Question

Given the following reaction half reaction:

Cd2+ + 2e ® Cd    E° = - 0.403 V

Pb2+ + 2e-®Pb     E° = - 0.126 V

where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M

construct a galvanic cell using half cell reactions and find

i) Which half cell –reaction is the cathode?

ii) Write the cell reaction.

iii) Calculate E_{cell}^{\circ} of this galvanic cell.

iv) Calculate Ecell of this galvanic cell

v) Calculate equilibrium constant using E°cell potentials of this galvanic cell based on the construction of ions.

06 Apr 2018
Answer :
Word Count : 389

The first step is to determine the cell reaction and total cell potential.

In order for the cell to be galvanic, E0cell > 0.

For this reaction to be galvanic, the cadmium reaction must be the oxidation reaction. Cd → Cd2+ + 2 e- E0 = +0.403 V
Pb2+ + 2 e- → Pb E0 = -0.126 V

The total cell reaction is:

Pb2+(aq) + Cd(s) → Cd2+(aq) + Pb(s)

and E0cell = 0.403 V + -0.126 V = 0.277 V

The Nernst equation is:

Ecell = E0cell - (RT/nF) x lnQ

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