Asked by Erika
                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.
            
            
        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.
Answers
                    Answered by
            DrBob222
            
    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.
    
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.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.