A zinc-copper battery is constructed as follows:
Zn | Zn+2(0.10 M) || Cu+2 (2.50 M)| Cu
The mass of each electrode is 200.0 g. Each half cell contains 1.00 liter of solution.
a) Calculate the cell potential when this battery is first connected.
b) Calculate the cell potential after a current of 10.0 amperes has flowed for 10.0 hours.
c) Calculate the mass of each electrode after 10.0 hours.
d) What is the total life span of this battery, delivering a current of 10.0 amperes, before it goes dead. (Hint: You need to know the limiting reagent. Is it Zn or Cu+2)
Perhaps I can get you started but that is it.
Since the concns are not standard, first calculate the Eo value for each half cell but BOTH as reductions.
For Zn^+ + 2e = Zn we have
E = Eo-(0.0592/n)*log[(Zn)/(Zn^+2)] =
-.763-(0.0592/2)*log[1/0.1)] = -0.793v. (check my work on this). This will be used as an oxidation reaction; therefore, reverse the equation and change the sign.
Eox = +0.793 v.
For the Cu, as a reduction.
E = Eo-(0.0592/n)*log[(Cu)/(Cu^+2)] =
+0.337 - (0.0592/2)*log(1/2.50) = 0.348v. Check this arithmetic, too.
Ered = 0.348 v.
Ecell = Eox + Ered = 0.793+0.348 = 1.14 v. Check this arithmetic (and chemistry).
The above is part a in DETAIL.
Post your work on the others if you get stuck and someone will help. Some hints:
amperes x seconds = coulombs.
96,485 coulombs will deposit one equivalent of a metal (1 equivalent = 1/2 mole for both Cu and Zn). For (b), you should start by calculating the amount of Zn that goes into solution and how much Cu^+2 comes out of solution, use those new concns to recalculate the half cell voltages just as I did in part (a), then calculate the new cell potential. Part (c) will follow from part (b). Follow the hint for part (d).
To calculate the cell potential for the given battery configuration, we can use the Nernst equation:
E = E° - (RT / (nF)) ln(Q)
Where:
E is the cell potential
E° is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of electrons transferred in the balanced equation
F is the Faraday's constant (96485 C/mol)
Q is the reaction quotient, which can be calculated using the concentrations of the species involved in the redox reaction.
a) Calculate the cell potential when this battery is first connected.
1. Identify the oxidation and reduction half-reactions:
Oxidation half-reaction: Zn → Zn+2 + 2e- (Zinc loses electrons and is oxidized)
Reduction half-reaction: Cu+2 + 2e- → Cu (Copper ions gain electrons and are reduced)
2. Write the balanced overall reaction by summing the half-reactions:
Zn + Cu+2 → Zn+2 + Cu
3. Calculate the reaction quotient (Q):
Q = [Zn+2] / [Cu+2]
Since the concentrations are not given, we assume both half-cells have 0.10 M and 2.50 M concentrations respectively:
Q = (0.10 M) / (2.50 M) = 0.04
4. Determine the standard cell potential (E°):
Consult a table of standard reduction potentials and find the reduction potentials for the Cu+2|Cu and Zn+2|Zn half-reactions.
E° for Cu+2|Cu = +0.34 V
E° for Zn+2|Zn = -0.76 V
The standard cell potential (E°) is the difference between the reduction and oxidation potentials:
E° = E°(cathode) - E°(anode)
E° = +0.34 V - (-0.76 V)
E° = +1.10 V
5. Substitute the values into the Nernst equation and solve for E:
E = E° - (RT / (nF)) ln(Q)
E = 1.10 V - [(8.314 J/(mol·K)) (298 K) / (2(96485 C/mol))] ln(0.04)
E ≈ 1.10 V
Therefore, the cell potential when this battery is first connected is approximately 1.10 V.
b) To calculate the cell potential after a current of 10.0 amperes has flowed for 10.0 hours, we need to consider the effect of Faraday's law of electrolysis.
According to Faraday's law, the amount of material (in moles) consumed or produced at an electrode is directly proportional to the quantity of electricity flowing through the cell.
To find the moles of reactants consumed, we can use the equation:
moles of reactant consumed = (current (A) x time (s)) / (nF)
Given:
current = 10.0 A
time = 10.0 hours = 36000 seconds (since 1 hour = 3600 seconds)
n = 2 (since 2 electrons are transferred in the half-reactions)
F = 96485 C/mol
moles of Zn consumed = (10.0 A x 36000 s) / (2 x 96485 C/mol)
Calculate this value.
c) To calculate the mass of each electrode after 10.0 hours, we need to use the moles of reactants consumed from part b) above.
Using the molar mass of Zn (65.38 g/mol) and Cu (63.55 g/mol), we can calculate the mass of each electrode.
mass of Zn electrode = moles of Zn consumed x molar mass of Zn
mass of Cu electrode = moles of Cu produced x molar mass of Cu
Calculate these values.
d) To determine the total lifespan of the battery delivering a current of 10.0 amperes before it goes dead, we need to identify the limiting reactant.
Since we've already calculated the moles of Zn and Cu consumed in part b), compare the moles to determine which reactant is limiting. The reactant with the lower moles is the limiting reactant.
The limiting reactant will be completely consumed first, and once it is depleted, the battery will stop working. The total lifespan can be determined by dividing the moles of the limiting reactant by its rate of consumption. The rate of consumption can be determined from the current value.
Using the balanced reaction equation and stoichiometry, we can determine the molar ratio between the limiting reactant and the electrons transferred.
Based on this information, calculate the total lifespan of the battery.
To answer these questions, we can use the Nernst equation and Faraday's laws of electrolysis. Let's break down each part step by step.
a) To calculate the cell potential when the battery is first connected, we need to find the standard cell potential and then determine if any changes occur due to concentration differences. The standard reduction potentials are given by the reduction half-reactions:
Zn+2 + 2e- → Zn E° = -0.76 V
Cu+2 + 2e- → Cu E° = +0.34 V
The overall cell potential (E°cell) can be calculated by subtracting the reduction potential of the anode (Zn) from the reduction potential of the cathode (Cu):
E°cell = E°(cathode) - E°(anode)
E°cell = 0.34 V - (-0.76 V)
E°cell = 1.10 V
However, we also need to consider concentration differences. The Nernst equation allows us to account for these differences:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
n is the number of electrons transferred in the balanced cell reaction
Q is the reaction quotient, which is the ratio of activities of products over reactants
In this case, n = 2 since 2 electrons are exchanged for both the Zn and Cu half-reactions.
Since [Zn+2] and [Cu+2] are given, we can substitute the values into the Nernst equation to find the cell potential (Ecell) when the battery is first connected.
b) To calculate the cell potential after a current of 10.0 amperes has flowed for 10.0 hours, we need to apply Faraday's laws of electrolysis. The equation relating electrolysis, current (I), time (t), and the quantity of electricity (Q) is:
Q = I * t
The quantity of electricity (Q) is directly proportional to the number of moles of electrons transferred (n). Since 1 Faraday (F) = 96,485 Coulombs (C) = n * 1 mol * 96,485 C, we can calculate the number of moles of electrons:
n = Q / (1 mol * 96,485 C)
We now know n, which can be substituted into the Nernst equation, along with the new concentrations at the given time.
c) To calculate the mass of each electrode after 10.0 hours, we need to calculate the moles of Zn and Cu ions that have been consumed or deposited due to the flow of current. This can be done by relating the number of electrons transferred to the moles of the substance involved in the half-reaction. From there, we can convert the moles to mass using the molar mass of each substance.
d) To determine the total life span of the battery delivering a current of 10.0 amperes, we need to identify the limiting reagent. We can do this by comparing the moles of electrons transferred (from Faraday's laws) to the stoichiometric coefficients in the balanced equation for each half-reaction. The reactant that has fewer moles per mole of electrons transferred is the limiting reagent. Once we know the limiting reagent, we can calculate the total charge consumed or deposited by it and use Faraday's laws to determine the total time it takes for that reactant to be completely consumed or deposited.
By following these steps, you should be able to successfully answer the questions.