I've been reading papers on Phosphided metals (CoP, NiP, FeP) as a catalyst for electrolysis. They increase the current density of the cell for a voltage. If it still requires 39.4 kilowatt-hours per kilogram of hydrogen (https://en.wikipedia.org/wiki/Electrolysis_of_water), why does it matter how fast the reaction occurs?

For example, with no catalyst, an electrolyzer runs at 0.6A/cm2 at 2.3V or with a catalyst it runs at 1A/cm2 at 1.4V. Both cases require about the same amount of power and should produce the same amount of hydrogen, correct?

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    $\begingroup$ The two examples you provide do not produce the same amount of hydrogen. Electrons are "stoichometric reagents/products" which can be indirectly measured in coloumbs ($\mathrm{1\ mol\ e^- =96485\ C}$). A higher current means more starting material is being consumed (or more product is being formed) for a given amount of time (recall that $\mathrm{1\ A=1\ C/s}$). This means that your second example would theoretically produce hydrogen $\mathrm{1/0.6=1.67}$ times faster. The lower voltage also decreases the chance of electrochemical side-reactions which would also consume/release electrons. $\endgroup$ Dec 8, 2019 at 14:41

1 Answer 1


Catalysts generally decrease the electrode overpotential, which leads to decreasing the needed voltage to perform electrolysis at given current densities toward its minimal theoretical value.

If higher current is possible at the same voltage due the catalyst, the lower voltage is needed for the same current, both leading to better efficiency.

As the energy needed to produce 1 mol of molecular hydrogen is

$$E=2 \cdot U \cdot F$$

where $U$ is the used voltage $F$ is the Faraday's constant cca 96490 C/mol

As the consequence, it has direct impact on efficiency of conversion of electricity to hydrogen for hydrogen based engines or just fuel.


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