Electrolysis

Lessons

Senior 5 · Chemistry

Electrolysis

4 lessons

Lesson 1 of 4

Electrolytic Cells, Selective Discharge and the Electrolysis of Water

25 min

Forcing a reaction to run backwards

Hydrogen burns in oxygen to form water, releasing energy; nobody has to push it. The reverse, splitting water into hydrogen and oxygen, never happens by itself. Yet connect two electrodes in slightly acidified water to a battery and bubbles of both gases stream from the electrodes. Electrical energy is driving a non-spontaneous reaction. This is electrolysis, the process that gives the world its aluminium, chlorine, sodium hydroxide and pure copper, and that may one day supply hydrogen fuel made with solar power.

Definition

Electrolysis: the decomposition of an electrolyte (molten or in solution) by a direct electric current.

Electrolytic cell: a cell in which electrical energy from an external supply drives a non-spontaneous redox reaction.

Cathode: the electrode connected to the negative terminal; reduction (gain of electrons) happens there. Anode: the electrode connected to the positive terminal; oxidation (loss of electrons) happens there.

Inert electrode: one that conducts but does not react, e.g. platinum or graphite. Active electrode: one that takes part in the reaction, e.g. a copper anode in copper(II) sulfate.

Discharge: the conversion of an ion into an uncharged atom or molecule at an electrode.

Inside an electrolytic cell

Electrons are pushed by the supply into the cathode. Cations move through the electrolyte towards the cathode and gain electrons there (reduction). Anions move towards the anode and lose electrons there (oxidation), and the electrons return to the supply. In the electrolyte the current is carried by ions; in the wires, by electrons.

Remember the pairs with "AN OX, RED CAT": anode oxidation, reduction cathode.

Which ion is discharged? Selective discharge

An aqueous electrolyte always contains at least two kinds of cation (the metal ion and $\mathrm{H^{+}}$ from water) and two kinds of anion (the anion of the salt and $\mathrm{OH^{-}}$ ). Three factors decide which is discharged:

Position in the electrochemical series. At the cathode, the ion that is most easily reduced is discharged first: $\mathrm{Ag^{+}}$ and $\mathrm{Cu^{2+}}$ before $\mathrm{H^{+}}$ , and $\mathrm{H^{+}}$ (from water) before $\mathrm{Na^{+}}$ , $\mathrm{Mg^{2+}}$ or $\mathrm{Al^{3+}}$ . At the anode, the ion most easily oxidised goes first: $\mathrm{I^{-}}$ before $\mathrm{Br^{-}}$ before $\mathrm{OH^{-}}$ (water); sulfate and nitrate are not oxidised at all, so water gives oxygen.
Concentration. A much higher concentration can make an ion discharge even if it is slightly harder to discharge. Chloride is the classic case (Lesson 2).
Nature of the electrode. An active anode such as copper may itself be oxidised instead of any anion (Lesson 3). Some electrodes also need an extra voltage (an overpotential) to release oxygen, which helps chloride compete.

Half-equation	$E^{\circ}$ / V
$\mathrm{Na^{+} + e^{-} \rightarrow Na}$	−2.71
$\mathrm{2H_2O + 2e^{-} \rightarrow H_2 + 2OH^{-}}$	−0.83
$\mathrm{2H^{+} + 2e^{-} \rightarrow H_2}$	0.00
$\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$	+0.34
$\mathrm{Ag^{+} + e^{-} \rightarrow Ag}$	+0.80
$\mathrm{Br_2 + 2e^{-} \rightarrow 2Br^{-}}$	+1.09
$\mathrm{O_2 + 4H^{+} + 4e^{-} \rightarrow 2H_2O}$	+1.23
$\mathrm{Cl_2 + 2e^{-} \rightarrow 2Cl^{-}}$	+1.36

Table 13.1: Standard electrode potentials at 25 °C (a preview of S6 Unit 11). At the cathode, the half-reaction higher in the table (more positive $E^{\circ}$ ) happens first; at the anode, the reverse of the lowest-placed (least positive) available half-reaction happens first.

The electrolysis of water

Pure water conducts too poorly to electrolyse, so a little dilute sulfuric acid (or sodium sulfate) is added to provide ions. Neither $\mathrm{SO_4^{2-}}$ nor $\mathrm{Na^{+}}$ is discharged, so only water reacts:

$\text{cathode: } \mathrm{4H^{+}(aq) + 4e^{-} \rightarrow 2H_2(g)} \qquad \text{anode: } \mathrm{2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-}}$

$\text{overall: } \mathrm{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)}$

Figure 13.1: A Hofmann voltameter. The same four electrons that make one $\mathrm{O_2}$ molecule make two $\mathrm{H_2}$ molecules, so the gas volumes are in the ratio 2 : 1.

The acid is not used up: $\mathrm{H^{+}}$ is consumed at the cathode and the same amount is made at the anode. In practice the oxygen volume is slightly less than half at first, because oxygen is a little more soluble in water than hydrogen.

Worked example

Problem. Predict the products at each inert electrode, and give the half-equations, for the electrolysis of (a) aqueous silver nitrate; (b) aqueous potassium iodide; (c) molten lead(II) bromide.

Solution.

Cathode: $\mathrm{Ag^{+}}$ is reduced more easily than water: $\mathrm{Ag^{+} + e^{-} \rightarrow Ag}$ (silver deposited). Anode: nitrate is not oxidised, so water is: $\mathrm{2H_2O \rightarrow O_2 + 4H^{+} + 4e^{-}}$ . The solution becomes acidic (nitric acid).
Cathode: $\mathrm{K^{+}}$ is very hard to reduce, so water is: $\mathrm{2H_2O + 2e^{-} \rightarrow H_2 + 2OH^{-}}$ . Anode: iodide is oxidised more easily than water: $\mathrm{2I^{-} \rightarrow I_2 + 2e^{-}}$ (brown solution). KOH is left in solution.
Molten: no water, so only $\mathrm{Pb^{2+}}$ and $\mathrm{Br^{-}}$ are present. Cathode: $\mathrm{Pb^{2+} + 2e^{-} \rightarrow Pb}$ ; anode: $\mathrm{2Br^{-} \rightarrow Br_2 + 2e^{-}}$ .

Worked example

Problem. In a Hofmann voltameter, 36.0 cm³ of hydrogen is collected. What volume of oxygen is expected at the same temperature and pressure, and how many moles of electrons have passed? (Molar gas volume 24.5 dm³ mol⁻¹ at 25 °C and 101.3 kPa.)

Solution. $\mathrm{2H_2O \rightarrow 2H_2 + O_2}$ : volume of $\mathrm{O_2} = \frac{36.0}{2} = 18.0\,\mathrm{cm^{3}}$ .

$n(\mathrm{H_2}) = \frac{36.0}{24\,500} = 1.47 \times 10^{-3}\,\mathrm{mol} \qquad n(e^{-}) = 2 \times 1.47 \times 10^{-3} = 2.94 \times 10^{-3}\,\mathrm{mol}$

Answer: 18.0 cm³ of oxygen; $2.94 \times 10^{-3}$ mol of electrons (each $\mathrm{H_2}$ needs 2 electrons).

Warning

Common errors: mixing up the electrodes (in an electrolytic cell the cathode is negative); writing $\mathrm{Na^{+}}$ as discharged from an aqueous solution; and forgetting that water itself can be oxidised or reduced.

Safety: a mixture of hydrogen and oxygen explodes if ignited. Test the gases in small amounts only, away from flames, and never collect the two gases together. Use low-voltage DC supplies only. If dilute acid gets in the eyes, rinse with water for at least 15 minutes (CDC/NIOSH first aid).

Tip

Electrons always flow into the cathode from the supply and out of the anode back to it. Whatever is reduced must be at the cathode; whatever is oxidised must be at the anode. If you are unsure of the sign, decide where reduction happens first.

Bristol ChemLabS: Teacher Training, Hofmann Voltameter. Watch how acidified water is split into oxygen and hydrogen, and compare the volumes of gas collected above the two electrodes.

✦

Practice

State the products at the cathode and anode (inert electrodes) for: (a) molten zinc chloride; (b) dilute sulfuric acid; (c) aqueous copper(II) bromide; (d) aqueous sodium sulfate.
Write the half-equations for the electrolysis of aqueous sodium sulfate, and explain why the solution near the cathode turns universal indicator purple.
Why is pure water a poor electrolyte, and why does adding sodium sulfate not change the products?
In a Hofmann voltameter 24.0 cm³ of oxygen is collected. What volume of hydrogen should be present, and how many moles of electrons passed? (24.5 dm³ mol⁻¹)
Explain why in aqueous silver nitrate silver, not hydrogen, is deposited at the cathode.
A student says "the cathode is always the negative electrode, so it must be where oxidation happens". Evaluate this statement.

Show answersHide answers

(a) zinc, chlorine; (b) hydrogen, oxygen; (c) copper, bromine; (d) hydrogen, oxygen.
Cathode: $\mathrm{2H_2O + 2e^{-} \rightarrow H_2 + 2OH^{-}}$ ; anode: $\mathrm{2H_2O \rightarrow O_2 + 4H^{+} + 4e^{-}}$ . $\mathrm{OH^{-}}$ forms at the cathode, so the solution there is alkaline.
It contains very few ions (only about $10^{-7}$ mol dm⁻³ of each); $\mathrm{Na^{+}}$ and $\mathrm{SO_4^{2-}}$ carry current but are harder to discharge than water, so water is still the substance reduced and oxidised.
48.0 cm³ of $\mathrm{H_2}$ ; $n(\mathrm{O_2}) = \frac{24.0}{24\,500} = 9.80 \times 10^{-4}$ mol, so $n(e^{-}) = 4 \times 9.80 \times 10^{-4} = 3.92 \times 10^{-3}$ mol.
$\mathrm{Ag^{+}}$ is much more easily reduced than water or $\mathrm{H^{+}}$ ( $E^{\circ} = +0.80$ V compared with 0.00 V or below), so it accepts electrons first.
Half right: in an electrolytic cell the cathode is negative, but it is where reduction happens, because electrons arrive there from the supply and are gained by cations. Oxidation happens at the anode.

Sources and credits

Standard electrode potentials: CRC Handbook of Chemistry and Physics (electrochemical series).
First aid: NIOSH Pocket Guide to Chemical Hazards: Sulfuric acid, US Centers for Disease Control and Prevention.
Video: Teacher Training, Hofmann Voltameter, Bristol ChemLabS, embedded from YouTube.
The text, Figure 13.1, the table, worked examples and Practice items were written and drawn by Ganzaa.

1Forcing a reaction to run backwards

2Inside an electrolytic cell

3Which ion is discharged? Selective discharge

4The electrolysis of water

Forcing a reaction to run backwards

Inside an electrolytic cell

Which ion is discharged? Selective discharge

The electrolysis of water