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
I did all the calculations, but I got 1.09V... not sure where I went wrong with calculations
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+?
Disregard of 1M solution, it's a typo. The volume of 1.00L is relevant
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+).
Yea I got 1.09 as well
To calculate the cell potential after 10.0 A of current has flowed for 10.0 hours, we need to consider the amount of charge that has passed through the circuit.
The formula for the amount of charge (Q) is given by:
Q = I * t
Where:
Q = amount of charge (in coulombs)
I = current (in amperes)
t = time (in seconds)
In this case, the current is given as 10.0 A and the time is given as 10.0 hours. Since 1 hour is equal to 3600 seconds, the time in seconds will be:
t = 10.0 hours * 3600 seconds/hour = 36000 seconds
Substituting these values into the formula, we have:
Q = 10.0 A * 36000 s = 360000 C
Next, we need to calculate the number of moles of electrons that have been transferred. Since 1 Faraday (F) is equal to 96500 C, we can use the formula:
n = Q / F
Where:
n = number of moles of electrons
Q = amount of charge (in coulombs)
F = Faraday's constant (in coulombs per mole of electrons)
Substituting the values, we have:
n = 360000 C / 96500 C/mol = 3.731 moles of electrons
Now we can calculate the change in cell potential (ΔE°) using the Nernst equation:
ΔE° = E°cell - (RT / (nF)) * ln(Q_prod / Q_react)
Where:
ΔE° = change in cell potential
E°cell = cell potential at standard conditions
R = gas constant (8.314 J/(mol K))
T = temperature (in Kelvin)
n = number of moles of electrons
F = Faraday's constant (in coulombs per mole of electrons)
ln = natural logarithm
Q_prod = concentration of products raised to their stoichiometric coefficients
Q_react = concentration of reactants raised to their stoichiometric coefficients
Since the temperature is not given, we assume it to be 298 K (room temperature). Plugging in the given standard reduction potentials and concentrations, we have:
ΔE° = (0.34 V) - ((8.314 J/(mol K)) * (298 K) / (3.731 mol) * ln((2.50 M)^1 / (0.10 M)^1)
Simplifying this:
ΔE° = 0.34 V - (22.843 J/mol * ln(25))
Now we need to convert the units of ΔE° from J/mol to V. Since 1 J/C is equal to 1 V, we can divide ΔE° by Faraday's constant (96500 C/mol) to give the cell potential. Therefore:
Cell potential (Ecell) = ΔE° / F
Substituting the values:
Ecell = (0.34 V - (22.843 J/mol * ln(25))) / 96500 C/mol
Calculating this value, we get:
Ecell ≈ 0.901 V
Therefore, the cell potential after 10.0 A of current has flowed for 10.0 hours is approximately 0.901 V.
Hence, the correct answer is A. 0.90 V.