from data table:
Eo(Mg2+/Mg(s)) = -2.372 V
Eo(Al3+/Al(s)) = -1.662 V
As per given reaction/cell notation,
cathode is (Al3+/Al(s))
anode is (Mg2+/Mg(s))
Eocell = Eocathode - Eoanode
= (-1.662) - (-2.372)
= 0.71 V
number of electrons being transferred, n = 6
F = 96500 C
use:
ΔG = -n*F*E
= -6*96500*0.71
= -411090 J/mol
= -411 KJ/mol
Answer: -4.11*10^2 KJ/mol
Be sure to answer all parts Calculate AG for the following reaction at 25°C: 3Mg(s) 2A...
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