Question

A voltaic cell is constructed that uses the following reaction and operates at 298 K.

Zn(s) + Ni2+(aq) Zn2+(aq) + Ni(s)

(a) What is the emf of this cell under standard conditions? _____V

b) What is the emf of this cell when [Ni2+] = 2.90 M and [Zn2+] = 0.193 M? _____ V

(c) What is the emf of the cell when [Ni2+] = 0.105 M and [Zn2+] = 0.931 M? ______ V

Answer 1

a)

Lets find Eo 1st
from data table:
Eo(Zn2+/Zn(s)) = -0.7618 V
Eo(Ni2+/Ni(s)) = -0.25 V

As per given reaction/cell notation,
cathode is (Ni2+/Ni(s))
anode is (Zn2+/Zn(s))

Eocell = Eocathode - Eoanode
= (-0.25) - (-0.7618)
= 0.5118 V
Answer: 0.512 V

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

use:
E = Eo - (2.303*RT/nF) log {[Zn2+]^1/[Ni2+]^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/[Ni2+]^1}
E = 0.5118 - (0.0591/2) log (0.193^1/2.9^1)
E = 0.5118-(-3.479*10^-2)
E = 0.5466 V
Answer: 0.547 V

c)

use:
E = Eo - (2.303*RT/nF) log {[Zn2+]^1/[Ni2+]^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/[Ni2+]^1}
E = 0.5118 - (0.0591/2) log (0.931^1/0.105^1)
E = 0.5118-(2.802*10^-2)
E = 0.4838 V
Answer: 0.484 V