Step 1: Write the overall decomposition of water. \[ 2\text{H}_2\text{O} \xrightarrow{\text{electricity}} 2\text{H}_2 + \text{O}_2 \] Reading the coefficients directly tells us how many molecules, and therefore how many volumes, of each gas are produced.
Step 2: Read off the mole ratio.
For every $2$ molecules of water broken down, we collect $2$ molecules of hydrogen gas but only $1$ molecule of oxygen gas, a ratio of $2:1$.
Step 3: Convert mole ratio to volume ratio.
Since equal moles of gas occupy equal volumes under the same conditions, this $2:1$ mole ratio becomes a $2:1$ volume ratio directly, meaning hydrogen collects twice as fast as oxygen.
Step 4: Assign each gas to its electrode.
Positive hydrogen ions are attracted to and reduced at the negative cathode, while oxygen is released at the positive anode, so the cathode collects the larger, double share and the anode collects the smaller, single share. \[ \boxed{\text{Cathode : 02-part hydrogen gas; Anode : 01-part oxygen gas}} \]