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Suppose pH of water is $6$, I think this means there is one $\text{H}^{+}$ ion for every $10^6$ water molecules.

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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 ?

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  • $\begingroup$ have a look at this hyperphysics.phy-astr.gsu.edu/hbase/electric/leadacid.html#c1 $\endgroup$ – anna v Jun 16 '17 at 7:27
  • $\begingroup$ Hey @annav my question is not much about batteries, its about how conduction occurs at water-metal interface $\endgroup$ – Hiiii Jun 16 '17 at 7:28
  • $\begingroup$ Like, in metals only electrons are mobile, but in water we don't have electrons, ions move. As ions move to the opposite plates, don't we run out of them ? $\endgroup$ – Hiiii Jun 16 '17 at 7:30
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    $\begingroup$ First I think that there is an equilibrium amount of ions in water so if you remove them the equilibrium will reestablish. Nevertheless, pure water is a quite bad conductor. Second, it is not necessarily the ion that moves, while there is a certain mobility of ions, its mostly electrons "hopping" making the ion virtually move. Finally this should create a gas at the electrodes, the typical "we make Hydrogen and Oxygen" experiment, so in that sense you use up the ions by actually using up the water. $\endgroup$ – mikuszefski Jun 16 '17 at 7:59
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    $\begingroup$ @Hiiii if the pH is not 7 initially, then it will deviate further from 7 as water is removed by electrolysis. $\endgroup$ – DavePhD Jun 16 '17 at 16:08
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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).

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    $\begingroup$ ... or until there isn't enough water to complete the circuit. $\endgroup$ – Dave Jun 16 '17 at 18:59
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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.

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  • $\begingroup$ isn't the battery being depleted and water being completely converted to gas the same thing in an ideal case? $\endgroup$ – Rishabh Jain Jun 16 '17 at 14:44
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    $\begingroup$ No, these are not the same thing. When I say depleted I mean fully depleted. The battery might be fully depleted before the water runs out, or the water might run out before the battery is fully depleted. $\endgroup$ – sammy gerbil Jun 16 '17 at 22:28
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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.

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  • $\begingroup$ Okay so its a molar volume, got it! Thank you :) May I ask how the charge flows outside the battery, on the other side at metal-water interface ? $\endgroup$ – Hiiii Jun 16 '17 at 7:58
  • $\begingroup$ That's the place where you want to use it.For example, you can connect a resistor (maybe a small fan) to it and current will flow through it . $\endgroup$ – Rishabh Jain Jun 16 '17 at 8:02
  • $\begingroup$ @ Rishabh Jain I think my question is not so clear; Actually in my question, I'm connecting the battery across water. Water is the load resistor in my setup. I'd like to know how charged ions move in water w/o getting used up $\endgroup$ – Hiiii Jun 16 '17 at 8:04
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    $\begingroup$ Did you mean battery?In that case, the ions will flow into the solution or away from it till both the solutions have the same voltage difference and then the current will stop(just like connecting two opposite batteries). $\endgroup$ – Rishabh Jain Jun 16 '17 at 8:08
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    $\begingroup$ @DavePhD The redox reaction reduces one electrode and oxidizes other.The inherent potential is drawn from the electrolyte. The lead acid battery you mentioned can even be reversed ,i.e. charged back if electricity is given the other way.For a very short and concise description,visit sciencing.com/do-batteries-go-flat-7787904.html $\endgroup$ – Rishabh Jain Jun 16 '17 at 14:37

protected by Qmechanic Jun 16 '17 at 8:28

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