Electrolysis uses electricity (energy) to drive a non-spontaneous reaction which is useful for recharging batteries, splitting water to make hydrogen and in producing chlorine from brine.
We came across electrolysis of molten and aqueous metal salts at GCSE, and the good news is nothing has changed 😀. If you need a quick refresher of the electrolysis of a molten ionic salt, click here!
For A level though, you do need to know the half equations for the oxidation and reduction of water as well as the rules predicting the products of electrolysis of a metal salt in solution.
Electrolysis of water
Water is stable with respect to hydrogen and oxygen under standard conditions. Splitting water to make hydrogen is not a feasible / spontaneous reaction as can be seen by the positive change in Gibbs energy and the negative Ecell value.
2H2O(l) ⇾ 2H2(g) + O2(g)
ΔrG⦵ = + 474 kJ mol-1 ; Ecell = -1.23V
However, if we supply more than +1.23V to the reaction, we can make it happen.
For electrolysis to be successful, the circuit must be completed by the movement of free charged ions through the electrolyte. Pure water has very few free H+ or OH– ions present so it is acidified (aqueous acids have a very high concentration of H+ ions) to increase the conductivity of the electrolyte.
Electrolysis of brine, aqueous sodium chloride
The salt for the solution comes from the mining of rock salt rather than sea water in most cases. This process is the basis of the chlor-alkali industry.
In the industrial set-up, the anode is titanium and the cathode is nickel.
The overall reaction is:
2H2O(l) + 2Cl–(aq) ⇾ Cl2(g) + H2(g) + 2OH–(aq)
E⦵cell = -2.19V which means this reaction is not spontaneous
We need to supply a potential difference of more than +2.19V to drive this reaction in a non-spontaneous direction.
Rules for predicting the products of the electrolysis of aqueous ionic salts
We can electrolyse the solutions of many metal salts. The products at each electrode can be predicted based on the following rules:
At the anode, either the anion or water will be oxidised
- If the salt is a halide, the halide ion is oxidised in preference to water
2Br–(aq) ⇾ Br2(l) + 2e–
- If the salt is a sulphate or a nitrate, water is oxidised
2H2O(l) ⇾ 4H+(aq) + O2(g) + 4e–
- If the salt is a hydroxide, the hydroxide ion is oxidised
4OH–(aq) ⇾ O2(g) + 2H2O(l) + 2e–
At the cathode, either the cation or water will be reduced
- If the metal is from group 1, group 2 or is aluminium, water is reduced and the metal ion remains in solution
2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq)
- For all other metals, the metal ion is reduced and deposited on the cathode
Cu2+(aq) + 2e– ⇾ Cu(s)
- If we are electrolysing an acid, the H+ ions are reduced
2H+(aq) + 2e– ⇾ H2(g)
Practice questions
1. Use the rules above to predict the products at the anode and cathode for the electrolysis of the following solutions with graphite electrodes. Write half equations for the reactions happening.
(a) KBr
(b) Al(NO3)3
(c) HNO3
(d) Mg(OH)2
(e) CuI2
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2. Which of the following is correct?
| Electrolyte | Product at anode | Product at cathode | |
| A | MgI2 (l) | iodine | magensium |
| B | AgNO3 (aq) | oxygen | hydrogen |
| C | NaCl (aq) | chlorine | sodium |
| D | PbS (l) | hydrogen sulphide | lead |
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3. The apparatus shown below was used to investigate the electrolysis of the sodium halides in solution.
(a) What would be observed at the anode and cathode during the electrolysis of aqueous sodium iodide?
(b) Write half equations for the reactions happening at each electrode during the electrolysis of sodium bromide.
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4. Chlorine is manufactured by the electrolysis of sodium chloride solution.
2NaCl(aq) + 2H2O(l) ⇾ Cl2(g) + H2(g) + NaOH(aq)
Give the half equation for the reaction at the negative electrode and explain why this is a reduction reaction.
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5. A steel object can be chromium plated using a solution of chromium(III) chloride and a graphite anode.
(a) Draw a labelled diagram of the apparatus needed.
(b) Write a half equation for the reaction happening at the cathode.
(c)Â The charge carried by 1 mole of electrons is 96500 coulombs and a coulomb is a current of 1 amp flowing for 1 second. Calculate the time in hours needed to deposit 26g of chromium at a current of 5 amps in your cell.
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6. The electrolysis of aqueous sodium chloride can be carried out using a membrane cell.Â
Explain why the products of this process are chlorine, hydrogen and aqueous sodium hydroxide with reference to the redox reactions that occur in the cell. Include relevant half equations in your answer.Â
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7. An experiment was carried out to electrolyse concentrated aqueous sulphuric acid using platinum electrodes.
(a) Write half equations for the reactions that occurred at the anode and the cathode.
(b) Calculate the volume of oxygen produced if the experiment ran for 18 minutes using a current of 0.80 amps at 298K.
1 Faraday = 96500 C mol-1 ; Q = I x t
Answers
1.
(a) 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) and 2Br–(aq) ⇾ Br2(l) + 2e–
(b) 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) and 2H2O(l) ⇾ 4H+(aq) + O2(g) + 4e–
(c) 2H+(aq) + 2e– ⇾ H2(g) and 2H2O(l) ⇾ 4H+(aq) + O2(g) + 4e–
(d) 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) and 4OH–(aq) ⇾ O2(g) + 2H2O(l) + 2e–
(e) Cu2+(aq) + 2e– ⇾ Cu(s) and 2I–(aq) ⇾ I2(aq) + 2e–
2. A
3. (a) at the anode you would see a yellow / orange / brown colour in solution
(b) at the anode: 2Br–(aq) ⇾ Br2(l) + 2e–
at the cathode: 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq)
4. 2H2O(l) + 2e– ⇾ H2(g) + 2OH–(aq) ; this is reduction because water gains two electrons
5. (a)
(b) Cr3+(aq) + 3e- ⇾ Cr(s)
(c) No. of mol of Cr = 26g / 52 gmol-1 = 0.5 mol
0.5 mol of Cr requires 0.5 x 3 mol of electrons = 1.5 mol
No. of coulombs needed = 1.5 x 96500 = 144750 C
time (s) = charge / current = 144750 / 5 = 28950 s = 8.04 hours
6. At the anode, Cl– ions are oxidised to Cl2.
2Cl–(aq) ⇾ Cl2(g) + 2e-
At the cathode, water is reduced to give OH– ions and H2.
2H2O(l) + 2e- ⇾ 2OH–(aq) + H2(g)
The hydroxide ions and sodium ions remain in solution, hence the third product is aqueous sodium hydroxide.Â
7. (a) Anode: 2H2O(l) ⇾ O2(g) + 4H+(aq) + 4e-
Cathode: 2H+(aq) + 2e- ⇾ H2(g)
(b) O2 : e- is a 1:4 ratio
no. of coulombs will be 4 x 96500 = 386000 C mol-1 for 1 mol of O2
Q = I x t = 0.8 x (18 x 60) = 864 C transferred during experiment
no. of mol of O2 produced = (864 / 386000) = 2.24 x 10-3 mol
volume of O2 = (2.24 x 10-3) x 24 dm3 = 0.054 dm3