Calculate the cell potential, the equilibrium constant, and the free-energy change for: Ca(s)+Mn2+(aq)(1M)⇌Ca2+(aq)(1M)+Mn(s) given the following Eo values: Ca2+(aq)+2e−→Ca(s) Eo = -1.59 V Mn2+(aq)+2e−→Mn(s) Eo = -0.54 V
1.) Calculate the equilibrium constant.
2.) Free-energy change?
In given reaction, Ca is getting oxidised
SO, this is anode.
Eo anode = -1.59 V
In given reaction, Mn2+ is getting oxidised
SO, this is Cathode.
Eo cathode = -0.54 V
use:
Eo cell = Eo cathode - Eo anode
= -0.54 V - (-1.59 V)
= 1.05 V
1)
here, number of electrons being transferred, n = 2
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
1.05 = (0.0592/2)*log Kc
log Kc = 35.473
Kc = 2.971*10^35
Answer: 2.97*10^35
2)
number of electrons being transferred, n = 2
F = 96500.0 C
use:
ΔG = -n*F*E
= -2*96500.0*1.05
= -202650 J/mol
= -202.65 KJ/mol
Answer: -202.65 KJ/mol
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