[confusedstud]

When a soluble ionic compound is added to water, does it cause a lot of the water molecules to ionize into H+ and OH- ions? Which thus allows then to conduct electricity better, because my textbook says "adding diluted NaCl allows the water to be a lot more conductive".

If that is true, then in a copper (ii) sulfate solution, there will be H+, Cu2+, OH- and SO4 2- ions, then won't the copper ion form a precipitate with the OH-? Similar to a sodium hydroxide+cooper (ii) sulfate solution reaction?

Thanks for the *delete me*

[Hunter2]

Water has its own equilibrium H₂O => H⁺ + OH^-. It has nothing to do with an addition of a salt.

Salt will dissociate: NaCl => Na⁺ + Cl^-

The solution gets conductive because more ions are present.

There wiil be no reaction between OH^- and cathions.

[confusedstud]

Water has its own equilibrium H₂O => H⁺ + OH^-. It has nothing to do with an addition of a salt.

Salt will dissociate: NaCl => Na⁺ + Cl^-

The solution gets conductive because more ions are present.

There wiil be no reaction between OH^- and cathions.

Oh, but if NaCl is the one conducting electricity, then won't sodium and chlorine gas be formed instead? I don't quite understand this  part..

[Darren]

Water has a very low dissociation, so it doesnt conduct electricity very well. Due to the low ionisation of water to form hydrogen and hydroxide ions. But its enough for you to get an electric shock when you touch an open circuit with wet hands. In electrolysis, you focus on water as a molecule, instead of the hydrogen or hydroxide ions that migrate to the cathod and anode respectively. So as water molecules, you use half equations to determine the potentials of the half equation to see which cation or anion is discharged from the solution. In dilute NaCl solution, you have water molecules and sodium ions and chloride ions. If you use electrode potentials, water will be discharged preferentially. And same goes for water being discharged instead of chloride ions. But in conc. NaCl, chlorine will be evolved instead of water as the concentration of anions plays a part in deciding which species is discharged, but only for the half equations which have electrode potentials close in values to each other, then will concentration become a factor to consider. And in copper sulfate solution, copper ions wont react with hydroxide ions because there are just too few hydroxide ions ionised from water. In fact, the concentration of hydrogen ions equals to concentration of hydroxide ions in water which is equal to 10^-7 moldm^-3.
Hope that helps you to understand better

[confusedstud]

Hi Darren, in my textbook it says that adding a little of NaCl makes the solution more conducive. But since the sodium cations and chlorine anions do not take part in the reaction then how does it make it more conductive? Since its the H+ and OH- ions that is reduced and oxidised respectively and not the Na+ or Cl-?

Also, does using a higher voltage battery cause more water to ionize? I think so because more H+ and OH- are discharged. If not how does increasing the voltage speed up the process?

Thanks for the *delete me*

[Darren]

Hi Darren, in my textbook it says that adding a little of NaCl makes the solution more conducive. But since the sodium cations and chlorine anions do not take part in the reaction then how does it make it more conductive? Since its the H+ and OH- ions that is reduced and oxidised respectively and not the Na+ or Cl-?

Also, does using a higher voltage battery cause more water to ionize? I think so because more H+ and OH- are discharged. If not how does increasing the voltage speed up the process?

Thanks for the *delete me*

Adding NaCl to water increases its conductivity because Na⁺ and Cl^- ions are free moving to conduct electricity as water molecules cant. Its not a reaction when you dissolve sodium chloride in water. Only electrolysis is the reaction here. Using a higher voltage Allows greater current to flow as V=IR. With more current, you can discharge a greater number of moles of products at the anode and cathode for the same amount of time used previously. Increasing the voltage just speeds us the process. I dont think it allows more water to ionise as the voltage only affects the speed or rate of discharge.

[Borek]

Water has a very low dissociation, so it doesnt conduct electricity very well. Due to the low ionisation of water to form hydrogen and hydroxide ions. But its enough for you to get an electric shock when you touch an open circuit with wet hands.

You are a relatively good conductor because of other ions, not because of water autodissociation.

In fact, the concentration of hydrogen ions equals to concentration of hydroxide ions in water which is equal to 10^-7 moldm^-3.

Only in neutral solution.

[Darren]

Water has a very low dissociation, so it doesnt conduct electricity very well. Due to the low ionisation of water to form hydrogen and hydroxide ions. But its enough for you to get an electric shock when you touch an open circuit with wet hands.

You are a relatively good conductor because of other ions, not because of water autodissociation.

In fact, the concentration of hydrogen ions equals to concentration of hydroxide ions in water which is equal to 10^-7 moldm^-3.

Only in neutral solution.

Oh but then why are you able to get an electric shock when you step on a puddle of water in the bathroom when a live wire from the hair dryer is touching another part of the puddle of water? Or is that only in the movies?

For the dissociation constant, if its not in neutral solution, then we would have to go by Le Chatelier's principle right? To determine how much of water is ionised in the non neutral solution?

[confusedstud]

oh so the presence of Na+ and Cl- allows the current to flow better? I still don't quite get this, even though there are more free mobile ions present in the electrolyte they do not take part in the reaction. Since only the H+ and OH- take part in the reaction how does having these addition ions allow the conductivity better?

As for the higher voltage, so the water doesn't auto ionize more than before? Only that the rate of oxidation and reduction of the OH-and H+ is increased because of the higher current/potential at the two electrodes (one will be more positive and the other more negative than before) so the process will be sped up? I heard that any one time there will be equal numbers of H+ and OH- that is given out during ionization (when some of them are given off in the electrochemical processes), is this true?

Lastly, when i have a solution of copper (ii) sulfate, will the amount of H+ and OH- in the solution (from water) be too insignificant for a precipitate to form? And only if another alkali is used, then it is able to precipitate because now the amount of OH- is more significant than before?

Thanks!

[Darren]

oh so the presence of Na+ and Cl- allows the current to flow better? I still don't quite get this, even though there are more free mobile ions present in the electrolyte they do not take part in the reaction. Since only the H+ and OH- take part in the reaction how does having these addition ions allow the conductivity better?

As for the higher voltage, so the water doesn't auto ionize more than before? Only that the rate of oxidation and reduction of the OH-and H+ is increased because of the higher current/potential at the two electrodes (one will be more positive and the other more negative than before) so the process will be sped up? I heard that any one time there will be equal numbers of H+ and OH- that is given out during ionization (when some of them are given off in the electrochemical processes), is this true?

Lastly, when i have a solution of copper (ii) sulfate, will the amount of H+ and OH- in the solution (from water) be too insignificant for a precipitate to form? And only if another alkali is used, then it is able to precipitate because now the amount of OH- is more significant than before?

Thanks!

The Na and Cl ions dont take part in the electrolysis reaction since they dont get discharged from dilute NaCl solution. Its their movement from the solution to the electrodes that carries the current to conduct electricity. Water does the same thing, the movement, but they take part in the reaction instead. So the movement of Na and Cl ions carry electricity when they move, but once they react the electrodes of opposite charges to their ionic charge, they remain around there without reaction happening to them. Thus, the movement of ions in solution generates the current.

You may want to know this formula, number of moles of product of electrolysis= (current x time) divided by (faraday's constant x number of moles of electrons transferred per mole of product)
So you can see that current plays a part in the amount of products generated if you keep the time constant for the process to take place.

Form the overall equation for electrolysis and you can see the mole ratios for water and the products used up or produced respectively, in the electrolysis. The equation will give you their mole ratios.

And for your last qn, you only get a ppt when you have a significant amount of hydroxide to react with copper ions. Water just doesnt produce enough of hydroxide ions to be available for reaction to form the ppt since it has a low dissociation.

[confusedstud]

Oh I get it more ions means better conductor despite it getting discharged pt not. Thanks for the help Darren!

[confusedstud]

oh so the presence of Na+ and Cl- allows the current to flow better? I still don't quite get this, even though there are more free mobile ions present in the electrolyte they do not take part in the reaction. Since only the H+ and OH- take part in the reaction how does having these addition ions allow the conductivity better?

As for the higher voltage, so the water doesn't auto ionize more than before? Only that the rate of oxidation and reduction of the OH-and H+ is increased because of the higher current/potential at the two electrodes (one will be more positive and the other more negative than before) so the process will be sped up? I heard that any one time there will be equal numbers of H+ and OH- that is given out during ionization (when some of them are given off in the electrochemical processes), is this true?

Lastly, when i have a solution of copper (ii) sulfate, will the amount of H+ and OH- in the solution (from water) be too insignificant for a precipitate to form? And only if another alkali is used, then it is able to precipitate because now the amount of OH- is more significant than before?

Thanks!

The Na and Cl ions dont take part in the electrolysis reaction since they dont get discharged from dilute NaCl solution. Its their movement from the solution to the electrodes that carries the current to conduct electricity. Water does the same thing, the movement, but they take part in the reaction instead. So the movement of Na and Cl ions carry electricity when they move, but once they react the electrodes of opposite charges to their ionic charge, they remain around there without reaction happening to them. Thus, the movement of ions in solution generates the current.

You may want to know this formula, number of moles of product of electrolysis= (current x time) divided by (faraday's constant x number of moles of electrons transferred per mole of product)
So you can see that current plays a part in the amount of products generated if you keep the time constant for the process to take place.

Form the overall equation for electrolysis and you can see the mole ratios for water and the products used up or produced respectively, in the electrolysis. The equation will give you their mole ratios.

And for your last qn, you only get a ppt when you have a significant amount of hydroxide to react with copper ions. Water just doesnt produce enough of hydroxide ions to be available for reaction to form the ppt since it has a low dissociation.

Hi Darren, according to Wikipedia it says that a higher potential means more autoionization.
Pure water is a fairly good insulator since it has a low autoionization, Kw = 1.0 x 10−14 at room temperature and thus pure water conducts current poorly, 0.055 µS·cm−1. Unless a very large potential is applied to cause an increase in the autoionization of water the electrolysis of pure water proceeds very slowly limited by the overall conductivity.

How does it work? Thanks Darren!

[Darren]

Hmm could that be the use of half equations for water to form hydrogen and oxygen in separate half epreactions? Because from the electrode potential, that is there i see the potential, or voltage to be coming from. But im not sure whether voltage actually affects the half equation, whether a higher voltage favours the forward reaction more or not. Ive only come across how concentrations have affected the half equation voltages at my level. But i will create a new topic somewhere else to ask about this before i get back to you about how voltage affects the ionisation of water it will require some time to get the discussions going

[confusedstud]

Hmm could that be the use of half equations for water to form hydrogen and oxygen in separate half epreactions? Because from the electrode potential, that is there i see the potential, or voltage to be coming from. But im not sure whether voltage actually affects the half equation, whether a higher voltage favours the forward reaction more or not. Ive only come across how concentrations have affected the half equation voltages at my level. But i will create a new topic somewhere else to ask about this before i get back to you about how voltage affects the ionisation of water it will require some time to get the discussions going

Okay! Thanks for the help, do let me know the link for the discussion. Thanks again!

[Darren]

Here's the link to the discussion we shall come to a conclusion after that whether voltage applied will affect water's dissociation or not

http://www.chemicalforums.com/index.php?topic=60050.msg214655#msg214655