The electrolysis of some other solutions using inert electrodes
3 marks · NormalA student electrolyses an aqueous solution of sodium sulfate using inert electrodes.
She collects 120 cm³ of gas at the anode over a period of time.
Only two types of negative ions are present: sulfate ions and hydroxide ions.
State which ion is discharged at the anode, write the ionic equation for the reaction, and calculate the volume of oxygen molecules produced in moles (to 2 significant figures). (1 mol of any gas occupies 24 000 cm³ at room temperature and pressure.)
State which ion is discharged at the anode.
Write the ionic half-equation for the reaction at the anode.
Calculate the amount of oxygen produced, in moles, to 2 significant figures.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) hydroxide ion (OH⁻) · must show hydroxide ion, OH-
(b) 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
(c) 0.0050 mol
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Identifies hydroxide ion (OH-) as the ion discharged at the anode.
B2 (b) · 1 mark · Gives correct ionic half-equation: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻.
B3 (c) · 1 mark · Converts volume of oxygen to moles correctly (120/24000 = 0.0050 mol; allow algebraic equivalent within tolerance).
Worked solution
(a) At the anode, the ions present are sulfate (SO₄²⁻) and hydroxide (OH⁻) ions. Sulfate ions are too difficult to oxidise, so hydroxide ions are discharged.
Answer: hydroxide ion (OH⁻)
(b) The ionic half-equation for discharge of hydroxide ions is:
(c) To find the moles of oxygen produced:
So, 0.0050 mol of oxygen (to 2 significant figures) is produced.