How would you start and finish these questions by a step by step process? Does Amp equal to Joules?
1. How many grams of aluminum can be produced by the electrolysis of aqueous aluminum chloride by a current of 5.00 amps for 5.00 hours?
2. Consider the following electrochemical cell: Cu(s) | Cu^2+(aq, 0.10 M) || Cu^2+(aq), ?? M | Cu(s)
A) What is the overall cell reaction and Ecell?
B) Determine the unknown [Cu^2+] if Ecell = 0.084 V at 25°C.
A cell can deposit atomic mass Al/3 (about 27/3 = approx 9g) Al when 96,485 coulombs is passed through a solution. How many coulombs do you have? That's coulombs = amperes (that's amps) x seconds.
So you have C = 5.00 x 5.00 hrs x (60 min/1 hr) x (60 sec/1 min) = about 9E4.
The problem boils down to this. 96,485 C will deposit 9g Al; how much Al will be deposited by 9E4 C.
That's 9g x (9E4/96,485) = ?
That 9 is a close estimate. You should recalculate these numbers.
2.
I don't know how Ecell can be calculated without knowing ??M concn Cu in the right hand cell. I suspect you meant Eocell.
B. If Ecell is 0.084, then
Ecell = Eocell - (0.0592/n)log (dilute)/(concd) and solve for dilute.
To start and finish these questions, you can follow these step-by-step processes:
1. Calculation of Aluminum Production:
Step 1: Write the balanced chemical equation for the electrolysis of aqueous aluminum chloride:
2AlCl₃ + 6H₂O → 2Al + 6HCl + 3O₂
Step 2: Determine the number of moles of electrons transferred in the reaction:
From the balanced equation, we can see that 6 moles of electrons are transferred for every 2 moles of aluminum produced.
Step 3: Calculate the total charge passed through the electrolysis cell:
Charge = Current (A) × Time (s)
Since the current is given in amps and the time is given in hours, convert the time to seconds.
Step 4: Convert the charge to moles of electrons using Faraday's constant:
1 Faraday = 96,485 coulombs = 1 mole of electrons
Step 5: Convert moles of electrons to moles of aluminum using the stoichiometry from the balanced equation:
2 moles of Al / 6 moles of e⁻ = Moles of Al / Charge (in Faradays)
Step 6: Convert moles of aluminum to grams using the molar mass of aluminum:
Molar mass of Al = 26.98 g/mol
Grams of Al = Moles of Al × Molar mass of Al
2. Electrochemical Cell Calculation:
A) Overall Cell Reaction and Ecell:
Step 1: Write the overall cell reaction by combining the half-reactions occurring at the anode and cathode. The anode half-reaction takes place on the left side (anode) and involves oxidation, while the cathode half-reaction takes place on the right side (cathode) and involves reduction.
Step 2: Determine the cell potential (Ecell) by subtracting the reduction potential of the anode (E°_anode) from the reduction potential of the cathode (E°_cathode).
B) Determination of Unknown [Cu^2+]:
Step 1: Look up the standard reduction potential of the Cu^2+(aq) | Cu(s) half-reaction.
Step 2: Use the Nernst equation to relate the cell potential (Ecell) to the concentration of Cu^2+ at the unknown side:
Ecell = E°cell - (0.0592 V/n) × log([Cu^2+])
Step 3: Rearrange the Nernst equation and solve for the unknown concentration [Cu^2+].
Substitute the given Ecell value and solve for [Cu^2+].
Following these step-by-step processes, you can start and finish these questions by calculating the desired values.