Electrolysis and Faraday's Law - AP Chemistry

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Question

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?

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Answer

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

$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$

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Question

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?

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Answer

$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$

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Question

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

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Answer

$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$

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Question

What current must be past through a solution of AlCl3 for 1.19 hours to produce 40g of solid Al?

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Answer

$40g , Al : x , $\frac{1, mol , Al}{26.982g}$, x, $\frac{3 , mol , e-}{1 , mol , Al}$ , x , $\frac{96485 C}{1 , mol , e^-}$ = 429108 , C$

$Current = $\frac{429108 C}{1.19hr , x , \frac{3600 s}${1hr}} = 100 A$

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Question

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

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Answer

$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$

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Question

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?

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Answer

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

$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$

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Question

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?

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Answer

$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$

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Question

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

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Answer

$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$

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Question

What current must be past through a solution of AlCl3 for 1.19 hours to produce 40g of solid Al?

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Answer

$40g , Al : x , $\frac{1, mol , Al}{26.982g}$, x, $\frac{3 , mol , e-}{1 , mol , Al}$ , x , $\frac{96485 C}{1 , mol , e^-}$ = 429108 , C$

$Current = $\frac{429108 C}{1.19hr , x , \frac{3600 s}${1hr}} = 100 A$

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Question

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

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Answer

$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$

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Question

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?

Tap to reveal answer

Answer

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

$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$

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Question

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?

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Answer

$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$

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Question

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

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Answer

$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$

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Question

What current must be past through a solution of AlCl3 for 1.19 hours to produce 40g of solid Al?

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Answer

$40g , Al : x , $\frac{1, mol , Al}{26.982g}$, x, $\frac{3 , mol , e-}{1 , mol , Al}$ , x , $\frac{96485 C}{1 , mol , e^-}$ = 429108 , C$

$Current = $\frac{429108 C}{1.19hr , x , \frac{3600 s}${1hr}} = 100 A$

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Question

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

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Answer

$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$

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Question

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?

Tap to reveal answer

Answer

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

$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$

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Question

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?

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Answer

$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$

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Question

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

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Answer

$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$

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Question

What current must be past through a solution of AlCl3 for 1.19 hours to produce 40g of solid Al?

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Answer

$40g , Al : x , $\frac{1, mol , Al}{26.982g}$, x, $\frac{3 , mol , e-}{1 , mol , Al}$ , x , $\frac{96485 C}{1 , mol , e^-}$ = 429108 , C$

$Current = $\frac{429108 C}{1.19hr , x , \frac{3600 s}${1hr}} = 100 A$

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Question

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

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Answer

$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$

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