(-1.097)=1.22 +0.0108=1.2308V`
`DeltaG= -nFE_("cell")= -6xx96500 xx 1.2308 = -712633J`
`= -712.633kJ`