12:22 AM ooo T-Mobile LTE 100% Close PROBLEM SET 10 due April 24, 2017 For each

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12:22 AM ooo T-Mobile LTE 100% Close PROBLEM SET 10 due April 24, 2017 For each question, write the letter of the correct answer inthe column onthe right. 1. What mass of chromium can be deposited byelectrolysis of an aqueous solution of Cra(soda for 160 min using a constant current of 15.0 A? (F 9648SC/mol AO 0431 g c) 232.8 g DO 0.187g 2. Copper is electroplated from Cusoesolution. A constant current of 439Ais applied by an external power supply. How long will ittake to deposit 1000gofcu? Al 19.2h B) 1317s C01.48 days D) 96 min 3. Use the information below to decide which ONE of the following statements of a 10M solution of sodium iodide INa0in water true conceming the electrolysis a-2.71, v Na(s: E 0.54V Em -083V AI The solution becomes more basic aselectrolysistakes place. B) Sodium metal is deposited lodine (la) is formed at the cathode. Dlonygen is evolved at the anode. E) Hydrogen is evolved at the anode. 4. When an aqueous solution of lithium sulfate u Solis electrolyzed, what are the expected products? Any of the half-reactions listed below maybe involved. -3.04 AI Lils and Haig) S. E of a molten salt with the formula MC, using a current of 386 Afor 162 min, deposits 1.52gofmetal. Identify the metal. A) Na C Ca DIRb EK

Explanation / Answer

What mass of chromoium can be deposited by electrolysis of an aqueous solution of cr2(so4)3 for 160 min

Q = it = 15 A * 160 min (60 sec / 1 min) = 144000 C

Q / F = 144000 / 96500 = 1.49 mole e-

Cr2(SO4)3 -- SO4 has a 2- charge, so together it has 6-. So Cr has to be in the 3+ oxidation state. Therefore, this is a 3-electron process.

1.49 mole e- (1 mol Cr / 3 moles e-) = 0.496 mole Cr (52.00 g/mol Cr) = 25.9 g

Anser 25.9 g

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Copper is electroplated from cuso4 solution . a constant current of 4.39 A is applied by an external power supply . how long will it take to deposit 100 g of Cu?

Answer =19.2 hours

the atomic mass of copper is 63.546

100 g * mol /63.546 =1.57 mol

Moles =it /Ef

where I = amps and t = time

1.57 mol =4.39 * t / 2*96500

T= (1.57 *2*96500     )/ 4.39

=69022.77

69022.7700012 seconds = 19.172991667 hours

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