Question

A zinc-copper battery is constructed as follows (standard reduction potentials given below):
Zn | Zn2+ (0.10 M) || Cu2+ (2.50 M) | Cu
Zn2+ + 2 e- → Zn(s) Eº = -0.76 V
Cu2+ + 2 e- → Cu(s) Eº = 0.34 V
The mass of each electrode is 200 g. Each half-cell contains 1.00 L of 1M solution. Calculate the cell
potential after 10.0 A of current has flowed for 10.0 hours.
A. 0.90V
B. 1.13V
C. –1.00V
D. 1.20V
E. 2.00V

Answer 1

I'm confused with the problem. How can the half cell contain 1.00 L of 1M solution when the cell is given as 0.1M Zn^2+ and 2.50M Cu^2+?

Answer 2

Disregard of 1M solution, it's a typo. The volume of 1.00L is relevant

Answer 3

I would do this.
Calculate g Zn that goes into solution and g Cu that are deposited from solution.
Coulombs = amperes x seconds = ?.
95,485 coulombs will use (65.4/2)g Zn and deposit (63.5/2)g Cu.
Calculate g Zn used, convert to mols, and add to the 0.1 mol already there.
Calculate g Cu deposited and convert to mols and subtract from the 2.50 mols there initially. Then
Ecell = Eocell - (0.0592/2)log(Zn^2+)/(Cu^2+).

Answer 4

I did all the calculations, but I got 1.09V... not sure where I went wrong with calculations

Answer 5

Yea I got 1.09 as well