Asked by Alex
What is the reaction that would occur in a galvanic cell between solutions of Al3+ and Pb2, where would oxidation and reduction take place, and what would the cell potentials be?
Answers
Answered by
bobpursley
Aluminum is much more prone to oxidation, so it will lose electrons (oxidation).
Half cell potentials would be
Al>>Al^3+3e -1.66 is reduction potential
Pb>>Pb^2 + 2e -.13 is reduction potential
If Eo means oxidation potential (-reduction potential)
Cell potential = EoCell= Eo<sub>Red,Cathode</sub>−Eo<sub>Red,Anode</sub>
then
Cell potential= 1.66-(-1.3)=1.79 volts.
check my thinking.
Half cell potentials would be
Al>>Al^3+3e -1.66 is reduction potential
Pb>>Pb^2 + 2e -.13 is reduction potential
If Eo means oxidation potential (-reduction potential)
Cell potential = EoCell= Eo<sub>Red,Cathode</sub>−Eo<sub>Red,Anode</sub>
then
Cell potential= 1.66-(-1.3)=1.79 volts.
check my thinking.
Answered by
buxtehude
bobpursley, I think the calculation is wrong.
Al is oxidation, Pb is reduction. Ecell= E reduction (Pb) - E oxidation
So the cell potential will be Ecell = -0.13 - (+1.68) = -1.81 V
Al is oxidation, Pb is reduction. Ecell= E reduction (Pb) - E oxidation
So the cell potential will be Ecell = -0.13 - (+1.68) = -1.81 V
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