5. Cathode Half Reaction:
Ni^+2 +2e^-1 ---------> Ni
(s)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
E0 = -0.23v
Anode Half Reaction
Al(s) ---------> Al^+3(aq) + 3e^-1Â Â Â Â E0
= 1.66V
2Al(s) ---------> 2Al^+3(aq) + 6e^-1Â Â Â
  E0 = 1.66V
3Ni^+2 +6e^-1 ---------> 3Ni
(s)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
E0 = -0.23v
----------------------------------------------------------------------------------
2Al(s) + 3Ni^+2 ---------> 2Al^+3 (aq) + 3Ni(s)Â Â
E0 = 1.43VÂ Â redox reaction or over all reaction
  n= 6
Ecell = E0cell-0.0592/n logQ
          Â
=1.43-0.0592/6 log[Al^+3]^2/[Ni^+2]^3
          Â
= 1.43-0.00986log(4)^2/(2*10^-5)^3
         =
1.43-0.00986*5.3010Â Â =1.377V >>>>answer
 Â
