Products of Electrolysis

What decides the products

What actually comes out at the two electrodes during electrolysis depends on

two things: the nature of the material being electrolysed, and the

type of electrode used.

• An inert electrode (platinum or gold, for instance) takes no part in the electrode reaction — it is only a source or sink for electrons.
• A reactive electrode does take part in the reaction, so the same electrolyte can give different products depending on whether the electrode is inert or reactive.

Beyond the electrode itself, when several oxidising or reducing species are

present together in the electrolytic cell, the one that actually reacts is

governed by their standard electrode potentials — the species offering

the thermodynamically more favourable half-reaction is expected to win out.

In practice this simple ranking can be overridden: some electrode reactions,

although thermodynamically feasible, are so sluggish that they barely proceed

at the voltage the potentials would suggest. An extra voltage, called the

overpotential, then has to be applied before that reaction becomes fast

enough to matter — and this can flip which reaction actually dominates.

Electrolysis of molten sodium chloride

Molten (fused) NaCl is the cleanest case, because only one cation and one

anion are present.

• Cathode (reduction): $\text{Na}^+ + e^- \rightarrow \text{Na}$
• Anode (oxidation): $\text{Cl}^- \rightarrow \tfrac{1}{2}\text{Cl}_2 + e^-$

With no competing species around, sodium metal is deposited at the cathode

and chlorine gas is liberated at the anode — exactly as the stoichiometry

suggests.

Electrolysis of aqueous sodium chloride

Once NaCl is dissolved in water rather than melted, the picture changes,

because water itself supplies extra ions ($\text{H}^+$, $\text{OH}^-$) and

molecules that can also react at the electrodes. Now there is a genuine

competition at each electrode.

At the cathode, two reductions are possible:

The reaction with the higher $E^\circ$ is the one that is preferred, so it

is the reduction of $\text{H}^+$ that actually occurs. Since the solution's

$\text{H}^+$ comes from the dissociation of water,

adding this to the reduction step gives the net cathode reaction:

At the anode, two oxidations are possible:

Here the reaction with the lower $E^\circ$ is the one favoured (an anode

reaction is an oxidation, so a lower standard potential means the species

gives up electrons more readily under these conditions). By that reasoning

water should be oxidised in preference to chloride. In practice, though, the

oxidation of water to oxygen suffers from a substantial overpotential, so

it is chloride that is oxidised instead.

Putting the two electrodes together, the net outcome of electrolysing

aqueous NaCl is:

• Dissolution: $\text{NaCl}(aq) \xrightarrow{\text{H}_2\text{O}} \text{Na}^+(aq) + \text{Cl}^-(aq)$
• Cathode: $\text{H}_2\text{O}(l) + e^- \rightarrow \tfrac{1}{2}\text{H}_2(g) + \text{OH}^-(aq)$
• Anode: $\text{Cl}^-(aq) \rightarrow \tfrac{1}{2}\text{Cl}_2(g) + e^-$
• Net: $\text{NaCl}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{Na}^+(aq) + \text{OH}^-(aq) + \tfrac{1}{2}\text{H}_2(g) + \tfrac{1}{2}\text{Cl}_2(g)$

So aqueous NaCl electrolysis gives NaOH, H₂, and Cl₂ — a completely

different outcome from the molten-salt case, purely because water introduces

new competing electrode reactions. (When concentrations rather than standard

states are involved, the plain standard electrode potentials are replaced by

the actual electrode potentials from the Nernst equation, so the comparison

above is really only a first approximation.)

Electrolysis of sulphuric acid

A similar competition decides what happens at the anode when sulphuric acid …