# How do you calculate the power (in watts) required per mole of product for or an electrochemical reduction?

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

Faraday’s law is usually summarized as

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

where
$m$ 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

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

where
$I$ 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}