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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 ionize. 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 ionizes 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).

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).

It is energetically unfavourable to split a water molecule into the two ions $H^+$ and $OH^-$ 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 $H^+$ and $OH^-$ ions.

When you apply a voltage to your electrodes in water, you convert the $H^+$ ions to hydrogen atoms and they bubble off as $H_2$. Likewise the $OH^-$ 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).

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 ionize. 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 ionizes 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).

It is energetically unfavourable to split a water molecule into the two ions $H^+$ and $OH^-$ i.e. you need to put in energy to do it. However at room temperature water molecules have a rage 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 $H^+$ and $OH^-$ ions.

When you apply a voltage to your electrodes in water you convert the $H^+$ ions to hydrogen atoms and they bubble off as $H_2$. Likewise the $OH^-$ ions are converted to water and oxygen molecules and the oxygen bubbles off. Th 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.

It is energetically unfavourable to split a water molecule into the two ions $H^+$ and $OH^-$ 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 $H^+$ and $OH^-$ ions.

When you apply a voltage to your electrodes in water, you convert the $H^+$ ions to hydrogen atoms and they bubble off as $H_2$. Likewise the $OH^-$ 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).