A cobalt electrode is immersed in 1.0M Co2+ and a lead electrode is immersed in 1.0 M Pb2+. Given are standard reduction potentials.
Co(2+) + 2e- -> Co ; E(reduction)= -0.28 V
Pb(2+) + 2e- -> Pb ; E(reduction)= -0.13 V
Which of the following describes the cell?
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Correct Answer: B. E(cell) = 0.15 V, galvanic
The standard electrode potential E(cell) for a cell is found using the equation:
E(cell) = E(oxidation) + E(reduction)
In this case, lead will be reduced because it has a higher reduction potential than cobalt. This means that cobalt is being oxidized. Remember that on the MCAT, you will always be given standard reduction potentials, so if we need to find an oxidation potential, we simply switch to the negative reduction potential. Our equations should now look like this:
Co -> Co2+ + 2e- ; E(oxidation)= +0.28 V
Pb2+ + 2e- -> Pb ; E(reduction)= -0.13 V
Thus, using the first equation, we arrive at
E(cell) = (+0.28 V) + (-0.13 V) = 0.15 V
Remember that if E(cell) > 0, the reaction is spontaneous and happens in a galvanic cell. (The opposite is a non-spontaneous, electrolytic cell). Also, lead cobalt batteries were used in some of the first electric cars!
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