Question

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

Pb²⁺(aq) + 2 e⁻ → Pb(s)

(a) If [Pb²⁺ ] = 2.20 M at the cathode and [Pb²⁺ ] = 0.00740 M at the anode, what is the cell potential of the concentration cell at 298.15 K?
V

(b) What is the cell potential of the cell when the concentration of Pb²⁺ ions is the same in the cathode and anode compartments?
V

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 {[Pb2+] at anode/[Pb2+]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 {[Pb2+] at anode/[Pb2+]at cathode}

E = 0 - (0.0592/2) log (0.0074/2.2)

E = 7.315*10^-2 V

Answer: 7.315*10^-2 V

b)

So, above expression becomes:

E = Eo - (0.0592/n) log {[Pb2+] at anode/[Pb2+]at cathode}

E = 0 - (0.0592/2) log (1/1)

E = 0 - (0.0592/2) *0

E = 0 V

Answer: 0 V