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Electrochemistry - Hydrogen fuel cells

Grade 11A LevelChemistry

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

🔑Concepts

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A hydrogen fuel cell is an electrochemical cell that converts the chemical energy of a fuel (hydrogen) and an oxidizing agent (oxygen) into electricity through a pair of redox reactions. Unlike batteries, they do not run down or need recharging as long as fuel and oxygen are supplied.

Diagram of a hydrogen fuel cell showing hydrogen entering the anode side and oxygen entering the cathode side.
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At the anode (negative electrode), hydrogen gas undergoes oxidation. The hydrogen molecules lose electrons to form hydrogen ions: 2H2(g)→4H+(aq)+4e−2H_2(g) \rightarrow 4H^+(aq) + 4e^-. These electrons flow through the external circuit to create an electric current.

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At the cathode (positive electrode), oxygen gas undergoes reduction. Oxygen molecules react with the hydrogen ions migrating through the electrolyte and the electrons arriving from the external circuit: O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l).

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The overall cell reaction is the combustion of hydrogen to form water: 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l). The only byproduct is water, making it a 'clean' energy source with no carbon emissions at the point of use.

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Advantages of fuel cells include high efficiency, silent operation, and lack of pollutants like CO2CO_2 or NOxNO_x. Disadvantages include the difficulty of storing hydrogen gas safely and the high cost of catalysts like platinum.

📐Formulae

2H2(g)→4H+(aq)+4e−2H_2(g) \rightarrow 4H^+(aq) + 4e^-

O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)

2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 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 (e−e^-). The total number of electrons is 4×6.02×1023=2.408×10244 \times 6.02 \times 10^{23} = 2.408 \times 10^{24} electrons.

Explanation:

The oxidation half-equation is 2H2→4H++4e−2H_2 \rightarrow 4H^+ + 4e^-. This shows a 1:21:2 molar ratio between H2H_2 and e−e^-. Therefore, 2.02.0 moles of H2H_2 release 4.04.0 moles of electrons. Using Avogadro's constant (L=6.02×1023L = 6.02 \times 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.

Problem 3:

In a specific hydrogen fuel cell design using an alkaline electrolyte, identify the reaction occurring at the anode and determine the volume of H2(g)H_2(g) required at r.t.p to produce 0.50.5 mol of electrons.

Alkaline hydrogen fuel cell diagram showing KOH electrolyte.

Solution:

  1. In alkaline conditions, the anode reaction is: H2(g)+2OH−(aq)→2H2O(l)+2e−H_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^-
  2. From the stoichiometry, 11 mol of H2H_2 produces 22 mol of electrons.
  3. To produce 0.50.5 mol of electrons: 0.52=0.25\frac{0.5}{2} = 0.25 mol of H2H_2.
  4. Volume at r.t.p: 0.25 mol×24 dm3/mol=6.0 dm30.25 \text{ mol} \times 24 \text{ dm}^3/\text{mol} = 6.0 \text{ dm}^3.

Explanation:

The relationship between moles of gas and moles of electrons is determined by the balanced half-equation. In all hydrogen fuel cells, 11 mole of H2H_2 always releases 22 moles of electrons.

Problem 4:

During the operation of the fuel cell shown, 3232 g of oxygen gas is consumed at the cathode. Calculate the mass of water produced during this process.

PEM fuel cell diagram highlighting the consumption of oxygen at the right electrode.

Solution:

  1. Molar mass of O2=32O_2 = 32 g/mol. Moles of O2O_2 used: 3232=1.0\frac{32}{32} = 1.0 mol.
  2. From the overall equation 2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O, the ratio of O2O_2 to H2OH_2O is 1:21:2.
  3. Moles of H2OH_2O produced: 1.0×2=2.01.0 \times 2 = 2.0 mol.
  4. Mass of H2OH_2O: 2.0 mol×18 g/mol=36 g2.0 \text{ mol} \times 18 \text{ g/mol} = 36 \text{ g}.

Explanation:

Using the balanced overall equation for the fuel cell, we apply stoichiometry to relate the mass of the reactant oxygen to the mass of the product water.

Hydrogen fuel cells Grade 11 Notes & Examples