Electrolysis and Faraday's Law

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AP Chemistry › Electrolysis and Faraday's Law

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How many grams of Cr can be obtained by the electrolysis of a Cr(NO3)3 if 10 amps are passed through the cell for 6 hours?

38.8 g

CORRECT

19.4 g

103 g

12.5 g

56.3 g

Explanation

$Cr^{3+} + 3 e- \rightarrow Cr(s)$

$mass = \frac{10 A , x, 6 hr , x , \frac{3600 s}{1 hr}}{ \frac{96485 C}{mol e-} } , x , \frac{1 mol , Cr}{3 mol e-} , x , \frac{51.996 g , Cr}{1 mol , Cr}= 38.8 g$

How long would it take to electroplate 28.3g of silver at a constant current of 2.0 amps from a solution of silver nitrate?

422 min

844 min

211 min

CORRECT

105 min

53 min

Explanation

$Ag^+ + 1 e- \rightarrow Ag(s)$

$28.3g : Ag : x : \frac{1 mol Ag}{107.87g} : x : \frac{1 mol e-}{1 mol Ag^+ } : x : \frac{96485 C}{1 mol e^-} = 25313 C$

$time = \frac{25313 C}{2 A}: x : \frac{1 min}{60 sec} = 211 min$

In the electrolysis of CaCl2 a 1.15 amp current is passed through the cell for 5.0 hours. How much Ca is produced?

4.3 g

CORRECT

8.6 g

18.3 g

2.2 g

1.1 g

Explanation

$Ca^{2+} + 2 e- \rightarrow Ca(s)$

$mass = \frac{1.15 A , x , 5 , hr , x, \frac{ 3600 , s}{1 , hr} }{\frac{96485, C}{mol , e-}}, x , \frac{1 mol Ca }{2, mol , e-}, x , \frac{40.078 g , Al}{1 mol , Al}= 4.3 g$

How much Al would be precipitated if an AlCl3 solution was electrolyzed for 1.00 hours with 5.00 amps?

3.36 g

1.68 g

CORRECT

6.72 g

0.84 g

1.25 g

Explanation

$Al^{3+} + 3 e- \rightarrow Al(s)$

$mass = \frac{5 A x 1 hr , x , \frac{3600 s}{1 hr} }{\frac{96485 C}{mol e-}} , x , \frac{1 mol , Al }{3 , mol , e-}, x, \frac{26.982 g Al}{1 mol Al} = 1.68 g$