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CHEM - Faraday and Mole Calculation Example 1

Chemistry, Faraday, Mole Calculations - IGCSE | AP Chemistry

· chemistry,question analysis,mole calculations,Faraday

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

Delving into the example

🤓Here is an example:

mole, Faraday, Chemistry, tuttee, dse, gce, ib, igcse,ap

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

Done!

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