Half cell reactions for oxidation of water by acidified solution of potassium dichromate

From Chemguide:

I don't understand the first equation. The chromium atoms in the dichromate ion receive 6 electrons, and their oxidation state declines from 6+ to 3+. But where do the 14 hydrogen cations get their 14 electrons in order to form water with the seven oxygens?

Look closely at the oxidation state of all atoms in the equation. $$\ce{Cr2O7^2- (aq) + 14 H+ (aq) + 6e- <=> 2Cr^3+ (aq) + 7 H2O (l)}$$
In the other equation $$\ce{O2 (g) + 4H+ (aq) +4e- <=> 2H2O (l)}$$
The total equation for this process would therefore be $$\ce{Cr2O7^2- (aq) + 14 H+ (aq) <=> 4Cr^3+ (aq) + 8H2O (l) + 3O2}.$$