Question

An electrochemical cell is composed of pure copper and pure lead electrodes immersed in solutions of their respective divalent ions. For a 0.7 M concentration of Cu²⁺, the lead electrode is oxidized yielding a cell potential of 0.520 V. Calculate the concentration of Pb²⁺ ions if the temperature is 25˚C. The standard potentials for Cu and Pb are +0.340 V and -0.126 V.

Please show your work!

Answer 1

The cell can be represented as below
Pb/Pb2+//Cu2+/Cu
At anode :
Pb ---> Pb2+ +2e- oxidation takes place

At cathode:

Cu2+ + 2e- ---> Cu reduction tahkes place
overall cell reaction is
Pb(s) + Cu2+(aq) -->Pb2+(aq) +Cu(s)

firstly let us find out standard cell potential of the cell
E0cell= E0Cathode-Eoanode
- E0Cu2+/Cu- E0pb2+/Pb
= 0.34 V- [-0.126 V] = 0.466 V
Also number of electrons involved in the redox process is =2
n=2***

given cell potential Ecell= 0.52 V

concentration of Cu2+, [Cu2+] = 0.7 M

Using Nernst equation let us find out concentration of Pb2+

Ecell = E0cell-0.0591/n logQ

where Q is the reaction qutient Q= [product concentration]/reactant concentration

In the above Nernst equation we know the values of all terms
other than Q ,So let us plug in all to get    Q

Ecell = E0cell-0.0591/n logQ
0.52 V= 0.466 V - 0.0591/2 logQ
0.52 V- 0.466 V = -0.02955logQ
0.054 V = -0.02955 logQ
log Q= 0.054 /-0.02955 =-1.8

Q= 10^(-1.8) = 0.01585

Q is the reaction qutient Q= [product concentration]/reactant concentration

[Pb2+]/[Cu2+] = 0.01585
[Pb2+] = 0.7 M * 0.01585 = 0.01109 M

Thus concentration of Pb2+ = 0.01109 M
********************************************

The cell can be represented as below
Pb/Pb2+//Cu2+/Cu
At anode :
Pb ---> Pb2+ +2e- oxidation takes place

At cathode:

Cu2+ + 2e- ---> Cu reduction tahkes place
overall cell reaction is
Pb(s) + Cu2+(aq) -->Pb2+(aq) +Cu(s)

firstly let us find out standard cell potential of the cell
E0cell= E0Cathode-Eoanode
- E0Cu2+/Cu- E0pb2+/Pb
= 0.34 V- [-0.126 V] = 0.466 V
Also number of electrons involved in the redox process is =2
n=2***

given cell potential Ecell= 0.52 V

concentration of Cu2+, [Cu2+] = 0.7 M

Using Nernst equation let us find out concentration of Pb2+

Ecell = E0cell-0.0591/n logQ

where Q is the reaction quotient Q= [product concentration]/reactant concentration

In the above Nernst equation we know the values of all terms
other than Q ,So let us plug in all to get    Q

Ecell = E0cell-0.0591/n [logQ]
0.52 V= 0.466 V - 0.0591/2 logQ
0.52 V- 0.466 V = -0.02955logQ
0.054 V = -0.02955 logQ
log Q= 0.054 /-0.02955 =-1.8

Q= 10^(-1.8) = 0.01585

Q is the reaction quotient Q= [product concentration]/reactant concentration

[Pb2+]/[Cu2+] = 0.01585
[Pb2+] = 0.7 M * 0.01585 = 0.01109 M

Thus concentration of Pb2+ = 0.01109 M
********************************************