Electrolysis: Calculations (CIE A Level Chemistry)

Calculate the amount of magnesium deposited when a current of 2.20 A flows through the molten bromide for 15 minutes.

Answer:

• The magnesium (Mg²⁺) ion is a positively charged cation that will move towards the cathode.
• Step 1: Write the half-equation at the cathode

Mg²⁺(aq)          +          2e^-       →         Mg(s)

1 mol                           2 mol               1 mol

• Step 2: Determine the number of coulombs required to deposit one mole of magnesium at the cathode
  • For every one mole of electrons, the number of coulombs needed is 96 500 C mol^-1
  • In this case, there are two moles of electrons required
  • So, the number of coulombs needed is:
  • F = 2 x 96 500
  • F = 193 000 C mol^-1
• Step 3: Calculate the charge transferred during the electrolysis
  • Q = I x t
  • Q = 2.20 x (60 x 15) = 1980 C 
• Step 4: Calculate the mass of magnesium deposited by simple proportion using the relative atomic mass of Mg

Charge (C)	Amount of Mg deposited (mol)	Amount of Mg deposited (g)
193 000	1	24.3
1980	= 0.0103	0.0103 x 24.3 = 0.25

• Therefore, 0.25 g of magnesium is deposited at the cathode

Calculate the volume of oxygen gas produced at room temperature, when a concentrated aqueous solution of sulfuric acid, H₂SO₄, is electrolysed for 35.0 minutes using a current of 0.75 A.

Answer:

• The oxygen gas is formed from the oxidation of negatively charged hydroxide (OH^-) ions at the anode-
• Step 1: Write the half-equation at the anode

   4OH^-(aq)         →         O₂(g)               +          2H₂O(l)            +          4e^-

   4 mol                           1 mol                           2 mol                           4 mol

• Step 2: Determine the number of coulombs required to form one mole of oxygen gas at the anode
  • For every one mole of electrons, the number of coulombs needed is 96 500 C mol^-1
  • So, for four moles of electrons, the number of coulombs needed is:
  • F = 4 x 96 500
  • F = 386 000 C mol^-1
• Step 3: Calculate the charge transferred during the electrolysis
  • Q = I x t
  • Q = 0.75 x (60 x 35) = 1575 C
• Step 4: Calculate the volume of oxygen liberated by simple proportion using the relationship 1 mol of gas occupies 24.0 dm³ at room temperature

Charge (C)	Amount of O2 liberated (mol)	Amount of O2 liberated (dm3)
386 000	1	24
1575	= 4.080 x 10-3	4.080 x 10-3 x 24.0 = 0.0979

• Therefore, 0.0979 dm³  of oxygen is formed at the anode

Calculating the volume of hydrogen gas produced at room temperature, when a concentrated aqueous solution of sodium sulfate, Na₂SO₄, is electrolysed for 17.5 minutes using a current of 3.25 A.

The hydrogen gas is formed from the reduction of positively charged hydrogen ions, H⁺ at the cathode.

Answer:

• Step 1: Write the half-equation at the cathode

2H⁺ (aq)   +     2e^-       →         H₂ (g)              

2 mols             2 mols             1 mol   

• Step 2: Determine the number of coulombs required to form one mole of hydrogen gas at the cathode
  • For every one mole of electrons, the number of coulombs needed is:
    • F = 96 500 C mol^-1
    • F = 1 x 96 500
    • F = 96 500 C
  • So, for two moles of electrons, the number of coulombs needed is:
    • F = 2 x 96 500
    • F = 193 000 C
• Step 3: Calculate the charge transferred during the electrolysis
  • Q = I x t
  • Q = 3.25 x (60 x 17.5) = 3 413 C
• Step 4: Calculate the volume of hydrogen liberated by simple proportion using the relationship 1 mol of gas occupies 24.0 dm³ at room temperature

Charge (C)	Amount of H2 liberated (mol)	Amount of H2 liberated (dm3)
193 000	1	24
3413	= 1.76 x 10-2	1.76 x 10-2 x 24 = 0.42

• Therefore, 0.42 dm³ of hydrogen is formed at the cathode