Electrolytic Reactions (1 Viewer)

[Narelle]

How do u work out which half reaction is oxidation and which is reduction...

For instance... If they tell u an molten salt solution undergoes electrolysis for eg. NaCl(l) , u look at the standard reduction table for the half equations but how do work out which becomes oxidised and which becomes reduced?

I know the answer is that Na becomes reduced and so Na solid forms at the cathode while Cl gas forms at the anode... but why and how do u work it out?


Anyone know an easy way to work it out or remember

 

[~*HSC 4 life*~]

How do u work out which half reaction is oxidation and which is reduction...

For instance... If they tell u an molten salt solution undergoes electrolysis for eg. NaCl(l) , u look at the standard reduction table for the half equations but how do work out which becomes oxidised and which becomes reduced?

I know the answer is that Na becomes reduced and so Na solid forms at the cathode while Cl gas forms at the anode... but why and how do u work it out?


Anyone know an easy way to work it out or remember

look at each species you've got in your beaker, then cosnider all the possible oxidation and reduction reactions. If the solution is molten, water is not involved, so you don't need to cnsider the redox reactions involving water..which makes life easier.

using your standard reduction table, compare electrode potential values for each POSSIBLE redox eqn, which ever is more positive is the one you should pick, i guess thats the easiest way to remember it

remember it can still be negative, because it is an electrolytic reaction, where a current is used to drive the reaction, but pick the one which is MORE positive like -0.5 more posive than -0.9 etc

 

[Xayma]

The oxidation reaction, is the most positive oxidiation potential (ie most negative reduction potential)

The reduction reactions, is the most positive reduction potential.

 

[tina_goes_doo]

Yep. Xayma basically summarises that up nicely.

For example, with the electrolysis of a dilute sodium chloride solution, there are two possible reactions which could undergo oxidation:

2Cl- -> Cl2(g) + 2e- Eo = -1.36V
2H2O(l) -> O2(g) + 4H+ + 4e- Eo = -1.23V

Similarly for reduction there are two possibilities:

Na2+ + 2e- -> Na(s) Eo = -2.74V
2H2O(l) + 2e- -> H2(g) + 2OH- Eo = -0.83V

This can be seen as competition between the sodium ions and the water molecules for electrons. From this we see that water is reduced rather than the sodium ions.

***Hence, the more positive value is the one most likely to undergo oxidation/reduction.***

So the overall reaction becomes:

2H2O(l) -> O2(g) + 2H2(g)

The sodium ions and the chlorine ions do not react during this electrolysis. They do, however, carry charge through the electrolyte so that decomposition of water can occur.

 

[mr EaZy]

thnx for that, but in the equation
2H20 (l)-> O2(g) + 4H+ + 4e- Eo = -1.23V

assuming the H+ ions dont get used up, does the solution get acidic?>

 

[Narelle]

Ok i know why i was so confused before... but i get it now

Say the situation is electrolysis with inert electrodes, molten NaCl as electrolyte and ur trying to work out what happens...

well the electrodes are inert so they simply act as conductors

as the others said, its molten not aqueous so u don't have to worry about the H2O equations...

and how do u work if Na and Cl undergo oxidation or reduction?

well you have Na IONS in electrolyte and Cl IONs in electrolyte... so then u look at the standard reduction table...

if u look at the table there's only one equation for Na ... and the Na ion (Na+) is on the left hand side of the equation. Therefore the Na ions that u have in the electrolyte undergo reduction (remember OIL RIG... or reduction reaction on left ect)

the Cl ion (Cl-) is on the right hand side of the equation so it undergoes oxidation.
there are two equations for Cl but like the others said u use the most positive one (u have to remember that with oxidation u have to reverse the sign of potential. So then 1.36V becomes -1.36V and 1.40 becomes -1.40 As u can see -1.36V is more positive.)

So now that u know in this electrolysis Na undergoes reduction and Cl undergoes oxidation, u remember An Ox Red Cat .... and so then u work out that Na(s) forms at the cathode (since it undergoes reduction) while Cl(g) forms at the anode (since it undergoes oxidation)

That was really simple, but it got me lost for a while, esp since i didn't do electrolytic equations properly before and its been a while since i studied it. You all probably knew all this but who knows maybe it might help someone. (oh by the way this is all under the SHIPWRECKS option, i should of mentioned that before)

Thanx everyone for ur other answers to this question too, they help out as well, esp for other situations.

In other situations where u might have more than equation possible than remember what xayma said.

and when u have aqueous solution (which is different to molten) then follow what tina said.

Remember also that other factors can affect electrolytic cells like nature of the electrodes, nature of electrolytes and concentration of electrolytes.

Temperature, the size of the applied voltage and the area of electrode immerse affect the rate and products of an electrolytic cell...

As for mr Ezy.... i don't know

 

[tennille]

Sorry to change the subject here, but when you work out the minimal voltage of an electrolytic cell, is it negative or positive? In conquering Chemistry, it is a positive an swer (exactly the same as a galvanic cell), but really, electrolytic cells need negative voltages... I'm really confused...could someone give me an example of how to work it out?

Thanks.

 

[Xayma]

Electrolytic cells have a negative potential. You need to supply a potential difference equal in magnitude but positive to bring it above 0 to get the reaction to occur.

Hence if an electrolytic cell had a potential of -1V you need a 1V potential difference to force the reaction to go.

 

[Tommy_Lamp]

so the minimum voltage in an electrolytic cell is the x-1 of the potential?

 

[tennille]

Thanks for that

 

[Narelle]

If they ask to describe what experiment u carried out to test factors effecting electrolysis, what one do u suggest we use?

For Concentration of the Electrolyte: tina already pointed out the NaCl one we could use. (compare the electrolysis of dilute NaCl to that of concentrated one... with dilute NaCl, water ends up being oxidised... with the latter Cl is produced at anode)

For Nature of Electrodes i think i'll use the one about copper sulfate ions with inert electrodes and then copper sulfate ions with copper electrodes.

What experiment did u use for NATURE OF THE ELECTROLYTE?... we didn't carry out this experiment because my school couldn't get the molten substances... i'm guessing cause its expensive and hard to obtain... Can anyone tell me what they used?

 

[tennille]

Basically, we tested all the factors using copper electrodes along with CUSO4 solution.

Changing voltage using 2v, 4v and 6v, but keeping the surface area, distance of the electrodes and concentration of electrolyte constant.

Changing surface area but keeping everyhting else constant.

Changing distance but keeping everything else constant.

Changing concentration from 0.1mol/L to 1mol/L but keeping everything else constant.

We didnt change the electrodes...we just did the main factors that affect the rate of electrolysis.

What do you mean by nature of electrolyte... do you mean a totally different electrolyte or differnet concentration?

 

[Narelle]

thanx..

umm as in molten or aqueous....i think thats wat the syllabus point means...

different concentration is another factor

if i then were to electrolyse aqueous copper sulfate, the copper would reduce right and form at the cathode... what happens to the sulfate? does water oxidise instead... or am i completely wrong

 

[tennille]

You might be right about water being oxidised at the anode...if it was a zinc anode, then zinc would go in to solution since it is more reactive but as it is a copper electrode in a CuSO4 solution, then it would remain as a metal. Copper would form on the cathode as a reduction reaction...

But I'm in the same position as you...I'm not completely sure about the anode reaction...when we did our experiment, we noticed that the cathode had gained mass and the anode had lost mass...so maybe Cu does go into solution...but that doesnt make sense...I'm sorry...

 

[tennille]

How has your study been going? I looked at your profile and its nice to see a girl doing science subjects...at least im not the only one...hehehe

 

[Narelle]

yeah this crap is so confusing sometimes... thanks anyway

study?... eh i'm wasting so much time... as in spending to much time on pointless stuff which most probably wont be tested on... i should move on so i can cover everything i need. Lucky did a bit before the trials even though results weren't the best.

yeah y the hell did i pick science?... well its not that bad, i find it interesting when i'm not confused and frustrated, teachers are great too even though im finding it annoying right now, feeling like i haven't got enough info. (its prob my fault anyway)

but yeah actually a lot of girls in my school do science (oh and no i dont go to an all gurls school)... and actually the gurls are like the top most of the science subject... except for physics, but then again there's only 4 gurls in our class and one of them is coming second... not me though... ) eh my rankings aren't that bad...but yeah science is good... i just dream about doin drama again... should of picked that... oh well...

so yeah.. to sum up... no ur not the only one...

ps: i hope they don't ask us to much on this thing, cause i'm not sure on it right now

 

[Narelle]

some answers

I emailed my beautiful chem teacher with some questions on electrolysis and here are her replies... i hope this helps ppl:

With any redox cell that you need to work out what is happening:
1. Write down the 'species' you have present. If it is molten NaCl then you have
Na+ ions and Cl- ions since it is ionic.
2. With these species written on the left hand edge of your page, then go to the
redox table and find equations that have these species in them:
Na+ + 1e- gives Na(s) = -2.71V
Cl2(g) + 2 e- gives 2Cl- = +1.36V
however this second equation is backwards since you have Cl- ions so :
2Cl- gives Cl2(g) + 2 e- = - 1.36 V
3. Look at both equations:
Na+ + 1e- gives Na(s) = -2.71V
2Cl- gives Cl2(g) + 2 e- = - 1.36 V
these are a matched pair since the electrons are on opposite sides of the
equation. this tells us that to produce Na (s) and Cl2(g) from molten NaCl we
need to supply - 4.07V (adding the two values together).
4. if you had aqueous NaCl then you do the same thing however you also have
water present in the solution and so you need to write down the water equations
also. you will find that there are two equations with water and that these will
have values that are either positive or closer to zero thatn the Na and Cl
equations. always choose the most positive or the negative value that is closer
to zero na dalways make sure you ahve a pair of equations with electrons on
oipposite sides and that you keep the correcxt species on the left hand side.

**********

For nature of the electrolyte compare 2 cells - both with inert electrodes one
with CuSo4 solution and the other with MgSO4 solution same concentrations. when
you write your equations don't use sulphate because it is very stable in
comparioson to other things and will not be involved. so for this example write
down 2 equations fro water, one producing H2(g) and one producing O2(g) - look
hard. Also write down Mg2+ and Cu2+.
you should have 3 reduction reactions and one oxidation reaction (H2O forming
O2).
The oxidation reaction must occur since it is the only one.
to select the reduction reaction look at the voltage values.
for MgSO4 the Mg2+ reaction is further from than the water and so Mg2+ will not
turn into Mg(s), but instead water will turn into H2(g) with O2 froming at the
other electrode.
For CuSO4 look at the same water equation and compare it with the Cu2+ one.
The Cu2+ equation has a positive value and is therefore more likely to occur
than the water one and so this cell will produce Cu(s) and O2(g)
I have suggested a different experiment but you can use the information in to work out this situation in which you use zinc sulfate and sodium
sulfate.