2. An industrial electrolytic process for producing aluminum uses the reactions

Question

2. An industrial electrolytic process for producing aluminum uses the reactions shown in the following equations. Anode (oxidation): C(s) + 2 O2–(aq) CO2(g) + 4 e– Cathode (reduction): Al3+() + 3 e– Al(s) How long would it take to produce enough aluminum to make a case (24 cans) of aluminum soft drink cans if each can used 5.00 g of aluminum, a current of 5.00 x 104 A was employed, and the current efficiency was 91.1%? 2. An industrial electrolytic process for producing aluminum uses the reactions shown in the following equations. Anode (oxidation): C(s) + 2 O2–(aq) CO2(g) + 4 e– Cathode (reduction): Al3+() + 3 e– Al(s) How long would it take to produce enough aluminum to make a case (24 cans) of aluminum soft drink cans if each can used 5.00 g of aluminum, a current of 5.00 x 104 A was employed, and the current efficiency was 91.1%? 2. An industrial electrolytic process for producing aluminum uses the reactions shown in the following equations. Anode (oxidation): C(s) + 2 O2–(aq) CO2(g) + 4 e– Cathode (reduction): Al3+() + 3 e– Al(s) How long would it take to produce enough aluminum to make a case (24 cans) of aluminum soft drink cans if each can used 5.00 g of aluminum, a current of 5.00 x 104 A was employed, and the current efficiency was 91.1%?

Explanation / Answer

current used = 5 x 10^4 x 0.911 = 45550 A

Coulombs = I x t = 45550 x t

t = time taken for depositing metal

for 5 g Al

time required = 5 x 96485 x 3/45550 x 27

                      = 1.2 s for 5 g

For 24 cans = 28.24 s is required

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