Does current run forever in water? (assuming the supply voltage is there forever) Suppose pH of water is $6$, I think this means there is one $\text{H}^{+}$ ion for every $10^6$ water molecules. 

When we plug in the battery, I believe we see a current as the $\text{H}^{+}$ ions drift to the $-ve$ side of the battery and suck the electrons injected by the negative plate of the battery. Similarly, $\text{OH}^{-}$ ions drift to the $+ve$ plate of the battery and give an electron away to the positive plate of the battery. This way $\text{H}^{+}$ and $\text{OH}^{-}$ ions neutralize themselves as they contribute to the current. Since the ions were neutralizing themselves, would the current cease to exist after some time when all the $\text{H}^+$ ions in the water were used up ? 
 A: Fist of all, saying that there is one hydrogen ion for $10^6$ ions of water is wrong.ph is the negative of log of concentration of hydrogen ions. pH 6 means that there is $10^{-6}$ moles of hydrogen in a litre of the solution. Moreover, in a battery the electrolyte gets depleted not the electrodes. So, the voltage will be constant till the electrolyte can provide the potential ideally.
In reality, the voltage curve drops at the end of the life of the battery.
A: It is energetically unfavourable to split a water molecule into the two ions $\text{H}^+$ and $\text{OH}^-$ i.e. you need to put in energy to do it. However at room temperature water molecules have a range of energies and there are always a few molecules with enough energy to ionise. So any sample of pure water at everyday temperatures always contains a few $\text{H}^+$ and $\text{OH}^-$ ions. 
When you apply a voltage to your electrodes in water, you convert the $\text{H}^+$ ions to hydrogen atoms and they bubble off as $\text{H}_2$. Likewise the $\text{OH}^-$ ions are converted to water and oxygen molecules and the oxygen bubbles off. The net result is to remove water from your container.
But as fast as the ion concentration is lowered by electrolysis, the remaining water ionises again to keep it constant. So electrolysis of pure water does not affect the ion concentration. You are correct that the current will continue to flow until all the water has gone (i.e. converted to hydrogen and oxygen).
A: The ions are converted into gases $H_2$ and $O_2$ at the electrodes, so water is gradually being removed from the container - but it requires a very very large amount of charge to flow in order to convert all of the water into gases.
The $H^+$ ions exist as hydronium ions $H_3 O^+$. There is a reversible equilibrium in the water :
$2H_2O <=> H_3O^+ + HO^-$
The product of concentrations of the ions is fixed at a certain value which depends on the temperature of the water. If one or both types of ions are removed by electrolysis, more water molecules dissociate to keep the product of concentrations constant :
$K_w = [H_3O^+] [HO^-]$.
At 25$^{\circ}$C the product $K_w \approx 1.0 \times 10^{-14}$ when concentrations are measured in moles per litre.
In ideal conditions there will always be ions available to conduct electricity through the water. It will keep flowing until either the battery is fully depleted or the water is completely converted to gases. 
In practice, an insulating layer may build up at the electrodes which prevents electrolysis from taking place. This layer is caused by other reactions at the electrodes due to impurities in the electrodes or the water.
See wikipedia articles Electrolysis of Water and Self-ionization of Water.
