What is the relationship between current, voltage and hydrogen quantity? And how are these quantities related to the efficiency of electrolysis?<\/p>\n\n\n\n
A simple way to determine the amount of hydrogen produced during electrolysis is to apply Faraday's laws. These generally describe the relationship between electric charge and material conversion in electrochemical reactions such as electrolysis.<\/p>\n\n\n\n
The charge required to split water corresponds to the multiplication of the charge number z, the amount of substance n in moles, and the Faraday constant F:<\/p>\n\n\n\n
Q = n \u00d7 z \u00d7 F<\/em><\/p>\n\n\n\n If the charge Q is now replaced by current I per time t, the following formula can be used after rearrangement:<\/p>\n\n\n\n n = ( I \u00d7 t ) \/ ( z \u00d7 F )<\/em><\/p>\n\n\n\n Now all missing values and constants can be substituted, giving:<\/p>\n\n\n\n ( 1 A \u00d7 1 s ) \/ ( 2 \u00d7 96,485 As\/mol ) = 5.182 \u00d7 10\u207b\u2076-6<\/sup> mol\/As<\/em><\/p>\n\n\n\n So per ampere-hour, a quantity of 0.418 liters of hydrogen is produced:<\/p>\n\n\n\n 5,182 * 10-6<\/sup> mol\/As \u00d7 22.414 Nl\/mol \u00d7 3600 s\/h = 0.418 Nl\/Ah<\/em><\/p>\n\n\n\n For one kilogram of hydrogen, approximately 4,650 Ah are therefore required.<\/p>\n\n\n\n ( 1000 g \/ 0.0899 g\/l ) \u00d7 0.418 Nl\/Ah = 4,649.61 Ah<\/em><\/p>\n\n\n\n The calculation here applies to one cell of a stack. Theoretically, a minimum thermodynamic cell voltage of 1.23 volts must be applied per cell. In practice, a higher voltage is often required. In PEM electrolysis this is typically between 1.7 V and 2.2 V per cell. Simplified, this overvoltage can be changed\/caused by four factors:<\/p>\n\n\n\n Unfortunately, the additional overvoltage energy cannot contribute to material conversion, as it is lost as heat \u2014 and therefore the efficiency of the electrolysis cell decreases.<\/p>\n\n\n\n The amount of charge that has flowed through the cell <\/strong>(e.g. in ampere-hours) is proportional to the amount of hydrogen produced, largely independent of efficiency. The voltage applied to the cell <\/strong>(Volt), and thus also the energy consumed <\/strong>(e.g. in kWh) depends on various factors, which ultimately determines the efficiency of the electrolysis cell.<\/strong><\/p>\n\n\n\n The system efficiency of an industrial electrolysis plant is even lower than the pure stack efficiency, since the losses from additional system components (e.g. pumps, cooling\/heating, water treatment, instrumentation and control) must also be taken into account.<\/p>\n\n\n\n Note: The charge number of atomic hydrogen is 1; however, two H atoms are always needed to form one hydrogen molecule (H\u2082). Therefore, theoretically z = 1 should be used and the entire equation then divided by two. This step is performed directly here for simplification.<\/p>\n\n\n\nVoltage<\/h2>\n\n\n\n
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Conclusion<\/h2>\n\n\n\n