Question

A battery is constructed based on the oxidation of magnesium and the reduction of Cu^2+. The initial concentrations of Mg^2+ and Cu^2+ are 1.2*10^-4M and 1.5M, respectively, in 1.0-L half-cells.

The initial voltage of the battery is 2.83V with the standard voltage at 2.71V.

QUESTION: What is the voltage of the battery after delivering 5.0A for 7.0hr?

Answer 1

E = E_cell^o - (0.0592/n)*log Q. [See the last line for the equation from which you can get the expression for Q).
You know E_cell^o = 2.71 v from the problem.
What you must do is calculate the new concns of Cu^+2 (which is being used up) and Mg^+2 (which is being increased) after the cell has run 5.0 amperes for 7 hours. This is how you do it.
Coulombs = amperes x time = 5.0 x 7 hrs x (3600 sec/hr) = ??
You know that Cu^+2 will be removed. How much? 96,485 coulombs will remove 1/2 mol Cu^+2 which is (63.54g/2). How much was there originally. 1.5 M, convert to grams. Subtract what you started with minus the final and that is the grams remaining (and it was in 1 L) so convert that to moles and that is the molarity. Do the same process for Mg. Now that you have the NEW concns of Cu^+2 and Mg^+2, substitute those into Q and calculate the new E for the cell.
By the way, the cell you are talking about is
Mg(s) + Cu^+2 ==<> Cu(s) + Mg^+2

Answer 2

bad explanation