Here we consider instead dissolving guns (which we will assume are pure iron) with sulfuric acid.
Fe+H2SO4->FeSO4+H2
It is apparent that they must use an outside source of electrical power to drive the dissolution of the iron. They use a small current so that the dissolution proceeds with the minimum voltage required. Assume standard values for the electrochemical potentials.
1. How much electrical energy supplied this way is thus required to dissolve an additional 1 kg of iron? Give your answer in kJ.
2. The characters realize that their hideout has been discovered by the police and they still have a last handgun that weighs 0.25 kg to dissolve. The cops will get there in an hour, so they have to speed up the reaction, by driving it at a higher current. What is the minimum total voltage in volts they'll need to drive the reaction at to get rid of the gun in time? Consider excess potential because of activation losses only and the exchange current I0 to be 1 A for the reaction over the surface of the entire tank (not a current density). α is 0.5 and everything is being done at room temperature. Assume standard electrochemical potentials.
To answer these questions, we need to use the electrochemical potentials and consider the amount of iron being dissolved. Here's how we can approach each question:
1. To determine the electrical energy required to dissolve 1 kg of iron, we need to calculate the amount of charge required and then multiply it by the voltage. Given the balanced equation: Fe + H2SO4 -> FeSO4 + H2, we can see that the stoichiometric ratio between iron and charge is 1:2. This means that it takes 2 Faradays of charge to dissolve 1 mole of iron.
First, we calculate the number of moles of iron in 1 kg:
Number of moles of Fe = (Mass of Fe) / (Molar mass of Fe)
= 1000 g / (55.85 g/mol)
≈ 17.90 mol
Now, we can calculate the number of Faradays of charge required:
Number of Faradays = (Number of moles of Fe) * (2 Faradays/mol)
= 17.90 * 2
= 35.80 Faradays
Since 1 Faraday is equivalent to 96,485 Coulombs (C), we can calculate the total charge in Coulombs:
Total charge = (Number of Faradays) * (96,485 C/Faraday)
= 35.80 * 96,485
≈ 3,447,518 C
Finally, to calculate the electrical energy in kJ, we use the formula:
Electrical energy = (Total charge) * (Voltage)
= (3,447,518 C) * (Voltage)
= 3,447,518 * Voltage kJ
2. In this question, we need to determine the minimum total voltage required to dissolve a 0.25 kg handgun in one hour. We know the exchange current (I0) is 1 A, α is 0.5, and consider excess potential due to activation losses.
To find the minimum total voltage, we can use the Butler-Volmer equation for the anodic (oxidation) reaction:
i = I0 * [exp(alpha * (eta - E0)) - exp(-(1-alpha) * (eta - E0))]
Here, i represents the current density, I0 is the exchange current, alpha is the charge transfer coefficient (0.5 in this case), eta is the overpotential, and E0 is the standard electrochemical potential.
To simplify the problem, we will assume the handgun is a perfect sphere with a uniform surface area.
Given:
Mass of handgun (m) = 0.25 kg
Time (t) = 1 hour = 3600 seconds
First, we calculate the number of moles of iron in the handgun:
Number of moles of Fe = (Mass of handgun) / (Molar mass of Fe)
= 0.25 kg / (55.85 g/mol)
≈ 0.00448 mol
Next, we calculate the total charge required for the dissolution:
Total charge = (Number of moles of Fe) * (2 Faradays/mol)
= 0.00448 * 2
= 0.00896 Faradays
To determine the current density, we divide the charge by the time:
Current density = (Total charge) / (Time)
= 0.00896 Faradays / 3600 seconds
≈ 2.49 × 10^-6 A/cm^2
The current density (i) is given by:
i = I0 * [exp(alpha * (eta - E0)) - exp(-(1-alpha) * (eta - E0))]
Since the current density given is the average current density over the entire tank, we can calculate the voltage required using the Tafel equation:
Voltage = [ln(i/I0)] / [alpha * (2.303 * RT/F)]
Where:
R = Gas constant = 8.314 J/(mol * K)
T = Temperature in Kelvin (room temperature)
F = Faraday constant = 96,485 C/mol
Using the equation above, we can solve for the voltage required.