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.
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) + 6e- &lt;=&gt; 2Cr^3+ (aq) + 7 H2O (l)}$$

The oxidation state in chromium changes from +6 to +3. This is what the electrons in the equation are necessary for. Oxygen retains its oxidation state of -2 in both sides of the equation, as well as hydrogen which remains its +1 oxidation state.

In the other equation
$$\ce{O2 (g) + 4H+ (aq) +4e- &lt;=&gt; 2H2O (l)}$$

the oxygen gets reduced by changing its oxidation state form 0 to -2. Again for hydrogen, the oxidation state of +1 is retained. 

The total equation for this process would therefore be
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) &lt;=&gt; 4Cr^3+ (aq) + 8H2O (l) + 3O2}.$$

From the standard potentials you can see, that this reaction may happen, it does not tell you anything about the conditions under which this reaction will happen. There are a lot more factors to consider.