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Unformatted text preview: Electrolysis of Water: A Thermodynamic View From the tables of standard electrochemical potentials, the voltage for the reaction H2O H2 + ½ O2 is ‐1.23 volts under standard conditions (note that the voltage is negative in the direction written, i.e. it is uphill). We will approximate the free energy change as ΔG ~ ΔG0 i.e. the free energy change is approximately the same as the standard free energy change (more on this later). We have seen that ΔG0 = ‐n F Δ E 0 where n is the number of electrons transferred, F is the Faraday, which is the number of coulombs per mole, 96,500, and ΔE0 is the standard potential of ‐1.23 volts. Thus ΔG ~ ΔG0 = ‐ 2 (96,500)(‐1.23) = 237.1 KJ (volt coulombs = joules) However we also have ΔG = ΔH – T ΔS and the values for the absolute entropies of each substance are available from tables, (calculated from measured heat capacities) and ΔH0, the standard heat of formation of water is ‐285.8 KJ. The standard heats for the elements H2 and O2 are zero by definition. So, SH2O = 69.9 J/K , SH2 = 130.7 J/K, SO2 = (0.5 x 205.1) = 102.6 J/K. Summing the heats for products, and subtracting the heat for the reactant gives ΔH = 285.8 KJ. Similarly summing the entropies for products and subtracting the reactant entropy gives: S = 163.4 J/K. and ΔG = 285.8 – (298) (163.4) = 285.8 – 48.7 = 237.1 KJ It checks!! Thermo rules! Note that since TΔS is negative, entropy is actually helping this reaction, which is “obvious” since the number of moles of gas is greater on the right side. As the electrolysis proceeds the electrolysis vessel cools and heat spontaneously flows into it from the surroundings. The Reverse Reaction Runs a Fuel Cell For the reaction H2 + O2 H2O the voltage is now +1.23 volts which is equal to electrical energy of ‐237.1 KJ using the same equation shown above. However now TΔS = ‐48.7 KJ is hindering the reaction, and this number of joules of waste heat shows up inside the cell and has to be gotten rid of. Also, the overall efficiency can be calculated as ‐237/‐285 = 83% since ΔG = ΔH – T ΔS = ‐285.8 + 48.7 = ‐237. This high apparent efficiency is not attainable since we are measuring the voltage more or less reversibly under standard conditions. As soon as we add a load, and draw some current, the voltage and actual efficiency drop and more heat is produced inside. Drawing current also changes the concentrations of reagents, and the approximation of ΔG ~ ΔG0 is no longer so good. However the 83% efficiency number is the thermodynamic upper limit and this whole calculation shows the thermodynamic minimum requirements for making hydrogen and oxygen from water. How Much Power does it take to make Hydrogen from Water? We have calculated a free energy input of 237 KJ to electrolyze one mole of water, which weighs 18 g. This process makes one mole or 2 g of H2 . We noted above that joules are equal to volt coulombs. If we want to produce a mole of H2 every hour, (assuming maximum thermodynamic efficiency) it would require 237, 000 joules /3600 sec. or a power expenditure of 66 W for an hour. (coulombs/sec = amps, volt x amp = watts) This doesn’t seem so effective, and on a weight basis it certainly isn’t impressive since hydrogen is so light. On the other hand, a molar volume at and 1 atmosphere pressure) is 24 liters (ideal gas law) which doesn’t seem so bad. This is a volume of about 1 cubic foot. There is of course also 12 liters of oxygen produced during the electrolysis, and although pure oxygen has some market value currently, it would be less if widespread electrolysis of water was being carried out. ...
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 Fall '11
 Rasmussen

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