Question #b278f

1 Answer
Feb 27, 2017

Answer:

You have passed 18000 coulombs of charge through the cell, with the result the 0.0935 mol (5.93 g) of Cu is produced.

Explanation:

A 5-amp current is the result of 5 coulombs of charge passing through the cell every second. In one hour (3600 s) this will amount to

#5 xx 3600 = 18000 C#

To change this into "Chemistry units" we need to used the quantity known as a faraday of charge. One faraday of charge is the total charge carried by one mole of electrons, and is equal to 96 350 coulombs.

So, the charge determined above is

#(18000 C) / (96350 (C/"mol")) = 0.187# mol of electrons

The copper ion in #CuSO_4# is a #Cu^(2+)# ion. To reduce one mole of #Cu^(2+)# ions, you must deliver two moles of electrons to the cathode of the cell:

#Cu^(2+) + 2e^-##rarr Cu#

Therefore, the 0.187 mol of electrons devilered during the hour mentioned above will reduce #0.187 -: 2 = 0.0935# mol of #Cu^(2+)# into 0.0935 mol (5.93 g) of Cu.