Is Twice as Much Hydrogen Produced as Chlorine in Brine Electrolysis?

No: in the ideal electrolysis of concentrated brine, hydrogen and chlorine form in equal numbers of molecules. That means equal mole amounts and equal ideal-gas volumes when measured at the same temperature and pressure. The often-remembered 2:1 ratio is for hydrogen compared with oxygen during water electrolysis, not hydrogen compared with chlorine.

The half-reactions explain the ratio

At the cathode, water gains electrons and produces hydrogen gas:

$2\mathrm{H_2O} + 2e^- \rightarrow \mathrm{H_2} + 2\mathrm{OH^-}$

At the anode, chloride ions lose electrons and produce chlorine gas:

$2\mathrm{Cl^-} \rightarrow \mathrm{Cl_2} + 2e^-$

Each reaction uses or produces two electrons for every one molecule of gas. The two electrons released at the anode can therefore support the production of one hydrogen molecule at the cathode. Adding the half-reactions cancels the electrons:

$2\mathrm{Cl^-} + 2\mathrm{H_2O} \rightarrow \mathrm{Cl_2} + \mathrm{H_2} + 2\mathrm{OH^-}$

Sodium ions in the brine are spectator ions in this equation. Including them gives the overall chlor-alkali reaction:

$2\mathrm{NaCl} + 2\mathrm{H_2O} \rightarrow \mathrm{Cl_2} + \mathrm{H_2} + 2\mathrm{NaOH}$

The coefficients for hydrogen and chlorine are both 1. Thus, the ideal amount ratio is $n(\mathrm{H_2}):n(\mathrm{Cl_2})=1:1$.

Worked example using electric charge

Suppose a brine cell passes $19{,}300\ \mathrm{C}$ of charge, and assume the current produces only the stated products. Faraday’s constant is approximately $96{,}485\ \mathrm{C\,mol^{-1}}$, so the amount of electrons transferred is:

$n(e^-) = \frac{19{,}300}{96{,}485} \approx 0.200\ \mathrm{mol}$

Both gas-forming half-reactions require two moles of electrons per mole of gas. Therefore:

$n(\mathrm{H_2}) = n(\mathrm{Cl_2}) = \frac{0.200}{2} = 0.100\ \mathrm{mol}$

At $25^\circ\mathrm{C}$ and $1\ \mathrm{atm}$, an ideal gas has a molar volume of about $24.5\ \mathrm{L\,mol^{-1}}$. Each gas would occupy approximately $0.100\times24.5=2.45\ \mathrm{L}$ under those conditions. The volumes match because the mole amounts match—not because the gases have the same mass.

Why the “twice as much” idea appears

During electrolysis of water, the net reaction is:

$2\mathrm{H_2O} \rightarrow 2\mathrm{H_2} + \mathrm{O_2}$

This reaction makes two moles of hydrogen for every mole of oxygen. If you remember that ratio and apply it to brine, it is easy to mistakenly expect twice as much hydrogen as chlorine. But chlorine formation has a different anode half-reaction from oxygen formation, and the electron balance gives a 1:1 ratio for hydrogen and chlorine.

The word “amount” can also mean different things. Equal moles of hydrogen and chlorine do not mean equal masses. One mole of hydrogen gas has a mass of about $2.016\ \mathrm{g}$, while one mole of chlorine gas has a mass of about $70.90\ \mathrm{g}$. So the hydrogen mass is much smaller, even though the gases are produced in equal mole amounts and have equal ideal volumes under matching conditions.

When the ratio can differ in practice

The 1:1 result applies to the ideal reaction when chloride is oxidized to chlorine and water is reduced to hydrogen. In real cells, side reactions and gas losses can affect what you collect. For example, oxygen can also form at the anode, particularly when chloride concentration or operating conditions favor oxygen evolution. If oxygen is produced, some of the charge goes to making oxygen instead of chlorine; the amount of collected chlorine may then be less than the amount predicted for a chlorine-only anode reaction.

Current efficiency, gas dissolution in the liquid, leaks, and unequal collection conditions can also make measured volumes differ. These effects do not change the balanced equation; they mean the cell did not convert all the charge into the two measured gases in the ideal proportions.

Common mistakes

• Using the water-electrolysis ratio: The 2:1 ratio compares hydrogen with oxygen, not hydrogen with chlorine.
• Confusing mass with gas volume: Equal moles give equal ideal gas volumes at the same temperature and pressure, but hydrogen and chlorine have very different molar masses.
• Ignoring side reactions: The 1:1 prediction assumes chlorine is the anode product and hydrogen is the cathode product. Oxygen evolution or collection losses can change experimental readings.
• Treating brine like molten salt: In aqueous brine, water is reduced at the cathode to make hydrogen; sodium metal is not the cathode product.