Is it the transfer current density of the product multiplied by the overpotential?


Faraday’s law is usually summarized as

$$m=\frac{M\cdot Q}{z\cdot F}\tag{1}$$

$m$ is mass,
$M$ is molar mass,
$Q$ is electric charge,
$z$ is charge number, and
$F=96\,485.332\,89(59)\ \mathrm{C\ mol^{-1}}$ is the Faraday constant.

It essentially consists of two parts:

$$Q=n\cdot z\cdot F\tag{2}$$

and the definition of molar mass $M$

$$m=M\cdot n\tag{3}$$

Since you want to calculate values per amount of substance $n$ and not per mass $m$, you only need equation $\text(2)$.

In case of electrolysis at a constant current, the electric charge $Q$ is given by

$$Q=I\cdot t\tag{4}$$

$I$ is electric current, and
$t$ is time.

The power $P$ is given by $$P=U\cdot I\tag{5}$$ where
$U$ is voltage.

Inserting $\text(4)$ into $\text(5)$ yields $$P=U\cdot\frac Qt\tag{6}$$

And inserting $\text(2)$ into $\text(6)$ yields $$P=U\cdot\frac nt\cdot z\cdot F\tag{7}$$

Now you can see why your question doesn’t make sense. The power $P$ (e.g. in watt) does not correspond to the deposited amount of substance $n$ (e.g. in mol); it actually corresponds to a deposition rate $n/t$ (e.g. in mol per second).

However, you could calculate the energy $E$ (e.g. in joule, watt seconds, or kilowatt hours) per amount of substance $n$ since

$$P=\frac Et\tag{8}$$

and thus

$$\begin{align} E&=P\cdot t\\[6pt] &=U\cdot n\cdot z\cdot F\tag{9} \end{align}$$


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