Question

Calculate K and ∆G for the redox reaction between Cu2+ (0.15 V) and Al3+ (-1.66 V)

Answer 1

Lets find Eo 1st
from data table:
Eo(Al3+/Al(s)) = -1.66 V
Eo(Cu2+/Cu(s)) = 0.15 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:
cathode is (Cu2+/Cu(s))
anode is (Al3+/Al(s))

The chemical reaction taking place is
3 Cu2+(aq) + 2 Al(s)   -->   3 Cu(s) + 2 Al3+(aq)

Eocell = Eocathode - Eoanode
= (0.15) - (-1.66)
= 1.81 V

a)
here, number of electrons being transferred, n = 6
Eo = (2.303*R*T)/(n*F) log Kc
At 25 oC or 298 K, R*T/F = 0.0592
So, Eo = (0.0592/n)*log Kc
1.81 = (0.0592/6)*log Kc
log Kc = 183.4459
Kc = 2.79*10^183
Answer: 2.79*10^183

b)
number of electrons being transferred, n = 6
F = 96500 C
use:
ΔG = -n*F*E
= -6*96500.0*1.81
= -1047990 J/mol
= -1048 KJ/mol
Answer: -1048 KJ/mol