Electrolysis of NaCl (1 Viewer)

[richz]

hello there,

my teacher recently performed this prac, but im confused at some things. He just put it up, while we did another exp and i didnt see much of it. So i got a few qs.

a) What voltage was used?
b) What colour was the NaCl initially and after electroylsis, Cl2 and H2?

thnx in advance

 

[richz]

very quiet here.....

 

[fantasy27]

i think its because not many people are up to this topic yet.. just a guess

 

[mitochondria]

Hello there I assume you were talking about the electrolysis of NaCl solution.. as opposed to the electrolysis of NaCl - simply bcause the electrolysis of NaCl is quite dangerous, first of all, the melting point of NaCl is ~800^oC and if I remember correctly a normal bunsen flame only goes up to aroudn 600^oC, not to mention that you need a container which has a melting point higher than 800^oC so equipment-wise it's quite twoublesome... well, if that's being taken care of, consider the product of the the electrolysis of molten NaCl:

NaCl_(l) ----> Na_(s) + Cl_2(g)

I assure you that the products are not mr and mrs nice..

Soooo, anyway, back to your original question, when you do electrolysis of NaCl solution, what you're essentially doing is an electrolysis of dihydrogen monoxide.. I mean water having said that, immediately you should think about:

REDOX!

and you should rush to get your table of standard reduction potentials (or simply follow this link http://www.jesuitnola.org/upload/clark/Refs/red_pot.htm) The two equations we want are:

2H₂O ---> O₂ + 4H⁺ + 4e^- = -1.23V and
2H₂O + 2e^----> H₂ + 2OH^- = -0.83V

solve the equations simultaneously and you should get an overall voltage of -2.89V (please check my calculation, and you can write out the balanced equation in all the glory details yourself )

recall that a negative potential means the reaction is not spontaneous, so in order to do an electrolysis (which is essentially reversing the reaction) you will need to supply a voltage of greater than 2.89V

ummm.. I'm confused by (ii).. but Cl₂ is green-yellow and H₂ is colourless

hope that helps!


P.S. subjected to error... please yell at me if you find any mistakes

 

[Templar]

ummm.. I'm confused by (ii).. but Cl₂ is green-yellow and H₂ is colourless

Perhaps they added indicator to show the creation of H and OH ions at the electrodes.

 

[Jumbo Cactuar]

Hello there I assume you were talking about the electrolysis of NaCl solution.. as opposed to the electrolysis of NaCl - simply bcause the electrolysis of NaCl is quite dangerous, first of all, the melting point of NaCl is ~800^oC and if I remember correctly a normal bunsen flame only goes up to aroudn 600^oC, not to mention that you need a container which has a melting point higher than 800^oC so equipment-wise it's quite twoublesome... well, if that's being taken care of, consider the product of the the electrolysis of molten NaCl:

NaCl_(l) ----> Na_(s) + Cl_2(g)

I assure you that the products are not mr and mrs nice..

Soooo, anyway, back to your original question, when you do electrolysis of NaCl solution, what you're essentially doing is an electrolysis of dihydrogen monoxide.. I mean water having said that, immediately you should think about:

REDOX!

and you should rush to get your table of standard reduction potentials (or simply follow this link http://www.jesuitnola.org/upload/clark/Refs/red_pot.htm) The two equations we want are:

2H₂O ---> O₂ + 4H⁺ + 4e^- = -1.23V and
2H₂O + 2e^----> H₂ + 2OH^- = -0.83V

solve the equations simultaneously and you should get an overall voltage of -2.89V (please check my calculation, and you can write out the balanced equation in all the glory details yourself )

recall that a negative potential means the reaction is not spontaneous, so in order to do an electrolysis (which is essentially reversing the reaction) you will need to supply a voltage of greater than 2.89V

ummm.. I'm confused by (ii).. but Cl₂ is green-yellow and H₂ is colourless

hope that helps!


P.S. subjected to error... please yell at me if you find any mistakes

For some critical Cl- concentration L :

if [Cl-] < L

2H₂O ---> O₂ + 4H⁺ + 4e^- E^o = -1.23V
2H₂O + 2e^----> H₂ + 2OH^- E^o = -0.83V
2H₂O ---> O₂ + 2H₂ E^o = -2.89V

Now E = E^o - RT/nF ln (1)
E = -2.89V

if [Cl-] > L

2Cl^----> Cl_2(g) + 2e^- E^o = -1.36V
2H₂O + 2e^----> H₂ + 2OH^- E^o = -0.83V
2H₂O + 2Cl^- ---> H₂ + 2OH^- + Cl_2(g) E^o = -2.19V

Now E = E^o - RT/nF ln (^[OH-]2/_[Cl-]2)

n=2, {R,F = usual constants}

E = -2.19 - 8.617 x 10^-5 T ln (^[OH-]/_[Cl-]), say T=25C=298.15K
E = -2.19 - 2.569 x 10^-2 ln (^[OH-]/_[Cl-]) V

Say initially pH = 7.0
pOH = 7.0,
[OH-] = 1 x 10^-7

E = -2.19 - 2.569 x 10^-2 (ln [OH-] - ln[Cl-]) V
E = -1.776 + 2.569 x 10^-2 ln[Cl-] V


**gasp**

But L = [Cl-] when E_a - E_b = 0
0 = -1.776 + 2.569 x 10^-2 ln(L) + 2.89
-1.114 = 2.569 x 10^-2 ln(L)
ln(L) = -43.36
L = 1.476 x 10^-19 M

Therefore the chloride ions will be oxidised in a pH neutral solution if in a concentration of 1.476 x 10^-19 M or greater given sufficient electrical potential.


xrtzx the reaction contains many many variables. The electrolysis reaction is definately:
2H₂O + 2Cl^- ---> H₂ + 2OH^- + Cl_2(g) E^o = -2.19V
The concentration is very unlikely to be a limiting one, so ask your peers.
The voltage must be larger than the limiting voltage dependent on pH and Cl- concentrations (also temperature).

Sorry for any errors.

Think I'm done with electrochemistry after that marathon

 

[richz]

cool thnx guys