Calculating mass and volume

In order to do these calculations you must know that:

• the charge associated with 1 mole of electrons is 96,500 C.
• the production of one mole of an element from its ion always requires n x 96,500 C, where n is the number of moles of electrons in the ion-electron equation.

Example 1

Calculate the mass of aluminium produced when molten aluminium oxide is electrolysed using a current (I) of 20,000 A for 5 hours, 21 min and 40s.

Step 1: Work out the quantity of charge (Q) passed in the time (t) given.

I = 20000 A

t = 19300 s

Q = I × t

= 20000 × 19300

= 386000000 C

= 3.86 × 10⁸ C

Step 2: From the ion-electron equation, work out the relationship between the number of moles of electrons and number of moles of product.

Al³⁺ + 3e^-→Al (from data book)

1 mol + 3 mol→1 mol

Step 3: Use the information in steps 1 and 2 to calculate the mass of aluminium produced.

Find: mass of Al (call this m grams) from 3.86 × 10⁸ C (from Step 1)

Link: 1 mol of Al ↔3 mol e^- (from Step 2)

Convert: 27g↔3 × 96500

Proportion: m ÷ 27 = (3.86 × 10⁸) ÷ (3 × 96500)

m = (27 × 3.86 × 10⁸) ÷ (2.895 × 10⁵)

m = 36 × 10³ g

= 36 kg

Answer: Mass of aluminium produced is 36 kg

The volume of gas produced during electrolysis can be calculated if the molar volume is known.

Example 2

In the industrial production of chlorine gas, a current of 50,000 A was passed through a salt solution for 1 hour. Calculate the volume of gas which would be produced. (Take the molar volume to be 25.0 l mol-1)

Step 1: Work out the quantity of charge (Q) passed in the time given:

I = 50000 A

t = 3600 s

Q = I × t

= 50000 × 3600

= 180000000

= 1.8 × 10⁸ C

Step 2:From the ion-electron equation, work out the relationship between the number of moles of electrons and number of moles of product gas:

2Cl^-(aq)→Cl₂(g) + 2e^- (from data book)

2 mol → 1 mol + 2 mol

Step 3:Use the information from steps 1 and 2 and the molar volume to calculate the volume of gas produced:

Find: volume of Cl₂ (call this v litres) from 1.8 × 10⁸ C (from Step 1)

Link: 1 mol of Cl₂↔2 mol e^- (from Step 2)

Convert: 25 l↔2 × 96500

Proportion: v ÷ 25 = (1.8 × 10⁸) ÷ (2 × 96500)

v = (25 × 1.8 × 10⁸) ÷ (1.93 × 10⁵)

= 2.33 × 10⁴ litres

Answer: The volume of chlorine produced is 2.33 x 10⁴ litres.