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Electrochemistry - Hydrogen fuel cells

Grade 11IGCSEChemistry

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

🔑Concepts

A hydrogen fuel cell is an electrochemical cell that converts the chemical energy of a fuel, hydrogen (H2H_2), and an oxidizing agent, oxygen (O2O_2), into electricity through a pair of redox reactions.

Unlike a battery, which stores a finite amount of energy, a fuel cell produces electricity continuously as long as fuel (H2H_2) and oxidant (O2O_2) are supplied.

At the anode (negative electrode), oxidation occurs. Hydrogen gas molecules lose electrons to form hydrogen ions (H+H^+).

At the cathode (positive electrode), reduction occurs. Oxygen gas reacts with the hydrogen ions and the electrons arriving from the external circuit to form water (H2OH_2O).

The movement of electrons from the anode to the cathode through an external circuit creates the electric current used to power devices.

The only product of the reaction is water (H2OH_2O), meaning there are no pollutant emissions like carbon dioxide (CO2CO_2) or nitrogen oxides (NOxNO_x) at the point of use.

Major disadvantages include the high cost of catalysts like platinum, the difficulty of storing highly flammable hydrogen gas (H2H_2) at high pressure, and the fact that most hydrogen is currently produced from fossil fuels like methane (CH4CH_4).

📐Formulae

ightarrow 4H^+(aq) + 4e^-$$
ightarrow 2H_2O(l)$$
ightarrow 2H_2O(l)$$

💡Examples

Problem 1:

Calculate the total number of electrons transferred when 2.02.0 mol of hydrogen gas (H2H_2) is completely reacted in a fuel cell.

Solution:

From the anode half-equation, 22 moles of H2H_2 produce 44 moles of electrons (ee^-). The total number of electrons is 4imes6.02imes1023=2.408imes10244 imes 6.02 imes 10^{23} = 2.408 imes 10^{24} electrons.

Explanation:

The oxidation half-equation is 2H2ightarrow4H++4e2H_2 ightarrow 4H^+ + 4e^-. This shows a 1:21:2 molar ratio between H2H_2 and ee^-. Therefore, 2.02.0 moles of H2H_2 release 4.04.0 moles of electrons. Using Avogadro's constant (L=6.02imes1023L = 6.02 imes 10^{23}), we find the total particle count.

Problem 2:

State the change in the oxidation state of oxygen in a hydrogen fuel cell reaction.

Solution:

The oxidation state of oxygen changes from 00 in O2(g)O_2(g) to 2-2 in H2O(l)H_2O(l).

Explanation:

In its elemental form (O2O_2), the oxidation state of oxygen is 00. In the product water (H2OH_2O), oxygen is more electronegative than hydrogen and takes an oxidation state of 2-2. Because the oxidation state decreases, oxygen undergoes reduction.

Hydrogen fuel cells - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry