percent ionization explanation
For $\ce{HA &lt;=&gt; A- + H+}$  with a particular $K_a$ value at a particular temperature (Also, remember it usually applies to weak acid as $K_a$ of strong acid is rather meaningless),
Consider its equilibrium expression: $$K_a=\frac{\ce{[A^{-}][H+]}}{\ce{[HA]}}$$ when water (of the same temperature) is added, all three component concentration will decrease with the same extent. 

Think about whether the numerator and denominator has a bigger decrease and you will find the answer (temperature is relatively constant). And for your free energy concern, being out of equilibrium necessarily means there is a favorable driving force pushing the reaction in one direction to reach back eqm.



Acquire a physical picture about the initial and equilibrium state and the dynamic state of a reaction in terms of collision theory. Think equilibrium in the form of competing forward and backward reaction and equilibrium constant is actually derived from rate constant.



Another reminder: Temperature and concentration/pressure all affect free energy. At the initial state (when you just add water and no reaction has taken place yet), there is only change in concentration but not temperature. So even if the temperature is changed upon reaction, the percentage of ionization will also be higher in all circumstances.