Electrolysis and Its Applications

Lessons

Senior 3 · Chemistry

Electrolysis and Its Applications

5 lessons

Lesson 2 of 5

Electrolysis of Acidified Water and Sodium Chloride Solution

25 min

Twice as much at one side

Pure water hardly conducts electricity. Add a little dilute sulfuric acid and switch on, and bubbles stream from both electrodes, but one tube fills twice as fast as the other. The gases are hydrogen and oxygen, in the ratio of the formula $\mathrm{H_2O}$ : the current is splitting water. In 1866 the German chemist August Wilhelm von Hofmann designed the three-tube apparatus still used to show this (Figure 2.1).

Hofmann's own drawing of his apparatus, from his 1866 book of chemistry lectures: two battery cells on the left are wired to the electrodes at the foot of the two side tubes, and the tall central tube is the reservoir. Image: A. W. von Hofmann, public domain, via Wikimedia Commons

Ions from the water too

In an aqueous solution the electrolyte's ions are not alone. Water ionises very slightly, so hydrogen ions and hydroxide ions are always present:

$\mathrm{H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)}$

So at each electrode two kinds of ion compete, and only one is discharged. This is selective discharge, and with inert electrodes it follows two rules.

Table 2.1: Which ion is discharged (inert electrodes)

Electrode	Competing ions	Rule	Product
cathode (−)	metal ion or $\mathrm{H^+}$	a metal less reactive than hydrogen (copper, silver) is deposited; for a more reactive metal (potassium, sodium, calcium, magnesium, aluminium), hydrogen forms instead	metal, or hydrogen
anode (+)	$\mathrm{OH^-}$ , and $\mathrm{Cl^-}$ , $\mathrm{Br^-}$ , $\mathrm{I^-}$ , $\mathrm{SO_4^{2-}}$ or $\mathrm{NO_3^-}$	a concentrated halide gives the halogen; otherwise $\mathrm{OH^-}$ is discharged and oxygen forms. Sulfate and nitrate ions are never discharged	halogen, or oxygen

The two half-equations for water's ions are:

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

Definition

Selective discharge: when several ions are attracted to an electrode, only the one that gains or loses electrons most easily is discharged (turned into an atom or molecule). Example: in sodium chloride solution, $\mathrm{H^+}$ rather than $\mathrm{Na^+}$ is discharged at the cathode, so hydrogen, not sodium, forms.

Electrolysis of acidified water

Dilute sulfuric acid contains $\mathrm{H^+}$ , $\mathrm{SO_4^{2-}}$ and a few $\mathrm{OH^-}$ ions. Hydrogen ions are discharged at the cathode; at the anode, hydroxide ions are discharged because sulfate ions never are.

Four electrons give two molecules of hydrogen but only one of oxygen, so the volumes are in the ratio 2 : 1. The overall change is the decomposition of water:

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

The sulfuric acid is not used up; as water is removed, the acid slowly becomes more concentrated. Tests: hydrogen burns with a squeaky pop when a lighted splint is held to it; oxygen relights a glowing splint.

Worked example

Problem. In a Hofmann voltameter, $36\,\mathrm{cm^3}$ of hydrogen is collected. (a) What volume of oxygen forms at the same time? (b) How many moles of electrons passed? (Molar volume $24\,000\,\mathrm{cm^3\,mol^{-1}}$ .)

Solution. (a) Hydrogen and oxygen form in the ratio 2 : 1, so oxygen $= \frac{36}{2} = 18\,\mathrm{cm^3}$ .

(b) $n(\mathrm{H_2}) = \frac{36}{24\,000} = 0.0015\,\mathrm{mol}$ . Each $\mathrm{H_2}$ needs 2 electrons ( $\mathrm{2H^+ + 2e^- \rightarrow H_2}$ ), so $n(\mathrm{e^-}) = 2 \times 0.0015 = 0.0030\,\mathrm{mol}$ .

Sodium chloride solution: dilute and concentrated

Sodium chloride solution contains $\mathrm{Na^+}$ , $\mathrm{Cl^-}$ , $\mathrm{H^+}$ and $\mathrm{OH^-}$ . Sodium is far more reactive than hydrogen, so hydrogen always forms at the cathode, whatever the concentration. At the anode, the product depends on the concentration of chloride ions.

Table 2.2: Electrolysis of sodium chloride solution (inert electrodes)

Solution	Cathode (−)	Anode (+)	Left in solution
dilute NaCl(aq)	hydrogen: $\mathrm{2H^+ + 2e^- \rightarrow H_2}$	mainly oxygen: $\mathrm{4OH^- \rightarrow 2H_2O + O_2 + 4e^-}$	sodium chloride, becoming more concentrated
concentrated NaCl(aq), brine	hydrogen: $\mathrm{2H^+ + 2e^- \rightarrow H_2}$	chlorine: $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$	sodium hydroxide

With brine, $\mathrm{H^+}$ and $\mathrm{Cl^-}$ are removed, so $\mathrm{Na^+}$ and $\mathrm{OH^-}$ are left behind: the solution turns into sodium hydroxide. A few drops of universal indicator turn blue-purple around the cathode, and damp blue litmus held in the anode gas turns red and then white, the test for chlorine from Unit 4. Overall:

$\mathrm{2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)}$

Worked example

Problem. Predict the products at each inert electrode when these solutions are electrolysed: (a) concentrated potassium bromide; (b) potassium nitrate; (c) concentrated copper(II) chloride.

Solution.

(a) Cathode: potassium is more reactive than hydrogen, so hydrogen. Anode: concentrated halide, so bromine.

(b) Cathode: hydrogen. Anode: nitrate is never discharged, so $\mathrm{OH^-}$ is, giving oxygen. The potassium nitrate is unchanged; only water is decomposed.

(c) Cathode: copper is less reactive than hydrogen, so copper is deposited, $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$ . Anode: concentrated chloride, so chlorine.

Warning

Common mistakes and safety. (1) Sodium is never formed from sodium chloride solution: water's hydrogen ions are discharged instead. Sodium metal comes only from molten sodium chloride. (2) The anode gives chlorine only when the chloride is concentrated. (3) Chlorine is toxic and a mixture of hydrogen and chlorine can explode in bright light, so electrolyse brine only on a small scale in a well-ventilated room or fume cupboard, for a short time. If the solution touches your skin, rinse with plenty of running water for at least 15 minutes.

Tip

Cathode: below hydrogen, metal; above hydrogen, hydrogen. Anode: concentrated halide, halogen; otherwise, oxygen. Hydrogen and oxygen from water always come out 2 : 1 by volume.

Cognito: Electrolysis Part 3, Aqueous Solutions. Watch how the reactivity of the metal decides the cathode product and how halide ions decide the anode product.

✦

Practice

Name the four ions present in sodium chloride solution, and say where each comes from.
Write the half-equations at each electrode for the electrolysis of dilute sulfuric acid. Why is the volume of hydrogen twice that of oxygen?
Describe a test for each gas formed in the electrolysis of acidified water.
In a Hofmann voltameter, $15\,\mathrm{cm^3}$ of oxygen is collected. What volume of hydrogen is collected, and how many moles of electrons have passed? (Molar volume $24\,000\,\mathrm{cm^3\,mol^{-1}}$ .)
Concentrated sodium chloride solution is electrolysed with graphite electrodes. Name the product at each electrode and the substance left in solution. How could you show that the solution near the cathode has become alkaline?
Predict the products at each inert electrode for (a) magnesium sulfate solution; (b) silver nitrate solution; (c) concentrated hydrochloric acid.
Why does the electrolysis of very dilute sodium chloride solution give mainly oxygen, not chlorine, at the anode?
A student says: "When salt water is electrolysed, sodium metal collects on the cathode." Evaluate this statement.

Show answersHide answers

$\mathrm{Na^+}$ and $\mathrm{Cl^-}$ from the salt; $\mathrm{H^+}$ and $\mathrm{OH^-}$ from the slight ionisation of water.
Cathode: $\mathrm{2H^+(aq) + 2e^- \rightarrow H_2(g)}$ . Anode: $\mathrm{4OH^-(aq) \rightarrow 2H_2O(l) + O_2(g) + 4e^-}$ . For every 4 electrons, 2 molecules of hydrogen but only 1 of oxygen form, and equal amounts of gas have equal volumes.
Hydrogen: a lighted splint gives a squeaky pop. Oxygen: a glowing splint relights.
$2 \times 15 = 30\,\mathrm{cm^3}$ of hydrogen. $n(\mathrm{O_2}) = \frac{15}{24\,000} = 0.000625\,\mathrm{mol}$ ; each $\mathrm{O_2}$ needs 4 electrons, so $n(\mathrm{e^-}) = 0.0025\,\mathrm{mol}$ .
Hydrogen at the cathode, chlorine at the anode, sodium hydroxide left. Add universal indicator (or phenolphthalein): it turns blue-purple (or pink) around the cathode.
(a) Hydrogen and oxygen. (b) Silver at the cathode, oxygen at the anode. (c) Hydrogen at the cathode, chlorine at the anode.
In very dilute solution there are few chloride ions near the anode compared with hydroxide ions (and water), so hydroxide ions are discharged and oxygen forms.
Wrong. Sodium is much more reactive than hydrogen, so the hydrogen ions from water are discharged instead and hydrogen gas forms. (Any sodium formed would react at once with the water.) Sodium is made only from molten sodium chloride.

Sources and credits

Image: Hofmann voltameter 1866.jpg, from A. W. von Hofmann, Introduction to Modern Chemistry (1866), public domain, via Wikimedia Commons.
Video: Electrolysis Part 3, Aqueous Solutions, Cognito, embedded from YouTube.
Written by Ganzaa. The diagram, tables, worked examples and Practice items were drawn and written by Ganzaa.

1Twice as much at one side

2Ions from the water too

3Electrolysis of acidified water

4Sodium chloride solution: dilute and concentrated

Twice as much at one side

Ions from the water too

Electrolysis of acidified water

Sodium chloride solution: dilute and concentrated