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