Question

Enter your answer in the provided box.

Calculate the emf of the following concentration cell:

Mg(s) | Mg²⁺(0.33 M) || Mg²⁺(0.54 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*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Mg2+] at anode/[Mg2+]at cathode}

E = 0 - (0.0591/2) log (0.33/0.54)

E = 6.323*10^-3 V

Answer: 6.32*10^-3 V