Upsilon
Hazard to Others
Posts: 392
Registered: 6-10-2013
Member Is Offline
Mood: No Mood
|
|
Electrolysis involving polyatmoic ions
So I understand how electrolysis works - the positively charged cathode attracts the negative ions in solution, takes an electron from the ion, and
leaves the elemental form of the ion. Same thing with the anode but the other way around. But my question is, what in the world happens with
polyatomic ions, like nitrate, carbonate, manganate, chromate, etc.? Basically, what happens in reactions like these:
NO3- -> ? + e-
MnO42- -> ? + e2-
NH4+ + e- -> ?
Is there some general rule as to what happens so that you can predict the outcome for all polyatomic ions?
[Edited on 14-9-2015 by Upsilon]
[Edited on 14-9-2015 by Upsilon]
|
|
|
j_sum1
Administrator
Posts: 6320
Registered: 4-10-2014
Location: At home
Member Is Offline
Mood: Most of the ducks are in a row
|
|
What you need is one of these.
You list all of the half reactions that might apply to your situation, considering as reactants all species in your electrolyte as well as whatever
materials your anode and cathode are made from.
Remember that for oxidation reactions you need to switch the sign of the electrode potentials.
Select from your list the most positive reduction reaction and the most positive oxidation reaction. (One or both may still be negative.) All other
things being equal, these are the two half reactions that will occur preferentially to everything else. Add the two reduction potentials. (After the
oxidation sign switch of course.) If the result is negative, this tells you the minimum voltage that you need to apply. If you get a positive number
when you add then the reaction will occur spontaneously.
Now, it does not always work out as easily as this. The standard reduction potentials are for solutions at 1M and 25°C. Changes to your
concentration in particular may have an effect. Similarly, high current densities or high overpotential may cause reactions to occur simultaneously.
Lack of stirring may prevent some reactions from occurring since the local concentration may drop. Also, build up of passive oxide coatings may
impede a reaction or cause it to cease altogether. Finally, use or non-use of a membrane may prevent/allow mixing of reaction products and secondary
reactions. Some of these complications may be used to your advantage depending on your application.
But to answer your basic question, polyatomic ions behave pretty much in the same way. You merely need to consult the appropriate tables to see what
the possible reactions will be, and go from there.
|
|
|
Detonationology
Hazard to Others
Posts: 362
Registered: 5-5-2015
Location: Deep South
Member Is Offline
Mood: Electrophillic
|
|
When most polyatomic ions are reduced, they can break down further into smaller polyatomics, if possible. For example, when nitrate is reduced, it
forms nitrite.
[Edited on 9-14-2015 by Detonationology]
|
|
|
Upsilon
Hazard to Others
Posts: 392
Registered: 6-10-2013
Member Is Offline
Mood: No Mood
|
|
It's been a couple of years since I learned electrochemistry, so I apologize for me being a little rusty on this topic. The simpler half-reactions in
the table make perfect sense to me:
Ca2+ (aq) + 2e- -> Ca (s)
Meaning that electrolyzing a solution of a calcium salt with inert electrodes will cause calcium metal to evolve at the cathode. This is as I would
have expected, since the negative cathode attracts the positive calcium ions, and donates electrons to the calcium ion until its charge balances,
causing it to revert to elemental form. However, the more complicated half-reactions involving polyatomic ions are confusing to me. For example, take
the half-reaction listed for nitrate:
NO3- (aq) + 2H+ + e- -> NO2 (g) + H2O
From this, I'm reading that this only occurs at the anode in an acidic solution. But this only shows the reduction of nitrate; what of its oxidation?
When the nitrate ion comes in contact with the positive anode and gives up an electron?
[Edited on 15-9-2015 by Upsilon]
|
|
|
j_sum1
Administrator
Posts: 6320
Registered: 4-10-2014
Location: At home
Member Is Offline
Mood: Most of the ducks are in a row
|
|
Coupla things.
1. Poor example for your simple half reaction. Ca2+ ain't going nowhere in aqueous solution. The electrode potential for that is -2.87V You are
going to produce hydrogen from your water at -0.83V and so that reaction will take precedence. At least until all of the water is gone. And then you
are in the realm of molten salt electrolysis (which is conceptually simpler but practically a whole lot more difficult.)
2. If you look at your nitrate reaction carefully you will note that it begins in an oxidation state of +5 and ends up in an oxidation state of +4.
So, it is reduction and will occur at the cathode. +5 is a reasonably high oxidation state. I don't think N is going to oxidise any higher than
that.
The usual caveats on oxidation numbers apply -- that is, they refer to almost nothing in the real world. They are merely an accounting term so that
you can see what is going on in your process. As such, they are darned useful -- at least for a beginner.
|
|
|
Upsilon
Hazard to Others
Posts: 392
Registered: 6-10-2013
Member Is Offline
Mood: No Mood
|
|
But the reduction potentials alone aren't the only thing at play here, right? You said yourself that it also depends on a number of other factors, a
major one being concentration. H+ is only about one part per 10 million in pure water due to its self ionization. Compared to what the concentration
of Ca2+ would be in the solution, it seems to me that it would be much more likely that it would get reduced instead of the H+. The H2 gas formed
could be explained by the calcium metal instantaneously reacting with the water to produce hydrogen gas and calcium hydroxide.
Am I missing something here?
|
|
|
j_sum1
Administrator
Posts: 6320
Registered: 4-10-2014
Location: At home
Member Is Offline
Mood: Most of the ducks are in a row
|
|
Reduction of water
In acid conditions
2H+ + 2e- --> H2 V=0.00
In neutral or basic conditions
2H2O + 2e- --> H2 + 2OH- V=-0.83
Either way, you are going to produce H2 preferentially to Ca metal.
This makes the -0.83 mark on the table very significant. It marks the dividing line between metals that can be reduced via aqueous chemistry and
those that cannot. You can deposit lots of metals by electrolysis: gold, silver, copper, tin, lead, chromium and zinc. But you never heard of
magnesium plating, sodium plating, aluminium plating or lithium plating. It just doesn't happen. And if it did, the metal would immediately begin to
react with the water.
You are right however that making your solution acidic will increase the chance of liberating H2. You may even prevent some of your other reduction
reactions from occurring.
|
|
|
Upsilon
Hazard to Others
Posts: 392
Registered: 6-10-2013
Member Is Offline
Mood: No Mood
|
|
Ok, gotcha. But I'm still slightly confused about that nitrate half-reaction. Say I have an acidic solution of calcium nitrate. The nitrate
half-reaction given is a reduction. The reduction potential given for that half-reaction is higher than the reduction potential of H+, so does that
mean nitrogen dioxide will evolve at the cathode instead of hydrogen gas? And what of at the anode, just oxygen gas I'm assuming because there's
nothing else in there to be oxidized?
And that is only if the solution is acidic. Say I just have calcium nitrate in pure water. There is no reaction given for the oxidation of nitrate;
therefore it cannot be oxidized? Meaning oxygen will appear at the anode and hydrogen at the cathode?
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
