# Question #b278f

Feb 27, 2017

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

$5 \times 3600 = 18000 C$

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

$\frac{18000 C}{96350 \left(\frac{C}{\text{mol}}\right)} = 0.187$ mol of electrons

The copper ion in $C u S {O}_{4}$ is a $C {u}^{2 +}$ ion. To reduce one mole of $C {u}^{2 +}$ ions, you must deliver two moles of electrons to the cathode of the cell:

$C {u}^{2 +} + 2 {e}^{-}$$\rightarrow C u$

Therefore, the 0.187 mol of electrons devilered during the hour mentioned above will reduce $0.187 \div 2 = 0.0935$ mol of $C {u}^{2 +}$ into 0.0935 mol (5.93 g) of Cu.