Question

Question #b278f

Answer 1

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.