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4.1.4 Charge transfer, aqueous electrolysis and half-equations

How charge moves in electrolysis

Definition

Oxidation

The loss of electrons by a substance.

Definition

Reduction

The gain of electrons by a substance.

  1. In an electrolytic cell, charge is carried by electrons in the external circuit and by ions in the electrolyte.
  2. Movement of electrons in the external circuit:
    1. The power supply pushes electrons out of its negative terminal, along the wire and onto the cathode.
    2. At the same time, electrons are drawn off the anode and flow along the wire back to the positive terminal.
    3. Electrons travel only through the wires and the electrodes, and never through the electrolyte itself.
  3. Loss or gain of electrons at the electrodes:
    1. At the cathode, positive ions gain electrons from the electrode, so reduction takes place.
    2. At the anode, negative ions, or sometimes the metal of the anode itself, lose electrons to the electrode, so oxidation takes place.
    3. Once the ions have gained or lost electrons they become neutral atoms or molecules, which is described as being discharged.
  4. Movement of ions in the electrolyte:
    1. Positive ions, or cations, move through the electrolyte towards the cathode.
    2. Negative ions, or anions, move through the electrolyte towards the anode.
    3. This movement of ions in opposite directions carries the charge between the two electrodes and completes the circuit.
  5. For every electron gained at the cathode, one electron is released at the anode, so the same amount of charge passes through every part of the circuit.
  6. The phrase OIL RIG, meaning Oxidation Is Loss and Reduction Is Gain of electrons, helps you link each electrode to the correct process.

Copper(II) sulfate with graphite electrodes

  1. Aqueous copper(II) sulfate contains copper(II) ions, Cu2+\text{Cu}^{2+}Cu2+, and sulfate ions, SO42−\text{SO}_4^{2-}SO42−​, together with hydrogen ions, H+\text{H}^{+}H+, and hydroxide ions, OH−\text{OH}^{-}OH−, from water.
  2. At the cathode, copper is less reactive than hydrogen, so copper(II) ions are discharged instead of hydrogen ions.
  3. A pink-brown layer of copper forms on the cathode: Cu2+(aq)+2e−→Cu(s)\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Cu}(\text{s})Cu2+(aq)+2e−→Cu(s).
  4. At the anode, sulfate ions are not discharged, so hydroxide ions lose electrons and oxygen is released instead.
  5. Bubbles of a colourless gas that relights a glowing splint form at the anode: 4OH−(aq)→O2(g)+2H2O(l)+4e−4\text{OH}^{-}(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^{-}4OH−(aq)→O2​(g)+2H2​O(l)+4e−.
  6. The blue colour of the solution fades, because copper(II) ions are removed at the cathode and nothing replaces them.
  7. The solution slowly becomes acidic, because the hydrogen ions and sulfate ions left behind make sulfuric acid.
  8. The graphite cathode gains mass as copper builds up on it, while the mass of the graphite anode does not change.

Copper(II) sulfate with copper electrodes

  1. When both electrodes are made of copper, the cathode reaction is unchanged, and copper is deposited on the cathode.
  2. At the anode, copper atoms lose electrons more readily than hydroxide ions do, so the anode itself dissolves: Cu(s)→Cu2+(aq)+2e−\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}Cu(s)→Cu2+(aq)+2e−.
  3. The copper anode loses mass and becomes thinner, while the copper cathode gains mass.
  4. No gas is given off at either electrode.
  5. The blue colour stays the same, because every copper(II) ion removed at the cathode is replaced by one released at the anode.
  6. When the anode is pure copper, the mass lost by the anode equals the mass gained by the cathode.
  7. The same arrangement is used to purify copper, with an impure copper anode dissolving and pure copper building up on the cathode.
Practical

Electrolysis of copper(II) sulfate

  • Aim: to compare the products and the changes in electrode mass when aqueous copper(II) sulfate is electrolysed with graphite electrodes and with copper electrodes.
  • Apparatus: a 250 cm3250\ \text{cm}^3250 cm3 beaker, aqueous copper(II) sulfate of concentration about 0.5 mol/dm30.5\ \text{mol/dm}^30.5 mol/dm3, two graphite rods, two strips of copper foil, a low-voltage d.c. power pack, an ammeter, a variable resistor, connecting leads with crocodile clips, emery paper, distilled water, paper towels, a balance reading to 0.01 g0.01\ \text{g}0.01 g, a small test tube, wooden splints and eye protection.
  • Method, graphite electrodes:
    • Weigh both graphite rods, then clamp them in the beaker of copper(II) sulfate solution about 3 cm3\ \text{cm}3 cm apart.
    • Fill a small test tube with the solution and invert it over the anode to collect any gas.
    • Connect the rods to the power pack through the ammeter and the variable resistor, switch on and set the current to about 0.3 A0.3\ \text{A}0.3 A.
    • Leave the cell running for 202020 minutes, keeping the current constant and watching both electrodes and the colour of the solution.
    • Test the gas collected at the anode with a glowing splint.
    • Switch off, rinse both rods gently with distilled water, let them dry and reweigh them.
  • Method, copper electrodes:
    • Clean two copper strips with emery paper, rinse and dry them, then weigh each one and label which will be the anode and which the cathode.
    • Repeat the electrolysis in fresh copper(II) sulfate solution using the same current for the same time.
    • Rinse, dry and reweigh both copper strips.
  • Results:
    • With graphite electrodes, pink-brown copper coats the cathode, a colourless gas that relights a glowing splint forms at the anode, and the blue colour fades.
    • With copper electrodes, the cathode gains mass, the anode loses about the same mass, no gas forms and the blue colour stays the same.
  • Watch out:
    • Keep the current and the time the same for both runs, because the mass of copper deposited depends on the total charge passed.
    • Copper deposited too quickly is loose and can fall off during rinsing, which makes the gain in mass of the cathode look too small.
    • Dab the electrodes dry or leave them to dry rather than rubbing them, so that no copper is wiped away.
  • Safety: wear eye protection, avoid skin contact with copper(II) sulfate solution because it is harmful and an irritant, and wash your hands at the end.

Electrolysis of aqueous halides

  1. An aqueous halide, such as a chloride, bromide or iodide solution, contains the ions of the dissolved compound together with hydrogen ions and hydroxide ions from water.
  2. There are two possible positive ions and two possible negative ions, so a separate decision is made at each electrode.
  3. At the cathode:
    1. If the metal is more reactive than hydrogen, such as sodium, potassium or magnesium, the metal ions stay in solution and hydrogen is produced.
    2. If the metal is less reactive than hydrogen, such as copper or silver, the metal itself is produced.
    3. The concentration of the solution has no effect on which product forms at the cathode.
  4. At the anode:
    1. In a concentrated halide solution, the halide ions are discharged and the halogen is produced, so a chloride gives chlorine, a bromide gives bromine and an iodide gives iodine.
    2. In a dilute halide solution, hydroxide ions are discharged instead and oxygen is produced.
    3. Chlorine appears as a pale yellow-green gas, bromine colours the solution around the anode orange-brown, and iodine colours it brown.
  5. Concentrated aqueous sodium chloride therefore gives hydrogen and chlorine, while dilute aqueous sodium chloride gives hydrogen and oxygen.
Example

Concentrated aqueous potassium bromide

  • The solution contains potassium ions, bromide ions, hydrogen ions and hydroxide ions.
  • Potassium is more reactive than hydrogen, so hydrogen is produced at the cathode.
2H+(aq)+2e−→H2(g) 2\text{H}^{+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{H}_2(\text{g}) 2H+(aq)+2e−→H2​(g)
  • The solution is concentrated, so bromide ions are discharged and bromine is produced at the anode.
2Br−(aq)→Br2(aq)+2e− 2\text{Br}^{-}(\text{aq}) \rightarrow \text{Br}_2(\text{aq}) + 2\text{e}^{-} 2Br−(aq)→Br2​(aq)+2e−
  • You see colourless bubbles at the cathode and an orange-brown colour spreading out from the anode.
Example

Dilute aqueous copper(II) chloride

  • The solution contains copper(II) ions, chloride ions, hydrogen ions and hydroxide ions.
  • Copper is less reactive than hydrogen, so copper is produced at the cathode.
Cu2+(aq)+2e−→Cu(s) \text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{Cu}(\text{s}) Cu2+(aq)+2e−→Cu(s)
  • The solution is dilute, so hydroxide ions are discharged and oxygen is produced at the anode.
4OH−(aq)→O2(g)+2H2O(l)+4e− 4\text{OH}^{-}(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^{-} 4OH−(aq)→O2​(g)+2H2​O(l)+4e−
  • You see a pink-brown coating on the cathode and colourless bubbles at the anode that relight a glowing splint.

Writing ionic half-equations

Definition

Ionic half-equation

An equation that shows the electrons gained or lost by one type of particle at one electrode.

  1. An ionic half-equation shows what happens to one type of particle at one electrode, with the electrons written in.
  2. Build each half-equation in four steps.
    1. Write the formula of the particle that reacts and of the product, remembering that hydrogen, oxygen and the halogens form diatomic molecules.
    2. Balance the atoms of each element by placing numbers in front of the formulae.
    3. Balance the charge by adding electrons, e−\text{e}^{-}e−, to the side that is more positive.
    4. Check that the total charge is the same on both sides, and add state symbols if the question asks for them.
  3. At the cathode, the electrons appear on the left-hand side because they are gained, so every cathode half-equation shows reduction.
  4. At the anode, the electrons appear on the right-hand side because they are lost, so every anode half-equation shows oxidation.
  5. The half-equations you need most often are listed below.
    1. Lead at the cathode: Pb2++2e−→Pb\text{Pb}^{2+} + 2\text{e}^{-} \rightarrow \text{Pb}Pb2++2e−→Pb
    2. Copper at the cathode: Cu2++2e−→Cu\text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu}Cu2++2e−→Cu
    3. Hydrogen at the cathode: 2H++2e−→H22\text{H}^{+} + 2\text{e}^{-} \rightarrow \text{H}_22H++2e−→H2​
    4. Sodium at the cathode from a molten compound: Na++e−→Na\text{Na}^{+} + \text{e}^{-} \rightarrow \text{Na}Na++e−→Na
    5. Chlorine at the anode: 2Cl−→Cl2+2e−2\text{Cl}^{-} \rightarrow \text{Cl}_2 + 2\text{e}^{-}2Cl−→Cl2​+2e−
    6. Bromine at the anode: 2Br−→Br2+2e−2\text{Br}^{-} \rightarrow \text{Br}_2 + 2\text{e}^{-}2Br−→Br2​+2e−
    7. Oxygen at the anode: 4OH−→O2+2H2O+4e−4\text{OH}^{-} \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^{-}4OH−→O2​+2H2​O+4e−
    8. Copper anode dissolving: Cu→Cu2++2e−\text{Cu} \rightarrow \text{Cu}^{2+} + 2\text{e}^{-}Cu→Cu2++2e−
Example

Building the oxygen half-equation

  • Hydroxide ions, OH−\text{OH}^{-}OH−, are discharged to form oxygen, O2\text{O}_2O2​, and water, H2O\text{H}_2\text{O}H2​O.
OH−→O2+H2O \text{OH}^{-} \rightarrow \text{O}_2 + \text{H}_2\text{O} OH−→O2​+H2​O
  • One O2\text{O}_2O2​ molecule and two H2O\text{H}_2\text{O}H2​O molecules contain four oxygen atoms and four hydrogen atoms, so four hydroxide ions are needed.
4OH−→O2+2H2O 4\text{OH}^{-} \rightarrow \text{O}_2 + 2\text{H}_2\text{O} 4OH−→O2​+2H2​O
  • The left-hand side has a total charge of 4−4-4− and the right-hand side has a charge of 000, so four electrons are added to the right.
4OH−→O2+2H2O+4e− 4\text{OH}^{-} \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^{-} 4OH−→O2​+2H2​O+4e−
  • The electrons are on the right, so the half-equation shows oxidation, which matches a reaction at the anode.
Common Mistake
  • Do not say that electrons flow through the electrolyte, because ions carry the charge through the electrolyte and electrons only move in the wires and electrodes.
  • Do not put the electrons on the wrong side, because they are gained on the left at the cathode and lost on the right at the anode.
  • Do not use concentration to decide the cathode product, because only the reactivity of the metal compared with hydrogen decides it.
  • Do not write Cl\text{Cl}Cl or H\text{H}H as products, because chlorine and hydrogen are released as diatomic molecules, Cl2\text{Cl}_2Cl2​ and H2\text{H}_2H2​.
Exam technique

Explaining charge transfer

  • For a charge-transfer question, describe three things in order: electrons moving in the external circuit, electrons being lost or gained at each electrode, and ions moving through the electrolyte.
  • When explaining the colour of copper(II) sulfate solution, link it to whether the copper(II) ions are replaced, so it fades with graphite and stays blue with copper.
  • Check every half-equation for both atoms and charge before moving on, because a missing electron loses the mark.
  • When asked why a change is oxidation or reduction, refer to loss or gain of electrons rather than to oxygen.
Self review
  • How is charge carried through the wires and through the electrolyte during electrolysis?
  • Why does the blue colour of copper(II) sulfate fade with graphite electrodes but not with copper electrodes?
  • Which products form at each electrode when dilute aqueous sodium chloride is electrolysed?
  • What is the ionic half-equation for the formation of bromine at the anode?
  • At which electrode does reduction take place, and why?
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4.1.4 Charge transfer, aqueous electrolysis and half-equations Revision Guide

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