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CHEM - Faraday and Mole Calculation Example 2

Chemistry, Faraday, mole calculations- IBID | DSE | GCE | IAL | AP Chemistry

· chemistry,question analysis,mole calculations,Faraday

Here is another example of mole calculation🧐

mole, Faraday, Chemistry, tuttee, dse, gce, ib, igcse,ap
Time to begin!

Step 1:

Start by writing down the equation

2Cl- > Cl2 + 2e-

Step 2:

Find the number of moles in the 18 cm3 of chlorine gas

Volume = Moles × Molar Volume

❗️​❗️​ Remember to use the same units (cm3)

By rearranging the above equation,

Moles = 18 cm3 / 24000 cm3 = 0.00075 moles

Step 3:

Note that electricity always refers to the flow of electrons

From the half equation, 2 electrons are needed to produce chlorine gas

Hence, 2e- = 96500 × 2 = 193000 C

Final Step💪

Don't forget about the moles!

Multiply the number of coulombs by the amount of moles calculated earlier

193000 × 0.00075 = 144.75 Coulombs

Therefore, 144.75 Coulombs are required to produce 18 cm3 of chlorine gas

Job done

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