Hydrogen by electrolysis

Electrolysis is the decomposition of a substance into simpler substances by passing an electric current through it. Three terms describe the setup:

• Electrolyte — the liquid being decomposed, which also conducts the current. Pure water barely conducts, so a little dilute sulphuric acid ($\text{H}_2\text{SO}_4$) or dilute hydrochloric acid is added; this "acidified water" is the electrolyte.
• Electrode — a metal or graphite rod dipped into the electrolyte and connected to the battery. The rod joined to the positive terminal is the anode; the rod joined to the negative terminal is the cathode.
• Electrolysis — the overall process: current flowing through the electrolyte splits it apart, and gas bubbles rise from each electrode.

When current flows through acidified water, hydrogen gas collects at the cathode and oxygen gas collects at the anode. In words:

where: the arrow means "decomposes into under the action of electric current" — this is not a spontaneous reaction; it needs continuous electrical energy input. With formulas and state symbols, the balanced equation is:

where: $\text{H}_2\text{O}(l)$ is liquid water (the electrolyte, acidified), $\text{H}_2(g)$ is hydrogen gas collecting at the cathode, and $\text{O}_2(g)$ is oxygen gas collecting at the anode. The coefficients 2 : 2 : 1 tell you two water molecules yield two hydrogen molecules and one oxygen molecule — so hydrogen and oxygen are always produced in a 2 : 1 ratio by volume, whatever the scale of the experiment.

The goal is to show, using only the balanced equation and the idea that equal volumes of any gas at the same temperature and pressure contain equal numbers of molecules, that the volume of hydrogen collected must always be exactly twice the volume of oxygen collected, no matter how long the experiment runs or how large the apparatus is.

Step 1. Start from the balanced equation, which fixes the number ratio of molecules produced: $2\text{H}_2\text{O}(l) \to 2\text{H}_2(g) + \text{O}_2(g)$. For every 2 molecules of hydrogen gas formed, exactly 1 molecule of oxygen gas is formed — this is a fixed counting fact from the equation, true regardless of how much water is electrolysed.

Step 2. Bring in the rule that equal volumes of gases, measured at the same temperature and pressure, contain equal numbers of molecules. Both gases here are collected in the same apparatus, at the same time, at essentially the same temperature and pressure (the two test tubes sit side by side in the same beaker). So the volume of each gas is directly proportional to the number of molecules of that gas.

Step 3. Combine the two facts. Since molecules of $\text{H}_2$ and $\text{O}_2$ are produced in a fixed 2 : 1 number ratio, and volume is proportional to molecule number under matched conditions, the volumes collected must also be in a 2 : 1 ratio: $V(\text{H}_2) : V(\text{O}_2) = 2 : 1$.

Step 4. Check this against the experiment. If the cathode tube collects a volume $V$, the anode tube collects $V/2$ in the same time — exactly what is observed. This is why a student can identify which gas is which simply by comparing tube volumes, without testing either gas: the tube with twice the volume is hydrogen.

Question: In an electrolysis set-up, a student collects gas in two inverted test tubes over the electrodes. After 10 minutes, the tube over one electrode holds $30\ \text{cm}^3$ of gas and the tube over the other holds $15\ \text{cm}^3$. Identify each gas, state which electrode it collects over, and verify the volume ratio.

Step 1. Compare the two volumes: $30\ \text{cm}^3$ and $15\ \text{cm}^3$. Their ratio is $30 : 15 = 2 : 1$.

Step 2. From the balanced equation, hydrogen and oxygen are produced in a $2 : 1$ ratio by volume, with hydrogen being the larger amount. So the $30\ \text{cm}^3$ tube contains hydrogen and the $15\ \text{cm}^3$ tube contains oxygen.

Step 3. Match each gas to its electrode: hydrogen collects at the electrode joined to the negative terminal, the cathode; oxygen collects at the electrode joined to the positive terminal, the anode.

Step 4. Verify: $V(\text{H}_2)/V(\text{O}_2) = 30/15 = 2$, confirming the $2 : 1$ ratio predicted by $2\text{H}_2\text{O} \to 2\text{H}_2 + \text{O}_2$. Answer: $30\ \text{cm}^3$ of hydrogen collects at the cathode and $15\ \text{cm}^3$ of oxygen collects at the anode, in the expected 2:1 ratio.

Competitive stretch

If the same experiment ran twice as long, both volumes would double to $60\ \text{cm}^3$ hydrogen and $30\ \text{cm}^3$ oxygen — the ratio stays $2 : 1$ because it comes from the fixed molecule count in the equation, not from the time elapsed; only the absolute amounts scale with time.

• Thinking the acid added is "used up" as a reactant — the acid only helps the water conduct; the decomposition equation shows only water breaking into hydrogen and oxygen, not the acid.
• Mixing up cathode and anode — hydrogen collects at the cathode (negative electrode), oxygen at the anode (positive electrode).
• Assuming equal volumes of hydrogen and oxygen are produced instead of the correct 2:1 ratio.
• Trying to explain the ratio through the detailed ionic mechanism (movement and discharge of ions) — that is beyond Class 8; the ratio is explained here purely from the balanced equation and matched gas volumes, which is all the syllabus expects.

Ask the tutor: Want to see how this 2:1 ratio connects to the laboratory method of preparing hydrogen from metals and acids, or try a few more identify-the-gas questions?