Colourful Solutions > Electron transfer reactions > Electrolysis of aqueous solutions

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Higher-level only

Electrolysis of aqueous solutions is much less complicated technically as there is no need to heat the ionic solid to molten. However, analysis of aqueous solutions must take into account the ions ever-present in water that can compete in the reactions.

Syllabus ref: R3.2.15

Reactivity 3.2.15 - During electrolysis of aqueous solutions, competing reactions can occur at the anode and cathode, including the oxidation and reduction of water. (HL)

  • Deduce from standard electrode potentials the products of the electrolysis of aqueous solutions.

Guidance

  • Electrolytic processes should include the electrolysis of water and of aqueous solutions.
  • The effects of concentration and the nature of the electrode are limited to the electrolysis of NaCl(aq) and CuSO4(aq).

Tools and links

Electrolysis of solutions

Solutions of ionic compounds in water contain hydrogen and hydroxide ions, as well as the ions of the solute. These H+ and OH- ions compete at the electrodes with the solute ions.

The ions that are successfully released at the electrodes depend on three factors

  1. 1 The position of the ion in the electrochemical series
  2. 2 The concentration of the ion in the solution
  3. 3 The nature of the electrode

1 The position of the ion in the electrochemical series

This is probably better expressed as the position of the redox equilibrium in the electrochemical series. All the redox equilibria are expressed as reductions going from left to right, for example:

Cu2+ + 2e ⇋ Cu

However, going from right to left this would be an oxidation. Hence it may be called a redox equilibrium.

The Copper ion || Copper equilibrium has an electrode potential of +0.34 V. This is more positive than hydrogen's electrode potential of 0.00 V. This means that copper ions are more easily oxidised than hydrogen ions (remember that as we go down the series the redox equilibria have better oxidising properties, i.e they can remove electrons from other things more easily - including electrodes). Consequently in a competition between the two ions, the copper ions will preferentially pick up the electrons.

As a rule of thumb, if the metal appears below hydrogen in the electrochemical series it will be preferentially deposited.

2 The concentration of the ions

When two ions with similar reactivity are in competition then the relative concentration of the two ions becomes an important factor.

A good example of this is the electrolysis of aqueous sodium chloride. When the chloride concentration is large, the chloride ions lose electrons and chlorine gas is released at the electrode, but when it is in low concentration, the hydroxide ions from the water are preferentially released.

3 The nature of the electrode

Usually, inert electrodes such as graphite or platinum are used for electrolysis. These electrodes do not interfere with the reactions occuring at the surface of the electrode, they simply act as a point of contact between the electrical circuit and the solution.

However, if metal electrodes are used in metal ion solutions they can get involved in the reactions by dissolving as ions, leaving their electrons behind (this can only happen when the metal takes the place of the anode, the positive electrode) - this is called electrode participation, it is discussed in greater detail in section 9.73.

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Electrolysis of sodium chloride solution

The ions present in the solution are: sodium ions, chloride ions, hydrogen ions, hydroxide ions

At the cathode

The positive ions are attracted to the negative cathode. There is competition between the sodium ions, Na+(aq), and the hydrogen ions, H+(aq). As the hydrogen ion || hydrogen redox equilibrium is lower in the electrochemical series than the sodium ion || sodium redox equilibrium, the hydrogen ions are preferentially reduced (hydrogen ions are a better oxidising agent than sodium ions) and hydrogen gas is produced at the electrode (bubbles are seen)

2H+ + 2e → H2

One way to remember this is to consider what would happen if sodium were to be liberated in water. Sodium is far too reactive and would react immediately with the water producing hydrogen gas. The end result would be the same - hydrogen gas is produced

At the anode

There is competition between the negative ions at the positive anode. The chloride ions compete with the hydroxide ions to release their electrons to the anode. When the solution is fairly concentrated the chloride ions preferentially lose electrons to become chlorine atoms (and then molecules)

2Cl- - 2e → Cl2

However, in dilute solutions of sodium chloride the hydroxide ions are preferentially oxidised. This is a concentration effect.

Ions remaining in solution

The ions that are removed from the solution, then, are the hydrogen ions and the chloride ions. This means that the sodium ions and the hydroxide ions remain in the solution - i.e sodium hydroxide is also produced.

Note: When the solution of chloride ions is dilute then OH-. ions are preferentially released at the anode.

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Electrolysis of copper(II) sulfate solution

The ions present in the solution are: copper ions, Cu2+; sulfate ions,SO42-; hydrogen ions, H+; and hydroxide ions, OH-.

At the cathode

The positive ions are attracted to the negative cathode. There is competition between the copper ions and the hydrogen ions. As the hydrogen ion | hydrogen redox equilibrium appears higher in the electrochemical series than the copper ion | copper equilibrium, then the copper ions are preferentially reduced and copper metal is deposited at the electrode (a pink layer is observed)

Cu2+ + 2e → Cu

At the anode

There is competition between the negative ions at the positive anode. The sulfate ions compete with the hydroxide ions to release their electrons to the anode. The hydroxide ions are much better reducing agents and are preferentially released AS OXYGEN GAS and water

4OH- - 4e → 2H2O + O2

Ions remaining in solution

The ions that are removed from the solution, then, are the copper ions and the hydroxide ions. This means that the hydrogen ions and the sulfate ions remain in the solution - i.e sulfuric acid is also produced.

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Participating electrodes

When non-inert electrodes are used in electrolysis they can interact with the solutions being electrolysed. This specifically is important when electrolysis of ions of relatively unreactive metals are electrolysed using the same metal as an electrode. In this case the anode can participate in the reaction by losing atoms in the form of ions to the solution, leaving behind their own electrons.

This is probably best understood by looking at an example. In the electrolysis of copper(II) sulfate solution using copper electrodes. The reaction that would proceed at the anode with inert electrodes would be:

4OH-(aq) - 4e → 2H2O(l) + O2(g)

This requires migration of hydroxide ions, which are at a fairly low concentration, to the anode for liberation. The alternative possibility in the case of a copper anode, is for the following reaction to take place:

Cu(s) - 2e → Cu2+(aq)

This second reaction has the advantage of not requiring any migration (the copper atoms are already there, they make up the electrode), and the reaction itself is energetically more advantageous. For both of these reaons, the second reaction takes preference.

Consequently, the following reactions occur:

At the anode (positive electrode):

Cu(s) - 2e → Cu2+(aq)

At the cathode: (negative electrode)

Cu2+(aq) + 2e → Cu(s)

The overall effect is for copper ions to be liberated as copper at one electrode at the same time as they are being produced by the other electrode. The concentration of copper(II) in solution does not change. However, the mass of copper at the anode decreases and the mass of copper at the cathode increases.

This is summarised in the following table:

Electrolysis of a copper(II) solution using copper electrodes
solution anode cathode
[Cu2+(aq)] unchanged mass Cu decreases mass Cu increases
  Cu  →  Cu2+ + 2e Cu2+ + 2e  →  Cu

Electrolysis of dilute sulfuric acid

Ions present: hydrogen ions, H+; sulfate ions, SO42-; hydroxide ions, OH-.

In this case there is no choice at the cathode; hydrogen ions must be released as hydrogen gas.

2H+ + 2e → H2

At the anode

There is competition between the sulfate ions and the hydroxide ions. sulfate ions are too big and bulky to ever be released, therefore it must be the hydroxide ions:

4OH- - 4e → 2H2O + O2

Ions remaining in solution

The only ions removed are hydrogen ions (not all of them) and hydroxide ions. Therefore the ions remaining are hydrogen ions and sulfate ions.

These are the ions of sulfuric acid. So the net result of electrolysing sulfuric acid is that the sulfuric acid gets more concentrated and the water solvent is electrolysed producing hydrogen oxygen gas at the electrodes.

For this reason the process is often known as the electrolysis of water - it is only the elements of water that are released.

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Summary

Reactive metals (more reactive than hydrogen) are never deposited during electrolysis of aqueous solutions. If the metal ion comes from a metal more reactive than hydrogen, then hydrogen gas is liberated at the cathode.

Halide ions (chloride, bromide, iodide) are released preferentially and if these are not present, the hydroxide ions from the water are released at the anode.

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Worked examples

Q971-01 Which statement is correct about the electrolysis of copper(II) sulfate solution using graphite electrodes?
  1. A colourless gas is produced at the negative electrode
  2. The electrolyte does not change colour
  3. The negative electrode decreases in mass
  4. A colourless gas is produced at the positive electrode
Answer

copper(II) sulfate solution contains the following ions:

Cu2+(aq); SO42-(aq);H+(aq); OH-(aq)

At the cathode: Cu2+(aq) + 2e → Cu(s)

At the anode: 4OH-(aq) - 4e → 2H2O(l) + O2(g)

Therefore a colourless gas is produced at the positive electrode (anode)
Q971-02 Consider the following statements regarding electrolysis of molten lead(II) bromide.
  1. I.     Oxidation takes place at the anode where lead ions gain electrons
  2. II.    Reduction takes place at the cathode where lead ions gain electrons
  3. III.   Oxidation takes place at the anode where bromide ions lose electrons
  4. IV.   Reduction takes place at the cathode where bromide ions lose electrons

Which of the above statements are correct?

  1. I and II only
  2. I and IV only
  3. II and III only
  4. II and IV only
Answer

Molten lead(II) bromide contains only lead ions and bromide ions.

Pb2+(l); Br-(l)

At the cathode: Pb2+(l) + 2e → Pb(l)

At the anode: 2Br-(l) - 2e → Br2(g)

Reduction takes place at the cathode, oxidation takes place at the anode. Statements II and III are correct.
Q971-03 Two moles of electrons are passed through an electrolytic cell containing molten sodium chloride. The same charge is passed through a second cell containing aqueous sodium chloride. In both cells the electrodes are made of platinum. Which statement is correct?
  1. One mole of sodium metal will be formed in the first cell
  2. Chlorine gas will be formed at the cathodes of both cells
  3. One mole of hydrogen gas will be formed in the second cell
  4. One mole of oxygen gas will be formed at the anode of the second cell
Answer

Molten sodium chloride contains only sodium ions and chloride ions. Na+(l); Cl-(l)

At the cathode: Na+(l) + 1e → Na(l)

At the anode: 2Cl-(l) - 2e → Cl2(g)

The cathode reaction requires 1 mole of electrons per mole of sodium formed. The anode reaction requires 2 moles of electrons per mole of chlorine gas formed.

Aqueous sodium chloride solution contains sodium ions, chloride ions, hydrogen ions and hydroxide ions. Na+(l); Cl-(l); H+(aq); OH-(aq)

At the cathode: 2H+(aq) + 2e → H2(g)

At the anode: 2Cl-(aq) - 2e → Cl2(g)

The cathode reaction requires 2 mole of electrons per mole of hydrogen formed. The anode reaction requires 2 moles of electrons per mole of chlorine gas formed.

The only correct statement is that one mole of hydrogen gas will be formed in the second cell.
Q971-04 Which statement is correct?
  1. Spontaneous redox reactions produce electricity in an electrolytic cell.
  2. Electricity is used to carry out a non-spontaneous redox reaction in a voltaic cell
  3. Oxidation takes place at the negative electrode in a voltaic cell and the positive electrode in an electrolytic cell
  4. Oxidation takes place at the negative electrode in a voltaic cell and reduction takes place at the positive electrode in an electrolytic cell
Answer

Electrolytic cells use electricity and voltaic (electrochemical) cells create electricity. In a voltaic cell the negative electrode produces electrons by an oxidation process, eg: M(s) → M2+(aq) + 2e. At the positive electrode in an electrolytic cell the negative ions have their electrons removed, eg: X2-(aq) → X2(g) + 2e. Hence, statement C is correct.
Q971-05 Aqueous solutions of AgNO3, Cu(NO3)2 and Cr(NO3)3 are electrolysed using the same quantity of electricity. How do the number of moles of metals formed compare?
  1. Ag = Cu = Cr
  2. Ag > Cu > Cr
  3. Ag < Cu < Cr
  4. Cu > Ag > Cr
Answer

Inspection of the equations for the release of the metals at the cathode in each cell:

  • Ag+(aq) + 1e → Ag(s)
  • Cu2+(aq) + 2e → Cu(s)
  • Cr3+(aq) + 3e → Cr(s)

Shows that the moles of electrons required to make 1 mole of metal are 1, 2 and 3 moles respectively. Hence if the same numer of moles of electrons are used the moles of metal formed in the three cells is in the order: Ag > Cu > Cr

Q971-06 What are the products formed at the cathode and anode respectively when dilute hydrochloric acid and dlute nitric acid are electrolysed?
  1. nitric acid: nitrogen and oxygen. hydrochloric acid: hydrogen and oxygen
  2. nitric acid: hydrogen and oxygen. hydrochloric acid: hydrogen and oxygen
  3. nitric acid: hydrogen and oxygen. hydrochloric acid: hydrogen and chlorine
  4. nitric acid: nitrogen and oxygen. hydrochloric acid: hydrogen and chlorine
Answer

Dilute hydrochloric acid, ions present: H+(aq); Cl-(aq), OH-(aq)

At the cathode: H+(aq) + 2e → H2(g)

At the anode: 2Cl-(aq) - 2e → Cl2(g)

Dilute nitric acid, ions present: H+(aq); Cl-(aq), NO3-(aq)

At the cathode: H+(aq) + 2e → H2(g)

At the anode: 4OH-(aq) - 4e → 2H2O(l) + O2(g)

Hence, response C is correct
Q971-07 Suggest three observations that may be made when aqueous copper(II) chloride is electrolysed using graphite electrodes.
Answer

copper(II) chloride solution, ions contained Cu2+(aq); Cl-(aq), H+(aq), OH-(aq)

At the cathode: Cu2+(aq) + 2e → Cu(s)

At the anode: 2Cl-(aq) - 2e → Cl2(g)

Observation 1: a red/ brown deposit is seen at the cathode.

Observation 2: bubbles of gas are observed at the anode

Observation 3: the colour of the copper(II) chloride solution fades.

Observation 4: a smell of chlorine (swimming pools) is noted.
Q971-08 What is left behind in the solution when a solution of copper(II) sulfate(aq) is electrolysed until there is no further deposition of copper at the cathode?
Answer

copper(II) sulfate solution, ions contained are Cu2+(aq); SO42-(aq), H+(aq), OH-(aq).

At the cathode: Cu2+(aq) + 2e → Cu(s)

At the anode: 4OH-(aq) - 4e → 2H2O(l) + O2(g)

The ions remaining in solution are SO42-(aq) and H+(aq), this is sulfuric acid
Q971-09 In the electrolysis of dilute sulfuric acid, if 25cm3 of oxygen are liberated at the anode. How much hydrogen is liberated at the cathode?
Answer

sulfuric acid (aq) contains the ions SO42-(aq), H+(aq) and OH-(aq).

At the cathode: 2H+(aq) + 2e → H2(g)

At the anode: 4OH-(aq) - 4e → 2H2O(l) + O2(g)

By inspection, we see that twice as many moles of electrons are required to make one mole of oxygen than for one mole of hydrogen. During an electrolysis the same number of moles of electrons must pass through both electrodes, therefore the volume of oxygen created is always half the volume of hydrogen. Thus 50cm3 of hydrogen are liberated.
Q971-10 During the electrolysis of potassium iodide solution, some universal indicator is added to the solution. Suggest any colour changes that may be observed around the electrodes.
Answer

Potassium iodide (aq) contains the ions K+(aq), I-(aq), H+(aq) and OH-(aq).

At the cathode: 2H+(aq) + 2e → H2(g)

At the anode: 2I-(aq) - 4e → I2(aq)

Around the cathode, the hydrogen ions are being removed leaving an excess of hydroxide ions. This will produce a basic solution and the phenolphthalein will turn from colourless to red.

Around the anode, the iodine produced is brownish/red in colour.

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