Question

Posted by Ana on Saturday, May 3, 2014 at 11:23am.


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

chemistry - Science - DrBob2222, Saturday, May 3, 2014 at 12:15pm

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+?

chemistry - Science - Ana, Saturday, May 3, 2014 at 12:19pm
Disregard of 1M solution, it's a typo. The volume of 1.00L is relevant

chemistry - Science - DrBob222, Saturday, May 3, 2014 at 11:36pm
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+).

chemistry - Science - Ana, Monday, May 5, 2014 at 9:16pm
I did all the calculations, but I got 1.09V... not sure where I went wrong with calculations

Answer 1

I worked through it and came up with 1.09v also. 1.13 seems to be the closest ad I would go with that. Sometimes answer sheets give approximate answers and this could be one of them. If I'm missing something I don't see it.