When 1 mole of Cu2+(aq) is reduced to Cu (s), how many faraday of charge are required?
A) 1
B) 2
C) 6.02 x 1023
D) none, charge is measured in Coulombs (C)
Explanation
For the MCAT, an electrolytic cell will always have a negative standard-state cell potential. A power supply must be used to drive the non-spontaneous reaction. Here, we have two half-reactions:
Ag+(aq) + e– → Ag(s) E° = 0.80
Cu2+(aq) + 2e– → Cu(s) E° = 0.34
One of the reactions must run in reverse in order to pair a reduction with an oxidation (which is necessary for a redox reaction). If the copper reaction runs in reverse, the standard reduction potential will be -0.34 + 0.8 = 0.46 V. This cannot be correct as the voltage is positive. Try running the Ag reaction in reverse and the copper forward: the standard-state cell potential is -0.80 + 0.34 = -0.46 V. (Note that we should not multiply the Ag potential by two, even though it only contains one electron while the other reaction contains two. This is important to remember for electrochemistry problems!) This could work for an electrolytic cell being driven by a power source. Both half-reactions cannot run in reverse, as we must always pair an oxidation with a reduction, and both half-reactions cannot run forward for the same reason.
a) -1.14 V, incorrect, This answer has both half-reactions running in reverse.
b) -0.46 V, correct. This answer has a negative standard-state cell potential, and one reaction runs forward (or reduces) while the other runs in reverse (or oxidizes).
c) 0.46 V, incorrect, An electrolytic cell on the MCAT will have a negative standard-state cell potential.
d) 1.14 V, incorrect, Both reactions cannot run forward, and the standard-state cell potential must be negative for an electrolytic cell.
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