Electrolysis and the products at the electrodes

Pushing a reaction the wrong way

A cheap spoon can have a bright silver skin. A car tap can have a shiny layer of chromium. A bar of copper from a refinery can be almost pure. All three were made by passing a current through a solution.

In a galvanic cell, such as the Daniell cell, a reaction runs by itself and pushes electrons through a wire. Electrolysis is the opposite. We connect a battery and push electrons into a cell. The battery forces a reaction that would not happen on its own.

The two kinds of cell differ in the signs of their electrodes:

• Galvanic cell: the anode is negative and the cathode is positive. The cell makes electricity.
• Electrolytic cell: the anode is joined to the positive terminal of the battery, so it is positive. The cathode is joined to the negative terminal, so it is negative. The cell uses electricity.

Two things stay the same in both. Oxidation happens at the anode and reduction at the cathode. Electrons leave the cell at the anode and enter it at the cathode.

For an electrolysis to start, the battery must give more voltage than the cell would give if it ran the other way. Think of rolling a ball uphill: the push must beat the slope.

An electrolytic cell

Look at the picture. Two platinum electrodes dip into a solution of copper sulphate. A battery is joined to them.

The battery pulls electrons out of the left electrode and pushes them into the right one. So the left electrode is the anode and the right one is the cathode. The ions in the solution move in answer to the charges:

• Positive ions, the cations, move towards the negative cathode.
• Negative ions, the anions, move towards the positive anode.

At the cathode, copper ions take electrons and become copper metal: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$. At the anode, water gives up electrons and oxygen gas bubbles off. The sulphate ions do not react at all.

Why did sulphate ions and water not behave the same way as copper ions? In a solution there is a choice at each electrode. We need a rule for it.

The simple case: molten sodium chloride

When sodium chloride is melted, only two kinds of ion are present, $\text{Na}^+$ and $\text{Cl}^-$. There is no choice.

• Cathode: $\text{Na}^+ + e^- \rightarrow \text{Na}(l)$
• Anode: $2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$

The overall change is $2\text{NaCl}(l) \rightarrow 2\text{Na}(l) + \text{Cl}_2(g)$. This is how sodium metal is made. Sodium is far too reactive to be found as a metal, so we cannot make it any other way.

A solution: water joins in

Now dissolve the sodium chloride in water and electrolyse with platinum electrodes. There are four species around: $\text{Na}^+$, $\text{Cl}^-$ and the water molecules. Each electrode has two candidates.

At the cathode the choice is between two reductions:

• $\text{Na}^+ + e^- \rightarrow \text{Na}$, with $E^\circ = -2.71\ \text{V}$
• $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$, with $E^\circ = -0.83\ \text{V}$

The reduction with the higher potential happens first, because it needs less push. Water wins. Hydrogen gas bubbles off, and the solution around the cathode fills with $\text{OH}^-$ ions.

At the anode the choice is between two oxidations:

• $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$, which belongs to the couple with $E^\circ = +1.36\ \text{V}$
• $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$, which belongs to the couple with $E^\circ = +1.23\ \text{V}$

The oxidation of the couple with the lower potential happens first. By this rule water should win, but in a strong solution of sodium chloride chlorine is what we get.

The reason is overpotential. This is the extra voltage that an electrode reaction needs beyond what $E^\circ$ predicts. Oxygen is slow to form on an electrode and needs about $0.4\ \text{V}$ more. So the working value for water at the anode is about $1.60\ \text{V}$, which is above chlorine's $1.36\ \text{V}$. Chloride ions go first.

The result for strong sodium chloride solution:

• Cathode: hydrogen gas.
• Anode: chlorine gas.
• In the solution: sodium ions and hydroxide ions, which is sodium hydroxide.

The overall equation is $2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g) + \text{Cl}_2(g)$. Industry makes chlorine and sodium hydroxide this way.

The rules for inert electrodes

An inert electrode, such as platinum or graphite, only carries electrons and does not react. For a solution of a salt in water, use these rules.

At the cathode:

• The species with the higher reduction potential is reduced first.
• Water is a rival, at $-0.83\ \text{V}$. Ions of very active metals, such as sodium, potassium, calcium, magnesium and aluminium, have a lower potential. They stay in solution and hydrogen forms.
• Ions of less active metals, such as $\text{Cu}^{2+}$ and $\text{Ag}^+$, are reduced to the metal. In an acid, $\text{H}^+$ ions are reduced to hydrogen.

At the anode:

• The species with the lower potential is oxidised first.
• Water is a rival, at a working value of about $1.60\ \text{V}$. Chloride, bromide and iodide ions come before it, and the halogen forms.
• Sulphate, nitrate and fluoride ions are not oxidised in water. Water is, and oxygen forms.

Four worked cases

1. Dilute sulphuric acid, platinum electrodes. Cathode: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$. Anode: water gives oxygen, $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$. The overall change is that water breaks up: $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$. Twice as many moles of hydrogen as of oxygen form.
2. Aqueous sodium sulphate, platinum electrodes. Neither $\text{Na}^+$ nor $\text{SO}_4^{2-}$ reacts. Hydrogen forms at the cathode and oxygen at the anode, so again only the water is used up. The salt is only there to carry the current.
3. Aqueous copper sulphate, platinum electrodes. Copper forms at the cathode and oxygen at the anode. The solution loses $\text{Cu}^{2+}$ ions and gains $\text{H}^+$ ions, so the blue colour fades and the solution becomes acidic.
4. Aqueous copper sulphate, copper electrodes. Now the electrode itself can react. Copper has a low potential, $0.34\ \text{V}$, so the copper anode dissolves before water is touched: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$. Copper deposits on the cathode. The solution stays the same, and copper moves from one electrode to the other.

An electrode that takes part in the reaction is an active electrode. Case 4 is the idea behind plating and refining.

Electrolysis at work

Electroplating. To plate a spoon with silver, make the spoon the cathode and a bar of silver the anode. The solution holds a silver salt. Silver dissolves from the bar and deposits on the spoon, so the solution stays at the same concentration.

Refining copper.

The anode is a thick block of impure copper and the cathode is a thin sheet of pure copper. Copper dissolves from the anode and deposits on the cathode. Silver, gold and platinum do not dissolve and fall to the bottom as anode mud, which is worth collecting. Metals more active than copper dissolve but are not deposited, so they stay in the solution.

Making metals and chemicals. Sodium, and aluminium from its oxide dissolved in molten cryolite, are made by electrolysis. So are chlorine and sodium hydroxide, as we saw above.

Start with copper and chloride, 3 and 2. Then try zinc with sulphate, 2 and 1, and sodium with sulphate, 1 and 1. When does hydrogen form?

Practise with new numbers

Try it in Python

The program makes the choice at each electrode and works out the least voltage.

Common mistakes

• Treating the anode as negative in every cell. In an electrolytic cell it is positive, because it is joined to the positive terminal.
• Forgetting water. In a solution, water competes at both electrodes, and sodium ions are never reduced there.
• Using the wrong comparison at the anode. At the cathode the higher potential wins. At the anode the lower potential wins.
• Calling every electrode inert. A copper anode in copper sulphate dissolves.
• Saying that sulphate ions turn into oxygen. The oxygen comes from the water.

Remember these

• Electrolysis uses electricity to force a reaction. Anode: positive, oxidation. Cathode: negative, reduction.
• Cathode: the higher reduction potential is reduced. Water, at $-0.83\ \text{V}$, wins over very active metal ions.
• Anode: the lower potential is oxidised. Water, at about $1.60\ \text{V}$ in practice, loses to the halogens and wins over sulphate and nitrate.
• An active electrode can dissolve. An inert one only carries electrons.
• Overpotential is the extra voltage a reaction needs beyond what $E^\circ$ predicts.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. At which electrode do electrons enter an electrolytic cell, and what is the sign of that electrode?

2. What forms at each platinum electrode when potassium nitrate solution is electrolysed?

3. Copper sulphate solution is electrolysed with platinum electrodes. Why does the solution become acidic?

4. Why does chlorine form from concentrated brine, when water has a lower potential than chloride ions?