Question

Calculate the emf of the following concentration cell: Mg(s) | Mg2+(0.32 M) || Mg2+(0.45 M) | Mg(s) V

Answer 1

For concentration cell, cathode and anode are same electrode
So, Eo = 0
Number of electron being transferred in balanced reaction is 2
So, n = 2
If E is positive anode will be the one with lower concentration

use:
E = Eo - (2.303*RT/nF) log {[Mg2+] at anode/[Mg2+]at cathode}

Here:
2.303*R*T/F
= 2.303*8.314*298.15/96500
= 0.0592

So, above expression becomes:
E = Eo - (0.0592/n) log {[Mg2+] at anode/[Mg2+]at cathode}
E = 0 - (0.0592/2) log (0.32/0.45)
E = 4.38*10^-3 V
Answer: 4.38*10^-3 V