A. What mass of aluminum metal can be produced per hour in the electrolysis of a

Question

A. What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21A ?
B. Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction.
Ag+(aq)+e??Ag(s) Part A What mass of silver would plate onto the cathode if a current of 8.5A flowed through the cell for 66min ? Express your answer using two significant figures. A. What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21A ?
B. Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction.
Ag+(aq)+e??Ag(s) Part A What mass of silver would plate onto the cathode if a current of 8.5A flowed through the cell for 66min ? Express your answer using two significant figures.
B. Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction.
Ag+(aq)+e??Ag(s) Part A What mass of silver would plate onto the cathode if a current of 8.5A flowed through the cell for 66min ? Express your answer using two significant figures. B. Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction.
Ag+(aq)+e??Ag(s) B. Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction.
Ag+(aq)+e??Ag(s) B. Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction.
Ag+(aq)+e??Ag(s) Part A What mass of silver would plate onto the cathode if a current of 8.5A flowed through the cell for 66min ? Express your answer using two significant figures. Part A What mass of silver would plate onto the cathode if a current of 8.5A flowed through the cell for 66min ? Express your answer using two significant figures. Part A What mass of silver would plate onto the cathode if a current of 8.5A flowed through the cell for 66min ? Express your answer using two significant figures.

Explanation / Answer

1. Given current, I = 21A

time, t = 1 hour = 60x60 s = 3600s

Hence total charge given during electrolysis, Q = It = 21Ax3600s = 75600C

Al3+  + 3e- ----------------> Al

3F charge----------->27g ((atomic mass of Al)

1F charge -------> 27/3 = 9g

96500 C(= 1F) of charge produces Al = Equivlent mass of Al = 27/3 = 9g

Hence 75600C will produce Al = 9g x (75600C/96500C) = 7.05g = 7.1g (upto 2 SF, answer)

2. We can also solve this problem in a similar way as

Given current, I = 8.5A

time, t = 66 min = 66x60 s = 3960s

Hence total charge given during electroplating, Q = It = 8.5Ax3960s = 33660C

Ag+ + e- ----------------> Ag

1F charge --------> 108 g (atomic mass of Ag)

96500 C(= 1F) of charge produces Ag = Equivlent mass of Ag = 108g

Hence 33660C will produce Ag = 108g x (33660C/96500C) = 37.67g = 38 g (upto 2 SF, answer)

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