What is the difference between N = Q/nF and m = Z*I*t in electrolysis?

[Shan43]

TL;DR Summary

Both of these equations seem to be revolving around Faraday's first law of electrolysis but would they be equal in the sense of what they are solving for?

N is equal to the number of moles of the electrolyzed species (in this case I am trying to find the amount of hydrogen produced) , Q is charge, n is stoichiometric number of electrons consumed in the electrode reaction (please explain what this means, and F is the Faraday constant. For the second equation, m is the mass of the substance undergoing electrolysis, Z is the electrochemical equivalent of the substance (also please explain what this is), I is current and t is time, I'm assuming in seconds. In other words, which of these equations would be the best when trying to find the amount of hydrogen produced during water electrolysis?

 

[ergospherical]

$n$ is just the number of electrons transferred in the electrode reaction equation (as written). For example, oxidation at the anode for water-electrolysis would have a half-reaction $\mathrm{4OH^{-}} \longrightarrow \mathrm{2H_2 O} + \mathrm{O_2} + \mathrm{4e^{-}}$ and $n=4$. A single mole of electrons has a charge of $F$, so $N$ moles of the reaction as written ("$N \ \mathrm{mol \ rxn}$") results in a charge transfer of $N\cdot nF$.

Meanwhile, the electrochemical equivalent $Z$ is pretty much defined as the ratio of the mass of some species used/evolved in the reaction to the charge transferred through the circuit, $Z \equiv m/Q$.

 

[Shan43]

Thank you for your clarification between the two variables. But which equation would be used to find the amount of hydrogen that will be produced from the electrolyzer? (This is not a homework question, more of a research question)

 

[ergospherical]

Well, at the cathode $2\mathrm{H^+} + \mathrm{2e^-} \longrightarrow \mathrm{H_2}$, i.e. $2$ moles of electrons liberates $1$ mole of $\mathrm{H_2}$ gas. If a current $I$ flows for a time $t$, then the charge through either electrode is $It$ and the number of moles of electrons transferred is $It/F$. Therefore you have $It/(2F)$ moles of $\mathrm{H_2}$.

 

[Shan43]

Well, at the cathode $2\mathrm{H^+} + \mathrm{2e^-} \longrightarrow \mathrm{H_2}$, i.e. $2$ moles of electrons liberates $1$ mole of $\mathrm{H_2}$ gas. If a current $I$ flows for a time $t$, then the charge through either electrode is $It$ and the number of moles of electrons transferred is $It/F$. Therefore you have $It/(2F)$ moles of $\mathrm{H_2}$.

Wow, thank you for the clear and concise explanation of faraday's law! But, just to be sure, the F is multiplied by the 2 because it takes 2 moles of electrons to produce one mole of H₂ gas correct?