Electrolysis of Aqueous Solutions

Electrolysing a solution is more complicated than electrolysing a molten compound, because the water adds extra ions to the mix.

When an ionic compound dissolves in water, its own ions separate and become free to move. On top of that, a tiny fraction of the water molecules break down to release hydrogen ions ($H^{+}$) and hydroxide ions ($O{H}^{-}$):

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$

So the electrolyte contains four kinds of ion, not two. For example, copper sulfate solution contains $C{u}^{2+}$ and $S{O}_4^{2-}$ from the salt, plus $H^{+}$ and $O{H}^{-}$ from the water. At each electrode there is now a choice about which ion is discharged, and simple rules decide the winner.

In an aqueous solution the water provides $H^{+}$ and $O{H}^{-}$ ions in addition to the ions from the dissolved compound.

At the negative cathode, two positive ions compete: the metal ion from the salt and the $H^{+}$ from the water. The rule depends on the reactivity of the metal.

At the cathode, hydrogen is produced unless the metal is less reactive than hydrogen, in which case the metal is deposited.

The reason is that the less reactive ion is discharged more easily. Hydrogen sits between zinc and copper on the reactivity comparison used for this rule:

Metal in the compound	More or less reactive than hydrogen?	Product at cathode
Potassium, sodium, calcium, magnesium, aluminium, zinc, iron	More reactive	Hydrogen gas
Copper, silver, gold	Less reactive	The metal is deposited

So electrolysing copper sulfate solution deposits a layer of copper on the cathode, because copper is less reactive than hydrogen. Electrolysing sodium chloride solution gives hydrogen at the cathode, because sodium is more reactive than hydrogen and stays in solution.

At the positive anode, negative ions compete: usually the $O{H}^{-}$ from the water against a negative ion from the salt.

At the anode, oxygen is produced unless the solution contains a halide ion (chloride, bromide or iodide), in which case the halogen is produced.

If a halide ion (from Group 7) is present, the halogen is discharged. If there is no halide, the hydroxide ions are discharged to give oxygen instead. Ions such as sulfate ($S{O}_4^{2-}$) and nitrate stay in solution, so oxygen forms when they are the salt's negative ion.

Negative ion in the solution	Product at anode
Chloride ($C{l}^{-}$)	Chlorine gas
Bromide ($B{r}^{-}$)	Bromine
Iodide ($I^{-}$)	Iodine
Sulfate, nitrate (no halide)	Oxygen gas

So sodium chloride solution gives chlorine at the anode, while copper sulfate solution gives oxygen at the anode.

Predicting products is a two-step job: apply the cathode rule, then the anode rule.

Example 1 — copper sulfate solution, $CuS{O}_4(aq)$. Ions present: $C{u}^{2+}$, $S{O}_4^{2-}$, $H^{+}$, $O{H}^{-}$.

• Cathode: copper is less reactive than hydrogen, so copper is deposited.
• Anode: no halide is present (only sulfate), so oxygen is produced.

Result: copper coats the cathode, oxygen bubbles off the anode, and the solution slowly turns from blue towards colourless as $C{u}^{2+}$ ions are removed.

Example 2 — sodium chloride solution, $NaCl(aq)$. Ions present: $N{a}^{+}$, $C{l}^{-}$, $H^{+}$, $O{H}^{-}$.

• Cathode: sodium is more reactive than hydrogen, so hydrogen is produced.
• Anode: chloride is a halide, so chlorine is produced.

Result: hydrogen bubbles at the cathode, chlorine at the anode, leaving sodium hydroxide solution behind (from the leftover $N{a}^{+}$ and $O{H}^{-}$).

(Higher Tier only) Half equations for the electrode reactions are assessed only at Higher tier.

A half equation shows the electrons gained or lost at one electrode, and must balance for both atoms and charge.

At the cathode (reduction, ions gain electrons):

$2H^{+}+2e^{-}\to H_2$ $C{u}^{2+}+2e^{-}\to Cu$

At the anode (oxidation, ions lose electrons):

$2C{l}^{-}\to C{l}_2+2e^{-}$ $4O{H}^{-}\to O_2+2H_{2}O+4e^{-}$

The oxygen half equation is the trickiest to recall. Check its balance: the left side has 4 O, 4 H and a total charge of $4-$; the right side has 4 O (two in $O_2$, two in $2H_{2}O$), 4 H, and the $4e^{-}$ carry away a charge of $4-$, so both sides have 4 oxygen, 4 hydrogen and a charge of $4-$. You may be asked to complete or balance a half equation you are given rather than recall it from scratch.

This required practical investigates what is produced when different aqueous solutions are electrolysed using inert graphite electrodes. The skill being tested is developing a hypothesis and then checking it against the products you observe.

Apparatus. An electrolysis cell (a small beaker or a special cell) fitted with two carbon (graphite) electrodes, connected to a low-voltage d.c. power supply. Test tubes or a lid can collect gases over each electrode.

Method.

1. Half-fill the cell with the solution to be tested (for example copper chloride, copper sulfate or sodium chloride solution).
2. Place a carbon electrode in each side and connect them to the d.c. supply, noting which is positive (anode) and which is negative (cathode).
3. Switch on and observe both electrodes: look for gas bubbles, a solid coating, or a colour change.
4. Test any gas collected: a lit splint gives a squeaky pop for hydrogen; a glowing splint relights in oxygen; chlorine bleaches damp litmus paper.
5. Repeat with each solution, recording the product at each electrode.

Variables.

Variable type	In this practical
Independent	The solution being electrolysed
Dependent	The product formed at each electrode
Control	Inert (graphite) electrodes, same voltage, same time

Developing a hypothesis. Before testing each new solution, use the cathode and anode rules to predict the products, then run the experiment to see if your prediction holds. Using inert electrodes matters because graphite does not react, so any product comes from the ions in the solution and not from the electrode itself. This is what lets you draw a valid conclusion about the effect of the solution.

1. Forgetting that water adds ions

An aqueous solution always contains $H^{+}$ and $O{H}^{-}$ from the water as well as the salt's ions. These extra ions are why hydrogen or oxygen can form instead of the salt's own elements.

2. Applying the molten rule to a solution

For a molten compound the metal always forms at the cathode. In a solution, the metal only forms if it is less reactive than hydrogen; otherwise hydrogen is produced. Do not assume sodium forms from sodium chloride solution.

3. Missing a halide and predicting the wrong anode product

Scan the negative ion first. A halide (chloride, bromide, iodide) gives the halogen; sulfate or nitrate gives oxygen. Chloride solution gives chlorine, not oxygen.

4. Mixing up the gas tests

Hydrogen gives a squeaky pop with a lit splint; oxygen relights a glowing splint; chlorine bleaches damp litmus paper. Quoting the wrong test loses the identification mark.

5. Not saying why inert electrodes are used

The electrodes must be inert (graphite) so they do not react and change the products. This keeps the comparison fair and means any product comes from the solution's ions.

6. Half equations that do not balance

Both atoms and charge must balance, and electrons go on the side that equalises the charge. For $2C{l}^{-}\to C{l}_2+2e^{-}$, two chloride ions release two electrons; leaving out the 2 in front of $C{l}^{-}$ or the electrons loses the mark.