[Cupro Chlorous]

Ok, well I have this problem i've been trying to solve for along time.

Consider the formula:

NaCl_(aq) + H₂O_(l) + e^- -> NaOCl_(aq?) + H_2(g) + e^-

Suppose I run a 12 volt DC current (with a max of 1 Amp) through the solution.  How would I go about calculating the moles of NaOCl at any given time?

What really threw me off is the fact that I also read that OCl^- is easily reduced at the cathode.

[Borek]

NaCl_(aq) + H₂O_(l) + e^- -> NaOCl_(aq?) + H_2(g) + e^-

Looks like electrons cancel out.

[Cupro Chlorous]

Well they do cancel out, but thats because electricity is flowing through.

[plu]

From your equation, it seems that you need to move 1 mole of electrons to produce 1 mole of NaOCl.  If your current is 1 Ampere, the charge passed through the solution at any given time would be 1 x time.  The charge needed to move 1 mole of electrons is equal to the Faraday constant (approximately 96 500 Coulombs).  Dividing the calculated charge by the Faraday constant would give you the number of moles of electrons moved and thus the moles of NaOCl produced.

[Cupro Chlorous]

From your equation, it seems that you need to move 1 mole of electrons to produce 1 mole of NaOCl.  If your current is 1 Ampere, the charge passed through the solution at any given time would be 1 x time.  The charge needed to move 1 mole of electrons is equal to the Faraday constant (approximately 96 500 Coulombs).  Dividing the calculated charge by the Faraday constant would give you the number of moles of electrons moved and thus the moles of NaOCl produced.

Hmm, so 3600 coulombs/hr with a 1 amp current, so 3600/96500 = about 0.037 mols NaOCl/hour.
But do I have to do some equillibrium calculation considering the following equation:

Cl₂ + 2OH^- -> OCl^- + Cl^- + H₂O

And while doing this I notinced that increasing the voltage made it bubble faster.  Does the voltage have an impact on current?

[plu]

But do I have to do some equillibrium calculation considering the following equation:

Cl₂ + 2OH^- -> OCl^- + Cl^- + H₂O

And while doing this I notinced that increasing the voltage made it bubble faster.  Does the voltage have an impact on current?

No equilibrium calculation is required.  Since you are running a charge through the solution, you are directly forcing the reaction to go in one direction.  And yes, voltage is directly related to current

[Cupro Chlorous]

No equilibrium calculation is required.

Oh, ok thanks for your *delete me*

Somehow the equations seem loopy.

2Na⁺Cl_- + 2H₂O + 2e -> 2Na⁺OH^- + Cl₂ + 2e^- -> Na⁺OCl^- + Na⁺Cl^- + H₂
O

I got these equations from Electrochemistry Encyclopedia, fyi.

And yes, voltage is directly related to current

Hmm...  how then would I be able to calculate the current?  I'm using a 12 volt 1 amp DC adapter.

[plu]

Somehow the equations seem loopy.

2Na⁺Cl_- + 2H₂O + 2e -> 2Na⁺OH^- + Cl₂ + 2e^- -> Na⁺OCl^- + Na⁺Cl^- + H₂
O

It would look a lot better if you got rid of all the Na⁺ ions.  They're spectator ions in this reaction so their presence in the equation is immaterial

I got these equations from Electrochemistry Encyclopedia, fyi.Hmm...  how then would I be able to calculate the current?  I'm using a 12 volt 1 amp DC adapter.

Unless my knowledge of batteries is mistaken, I believe the current produced by a 1 Amp DC adapter is, in fact, 1 Amp