Scientific Principles & Theory
Scientific Background: Electrolysis is the decomposition of an electrolyte (an ionic compound in molten state or dissolved in water) by the passage of a direct electric current.
In an aqueous solution, water molecules dissociate into H⁺ and OH⁻ ions alongside the electrolyte ions. The products formed at the inert electrodes depend on relative reactivity:
- Cathode (-): H⁺ gas is discharged unless the metal ion in solution is less reactive than hydrogen (e.g. Cu²⁺, Ag⁺, Au⁺).
- Anode (+): Halide ions (Cl⁻, Br⁻, I⁻) are discharged as halogens. If no halide is present, OH⁻ ions are discharged to form oxygen gas and water:
4OH⁻ → O₂ + 2H₂O + 4e⁻.
Experimental Variables
Independent Variable
The electrolyte solution (e.g. sodium chloride, copper(II) sulfate, dilute sulfuric acid).
Dependent Variable
The products observed/identified at the cathode (-) and anode (+).
Control Variables
Electrode material (graphite/inert), voltage (constant DC supply), concentration of solutions.
⚠️ Lab Risk Assessment
| Hazard | Associated Risk | Control Measure |
|---|---|---|
| Chlorine gas (produced from sodium chloride electrolysis) | Toxic and respiratory irritant | Use low concentrations; run for a short duration; perform in a fume cupboard or well-ventilated laboratory. |
| Electrical current / wet environment | Electric shock | Do not touch electrical connections with wet hands; keep liquids away from power pack units. |
Apparatus & Procedure
Required Apparatus
- DC Power supply (6 V)
- Graphite (carbon) electrodes
- Electrolysis cell (beaker or U-tube)
- Connecting leads and crocodile clips
- Aqueous electrolytes (1.0 mol/dm³ NaCl, CuSO₄, dilute H₂SO₄)
- Test tubes (for gas collection over electrodes)
- Litmus paper, splints (for gas tests)
Step-by-Step Procedure
- Pour the selected electrolyte solution (e.g. aqueous copper(II) sulfate) into a beaker to a depth of about 5 cm.
- Mount two graphite electrodes in a holder and place them into the solution. Ensure they do not touch each other.
- Connect the electrodes to the DC terminals of the power supply using connecting leads and crocodile clips.
- Turn on the power supply and set the voltage to 6 V DC.
- Observe any bubble formation, colour changes, or solid deposits at both the anode (connected to the positive terminal) and the cathode (connected to the negative terminal).
- If gases are produced, invert gas-filled test tubes filled with the electrolyte solution over each electrode to collect the gases by downward displacement of water.
- Test any gas collected at the cathode: test for hydrogen using a burning splint (squeaky pop).
- Test any gas collected at the anode: test for oxygen using a glowing splint (relights) or chlorine using damp blue litmus paper (bleaches white).
- Switch off the power supply. Record all observations.
- Clean the electrodes thoroughly with distilled water before changing to the next electrolyte solution.
Fig 1. Laboratory experimental setup for Core Practical 1.60C.
Sample Data & Calculations
This representative dataset illustrates the values typically obtained when carrying out this experiment in the laboratory:
| Electrolyte Solution | Product at Cathode (-) | Product at Anode (+) |
|---|---|---|
| Copper(II) sulfate (CuSO₄) | Red-brown copper metal coating on electrode | Bubbles of oxygen gas (relights glowing splint) |
| Sodium chloride (NaCl) | Bubbles of hydrogen gas (squeaky pop test) | Bubbles of chlorine gas (bleaches blue litmus) |
| Dilute sulfuric acid (H₂SO₄) | Bubbles of hydrogen gas (rapid rate) | Bubbles of oxygen gas (half the volume of hydrogen) |
Data Processing & Analysis
- Reaction at Cathode (for CuSO₄): Cu²⁺(aq) + 2e⁻ → Cu(s) [Reduction]
- Reaction at Anode (for CuSO₄): 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ [Oxidation]
- Note: H⁺ is discharged over Na⁺ at the cathode because hydrogen is less reactive than sodium. Cu²⁺ is discharged over H⁺ because copper is less reactive than hydrogen.
Conclusion & Evaluation
Chemical Explanation: Saturated solutions are heavily dependent on temperature. Heating shifts solubility limits, allowing more solute to form coordinate bonds or ion-dipole interactions with solvent molecules. When cooled, the reverse process happens and solute precipitates out.
Experimental Error Analysis
| Error Type & Source | Effect on Final Result | Mitigation Strategy |
|---|---|---|
| Systematic Error Impure electrodes / organic contamination |
Side reactions occur, leading to false observations, gas yields, or electrode corrosion. | Clean the graphite rods thoroughly with sand-paper and wash with distilled water before every run. |
| Random Error Gas leaking during collection over electrodes |
The volume of gas measured is lower than actual, leading to incorrect mole ratios of gases. | Ensure the inverted test tubes are completely full of solution at the start and rest directly over the graphite rods. |
Exam Practice
A student electrolyses dilute sulfuric acid. Describe an experiment to verify that the gases produced are hydrogen and oxygen, explaining the relative volumes of the gases collected in terms of the half-equations.
View Model Answer & Mark Scheme
Model Answer (6/6 Marks):
- Setup: Set up an electrolysis cell containing dilute sulfuric acid. Insert two inert graphite electrodes connected to a DC power supply (6 V).
- Collection: Fill two test tubes with sulfuric acid and invert them over the positive (anode) and negative (cathode) electrodes.
- Execution: Pass the current until a measurable volume of gas is collected in both tubes.
- Testing Gases:
- Take the tube from the cathode (-) and test with a lit splint: a squeaky pop confirms hydrogen.
- Take the tube from the anode (+) and test with a glowing splint: if it relights, it is oxygen.
- Explanation of Volumes: The volume of hydrogen gas collected is double the volume of oxygen gas.
- Half-Equations Link:
- At cathode:
2H⁺ + 2e⁻ → H₂. Producing 1 mole of H₂ requires 2 moles of electrons. - At anode:
4OH⁻ → O₂ + 2H₂O + 4e⁻. Producing 1 mole of O₂ releases 4 moles of electrons. - To balance the electron flow, for every 4 electrons, 2 moles of H₂ gas are formed for every 1 mole of O₂ gas, hence a 2:1 volume ratio.
- At cathode:
Examiner Tip:
Remember that gases occupy volume proportional to their moles (Avogadro's Law). Writing the balanced half-equations is the key to scoring high marks on the volume explanation.