krit.club logo

Electrochemistry - Hydrogen fuel cells

Grade 12IGCSEChemistry

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 (H2H_2) and an oxidizing agent (O2O_2) into electricity through a pair of redox reactions.

•

Unlike traditional batteries, fuel cells do not run down or need recharging; they produce electricity as long as fuel and oxygen are supplied.

•

In an alkaline fuel cell, the electrolyte used is typically hot potassium hydroxide (KOHKOH). At the anode, hydrogen gas reacts with hydroxide ions: 2H2(g)+4OH−(aq)→4H2O(l)+4e−2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-.

•

At the cathode of an alkaline fuel cell, oxygen gas reacts with water and electrons to produce hydroxide ions: O2(g)+2H2O(l)+4e−→4OH−(aq)O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq).

•

In a Proton Exchange Membrane (PEM) or acidic fuel cell, H+H^+ ions move across the membrane. Anode reaction: H2(g)→2H+(aq)+2e−H_2(g) \rightarrow 2H^+(aq) + 2e^-; Cathode reaction: 12O2(g)+2H+(aq)+2e−→H2O(l)\frac{1}{2}O_2(g) + 2H^+(aq) + 2e^- \rightarrow H_2O(l).

•

The only chemical byproduct of the reaction is water (H2OH_2O), making it an environmentally friendly 'zero-emission' energy source at the point of use.

•

Advantages include high efficiency compared to internal combustion engines and quiet operation. Disadvantages include the difficulty of storing H2H_2 gas and the high cost of catalysts like Platinum (PtPt).

📐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.