Question

Calculate the cell potential (Ecell) for the following lead concentration cell at 298 K. [Pb2+] = 0.00900 M || [Pb2+] = 2.25 M 1st attempt See HintSee Periodic Table Ecell = 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 {[Pb2+] at anode/[Pb2+]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 {[Pb2+] at anode/[Pb2+]at cathode}

E = 0 - (0.0591/2) log (0.009/2.25)

E = 7.089*10^-2 V

Answer: 7.09*10^-2 V