Mobility is of vital importance for our modern society. Electric cars are under active development t — Physical Chemistry — Kinetics Chemistry Question
Electrochemical Energy Conversion
Mobility is of vital importance for our modern society. Electric cars are under active development to ensure our future needs of transportation. One of the major problems for electrically driven vehicles is a supply of a suitable source of electricity. Batteries have the drawback that they must be recharged, thus the action radius is limited. The in situ generation of electricity in fuel cells is an attractive alternative. A fuel battery, or flow battery, is a galvanic cell for which the reactants are continuously supplied. Fuel cells utilize combustion reactions to produce electricity. The reactants undergo half-reactions at the electrodes, and the electrons are transferred through an external circuit. The electrodes are separated by an ionically conducting liquid or a molten or solid electrolyte.
The electrode half-reactions for a hydrogen-oxygen fuel cell with a concentrated potassium hydroxide electrolyte are:
O2(g) + 2 H2O + 4 e– → 4 OH– (aq) (1)
H2(g) + 2 OH– (aq) → 2 H2O(l) + 2 e– (2)
The fuel-cell reaction, after making electron loss equal to electron gain, is:
2 H2(g) + O2(g) → 2 H2O(l) (3)
The reaction product is water! and the efficiency is about 50 – 60 %.
Which reaction occurs at the cathode?
Model Answer
At the cathode oxygen is reduced to hydroxide, i.e. half-reaction (1).
Which reaction occurs at the anode?
Model Answer
At the anode hydrogen is oxidized to water, i.e. reaction (2).
Give the electrode reactions when the electrolyte is phosphoric acid.
Model Answer
Anode: 2 H2(g) → 4 H+ + 4 e–
Cathode: 4 H+ + O2(g) + 4 e– → 2 H2O(g)
Fuel cell reaction: 2 H2(g) + O2(g) → 2 H2O(g)
The change of Gibbs energy G 0 is a measure of the driving force of a reaction. The change of energy is given by:
G 0 = – n F E
where n is the number of electrons transferred in the reaction and F is the Faraday constant (96 487 C mol -1). The standard electrode potential for O2(g) at 25 °C is +1.23 V.
Calculate the G 0 of the fuel-cell reaction under acidic conditions (see 21.3).
Model Answer
The standard electrode potential of the reaction at the anode = 0 V
The standard electrode potential of the reaction at the cathode = + 1.23 V
The total number of electrons transferred in the reaction = 4
ΔG o = – n F E = – 4 × 96487 × (1.23 V – 0 V) = – 474.716 J
The production of usable energy by combustion of fuels is an extremely inefficient process. In The Netherlands natural gas is a highly attractive energy source as it is abundantly available. Modern electric power plants are able to furnish only 35-40 % of the energy theoretically available from natural gas. The exothermic reaction of natural gas (methane) with oxygen is:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) + energy
Usually the energy released from this reaction is used indirectly to heat houses or to run machines. However, in a high-temperature ceramic fuel cell based on a solid oxide-ion conducting electrolyte, natural gas can be utilized directly, without a catalyst and with a high efficiency of conversion (75 %). The net fuel-cell reaction is:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Give the reactions at the anode and the cathode.
Model Answer
Anode: CH4(g) + 2 O2– (electrolyte) → 2 H2O(g) + CO2 + 4e–
Cathode: O2(g) + 4 e– → 2 O2– (electrolyte)
Net reaction: CH4(g) + 2 O2(g) → 2 H2O(g) + CO2(g)
Another high-temperature fuel cell utilizes molten Carbonate as the ionically conducting electrolyte. Hydrogen is used as fuel, oxygen is mixed with CO2.
Give the half-reactions at the anode and cathode, and the net fuel-cell reaction.
Model Answer
Anode: 2 H2(g) + 2 CO32– (l) → 2 H2O(g) + 2 CO2(g) + 4 e–
Cathode: O2(g) + 2 CO2(g) + 4 e– → 2 CO32– (l) (cathode)
Fuel cell reaction: 2 H2(g) + O2(g) → 2 H2O(g)