Electrolysis

6 marks · Normal

Potassium iodide (KI) is an ionic compound. When heated until molten and electrolysed using inert electrodes, iodine gas is produced at one electrode and potassium metal at the other.

(a)

State the ions present in molten potassium iodide.

[1 mark]

(b)

Write the half-equation for the reaction occurring at the cathode.

[1 mark]

(c)

Write the half-equation for the reaction occurring at the anode.

[1 mark]

(d)

State, in terms of electron transfer, which reaction involves reduction and which involves oxidation.

[2 marks]

(e)

Explain how you can use the mnemonic OILRIG to identify which reaction at the electrodes is oxidation and which is reduction.

[1 mark]

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) Potassium ions (K^+) and iodide ions (I^-) · must show K^+, I^-

(b) K^+ + e^- → K · must show potassium ion gains one electron, potassium atom

(d) Reduction is gain of electrons, which happens at the cathode (K^+ gains e^-); oxidation is loss of electrons, which happens at the anode (I^- loses e^-). · must show Reduction = gain of electrons at cathode, Oxidation = loss of electrons at anode

(e) OILRIG stands for 'Oxidation Is Loss, Reduction Is Gain' (of electrons), so loss of electrons (I^-) is oxidation and gain (K^+) is reduction. · must show Explain OILRIG as meaning loss = oxidation, gain = reduction

Common mistakes: Writing aqueous ions instead of molten; Swapping cathode and anode equations; Confusing oxidation and reduction definitions

Show mark scheme and worked solution

Marking points

B1 (a) · 1 mark · States both potassium ions (K^+) and iodide ions (I^-).

B1 (b) · 1 mark · Correct balanced half-equation for potassium at the cathode: K^+ + e^- → K.

B1 (c) · 1 mark · Correct balanced half-equation for iodide at the anode: 2I^- → I_2 + 2e^-.

M1 (d) · 1 mark · Identifies reduction as gain of electrons and oxidation as loss of electrons.

M1 (d) · 1 mark · Correctly assigns reduction to cathode (potassium) and oxidation to anode (iodine/iodide).

B1 (e) · 1 mark · Uses OILRIG correctly to relate oxidation to loss, reduction to gain, and applies it to electrode reactions.

Worked solution

a) The ions present in molten potassium iodide are potassium ions, K^+, and iodide ions, I^-.

b) At the cathode, potassium ions gain electrons to form potassium atoms:

c) At the anode, iodide ions lose electrons to form iodine molecules:

d) At the cathode, K^+ ions gain electrons (reduction; gain of electrons). At the anode, I^- ions lose electrons (oxidation; loss of electrons).

e) OILRIG stands for 'Oxidation Is Loss, Reduction Is Gain (of electrons)'. Therefore, the reaction where electrons are lost (anode) is oxidation, and the reaction where electrons are gained (cathode) is reduction.