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Electrochemistry - Hydrogen fuel cells

Grade 12A 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, fuel cells require a continuous source of fuel and oxygen to sustain the chemical reaction.

Diagram of an alkaline hydrogen fuel cell showing porous electrodes and a central electrolyte.
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In an alkaline fuel cell, the electrolyte is typically an aqueous solution of potassium hydroxide (KOHKOH). At the anode (negative electrode), hydrogen gas reacts with hydroxide ions: 2H2(g)+4OH−(aq)→4H2O(l)+4e−2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-

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At the cathode (positive electrode) of an alkaline fuel cell, oxygen gas reacts with water and electrons to form hydroxide ions: O2(g)+2H2O(l)+4e−→4OH−(aq)O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) This maintains the concentration of the electrolyte while producing electrical work.

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The main advantages of hydrogen fuel cells include high efficiency compared to internal combustion engines and the fact that the only byproduct at the point of use is water (H2OH_2O), making them 'zero-emission' if the hydrogen is produced sustainably.

📐Formulae

2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l) (Overall Reaction)

Ecellθ=Ereductionθ−EoxidationθE_{cell}^{\theta} = E_{reduction}^{\theta} - E_{oxidation}^{\theta}

ΔGθ=−nFEcellθ\Delta G^{\theta} = -nFE_{cell}^{\theta}

H2(g)→2H+(aq)+2e−H_2(g) \rightarrow 2H^+(aq) + 2e^- (Anode in Acidic Conditions)

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

💡Examples

Problem 1:

Calculate the standard cell potential (EcellθE_{cell}^{\theta}) for a hydrogen-oxygen fuel cell operating under acidic conditions, given the standard reduction potentials: Eθ(O2/H2O)=+1.23 VE^{\theta}(O_2/H_2O) = +1.23\,V and Eθ(H+/H2)=0.00 VE^{\theta}(H^+/H_2) = 0.00\,V.

Solution:

Ecellθ=Ecathodeθ−Eanodeθ=1.23 V−0.00 V=+1.23 VE_{cell}^{\theta} = E_{cathode}^{\theta} - E_{anode}^{\theta} = 1.23\,V - 0.00\,V = +1.23\,V

Explanation:

In a fuel cell, oxygen is reduced at the cathode and hydrogen is oxidized at the anode. The standard electrode potential for the oxygen electrode is +1.23 V+1.23\,V and for the hydrogen electrode (SHE) is 0.00 V0.00\,V. Subtracting the anode potential from the cathode potential gives the total cell voltage.

Problem 2:

Write the overall balanced equation for an alkaline fuel cell and identify the species being oxidized.

Solution:

Overall equation: 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l). The species being oxidized is H2(g)H_2(g).

Explanation:

During the reaction, the oxidation state of hydrogen increases from 00 in H2H_2 to +1+1 in H2OH_2O. Since oxidation is the loss of electrons (or increase in oxidation state), hydrogen is the species being oxidized at the anode.

Problem 3:

Identify the direction of electron flow and the ions moving through the electrolyte in the alkaline fuel cell shown. Write the half-equation occurring at the cathode.

Hydrogen fuel cell diagram labeled with H2 and O2 inlets and KOH electrolyte.

Solution:

  1. Electrons flow from the Anode (left) to the Cathode (right) through the external circuit.
  2. In the electrolyte (KOHKOH), hydroxide ions (OH−OH^-) move from the cathode to the anode.
  3. Cathode half-equation: O2(g)+2H2O(l)+4e−→4OH−(aq)O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)

Explanation:

In any electrochemical cell, oxidation occurs at the anode, releasing electrons which travel through the external wire to the cathode. In alkaline fuel cells, the charge carrier in the electrolyte is the OH−OH^- ion, which migrates toward the anode to react with H2H_2.

Problem 4:

Calculate the mass of water produced when 4.04.0 g of hydrogen gas is completely reacted in a hydrogen-oxygen fuel cell.

Proton Exchange Membrane fuel cell diagram used for stoichiometric calculations.

Solution:

  1. The overall reaction is: 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l)
  2. Calculate moles of H2H_2: n(H2)=massMr=4.02.0=2.0 moln(H_2) = \frac{mass}{M_r} = \frac{4.0}{2.0} = 2.0\text{ mol}
  3. From the stoichiometry, 22 moles of H2H_2 produce 22 moles of H2OH_2O. So, n(H2O)=2.0 moln(H_2O) = 2.0\text{ mol}.
  4. Calculate mass of H2OH_2O: mass=n×Mr=2.0×18.0=36.0 gmass = n \times M_r = 2.0 \times 18.0 = 36.0\text{ g}

Explanation:

The overall stoichiometry of the fuel cell is identical to the combustion of hydrogen. Using the molar mass of hydrogen (2.0 g/mol2.0\text{ g/mol}) and water (18.0 g/mol18.0\text{ g/mol}), we find that 4.0 g4.0\text{ g} of hydrogen yields 36.0 g36.0\text{ g} of water.

Hydrogen fuel cells Grade 12 Notes & Examples