Calculate the electrical voltage of an electrochemical cell using Pb(s) and
Co(s) as electrodes along with their appropriate solutions.
Co2+(aq) + 2e- --> Co(s) -0.28 V
Pb2+(aq) + 2e- --> Pb(s) -0.13 V
The oxidation reaction is
Co(s) --> Co2+(aq) + 2e- +0.28V
E0 net(cell) = E0 ox + E0 red
= (0.28 V) + (-0.13 V)
= +0.15 V
Does this look right? I am not sure if I answered what the question is asking.
Yes, that look fine to me. And your calculations are above reproach.
Yes, your approach is correct, and you have correctly calculated the electrical voltage of the electrochemical cell. The question asks for the electrical voltage of the cell, which is the difference between the standard reduction potentials of the oxidation and reduction half-reactions involved.
To calculate the overall voltage of the electrochemical cell, you need to determine the standard reduction potentials for both half-reactions (given in the question) and subtract the reduction potential of the anode (oxidation) half-reaction from the reduction potential of the cathode (reduction) half-reaction.
In this case, the reduction potential for the Co2+(aq) + 2e- --> Co(s) half-reaction is -0.28 V, and the reduction potential for the Pb2+(aq) + 2e- --> Pb(s) half-reaction is -0.13 V. According to convention, the less positive value is taken as the anode potential. Therefore, the Co(s) --> Co2+(aq) + 2e- half-reaction has an oxidation potential of +0.28 V.
To obtain the electrical voltage of the cell, you add the oxidation potential to the reduction potential:
E0 net(cell) = E0 ox + E0 red
= (0.28 V) + (-0.13 V)
= +0.15 V
So, the electrical voltage of the electrochemical cell, using Pb(s) and Co(s) as electrodes with their appropriate solutions, is +0.15 V.