Question

A concentration cell has both electrodes made of magnesium and is based on the following half-cell reaction.

Mg2+(aq) + 2 e− → Mg(s)

(a) If [Mg2+ ] = 2.18 M at the cathode and [Mg2+ ] = 0.00744 M at the anode, what is the cell potential of the concentration cell at 298.15 K?

(b) What is the cell potential of the cell when the concentration of Mg²⁺ ions is the same in the cathode and anode compartments?

Answer 1

a)

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

we have below equation to be used:

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 (0.00744/2.18)

E = 7.293*10^-2 V

Answer: 7.29*10^-2 V

b)

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

E = Eo - (0.0591/n) log {1}

E = Eo - (0.0591/n) *0

E = Eo

E = 0

Answer: 0 V