Question

2. Calculate the voltage of the following cell: Zn (s)│Zn2+ (O.200 M) ││ Cu2+ (0.100 M)│Cu (s).

3. Calculate the cell potential, the equilibrium constant, and the free-energy change for the following reaction: Ca (s) + Mn2+ (1.00 M) ˂=˃ Ca2+ (1.00 M) + Mn (s)

Answer 1

2)

Lets find Eo 1st
from data table:
Eo(Zn2+/Zn(s)) = -0.7618 V
Eo(Cu2+/Cu(s)) = 0.337 V

As per given reaction/cell notation,
cathode is (Cu2+/Cu(s))
anode is (Zn2+/Zn(s))

Eocell = Eocathode - Eoanode
= (0.337) - (-0.7618)
= 1.0988 V

Number of electron being transferred in balanced reaction is 2
So, n = 2

use:
E = Eo - (2.303*RT/nF) log {[Zn2+]^1/[Cu2+]^1}

Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591

So, above expression becomes:
E = Eo - (0.0591/n) log {[Zn2+]^1/[Cu2+]^1}
E = 1.099 - (0.0591/2) log (0.2^1/0.1^1)
E = 1.099-(8.9*10^-3)
E = 1.09 V
Answer: 1.09 V