1. Assuming that the cell is operating at standard conditions, what would be the potential of the cell
in vol d. ts?
Solution
d)Â Â standard condition
anode : Al---------------------> Al3+ + 3e-Â Â Â Â , Eo = -1.66 V
cathode : Cd2+ + 2e- ----------------> Cd  , Eo = -0.403 V
overall :
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2 Al +Â Â 3 Cd2+ --------------------> 2 Al3+ + 3 Cd

2. Eocell = Eo cathode - Eo anode
= -0.403 + 1.66
= 1.26 V
potential of the cell = 1.26 V
e)
QÂ Â = [Al3+]^2 / [Cd2+]^3
Q = (2.0)^2 / (0.00010)^3
Q = 4 X 10^12
Ecell = Eo cell - 0.05916 / n * log Q
= 1.26 - 0.05916 / 6 * log (4 x10^12)
= 1.14 V
potential of the cell = 1.14 V