The electrolysis of molten AlCl_3 for 325 hr with an electrical current of 15.0

Question

The electrolysis of molten AlCl_3 for 325 hr with an electrical current of 15.0 A produces ____g of aluminum metal A) 49.I B) 0.606 C) 4.55 times 10^-3 D) 16.4 E) 147 How many seconds required to produce 4.00 g of aluminum metal from the electrolysis of molten AlCl_3 with an electrical current of 12.0 A? A) 3.57 times 10^3 B) 1.19 times 10^3 C) 2.90 times 10^5 D) 9.00 E) 27.0 Calculate the delta G degree_rxn using the following information. 4 HNO_3(g) + 5N_2 H_4(1) rightarrow 7 N_2(g) + 12 H_2O(1) delta G degree_rxn = ? delta H degree f(kJ/mol) -133.9 50.6 -285.8 S degree(J/mol-K) 266.9 121.2 191.6 70.0 A) -3.15 times 10^3 kJ B) -2.04 times 10^3 kJ C) -3.30 times 10^3 kJ D) +4.90 times 10^3 kJ E) +3.90 times 10^3 kJ Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 degree C. Sn(s) | Sn^2+(aq 0.022M) | Ag+(aq, 2.7M) | Ag(s) A) +0.01 V B) -0.66 V C) +1.01 V D) -0.83 V E) +1.31 V The gas OF_2 can be produced from the electrolysis of an aqueous solution of KF, as shown in the equation below. OF_2(g) + 2H^+(aq) + 4 e^- rightarrow H_2O(l) + 2F-(aq) E degree = +2.15 V Using the given standard reduction potential, calculate the amount of OF_2 that is produced, and the electrode at which the OF_2 is produced, upon the passage of 0.480 faradays through an aqueous KF solution. A) 6.48 g of OF_2 at the anode. B) 26.0 g of OF_2 at the anode. C) 26.0 g of OF_2 at the cathode. D) 6.48 g of OF_2 at the cathode.

Explanation / Answer

Question 23.

AlCl3 --< Al3+ + 3Cl-

so

t = 3.25 h = 3.25 h * 3600 s / h = 11700 seconds

I = 15 A = 15 C/s

total Charge = I*t = 11700*15 = 175500 C

note that

1 mol of e- = 96500 C

so

175500 C --> 175500/96500 = 1.818 mol of e-

now, remember that

1 mol of Al+3 require 3 mol of e-

so

1.818 mol of e- --> 1.818/3 = 0.606 mol of Al

Change to mass

1 mol of Al = 26.98 g

0.606 mol --> 0.606*26.98 = 16.35 g of Al

the best anaswer is D

16.4 g of Aluminium

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