Question

An electrochemical cell is set up with a lead metal electrode immersed in a 0.1393 M solution of Pb2+ joined, through a salt bridge, to a 0.0511 M solution of Cd2+ into which is placed a cadmium metal electrode. Calculate the potential of this cell in its galvanic configuration and write out the shorthand notation for the galvanic cell.

Answer 1

Pb2+(aq) + 2e- ---> Pb(s) ..... E° = - 0.127 V
Cd2+(aq) + 2e- ---> Cd(s) ..... E° = - 0.402 V

The reduction potential of Pb2+ is greater than that of Cd2+, Pb2+ will be reduced and Cd will be oxidized.

Anode half-reaction: Cd ----> Cd2+ + 2e- ..... E° = + 0.402 V
Cathode half-reaction: Pb2+ + 2e- ----> Pb .. E° = - 0.127 V

Cell reaction: Cd + Pb2+ ----> Cd2+ + Pb . E°cell = + 0.275 V

The shorthand notation for the galvanic cell:

Cd (s) | Cd²⁺ (aq, 0.0511 M) || Pb²⁺ (aq, 0.1393 M) | Pb (s)

The overall cell reaction is: Pb2+ + Cd ==> Cd2+ + Pb . . .Eo cell = +0.275 V.

Q cell = [Cd2+]/[Pb2+] = moles Cd2+/moles Pb2+

Q cell =0.0511 / 0.1393 = 0.367

the potential of this cell is as follows:

E cell = Eo cell - 0.059/2 log Q

E cell = 0.275 V - 0.059/2 log 0.367

E cell = 0.275 V - 0.059/2 (-0.436)

E cell = 0.275 V +0.0129V

E cell = 0.2879 V