How come diluting a cell decreases the voltage?
Your assertion the voltage in a cell is decreased by diluting the cell  is not necessarily true. According to the Nernst Equation:

$$E_{cell}=E^0-\frac{RT}{nF}\ln(Q)$$

Where $Q$ is the reaction quotient relating $\frac{[\mathrm{metal~ion~formed~by~oxidation}]}{[\mathrm{metal~ion~reduced}]}$. 

The value of $Q$ can be increased by: increasing the concentration of the metal ion formed by oxidation, or by decreasing the concentration of the metal ion that is reduced. By increasing the value of $Q$, the magnitude of the term $-\frac{RT}{nF}\ln(Q)$ is increased. This means that a larger number is subtracted from the $E^0$ of the cell, and that the cell voltage decreases.

The value of $Q$ can be decreased by: decreasing the concentration of the metal ion formed by oxidation, or by increasing the concentration of the metal ion that is reduced. By decreasing the value of $Q$ to a fraction, the value of the term $-\frac{RT}{nF}\ln(Q)$ is made positive. This means that this term is added to the $E^0$ of the cell, and that the cell voltage increases.

This is consistent with the Le Châtelier's Theorem. Essentially, increasing the relative concentration of your reactants drives the reaction forward (higher $\Delta V$), and increasing the relative concentration of your products drives the reaction backwards (lower $\Delta V$).