Question

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?

Answer 1

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