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Electrochemical cell experiments (Core Practical 12)
Edexcel IAL Chemistry · WCH16 Practical Skills in Chemistry II
Core Practical 12 measures the emf of cells. The theory of electrode potentials is in the Unit 5 lessons; here the focus is on setting up the cell and explaining readings that differ from the data book.
Setting up
- Clean metal electrodes with emery paper to remove the oxide layer.
- Use solutions of the metal ions at 1.00 mol dm-3 and record the temperature.
- For ions of the same element in solution (Fe3+/Fe2+), use an inert platinum electrode.
- Join the half-cells with a salt bridge of filter paper soaked in saturated potassium nitrate solution. KNO3 is used because K+ and NO3- do not form precipitates with the ions in the half-cells (KCl would precipitate AgCl with Ag+).
- Measure the emf with a high-resistance digital voltmeter, which draws almost no current, so the emf is measured without the concentrations changing.
- A negative reading means the leads are reversed: the half-cell joined to the black (COM) lead is actually the positive electrode.
Why readings differ from the data book
- Concentrations not exactly 1.00 mol dm-3, or temperature not 298 K.
- Oxide layers or impurities on the electrodes.
- A current drawn by the voltmeter, or a dried-out salt bridge with high resistance.
Lowering the concentration of the ion in the positive half-cell makes its potential less positive and the cell emf smaller (the Unit 5 lesson explains why).
Data skills
- E°cell = E°(positive) - E°(negative).
- Percentage difference = (expected - measured) / expected x 100.
- Voltmeter resolution (often 0.01 V) limits precision.
Worked example
A Cu/Ag cell gives 0.43 V. E°(Ag+/Ag) = +0.80 V and E°(Cu2+/Cu) = +0.34 V. Find the expected emf and the percentage difference.
- Expected = 0.80 - 0.34 = 0.46 V
- % difference = (0.46 - 0.43) / 0.46 x 100 = 6.5%
- Likely causes: non-standard concentrations or oxidised electrodes.
Answer: 0.46 V; 6.5%
Check yourself
- E°(Zn2+/Zn) = -0.76 V and E°(Cu2+/Cu) = +0.34 V. Calculate the emf of the zinc-copper cell.
Show answer
1.10 V. 0.34 - (-0.76) = 1.10 V.
- The expected emf of a cell is 0.46 V, but the measured value is 0.43 V. Calculate the percentage difference.
Show answer
6.5 %. (0.46 - 0.43) / 0.46 x 100 = 6.5%.
- Why is potassium nitrate, not potassium chloride, used in the salt bridge for a cell containing a silver half-cell?
- Chloride ions would precipitate silver chloride
- Potassium chloride does not conduct
- Nitrate ions are reduced at the electrode
- Potassium chloride is toxic
Show answer
Chloride ions would precipitate silver chloride. Nitrates are all soluble.
- Why is a high-resistance voltmeter used?
- It gives a larger reading
- It protects the salt bridge
- It is more robust
- It draws almost no current, so it measures the emf
Show answer
It draws almost no current, so it measures the emf. If current flows, the measured voltage falls below the emf.
- Why are the metal electrodes cleaned with emery paper?
- To make them shiny for observation
- To remove the oxide layer so the metal is in contact with the solution
- To make them thinner
- To roughen them so they react faster with air
Show answer
To remove the oxide layer so the metal is in contact with the solution. An oxide coating changes the electrode potential.
- Which electrode is used in a half-cell containing Fe3+ and Fe2+ ions?
- Iron
- Copper
- Zinc
- Platinum
Show answer
Platinum. Platinum is inert and conducts electrons.
- In a Cu/Ag cell (silver positive), the Ag+ solution is diluted. What happens to the measured emf?
- It increases
- It decreases
- It stays the same
- It becomes zero
Show answer
It decreases. Lower [Ag+] makes the silver potential less positive.
Sources
Last updated 8 November 2026. Spotted a mistake? Tell us.