Saltwater Electrolysis - Adding HCl

  • #1
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Main Question or Discussion Point

Greetings forum!

I'm experimenting with saltwater electrolysis and have a few questions. I'm using a salt solution I've prepared overnight. The solution has about 2 parts water, 6 parts salt, so I will essentially be electrolyzing brine at the beginning of the experiment (For safety, the experiment is done outside with a fume hood.). At this point, the solution is very dry and flaky, with a voltage of about 1V without current. I'll be using a current of about 1A with a voltage peaking around 5V. The overall reaction should be:

2 NaCl(aq) + 2 H2O(l) → 2 NaOH(aq) + H2(g) + Cl2(g)

However, I will be adding another prepared solution of HCl + H2O to my original solution. How will adding the protonated water to the brine during electrolysis change the reactions? Thank you for your assistance! :smile:
 

Answers and Replies

  • #2
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How do you apply/control the voltage/current?
 
  • #3
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I'm using a 6V lantern battery and a potentiometer to adjust the voltage. I'm electrolyzing brine. I'll be adding the H3O+ solution during electrolysis. What do you think the reactions will be?
 
  • #4
uby
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Looks like you're changing the concentrations of various ionic species, not changing the chemistry of the system.
 
  • #5
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Looks like you're changing the concentrations of various ionic species, not changing the chemistry of the system.
Thank you for the response! What do you think the reactions (if any) will be?
 
  • #6
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How are you going to protonate the solution? By adding acid? HCl? Or other acid?

You probably have salt slurry not solution by the sound of it.

I think you will not get chlorine; you have to electrolyse molten NaCl for that. Read/skim this:

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php

There are probably lot of descriptions for this experiment on internet. Read them?

PS: Just out of curiosity - why are you trying to do this experiment?
 
  • #7
Borek
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I think you will not get chlorine; you have to electrolyse molten NaCl for that.
Judging by standard reduction potentials one would expect oxygen to be the product, but oxygen oxidation is a very slow process, so there is usually some chlorine produced as well even when you electrolyse water solutions.
 
  • #8
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Thanks for your input Borek! Could you illustrate this with an equation? I'm trying to get a better understanding of how the hydronium ion behaves as a reactant.
 
  • #9
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How are you going to protonate the solution? By adding acid? HCl? Or other acid?
I'm adding HCl.

You probably have salt slurry not solution by the sound of it.
Yes, I'll be electrolyzing brine/salt slurry.

I think you will not get chlorine; you have to electrolyse molten NaCl for that.
Could you illustrate this in an equation? I think I'd have a better understanding of how the hydronium ion behaves as a reactant (if it indeed does).

There are probably lot of descriptions for this experiment on internet. Read them?
There are plenty of examples of water ionization, brine/water electrolysis, and NaOH production. As simple as this experiment is, I couldn't find any examples of the exact stipulations during electrolysis.

PS: Just out of curiosity - why are you trying to do this experiment?
Just to learn more about acids, bases, electrolysis, and electrolytes. I'm very interested in the hydronium ion and its potential.
 
  • #10
Borek
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Initially you have NaCl and H2O in your solution, so there are three possible reactions:

2H2O + 2e- -> H2 + 2OH-

4OH- -> O2 + 2H2O + 4e-

2Cl- -> Cl2 + 2e-

(the first two can written differently, but it won't change the overall stoichiometry nor products - namely hydrogen and oxygen).

As I wrote earlier, if you check standard potentials, third reaction should not occur - as long as there is water present chlorides should be not oxidized and the only product on the anode should be oxygen. But in reality oxygen evolution is very slow (as we say kinetically disfavored) and requires substantial overpotential (larger potential that speeds the reaction up) to proceed fast, which means we are moving in the potential range where chlorine starts to evolve.

Net effect is that the solution becomes slightly basic. Why? Well, as long as only water is being electrolysed, pH doesn't change - OH- is being produced and consumed on anode and cathode in exactly the same amount, so its concentration doesn't change, and constant OH- concentration means constant pH, see http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product]discussion[/PLAIN] [Broken] of water autoionization[url]). However, when some of the charge is being used to remove Cl- from the solution, there will be a slight excess of OH- accumulating, as it will be not all discharged at the anode.

When you add HCl to this slightly basic solution all you do is you return back to the starting point - with NaCl only. Sure, if you add excess HCl you can acidify the solution, making it even easier to produce chlorine (lower concentration of OH- makes the second reaction more difficult).
 
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  • #11
NascentOxygen
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I'm using a 6V lantern battery and a potentiometer to adjust the voltage. I'm electrolyzing brine. I'll be adding the H3O+ solution during electrolysis. What do you think the reactions will be?
What electrodes are you using?
 
  • #12
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I'll be switching between copper and nickel.
 
  • #13
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When you add HCl to this slightly basic solution all you do is you return back to the starting point - with NaCl only. Sure, if you add excess HCl you can acidify the solution, making it even easier to produce chlorine (lower concentration of OH- makes the second reaction more difficult).
Wow. That was a ton of useful information. Thanks, Borek! Could you illustrate what the reactions will be after adding the HCl? Will any of the reactions include the hydronium ion? I'm hoping to put the hydronium ion in a position where it will serve as both a reactant and a product. Not sure how though. Any suggestions would be greatly appreciated! :smile:
 
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  • #14
NascentOxygen
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I'll be switching between copper and nickel.
Copper will react with the Cl, giving bits of green floating around and a mucky electrode.
 
  • #15
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Any thoughts on the reactions? I'm interested in how the hydronium ion will behave during the electrolysis of brine.
 
  • #16
Borek
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Read any general chemistry book - you are looking for acid/base reactions in water (water ion product, acid/base equilibrium).
 
  • #17
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Thanks for the advice, Borek. I think I finally found the answer for what I was looking for regarding hydronium *reduction* during electrolysis:

2H3O+(aq) + 2e- --> H2(g) + 2H2O(l)
or
2H+(aq) + 2e- --> H2(g)

I should've focused on the word *reduction* instead of *reaction* in my original searches. But going over acid-base redox certainly cleared that up. Thanks again!
 
  • #18
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In the brine electrolysis, the cathode reaction will always give you hydrogen. Addition of acid will assist this reaction, and also provide your electrolyte with better conductivity.

The anode reaction is a fairly fine balance in a 3-way competition between
(i) emission of chlorine gas:

2 Cl(aq) --> Cl2(g) + 2 e

(ii) emission of oxygen gas:

2 H2O --> 4 H+ + O2(g) + 4 e

and (iii) erosion of the anode (depending on the anode material)

M(solid) --> Mn+(aq) + n e

Keeping the chloride ion concentration high, and choosing an anode material that is a good conductor but otherwise unreactive will help promote reaction (i)

Acidification using HCl is a good move for your electrolysis, but completely useless if you are hoping to obtain NaOH as a product.

Glassy carbon or graphite is probably the best anode material. Note that even unreactive metals -- silver, gold, platinum -- tend to react with nascent chlorine.
 

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