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    Evaluating the efficiency of electrolysis

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    The amount of product produced by electrolysis is limited by Faraday's law. Therefore, as I understand it, the amount of product produced depends on the amps rather than on the volts (although the minimum voltage for a given reaction plus overvoltage must be met).

    For example, lets say you have a reaction that requires at least 0.3 Volts (0.3 volts includes the necessary overvoltage) to run. Instead of using 0.3 Volts you decide to double the voltage to 0.6 Volts. Volts X Amps = Watts, and Watts can be converted directly into Joules. Therefore, according to my calculations, it would require twice as much energy (IN JOULES!!!) to produce the SAME AMOUNT of product at 0.6 Volts than at 0.3 Volts EVEN THOUGH YOU HAVE DOUBLED YOUR ENERGY INPUT.

    I want to know if this is really true. In other words, does increasing the volts beyond what is absolutely necessary really cause an increase in inefficiency (more energy input for the same amount of product)?

    Please do not answer unless you are sure that you understand my question and know the correct answer based on experience.

    Please understand I am NOT asking how to calculate the amount of current required to perform a given reaction. I am just asking about the efficiency of electrolysis IN JOULES, NOT CURRENT!!!

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    Solution Preview

    According to Faraday's law,

    amount of substance (mols) deposited is proportional to current * time

    (leave z and F as they are constants)

    When you use 0.3v, let I1 be the current flowing through the circuit and A1 be the amount deposited and T1 be the time.

    So, A1 proportional to I1 * T1 or take A1 = K*I1*T1 where K is a constant

    when you use, 0.6volts, the current also doulbes according to Ohm's law I = V/R, R is a constant ...

    Solution Summary

    The solution provides and easy to follow explanation of the concepts and gives all mathematical steps wherever necessary.