Cambridge AS & A Level9701

Electrolysis

Chemistry 9701 Chapter Notes

What this chapter covers

ElectrolysisStandard electrode potentials E⦵, standard cell potentials E⦵cell and the Nernst equation
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In this chapter

  1. 1Fundamentals of Electrolytic Cells
  2. 2Electrolysis of Molten Compounds
  3. 3Electrolysis of Aqueous Solutions
  4. 4Quantitative Electrolysis: Faraday's Laws
  5. 5Standard Electrode Potentials (E°)
  6. 6Using Electrode Potentials to Predict Reactions
  7. 7Advanced Applications and Fuel Cells

1. Fundamentals of Electrolytic Cells

Electrolysis is the process of using a direct electric current to decompose an ionic compound, either molten or in solution, into its constituent elements. This process occurs in an electrolytic cell. The cell contains two electrodes (conductors, often graphite or metal) dipped into an electrolyte (the substance being broken down). A power supply pushes electrons towards the negative electrode, called the cathode, and pulls them from the positive electrode, called the anode. Positive ions (cations) in the electrolyte are attracted to the cathode, where they gain electrons in a process called reduction. Negative ions (anions) are attracted to the anode, where they lose electrons in a process called oxidation. A simple way to remember this is 'OIL RIG' (Oxidation Is Loss, Reduction Is Gain) and 'PANIC' (Positive Anode, Negative Is Cathode).

Key term

Electrolysis: The decomposition of a chemical compound by passing an electric current through it, either in a molten state or dissolved in a suitable solvent.

Examiner insight

Examiners award marks for correctly identifying the products of oxidation at the anode and reduction at thecathode, and for writing the correct half-equations with balanced charges.

Common pitfall

Confusing the sign of the anode and cathode. Remember PANIC: Positive Anode, Negative Is Cathode. This is for electrolytic cells only.

Worked example 14 marks

The diagram shows a simple electrolytic cell for the electrolysis of molten lead(II) bromide, PbBr₂. Label the components A, B, C, and D, and state the direction of electron flow in the external wire.

  1. 1

    Step 1: Identify the power supply terminals. The longer line represents the positive terminal and the shorter line represents the negative terminal.

  2. 2

    Step 2: Identify the electrodes. The electrode connected to the positive terminal is the anode (A). The electrode connected to the negative terminal is the cathode (B).

  3. 3

    Step 3: Identify the electrolyte. The molten substance being electrolysed is the electrolyte (C), which is molten PbBr₂.

  4. 4

    Step 4: Identify the ions. In molten PbBr₂, the ions are lead(II) ions, Pb²⁺ (D - cations), and bromide ions, Br⁻ (D - anions).

  5. 5

    Step 5: Determine electron flow. The power supply pushes electrons from the anode to the cathode through the external circuit. Therefore, electrons flow from electrode A to electrode B.

Recap

  • Electrolysis uses electricity to break down ionic compounds.
  • The negative electrode is the cathode, where reduction (gain of electrons) occurs.
  • The positive electrode is the anode, where oxidation (loss of electrons) occurs.
  • Cations (positive ions) move to the cathode.
  • Anions (negative ions) move to the anode.
  • Electrons flow from anode to cathode in the external circuit.

Quick check

  1. What type of reaction occurs at the anode?1 mark
  2. What is the name given to the positive electrode in an electrolytic cell?1 mark

2. Electrolysis of Molten Compounds

When an ionic compound is heated until it melts, its ions become mobile and can move to the electrodes to be discharged. This is the simplest case of electrolysis because there are only two types of ions present: the metal cation and the non-metal anion. The process is straightforward: the metal cation is always reduced at the negative cathode to form the metal, and the non-metal anion is always oxidised at the positive anode. For example, in the electrolysis of molten sodium chloride (NaCl), Na⁺ ions move to the cathode and Cl⁻ ions move to the anode.

Key term

Electrolyte: A molten ionic compound or an aqueous solution of ions that is decomposed during electrolysis.

Fun fact

The industrial process for extracting aluminium via electrolysis of molten aluminium oxide (mixed with cryolite) is so energy-intensive that aluminium has been called 'congealed electricity'. It consumes about 4% of the USA's entire electricity supply.

Worked example 15 marks

Predict the products at the anode and cathode during the electrolysis of molten zinc chloride (ZnCl₂). Write the half-equation for the reaction at each electrode and the overall equation.

  1. 1

    Step 1: Identify the ions present. Molten ZnCl₂ contains zinc cations (Zn²⁺) and chloride anions (Cl⁻).

  2. 2

    Step 2: Determine ion movement. Positive Zn²⁺ ions are attracted to the negative cathode. Negative Cl⁻ ions are attracted to the positive anode.

  3. 3

    Step 3: Write the cathode half-equation (Reduction). At the cathode, Zn²⁺ ions gain two electrons to form zinc metal. Zn²⁺(l) + 2e⁻ → Zn(l).

  4. 4

    Step 4: Write the anode half-equation (Oxidation). At the anode, Cl⁻ ions lose one electron each to form chlorine gas. Two ions are needed to form a Cl₂ molecule. 2Cl⁻(l) → Cl₂(g) + 2e⁻.

  5. 5

    Step 5: Write the overall equation. Combine the two half-equations. The electrons on both sides cancel out. Zn²⁺(l) + 2Cl⁻(l) → Zn(l) + Cl₂(g). This is simply ZnCl₂(l) → Zn(l) + Cl₂(g).

Recap

  • In molten electrolysis, only the ions from the compound are present.
  • The positive metal ion (cation) is reduced at the cathode to form the metal.
  • The negative non-metal ion (anion) is oxidised at the anode.
  • The products are simply the elements that make up the ionic compound.
  • This process is used to extract reactive metals like aluminium and sodium.

Quick check

  1. What substance would be formed at the cathode during the electrolysis of molten magnesium oxide (MgO)?1 mark

3. Electrolysis of Aqueous Solutions

When an ionic compound is dissolved in water, the solution contains ions from the compound AND ions from the partial dissociation of water (H⁺ and OH⁻). This means there is competition at each electrode. The product depends on three factors: the position of the ions in the electrochemical series, the concentration of the ions, and the nature of the electrode.

At the Cathode (Negative): Two positive ions compete. The one that is easier to reduce (has a more positive E° value) will be discharged. Rule of thumb: If the metal ion is from a reactive metal (like Group 1, 2, or Al), hydrogen gas will be produced from the reduction of H⁺ ions (2H⁺ + 2e⁻ → H₂). If the metal is less reactive than hydrogen (like Cu, Ag, Au), the metal will be deposited.

At the Anode (Positive): Two negative ions compete. Rule of thumb: If the solution contains a high concentration of halide ions (Cl⁻, Br⁻, I⁻), the halogen will be produced. If the solution is dilute or contains no halides, oxygen gas is produced from the oxidation of OH⁻ ions (4OH⁻ → O₂ + 2H₂O + 4e⁻).

Cathode (Reduction): 2H⁺(aq) + 2e⁻ → H₂(g) or Mⁿ⁺(aq) + ne⁻ → M(s)

Anode (Oxidation): 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻ or 2X⁻(aq) → X₂(g) + 2e⁻

Key term

Selective Discharge: The process where, out of a mixture of ions, only one type of ion is discharged (oxidised or reduced) at an electrode.

Examiner insight

Marks are often awarded for explaining *why* a particular ion is discharged, by referring to its relative reactivity or position in the electrochemical series compared to H⁺ or OH⁻.

Common pitfall

Assuming the ions from the salt are always the ones discharged in aqueous solution. Always remember to consider H⁺ and OH⁻ from water as competitors.

Worked example 14 marks

Predict the products at the anode and cathode for the electrolysis of a concentrated aqueous solution of sodium chloride using inert electrodes. Write the relevant half-equations.

  1. 1

    Step 1: Identify all ions present. From NaCl: Na⁺ and Cl⁻. From water: H⁺ and OH⁻.

  2. 2

    Step 2: Consider the cathode (negative electrode). The positive ions Na⁺ and H⁺ are attracted. Sodium is a very reactive metal (more negative E°), so it is harder to reduce than H⁺. Therefore, H⁺ ions are preferentially discharged, producing hydrogen gas.

  3. 3

    Step 3: Write the cathode half-equation. 2H⁺(aq) + 2e⁻ → H₂(g).

  4. 4

    Step 4: Consider the anode (positive electrode). The negative ions Cl⁻ and OH⁻ are attracted. Because the solution is 'concentrated', the Cl⁻ ions are preferentially discharged over OH⁻ ions, producing chlorine gas.

  5. 5

    Step 5: Write the anode half-equation. 2Cl⁻(aq) → Cl₂(g) + 2e⁻.

Recap

  • In aqueous solutions, water provides H⁺ and OH⁻ ions which compete at the electrodes.
  • At the cathode, the cation with the more positive E° (less reactive metal) is discharged.
  • For very reactive metals (K, Na, Ca, Mg, Al), hydrogen is produced at the cathode.
  • At the anode, concentrated halide ions (Cl⁻, Br⁻, I⁻) are discharged to form halogens.
  • If no concentrated halides are present, oxygen is produced at the anode from OH⁻ ions.

Quick check

  1. What gas is produced at the anode during the electrolysis of dilute sulfuric acid, H₂SO₄(aq)?1 mark
  2. What is deposited at the cathode during the electrolysis of copper(II) sulfate solution, CuSO₄(aq)?1 mark

4. Quantitative Electrolysis: Faraday's Laws

The amount of substance produced during electrolysis is directly proportional to the total electric charge passed through the electrolyte. This relationship is quantified by Faraday's laws. The total charge (Q), measured in coulombs (C), is calculated by multiplying the current (I), in amperes (A), by the time (t), in seconds. The Faraday constant (F) is the charge carried by one mole of electrons, which is approximately 96500 C mol⁻¹. The relationship between the Faraday constant, Avogadro's constant (L), and the charge on a single electron(e) is F = Le. To solve quantitative electrolysis problems, you use a three-step process: 1. Calculate the total charge passed (Q = It). 2. Calculate the moles of electrons passed (moles of e⁻ = Q / F). 3. Use the stoichiometry of the relevant half-equation to relate moles of electrons to moles of the substance produced, and then find its mass or volume.

Q = I × t

F = L × e

Moles of electrons = Q / F

1 F = 96500 C mol⁻¹

Key term

Faraday Constant (F): The magnitude of electric charge per mole of electrons, with a value of approximately 96500 coulombs per mole (C mol⁻¹).

Examiner insight

Examiners look for a clear, step-by-step method. Writing the half-equation first is a crucial step that is often awarded a mark.

Worked example 14 marks

Calculate the mass of copper deposited at the cathode when a current of 1.50 A is passed through a copper(II) sulfate solution for 40.0 minutes. (Ar of Cu = 63.5)

  1. 1

    Step 1: Write the half-equation for the reaction at the cathode. Cu²⁺(aq) + 2e⁻ → Cu(s). This shows that 2 moles of electrons are needed to deposit 1 mole of copper.

  2. 2

    Step 2: Calculate the total charge (Q) passed. First, convert time to seconds: 40.0 min × 60 s/min = 2400 s. Then, Q = I × t = 1.50 A × 2400 s = 3600 C.

  3. 3

    Step 3: Calculate the moles of electrons. Moles of e⁻ = Q / F = 3600 C / 96500 C mol⁻¹ = 0.0373 mol.

  4. 4

    Step 4: Use the mole ratio from the half-equation to find the moles of copper. Moles of Cu = (moles of e⁻) / 2 = 0.0373 mol / 2 = 0.01865 mol.

  5. 5

    Step 5: Calculate the mass of copper. Mass = moles × Ar = 0.01865 mol × 63.5 g/mol = 1.18 g (to 3 s.f.).

Worked example 23 marks

What volume of oxygen gas, measured at room temperature and pressure (RTP), is produced at the anode when a charge of 1930 C is passed through dilute sulfuric acid? (Molar volume of gas at RTP = 24.0 dm³ mol⁻¹)

  1. 1

    Step 1: Write the half-equation for the reaction at the anode. 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻. This shows that 4 moles of electrons are produced for every 1 mole of O₂.

  2. 2

    Step 2: Calculate the moles of electrons. Moles of e⁻ = Q / F = 1930 C / 96500 C mol⁻¹ = 0.0200 mol.

  3. 3

    Step 3: Use the mole ratio to find the moles of oxygen. Moles of O₂ = (moles of e⁻) / 4 = 0.0200 mol / 4 = 0.00500 mol.

  4. 4

    Step 4: Calculate the volume of oxygen at RTP. Volume = moles × molar volume = 0.00500 mol × 24.0 dm³ mol⁻¹ = 0.120 dm³.

Recap

  • The amount of product is proportional to the charge passed.
  • Calculate charge using Q = I × t (time must be in seconds).
  • Calculate moles of electrons using moles = Q / 96500.
  • Use the half-equation mole ratio to find moles of product.
  • Convert moles of product to mass (using Mr) or gas volume (using molar volume).

Quick check

  1. How many coulombs of charge are passed when a current of 5 A flows for 10 minutes?2 marks
  2. How many moles of electrons correspond to 1 Faraday of charge?1 mark

5. Standard Electrode Potentials (E°)

While electrolysis uses energy to force a non-spontaneous reaction, electrochemical cells (batteries) generate electricity from spontaneous reactions. To compare the tendency of different substances to be reduced, we use standard electrode potentials (E°). The standard electrode potential of a half-cell is the voltage measured when it is connected to a Standard Hydrogen Electrode (SHE) under standard conditions (298 K, 1 atm pressure for gases, 1.0 mol dm⁻³ concentration for solutions). The SHE itself is assigned a potential of 0.00 V. It consists of a platinum electrode in a 1.0 mol dm⁻³ solution of H⁺ ions, with hydrogen gas bubbled over it at 1 atm pressure. A more positive E° value indicates a greater tendency for the species to be reduced (a stronger oxidising agent). A more negative E° value indicates a greater tendency for the species to be oxidised (a stronger reducing agent).

E°(Standard Hydrogen Electrode) = 0.00 V

Key term

Standard Electrode Potential (E°): The potential difference (voltage) of a half-cell connected to a standard hydrogen electrode, measured under standard conditions.

Examiner insight

Students must be able to describe the SHE accurately, as it is the universal reference standard for all electrode potential measurements.

Common pitfall

Forgetting any one of the three standard conditions (concentration, temperature, pressure). All three must be stated for a full mark.

Worked example 14 marks

Describe the key features of the Standard Hydrogen Electrode (SHE).

  1. 1

    Step 1: State the electrode material. It uses an inert platinum electrode, often coated in finely divided platinum (platinum black) to increase its surface area and catalyse the reaction.

  2. 2

    Step 2: State the gaseous component. Hydrogen gas (H₂) at a pressure of 1 atmosphere (or 100 kPa) is bubbled over the electrode.

  3. 3

    Step 3: State the solution component. The electrode is immersed in a solution containing hydrogen ions (H⁺) at a concentration of 1.0 mol dm⁻³.

  4. 4

    Step 4: State the temperature. The system is maintained at a standard temperature, usually 298 K (25 °C).

  5. 5

    Step 5: State its defined potential. By convention, the standard electrode potential of the SHE is defined as exactly 0.00 volts.

Recap

  • Standard electrode potential (E°) measures the tendency of a half-cell to undergo reduction.
  • All E° values are measured against the Standard Hydrogen Electrode (SHE).
  • The SHE is assigned a potential of 0.00 V by definition.
  • Standard conditions are 298 K, 1 atm pressure, and 1.0 mol dm⁻³ concentration.
  • A more positive E° means the substance is more easily reduced.

Quick check

  1. What are the three standard conditions required for measuring an E° value?3 marks
  2. What is the E° value of the standard hydrogen electrode?1 mark

6. Using Electrode Potentials to Predict Reactions

Standard electrode potentials are powerful tools for predicting the direction of electron flow and the feasibility of a redox reaction. When two half-cells are connected, they form an electrochemical cell. The standard cell potential (E°cell) is the overall voltage of this cell, calculated by subtracting the E° of the half-cell being oxidised from the E° of the half-cell being reduced. A simpler way is: E°cell = E°(more positive) - E°(more negative). A positive E°cell value indicates that the reaction is spontaneous (feasible) under standard conditions. The half-cell with the more negative E° value will always be the site of oxidation (the negative electrode), and the half-cell with the more positive E° value will be the site of reduction (the positive electrode). Electrons always flow from the more negative half-cell to the more positive half-cell in the external circuit.

E°cell = E°(reduction) - E°(oxidation)

E°cell = E°(more positive) - E°(more negative)

Key term

Standard Cell Potential (E°cell): The potential difference between two half-cells in an electrochemical cell under standard conditions, which indicates the feasibility of the overall redox reaction.

Examiner insight

When predicting feasibility, it is not enough to just state 'it is feasible'. You must support your answer by calculating a positive E°cell value.

Common pitfall

Incorrectly subtracting the E° values. Always remember the formula E°cell = E°(more positive) - E°(more negative) to avoid sign errors.

Worked example 14 marks

An electrochemical cell is constructed from a zinc half-cell (E°(Zn²⁺/Zn) = -0.76 V) and a copper half-cell (E°(Cu²⁺/Cu) = +0.34 V). Calculate the standard cell potential and deduce the direction of electron flow.

  1. 1

    Step 1: Identify the more positive and more negative E° values. E°(Cu²⁺/Cu) = +0.34 V (more positive). E°(Zn²⁺/Zn) = -0.76 V (more negative).

  2. 2

    Step 2: Calculate the standard cell potential, E°cell. E°cell = E°(more positive) - E°(more negative) = (+0.34 V) - (-0.76 V) = 1.10 V.

  3. 3

    Step 3: Determine the direction of reaction. Since E°cell is positive, the reaction is feasible. The half-reaction with the more negative E° is oxidation, and the one with the more positive E° is reduction.

  4. 4

    Step 4: Identify the electrodes and electron flow. Zinc is oxidised (Zn → Zn²⁺ + 2e⁻), so the zinc electrode is the negative pole. Copper ions are reduced (Cu²⁺ + 2e⁻ → Cu), so the copper electrode is the positive pole. Electrons flow from the negative zinc electrode to the positive copper electrode through the external wire.

Recap

  • E°cell is calculated by E°(more positive) - E°(more negative).
  • A positive E°cell indicates a spontaneous (feasible) reaction.
  • A negative E°cell indicates the reaction is not spontaneous in the forward direction.
  • The half-cell with the more negative E° undergoes oxidation.
  • The half-cell with the more positive E° undergoes reduction.
  • Electrons flow from the negative electrode (oxidation) to the positive electrode (reduction).

Quick check

  1. If E°(Fe²⁺/Fe) = -0.44 V and E°(Sn²⁺/Sn) = -0.14 V, what is the E°cell for a cell made of these two half-cells?2 marks

7. Advanced Applications and Fuel Cells

Electrolysis has precise applications, such as determining the Avogadro constant, L. By electrolysing a solution like silver nitrate and accurately measuring the mass of silver deposited (m), the current (I), and time (t), we can find the total charge passed (Q=It). From this, we find the moles of electrons and thus the total number of electrons that passed. Knowing the moles of silver (m/Ar), we can find the number of atoms deposited. The ratio of these two numbers gives the number of atoms per mole, which is L. Beyond traditional electrolysis, fuel cells are a key technology. A hydrogen-oxygen fuel cell generates electricity directly from the reaction between hydrogen and oxygen. At the anode, hydrogen is oxidised (H₂ → 2H⁺ + 2e⁻), and at the cathode, oxygen is reduced (O₂ + 4H⁺ + 4e⁻ → 2H₂O). The overall reaction is 2H₂ + O₂ → 2H₂O. The main advantages are high efficiency and producing only water as a product, unlike combustion engines.

Overall fuel cell reaction: 2H₂(g) + O₂(g) → 2H₂O(l)

Key term

Fuel Cell: An electrochemical cell that generates a continuous electric current by consuming a fuel (like hydrogen) and an oxidant (like oxygen).

Fun fact

Some new public transport buses run on hydrogen fuel cells. Their only emission is pure water vapour, which drips onto the road as they drive.

Worked example 15 marks

In an experiment to determine the Avogadro constant, a current of 0.200 A was passed through a solution of silver nitrate for 1 hour. A mass of 0.806 g of silver was deposited on the cathode. Given that the charge on an electron is 1.60 × 10⁻¹⁹ C and the relative atomic mass of silver is 107.9, calculate a value for the Avogadro constant.

  1. 1

    Step 1: Calculate the total charge passed, Q. Time = 1 hour = 3600 s. Q = I × t = 0.200 A × 3600 s = 720 C.

  2. 2

    Step 2: Calculate the number of electrons that passed. Number of electrons = Total charge / Charge per electron = 720 C / (1.60 × 10⁻¹⁹ C) = 4.50 × 10²¹ electrons.

  3. 3

    Step 3: Calculate the moles of silver deposited. Moles of Ag = mass / Ar = 0.806 g / 107.9 g/mol = 0.00747 mol.

  4. 4

    Step 4: Write the half-equation to find the mole ratio. Ag⁺(aq) + e⁻ → Ag(s). This is a 1:1 ratio of electrons to silver atoms.

  5. 5

    Step 5: Calculate the Avogadro constant, L. L is the number of particles per mole. L = (Number of particles) / (moles) = (4.50 × 10²¹ atoms) / (0.00747 mol) = 6.024 × 10²³ mol⁻¹.

Recap

  • Electrolysis can be used experimentally to determine the Avogadro constant.
  • A fuel cell converts chemical energy from a fuel directly into electrical energy.
  • The hydrogen-oxygen fuel cell produces only water as a waste product.
  • Advantages of fuel cells include high efficiency and low pollution.
  • The overall reaction in a hydrogen fuel cell is the formation of water.

Quick check

  1. State one major advantage of a hydrogen fuel cell over a petrol engine.1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Molten aluminium bromide (AlBr₃) is electrolysed. Write the half-equations for the reactions occurring at the anode and the cathode.3 marks
  2. Predict the products formed at the inert anode and cathode during the electrolysis of (i) dilute sodium sulfate solution and (ii) molten copper(II) oxide.4 marks
  3. A current of 2.5 A is passed through a solution of nickel(II) sulfate for 50 minutes. Calculate the mass of nickel deposited at the cathode. (Ar of Ni = 58.7, F = 96500 C mol⁻¹)4 marks
  4. An electrochemical cell is set up using the following half-cells under standard conditions: Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), E° = +0.77 V; and Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ ⇌ 2Cr³⁺(aq) + 7H₂O(l), E° = +1.33 V. Calculate the standard cell potential (E°cell) and identify the oxidising agent in the spontaneous reaction.3 marks
  5. Explain why different products are formed at the anode during the electrolysis of dilute aqueous sodium chloride compared to concentrated aqueous sodium chloride.3 marks
  6. Calculate the volume of chlorine gas, measured at RTP, produced when a current of 4.0 A flows through molten magnesium chloride for 30 minutes. (Molar volume of gas at RTP = 24.0 dm³ mol⁻¹; F = 96500 C mol⁻¹).4 marks
  7. Using the following standard electrode potentials: E°(Cl₂/Cl⁻) = +1.36 V and E°(Br₂/Br⁻) = +1.07 V, predict whether chlorine gas will oxidise bromide ions. Justify your answer with a calculation.3 marks
  8. Describe the standard hydrogen electrode, including a labelled diagram, and explain its role in electrochemistry.5 marks
  9. A student wants to silver-plate a brass spoon. Draw a labelled diagram of the electrolytic cell they should set up. Ensure you label the anode, cathode, electrolyte, and power supply polarity.4 marks
  10. A hydrogen-oxygen fuel cell operating in acidic conditions produces a current of 150 A. Calculate the mass of hydrogen gas consumed in 1 hour. (F = 96500 C mol⁻¹)5 marks

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