Cell emf, cell diagrams and fuel cells (Core Practical 12)

Cell emf (spec 16.7)

E°cell = E°(positive electrode) - E°(negative electrode), or E°(right) - E°(left) for a cell diagram.

The half-cell with the more positive E° is the positive electrode, where reduction happens. Electrons flow through the wire from the negative electrode (oxidation) to the positive electrode. For Zn and Cu: E°cell = +0.34 - (-0.76) = +1.10 V.

Cell diagrams (spec 16.8)

Write the oxidation (negative) half-cell on the left and the reduction half-cell on the right. A single vertical line is a phase boundary; a double line is the salt bridge; a comma separates species in the same phase. Write each half-cell so the species change in the direction of the reaction, with the electrodes on the outside:

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Pt(s) | H2(g) | H+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)

Core Practical 12 and non-standard conditions (spec 16.6, 16.9)

Build half-cells (for example Cu/Cu2+, Zn/Zn2+, Ag/Ag+, Fe3+/Fe2+ with Pt), link them with filter paper soaked in potassium nitrate and read the emf on a digital voltmeter. A Cu/Zn cell gives about 1.10 V. Electrode potentials depend on concentration and temperature, so conditions must be stated. Lowering [Ag+] shifts Ag+ + e- ⇌ Ag to the left, making E for the silver half-cell less positive, so the emf of a Cu/Ag cell falls.

Fuel cells (spec 16.18, 16.19)

A fuel cell turns the energy of a fuel reacting with oxygen directly into a voltage. In the hydrogen-oxygen cell:

• Acidic electrolyte: negative electrode H2 -> 2H+ + 2e-; positive electrode O2 + 4H+ + 4e- -> 2H2O
• Alkaline electrolyte: negative electrode H2 + 2OH- -> 2H2O + 2e-; positive electrode O2 + 2H2O + 4e- -> 4OH-

Overall 2H2 + O2 -> 2H2O in each case. Methanol and other hydrogen-rich fuels can be used: in acid, CH3OH + H2O -> CO2 + 6H+ + 6e- at the negative electrode. Fuel cells produce only water from hydrogen and are more efficient than burning fuel, but hydrogen is hard to store and is often made using fossil fuels.

Worked example

Calculate E°cell for a cell made from Mg/Mg2+ (E° = -2.37 V) and Ag/Ag+ (E° = +0.80 V), and write its cell diagram.

1. Silver is more positive, so it is the positive electrode.
2. E°cell = +0.80 - (-2.37)

Answer: E°cell = +3.17 V; Mg(s) | Mg2+(aq) || Ag+(aq) | Ag(s)

Check yourself

1. Calculate E°cell for a cell made from Zn2+/Zn (E° = -0.76 V) and Fe3+/Fe2+ (E° = +0.77 V).
   Show answer

   +1.53 V. E°cell = +0.77 - (-0.76) = +1.53 V.

2. Calculate E°cell for the cell Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s). (Cu2+/Cu +0.34 V; Ag+/Ag +0.80 V)
   Show answer

   +0.46 V. E°cell = E°right - E°left = +0.80 - (+0.34) = +0.46 V.

3. Which is the correct cell diagram for a cell in which zinc reduces copper(II) ions?
   • Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s)
   • Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
   • Zn2+(aq) | Zn(s) || Cu(s) | Cu2+(aq)
   • Zn(s) || Zn2+(aq) | Cu2+(aq) || Cu(s)
   Show answer

   Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s). Oxidation (Zn -> Zn2+) on the left, reduction on the right, with || for the salt bridge.

4. In a Zn/Cu cell, in which direction do electrons flow in the external circuit?
   • From copper to zinc
   • Through the salt bridge from zinc to copper
   • From zinc to copper
   • They do not flow in the external circuit
   Show answer

   From zinc to copper. Zinc is oxidised and releases electrons, which travel through the wire to the copper electrode.

5. In a Cu/Ag cell, the Ag+ concentration is reduced from 1.00 to 0.010 mol dm^-3. How does the cell emf change?
   • It decreases, because E for the silver half-cell becomes less positive
   • It increases, because more silver dissolves
   • It stays the same
   • It becomes exactly zero
   Show answer

   It decreases, because E for the silver half-cell becomes less positive. Lower [Ag+] shifts Ag+ + e- ⇌ Ag to the left, so silver is a weaker positive electrode.

6. What is the reaction at the negative electrode of a hydrogen-oxygen fuel cell with an alkaline electrolyte?
   • O2 + 2H2O + 4e- -> 4OH-
   • H2 -> 2H+ + 2e-
   • H2 + 2OH- -> 2H2O + 2e-
   • O2 + 4H+ + 4e- -> 2H2O
   Show answer

   H2 + 2OH- -> 2H2O + 2e-. Hydrogen is oxidised; in alkali it combines with OH- to form water.

7. What is the reaction at the positive electrode of a hydrogen-oxygen fuel cell with an acidic electrolyte?
   • O2 + 4H+ + 4e- -> 2H2O
   • H2 -> 2H+ + 2e-
   • O2 + 2H2O + 4e- -> 4OH-
   • 2H2O -> O2 + 4H+ + 4e-
   Show answer

   O2 + 4H+ + 4e- -> 2H2O. Oxygen is reduced at the positive electrode; in acid, H+ is used.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work
• Pearson IAL Chemistry teacher practical guide

Last updated 7 November 2026. Spotted a mistake? Tell us.