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Electrochemistry - Fuel cells

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Fuel cells are galvanic cells that convert the chemical energy produced during the combustion of fuels like H2H_2, CH4CH_4, or CH3OHCH_3OH directly into electrical energy.

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The most common example is the Hydrogen-Oxygen (H2−O2H_2 - O_2) fuel cell used in the Apollo space program. It uses porous carbon electrodes with catalysts like PtPt or PdPd and a concentrated aqueous KOHKOH or NaOHNaOH electrolyte.

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At the Anode (Oxidation): 2H2(g)+4OH−(aq)→4H2O(l)+4e−2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-

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At the Cathode (Reduction): O2(g)+2H2O(l)+4e−→4OH−(aq)O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)

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The overall cell reaction is: 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l). Note that the product is pure water, which is eco-friendly and was used for drinking by astronauts.

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Fuel cells are highly efficient (theoretical efficiency is nearly 100%100\%, practical efficiency is around 60−70%60-70\%) compared to thermal power plants which have an efficiency of about 40%40\%.

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Unlike ordinary batteries, fuel cells do not store energy; they provide electricity as long as the fuel and oxidant are continuously supplied.

📐Formulae

ΔG=−nFEcell\Delta G = -nFE_{cell}

Efficiency (η)=ΔGΔH×100\text{Efficiency } (\eta) = \frac{\Delta G}{\Delta H} \times 100

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Ecell=Ecell∘−2.303RTnFlog⁡QE_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log Q

💡Examples

Problem 1:

Calculate the thermodynamic efficiency of a H2−O2H_2 - O_2 fuel cell if the standard free energy of formation of liquid water ΔGf∘\Delta G_f^\circ is −237.2 kJ/mol-237.2 \text{ kJ/mol} and the standard enthalpy of formation ΔHf∘\Delta H_f^\circ is −285.8 kJ/mol-285.8 \text{ kJ/mol}.

Solution:

Given: ΔG=−237.2 kJ/mol\Delta G = -237.2 \text{ kJ/mol} and ΔH=−285.8 kJ/mol\Delta H = -285.8 \text{ kJ/mol}. Efficiency η=ΔGΔH×100=−237.2 kJ/mol−285.8 kJ/mol×100≈83.0%\eta = \frac{\Delta G}{\Delta H} \times 100 = \frac{-237.2 \text{ kJ/mol}}{-285.8 \text{ kJ/mol}} \times 100 \approx 83.0\%.

Explanation:

The efficiency of a fuel cell is defined as the ratio of the maximum useful work (Gibbs free energy change) to the total heat released during combustion (Enthalpy change).

Problem 2:

For the reaction 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l), what is the value of nn (number of moles of electrons) used in the ΔG=−nFEcell\Delta G = -nFE_{cell} formula?

Solution:

In the oxidation half-reaction: 2H2→4H++4e−2H_2 \rightarrow 4H^+ + 4e^-. Therefore, for the balanced overall reaction as written, n=4n = 4.

Explanation:

Each hydrogen molecule H2H_2 loses 22 electrons to form 2H+2H^+. Since the balanced equation uses 22 moles of H2H_2, a total of 44 moles of electrons are transferred.

Fuel cells - Revision Notes & Key Formulas | ICSE Class 12 Chemistry