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
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.
To calculate the cell potential after the given conditions, follow these steps:
1. Calculate the amount of zinc (Zn) that goes into solution:
- You are given that 10.0 A of current has flowed for 10.0 hours. Using the equation Coulombs = Amperes x Seconds, we can calculate the total number of coulombs: Coulombs = 10.0 A x (10.0 hours x 3600 seconds/hour) = 3.6 x 10^5 C.
- The amount of coulombs required to use 1 mol of electrons is 96,485 C. So, the amount of coulombs required to use (65.4/2) g Zn (molar mass of Zn = 65.4 g/mol) is (3.6 x 10^5 C) / (96485 C/mol) x (65.4/2 g Zn) = 47.63 g Zn.
2. Calculate the amount of copper (Cu) deposited:
- The molar mass of copper is 63.5 g/mol. So, the amount of copper deposited is (3.6 x 10^5 C) / (96485 C/mol) x (63.5/2 g Cu) = 43.85 g Cu.
3. Calculate the change in amount of Zn and Cu in moles:
- The change in moles of Zn is (47.63 g Zn) / (65.4 g/mol Zn) = 0.7278 mol Zn.
- The change in moles of Cu is (43.85 g Cu) / (63.5 g/mol Cu) = 0.6906 mol Cu.
4. Calculate the final concentrations of Zn^2+ and Cu^2+:
- Initially, the concentration of Zn^2+ is 0.10 M, and since 1.00 L is relevant, we have 0.10 mol Zn^2+.
- Initially, the concentration of Cu^2+ is 2.50 M.
Adding the change in moles to the initial moles, we get:
- For Zn^2+: 0.10 mol + 0.7278 mol = 0.8278 mol.
- For Cu^2+: 2.50 mol - 0.6906 mol = 1.8094 mol.
5. Calculate the cell potential:
- The given reduction potentials are Eº(Zn) = -0.76 V and Eº(Cu) = 0.34 V.
- Using the Nernst equation, we have Ecell = Eocell - (0.0592/2)log(Zn^2+)/(Cu^2+).
- Plugging in the values, we get Ecell = (0.34 V) - (0.0592/2)log(0.8278)/(1.8094).
- Calculating this expression, we find that Ecell = 1.13 V.
Therefore, the correct answer is B. 1.13V.