Describe how a hydrogen fuel cell produces a potential difference.
How to approach this question
1. What are the two fuels supplied to a hydrogen-oxygen fuel cell, and which electrode does each go to?
2. What happens to the hydrogen at its electrode (oxidation or reduction)? What does this mean in terms of electrons?
3. What happens to the oxygen at its electrode?
4. What is a potential difference (voltage) in terms of electrons?
Full Answer
Hydrogen is supplied to the negative electrode where it is oxidised / loses electrons. Oxygen is supplied to the positive electrode where it is reduced / gains electrons. This flow of electrons from the negative to the positive electrode constitutes an electric current / creates a potential difference.
A hydrogen fuel cell generates a potential difference (voltage) through an electrochemical reaction.
1. **At the negative electrode (anode):** Hydrogen gas (H₂) is supplied. It is oxidised, meaning it loses electrons. The hydrogen molecules are split into protons (H⁺) and electrons (e⁻). The half-equation is: 2H₂ → 4H⁺ + 4e⁻.
2. **Electron Flow:** The electrons cannot pass through the electrolyte, so they travel through an external circuit from the negative electrode to the positive electrode. This flow of electrons is an electric current, which creates the potential difference.
3. **At the positive electrode (cathode):** Oxygen gas (O₂) is supplied. It is reduced, meaning it gains electrons. The oxygen molecules react with the protons (which have travelled through the electrolyte) and the electrons (from the external circuit) to form water. The half-equation is: O₂ + 4H⁺ + 4e⁻ → 2H₂O.
The overall reaction is the combination of hydrogen and oxygen to form water: 2H₂ + O₂ → 2H₂O.
Common mistakes
✗ Confusing which fuel goes to which electrode.
✗ Confusing oxidation and reduction.
✗ Just saying "hydrogen and oxygen react" without mentioning the role of electrons and electrodes.
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