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CHEM - Faraday and Mole Calculation Example 1

Chemistry, Faraday, Mole Calculations - IGCSE | AP Chemistry

April 30, 2018

When dealing with questions related to mole calculations, following a few simple steps can make it easier to solve them. 🤔

🤓Here is an example:

1.57 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions

One faraday is 96500 coulombs. It is also one mole of electrons.

If current of 0.2 Amps is passed through copper(II) sulphate for two hours, how much copper do you get?

Step 1:

Start by writing out the equation

Cu2+ ​+ 2e →​ Cu​

Step 2:

Find out the coulombs of electrons flowing

Charge (coulomb, C) = Current (ampere, A) × Time (second, s).

*Remember to convert the time into seconds !

Therefore, Charge = 0.2 × 7200 = 1440 coulombs​

Step 3:

Convert Coulombs into moles of electrons

Moles = Coulombs / faraday

Moles = 1440/96500 = 0.015

Step 4:

Find the moles of the product using the scale factor 

*From the equation, for every 2 moles of electrons, there will be 1 copper.

Scale factor = moles of product/moles of electrons = 1/2

Thus, there ​are 0.015 × 1/2 = 0.0075 moles of Cu

LAST STEP👏​

Convert moles into mass

Mass = Moles × Molar Mass

The mass of Cu = 0.0075 × 63.5 = 0.48 g

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