This is a question from my Chemistry textbook:
A particular lithium cell has an operating voltage of 3.5 V. The overall reaction taking place when discharging is as shown by this equation: LiC
6 + CoO
2 C
6 + LiCoO
2. Calculate the increase in mass of solid LiCoO
2 if the cell delivers a 0.60 A current for 2.5 hours operating at 65% efficiency. Assume energy is lost as heat.
I thought that if a galvanic cell operates at an efficiency less than 100%, the increase in mass at the cathode would be less than the amount predicted by Faraday's laws. Below are my calculations:
Q = It = 0.60 x 2.5 x 60 x 60 = 5400C
n(e-) = 5400 / 96500 = 0.056mol
n(LiCoO
2) = 0.056mol
m(LiCoO
2) = 0.056x 97.8 = 5.47g
m(LiCoO
2) with 65% efficiency = 0.65 x 5.47g = 3.6g
However, the solutions give a different answer. Instead of multiplying the mass of 5.47g by 0.65, they multiply it by 100/65 to give 8.4g. This is their reasoning - "only 65% of the chemical energy released by the reaction is converted into electrical energy, so the reaction will have progressed "further" than expected, to produce the stated amount of electricity, so more reactant would be used and more solid LiCoO
2 produced than if it was running at 100%".
The part regarding more reactant being consumed to produce the current of 0.60A sort of makes sense but wouldn't that "extra" chemical energy just be converted to heat energy and not lead to any extra deposition? If so, wouldn't the amount of LiCoO
2 deposited be simply equal to that predicted by Faraday's laws?
Thank you.