Question 1
Find the cell potential of a galvanic cell, based on the following reduction half-reactions at 25 °C:
Cd2+ + 2 e– → Cd E0 = -0.403 V
Pb2+ + 2 e– → Pb E0 = -0.126 V
where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M.
Answer: 0.3V
Question 2
Consider the following cell at 25∘C, where the reaction is:
2Al(s)+3Mn2+(aq)→2Al3+(aq)+3Mn(s)
[Mn2+]=0.50M and [Al3+]=1.50M.
Find the cell potential of such a system.
Answer: 0.47V
Question 3
Use the Nernst equation to determine the potential difference at 25oCin the following galvanic cell systems:
Cd(s)│Cd2+(aq) (0.002M) ││Zn2+(aq) (0.075M) │Zn(s)
Sn(s) │Sn(aq) (0.01M) ││Fe3+(aq)(0.05M) │ Fe2+(aq)(0.95M)