Question #b278f

1 Answer
Dwight
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 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.

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