krit.club logo

Electrochemistry - Fuel cells

Grade 12ICSEChemistry

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

πŸ”‘Concepts

β€’

Fuel cells are electrochemical cells that convert the chemical energy of a fuel, such as hydrogen or methane, directly into electrical energy. Unlike conventional cells, the reactants are continuously supplied to the electrodes from an external source.

A Hydrogen-Oxygen fuel cell schematic showing porous carbon electrodes and an alkaline electrolyte.
β€’

In a H2βˆ’O2H_2 - O_2 fuel cell, the anode reaction involves the oxidation of hydrogen: 2H2(g)+4OHβˆ’(aq)β†’4H2O(l)+4eβˆ’2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^-

β€’

The cathode reaction in a H2βˆ’O2H_2 - O_2 fuel cell involves the reduction of oxygen: O2(g)+2H2O(l)+4eβˆ’β†’4OHβˆ’(aq)O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) resulting in the overall cell reaction: 2H2(g)+O2(g)β†’2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l) with a theoretical cell potential of 1.23V1.23V.

β€’

Fuel cells are preferred over thermal power plants because they are highly efficient (theoretical efficiency up to 70%βˆ’100%70\% - 100\%) and pollution-free, as the only byproduct is water.

πŸ“Formulae

Ξ”G=βˆ’nFEcell\Delta G = -nFE_{cell}

EfficiencyΒ (Ξ·)=Ξ”GΞ”HΓ—100\text{Efficiency } (\eta) = \frac{\Delta G}{\Delta H} \times 100

Ξ”G=Ξ”Hβˆ’TΞ”S\Delta G = \Delta H - T\Delta S

Ecell=Ecellβˆ˜βˆ’2.303RTnFlog⁑QE_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log Q

πŸ’‘Examples

Problem 1:

Calculate the thermodynamic efficiency of a H2βˆ’O2H_2 - O_2 fuel cell if the standard free energy of formation of liquid water Ξ”Gf∘\Delta G_f^\circ is βˆ’237.2Β kJ/mol-237.2 \text{ kJ/mol} and the standard enthalpy of formation Ξ”Hf∘\Delta H_f^\circ is βˆ’285.8Β kJ/mol-285.8 \text{ kJ/mol}.

Solution:

Given: Ξ”G=βˆ’237.2Β kJ/mol\Delta G = -237.2 \text{ kJ/mol} and Ξ”H=βˆ’285.8Β kJ/mol\Delta H = -285.8 \text{ kJ/mol}. Efficiency Ξ·=Ξ”GΞ”HΓ—100=βˆ’237.2Β kJ/molβˆ’285.8Β kJ/molΓ—100β‰ˆ83.0%\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)β†’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 Ξ”G=βˆ’nFEcell\Delta G = -nFE_{cell} formula?

Solution:

In the oxidation half-reaction: 2H2β†’4H++4eβˆ’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.

Problem 3:

Identify the species being oxidized and reduced in the H2βˆ’O2H_2 - O_2 fuel cell shown, and calculate the total number of electrons (nn) transferred for the production of one mole of liquid water.

Fuel cell diagram highlighting the inlet of H2 at the anode and O2 at the cathode.

Solution:

The half-cell reactions are: Anode: H2+2OHβˆ’β†’2H2O+2eβˆ’H_2 + 2OH^- \rightarrow 2H_2O + 2e^- (Oxidation) Cathode: 12O2+H2O+2eβˆ’β†’2OHβˆ’\frac{1}{2}O_2 + H_2O + 2e^- \rightarrow 2OH^- (Reduction) For the formation of 11 mole of H2OH_2O, the net reaction is H2+12O2β†’H2OH_2 + \frac{1}{2}O_2 \rightarrow H_2O. Summing the electrons transferred in this stoichiometric ratio, we find n=2n = 2.

Explanation:

Hydrogen gas loses electrons at the anode (oxidation state changes from 00 to +1+1), while Oxygen gas gains electrons at the cathode (oxidation state changes from 00 to βˆ’2-2).

Problem 4:

A fuel cell operates using the reaction CH4(g)+2O2(g)β†’CO2(g)+2H2O(l)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l). If the cell is set up with methane at the left electrode and oxygen at the right electrode in an acidic electrolyte, identify the anode reaction.

Methane fuel cell diagram with phosphoric acid electrolyte.

Solution:

In an acidic medium, methane is oxidized at the anode as follows: CH4(g)+2H2O(l)β†’CO2(g)+8H+(aq)+8eβˆ’CH_4(g) + 2H_2O(l) \rightarrow CO_2(g) + 8H^+(aq) + 8e^-

Explanation:

The fuel (methane) is always fed to the anode where oxidation occurs. Since the carbon in CH4CH_4 is in the βˆ’4-4 oxidation state and in CO2CO_2 it is +4+4, a total of 88 electrons are released per molecule of methane.