Sciencemadness Discussion Board - Handling Redox Equations

Handling Redox Equations

CHRIS25 - 2-5-2014 at 00:05

I thought I had understood reasonably well now the redox equations and how thing are balanced, but I am not so sure with this one:

2Al + 3H₂⁺¹SO₄^2- = Al₂⁺³(SO₄)₃^2- + 3H₂⁺¹

Al has lost 3 electrons and therefore is oxidized and has become more positive in its combination with the sulphate.
The sulphate has gone from one molecule to three molecules as it joins with the Al. But its charge has remained the same except that when combined with the Al it makes the whole compound more negative, Yet the sulphate has NOT been reduced, so if there is no reduction then there is no oxidation. So obviously I am forgetting something or missing something. Now my confidence is shaken.

Mistake found:
Yep it's the hydrogen sorry, that has become neutral by gaining an electron and being reduced

[Edited on 2-5-2014 by CHRIS25]

[Edited on 2-5-2014 by CHRIS25]

[Edited on 2-5-2014 by CHRIS25]

[Edited on 2-5-2014 by CHRIS25]

Mildronate - 2-5-2014 at 01:20

1)write half reactions
2)balance all other elements except H or O
3) Balance O by adding H2O
4) Balance H by adding H+
5)balance charge by adding e-
6) Sum half reactions, simplify and check

[Edited on 2-5-2014 by Mildronate]

CHRIS25 - 2-5-2014 at 01:49

)write half reactions).....
I have read about this and understand the theory, but believe it or not I am not really doing this part well, can't grasp it for some reason, it gets me all knotted up. Yet if I simply look at the Whole equation with all reactants balanced in their place and then go about with the electonegativity and which is oxidized and what has lost electrons or gained etc, I find that easier and when I have tested myself with some questionaires I found online I always seem to get those right, then I look at the half reactions and get the gist but it loses me to be honest. Yes I know I am amateur, and maybe there is something that needs to click into place here....

[Edited on 2-5-2014 by CHRIS25]

Mildronate - 2-5-2014 at 06:23

1) write half reactions:

Al0 => Al3+
H+ => H20

2) Balance all other elements than O or H

Ok, nothing to balance

3) Balance O, by adding H2O

Ok, nothing to balance

4) Balance H, by adding H+

2H+ => H20

5) Balance charge in half reactions by adding some electrons

Al0 => Al3+ + 3e-
2H+ + 2e- => H20

6) Add integers to half reactions

2|Al0 => Al3+ + 3e-
3|2H+ + 2e- => H20

7) Sum half reactions

2Al0 +6 H+ => 2Al3+ + 3H20

blogfast25 - 2-5-2014 at 09:29

Quote: Originally posted by CHRIS25

) Yes I know I am amateur, and maybe there is something that needs to click into place here....

[Edited on 2-5-2014 by CHRIS25]

Balancing redox reactions is quite different from reactions where no changes in oxidation states occur.

The key point is to identify which species gets oxidised (Al, in your reaction) and which gets reduced (H+ in your reaction; strictly speaking H3O+

. For the 'amateur' that can be the biggest challenge of the whole thing.

Mildronate's 'algorithm' is correct and never fails if you apply it correctly. "Practice makes perfect."

For your reaction we get:

Al(s) + 3 H3O+(aq) === > Al3+(aq) + 3/2 H2(g) + 3 H2O(l)

Note how the sulphate ions don't participate: we call them 'spectator ions'; they are present but don't do anything. But you can add them to eliminate the charges on both sides, if you wish. The reaction is the same with HCl: then the chloride ions play the part of spectators.

[Edited on 2-5-2014 by blogfast25]

CHRIS25 - 2-5-2014 at 10:12

Thanks both of you. I think my first post here demonstrates that I understand the oxidation reduction with regards to balancing the sharing/donating/losing of electrons. I am looking at the half reactions and am still puzzled how it is arranged, for instance you have added 3H₂O(l) and Mildronate threw in a 6H⁺ ; and 3|2H+ + 2e- => H20?? (I mean the 3/2H⁺ - which side of the equation does that come from, the reactant or the result after the reaction? I have yet to understand how to re-adjust my perfectly good balanced equation With its charges included, and then convert that to half reactions. Some of what you have written I fully understand, so I am busy looking at video tutorials on this yet again.

[Edited on 2-5-2014 by CHRIS25]

blogfast25 - 2-5-2014 at 11:37

Quote: Originally posted by CHRIS25

I thought I had understood reasonably well now the redox equations and how thing are balanced, but I am not so sure with this one:

2Al + 3H₂⁺¹SO₄^2- = Al₂⁺³(SO₄)₃^2- + 3H₂⁺¹

Let’s try again.

What you wrote above really makes little sense and would confuse even Einstein.

When H2SO4 dissolves in water, it completely (well, more or less!) dissociates as follows:

H2SO4(aq) + 2 H2O(l) === > SO42-(aq) + 2 H3O+(aq)

As the sulphate ions get neither oxidised nor reduced, they don't not take part in the reactions.

Oxidation half reaction:

Al(s) === > Al3+ + 3 e-

Reduction half reduction it its most primitive form:

H3O+(aq) === > H2(g)

Balance O by adding 1 H2O to the right:

H3O+(aq) === > H2(g) + H2O(l)

Balance H: there’s one too many on the right hand sight;

so replace H2(g) by ½ H2(g):

H3O+(aq) === > ½ H2(g) + H2O(l)

Makes this electrically neutral by adding 1 e- to the left:

H3O+(aq) + e- === > ½ H2(g) + H2O(l)

This is the half reaction for the reduction of the oxonium ion to hydrogen and water.

Home work: write the balanced redox reaction for the oxidation of oxalic acid to CO2 by permanganate in acid conditions. Permanganate is reduced to Mn2+.

[Edited on 2-5-2014 by blogfast25]

Mildronate - 2-5-2014 at 13:10

You can look at some general chemistry books how to write this equations. I like Brown LeMay Bursten book ''Chemistry the central science''. I and blogfast25 write same equation, only he uses H3O+ (actually more correct)instead of H+ like I and he calculate it for 1mole Al, I for 2 mole, because then I get round numbers.

[Edited on 2-5-2014 by Mildronate]

CHRIS25 - 2-5-2014 at 14:02

Ok. Blogfast, It's late, but I will take up your challenge and do my homework. As a side note, most of the redox lessons I followed simply showed you how to add the positive and negative charges and balance things out according to periodic table and redox rules. I thought I had done a good job especially since I had learned the electronegativity concepts and bond polarity - guess not

Homework I will do tomorrow, but for now, having gone through your demonstration with a fine tooth-coombe I am utterly mystified by the very last bit, Up until this point I fully understand everything, but this ....

Makes this electrically neutral by adding 1 e- to the left:

H3O+(aq) + e- === > ½ H2(g) + H2O(l)

The hydronium on the left is positively charged, the Hydrogen and water on the right are neutral. The right hand side equals 3 Hydrogen molecules and one oxygen molecule, the left has 3 hydrogen molecules and one oxygen molecule. They appear equal, so why have you added one electron on the left side, it's this "+ e-" that throws me off course. This electron is attached to what? It's not a part of the oxygen since that lost 2 electrons and became more positive, and is holding 3 hydrogen atoms together and is now a polar covalent bond (hope I have at least that rightly understood), and I do not see how one electron is added to 3 Hydrogens.

[Edited on 2-5-2014 by CHRIS25]

Metacelsus - 2-5-2014 at 16:16

This is what's known as a half-reaction. The other half is Al -> Al³⁺ + 3 e^-.

CHRIS25 - 3-5-2014 at 02:55

@ Cheddite: Thankyou but I knew that. Having watched a very good explanation I now understand the missing link in my brain!....it was the nomenclature, plus means plus, but in half reactions it has nothing to do with adding, and thanks to a video he used words like: combining, donating with regards to the + sign, at that moment I had an epthanafinithiny. Sometimes instructors instruct in abstract concepts and this is where it gets confusing.

@Blogfast: This was too easy

Ok My planning process - I have been exhaustive so that you may easily see where I am going wrong if indeed I am: First I write a balanced equation thus: and since it takes place in water I have to calculate that in:

1) C₂H₂O₄ + KMnO₄ +H₂O= CO₂ + K + Mn + 4H₂O

then I go about putting in the charges

2) (+1+1-2) + (+1+7-2) + (+2-2) = (+1-2) + (0) + (0) + (+1-2) (these are not Totalized charges)

then I see what has changed with regards to electronegativity:

Carbon has lost 1 electron and is oxidized and combined with Oxygen (although carbon is now more positive the whole bond is a neutral polar covalent bond);
Manganese has gained 7 electrons and is neutral and becomes elemental;
Potassium has gained 1 electron and is neutral and elemental;
Water has increased on the right, due to an extra hydrogen being liberated from the Oxalic acid and has bonded with the oxygen as a result of the reduction (disassembling) of the potassium permanganate.

Now the fun starts:

3)
C⁺¹ = C⁺² + e-
Mn⁺⁷ + 7e- = Mn⁰
K⁺¹ + e- = K⁰

[Edited on 3-5-2014 by CHRIS25]

blogfast25 - 3-5-2014 at 05:27

Quote: Originally posted by CHRIS25

The hydronium on the left is positively charged, the Hydrogen and water on the right are neutral. The right hand side equals 3 Hydrogen molecules and one oxygen molecule, the left has 3 hydrogen molecules and one oxygen molecule. They appear equal, so why have you added one electron on the left side, it's this "+ e-" that throws me off course. This electron is attached to what?

It's not attached to anything. Look at a redox reaction as an electron transfer phenomenon: here Al 'provides' the electrons needed to reduce THREE H3O+ to hydrogen and water. So the electrons get transferred from the oxidised species to the reduced species. That is the ESSENCE of a redox reaction.

Re. permanganate and oxalic acid, some of your reasoning is correct, some not but it's very hard to get to the correct equation using your method. You assume that K+ is reduced to K(0). Trust me: IT IS NOT. K+ is a spectator ion.

1) Reduction reaction: (I'm going to dispense with the 'sub/sup' tags for a minute and will use Mildronate's simplified algorithm)

MnO4(-) === > Mn(2+)

This lacks for O on the right so add four H2O:

MnO4(-) === > Mn(2+) + 4 H2O

Now the left is lacking 8 H, so add 8 H+:

MnO4(-) + 8 H+ === > Mn(2+) + 4 H2O

Now look at the charges: - 1 + 8 = + 7 on the left
+ 2 on the right. We need to add 5 e- on the left to balance that:

MnO4(-) + 8 H+ + 5 e- === > Mn(2+) + 4 H2O (Eq.1)

This is the balanced reduction half reaction for MnO4(-) to Mn(2+)

2) Oxidation reaction:

C2O4H2 === > CO2

Balance in C by adding 1 CO2 on the right:

C2O4H2 === > 2 CO2

Oxygen is now also balanced

Balance H by adding 2 H+ to the right:

C2O4H2 === > 2 CO2 + 2 H+

Balance charge by adding 2 e- to the right:

C2O4H2 === > 2 CO2 + 2 H+ + 2 e- (Eq. 2)

This is the balanced oxidation half reaction.

3) To obtain the complete redox reaction, multiply Eq. 1 by 2 and Eq. 2 by 5 and add everything up. The electrons will now be balanced and drop out. Some H+ will also be redundant and we obtain:

2 MnO4(-) + 5 C2O4H2 + 6 H+ === > 2 Mn(2+) +10 CO2 + 8 H2O

Perfectly balanced. You can now add some K+ and some sulphate, if you wish:

2 KMnO4 + 5 C2O4H2 + 3 H2SO4 === > 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4

[Edited on 3-5-2014 by blogfast25]

CHRIS25 - 3-5-2014 at 06:54

Well I must admit, I am certainly a million miles away from being able to do this sort of thing. For start off I got the balanced equation wrong right at the start. Second, I had no idea sulphuric acid was needed. So I did not have a chance in hell of getting it right. Thirdly, the video demonstrations make it look so simple and logical, yet this reaction is complex and does not come close to what I have learned. For example, I am derailed on just the first line.

MnO4(-) === > Mn(2+)

This lacks for O on the right so add four H2O:

MnO4(-) === > Mn(2+) + 4 H2O

I was taught to do this:
Take the KMnO4: K=+1 Mn=+7 O4 adds up to +8 and work from here and see what has changed on the right. I can understand this far and understand Redox, but half reactions are a lot more complex to write out than I thought.

What you have done baffles me. Sorry, I don't get it and I think doing half reactions is too advanced for me at the moment. Guess I will just plod along until something in the coming months clicks. I can follow your reasoning and see what is going on, but I could never do it myself as you saw. i will go through this though and use it as a foundation to try and study further so you have not wasted your time, and I appreciate the extensive explanations Blogfast.

[Edited on 3-5-2014 by CHRIS25]

blogfast25 - 3-5-2014 at 10:44

One minor point:

"Second, I had no idea sulphuric acid was needed."

2 MnO4(-) + 5 C2O4H2 + 6 H+ === > 2 Mn(2+) +10 CO2 + 8 H2O

... calls for 6 H+. These can only be provided by an acid. It doesn't have to be sulphuric, I just chose it to be that one because in practical executions of that reaction we normally use good old sulphuric acid.

CHRIS25 - 3-5-2014 at 11:19

I have so much to understand......

aga - 3-5-2014 at 11:41

Quote:

I have so much to understand......

Same here - i get the principles, but even beginning on this is still far ahead of where i am.

The scariest part is now knowing that 'Spectator Ions' exist.

How is it possible to tell that a particular ion will not take part in the reaction ?

blogfast25 - 3-5-2014 at 11:59

Quote: Originally posted by aga

How is it possible to tell that a particular ion will not take part in the reaction ?

That's an ace question and not so easy to answer.

Most chemists will go by general/knowledge/experience etc. But there are cases where most of us will have to scratch our heads to figure out whether a species will take part in oxidation/reduction reactions or not.

This is usually done by postulating a hypothesis: 'Can species A be reduced or oxidised by species B?' To find the answer we have to figure out whether ΔG < 0 or not. And to do this we use so-called 'reduction half potentials'. Combining the oxidation and reduction half potentials for the hypothesised reaction then tells you whether ΔG < 0 or ΔG > 0 for that reaction.

In the specific case of K+ and sulphate anions in the permanganate/oxalate redox reaction, we know both K and S are at their highest oxidation state, so they cannot be further oxidised. And there are no reducing agents present, only permanganate and that's an oxidiser, so no reduction is possible either for H+ or sulphate. So both just sit there, doing nothing...

[Edited on 3-5-2014 by blogfast25]

CHRIS25 - 3-5-2014 at 12:01

I don't feel qualified to say this Adrian but my confidence is building. All I can say is that I begin by balancing the equation and assigning charges. For example in the above situation: 2 KMnO4 + 5 C2O4H2 + 3 H2SO4 === > 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4 I can see now that on the left K (potassium) started off as having a charge of +1 and there are two of them, then on the right it still has a charge of +1 BUT there are Two of them in the K2SO4. So from this alone I see that potassium does not take any part in the oxidation or reduction, it simply got divorced from the Mn02 and they both went for a walk and found a new love - a sulphate. I do not know if you follow this, it's the most simplest explanation I can give at the moment.

[Edited on 3-5-2014 by CHRIS25]
Having just read Blogfast's response my confidence just went down AGAIN

[Edited on 3-5-2014 by CHRIS25]

Etaoin Shrdlu - 3-5-2014 at 12:08

Aga, you can tell because the spectator ions are the ones which do not change states during the reaction. A classic example is the reaction of silver nitrate with sodium chloride.

AgNO₃ + NaCl --> AgCl + NaNO₃

The silver chloride is insoluble and will precipitate out of aqueous solution, the other chemicals will stay in solution. So the ions behave like so:

Ag⁺(aq) + NO₃^-(aq) + Na⁺(aq) + Cl^-(aq) --> AgCl(s) + Na⁺(aq) + NO₃^-(aq)

NO₃^- and Na⁺ never change states. They're always aqueous. Therefore, they're the spectator ions.

EDIT: Seeing blogfast's post I'm suddenly uncertain whether the question was about parsing known reactions or understanding whether unknown ones will go, but I'll leave this up anyway.

[Edited on 5-3-2014 by Etaoin Shrdlu]

aga - 3-5-2014 at 12:09

@blogfast25 : Ah! I am glad that Head Scratching is involved. I can do that.

So see if the Gibbs says it is likely to react or not, then test by experimentation ?

@CHRIS25 : that also kinda makes sense. It just moved. It's overall charge/contribution remained the same.

@Etaoin Shrdlu: -

Quote:

The silver chloride is insoluble and will precipitate out of aqueous solution

Another thing 'known' ! how is that known : is it from a table somewhere, or is it a predictable phenomenon ?

[Edited on 3-5-2014 by aga]

Etaoin Shrdlu - 3-5-2014 at 12:22

Quote: Originally posted by aga

Another thing 'known' ! how is that known : is it from a table somewhere, or is it a predictable phenomenon ?[Edited on 3-5-2014 by aga]

*cough*Wikipedia*cough*

There's probably a way to predict it, and you can make decent guesses based on the ions, but I'm betting nearly every combination of common ions has had at least aqueous solubility figured out already.

There are some rules of thumb; silver salts tend to be insoluble, alkali hydroxides tend to be soluble, non-alkali hydroxides tend to be insoluble, halide salts tend to be soluble...but there are so many exceptions especially when a "generally insoluble" is combined with a "generally soluble." Much easier to check.

EDITED for missing word.

[Edited on 5-3-2014 by Etaoin Shrdlu]

blogfast25 - 3-5-2014 at 12:22

aga:

Always test by experiment.

For insolubility there are some broad guidelines. Predicting from theory is difficult, also because solubility is a relative thing. Even insoluble stuff like AgCl has some solubility.

aga - 3-5-2014 at 12:33

Quote: Originally posted by Etaoin Shrdlu

Wikipedia

So solubility has been tested by experimentation and tabulated already, but if in doubt, Test for yourself.

Many thanks. Now have a table (blush) ...

CHRIS25 - 3-5-2014 at 12:51

Adrian, I always keep Wolfram alpha and wikpedia as tabs on my safari so they load within seconds. Here is a guide for you, pick on Aluminium Hydroxide in wikpedia. Look on right side column down where it says solubility. Most of the time the solubility is given in grams per 100mLs at three different temperatures. Wolfram alpha is also a good source of detailed info. And one last link I always keep as a quick tab is this: http://www.ptable.com/ thoroughly recommend it. And this one http://www.webqc.org/balance.php?reaction. Though I always calculate first then check, since I want to quicken my brain and stay alert. You might also want this one http://www.webqc.org/molecular-weight-of-Na2(SiO2)7.html. Once again I calculate and double check, and until the day I am awarded the nobel prize for atomic chemistry I will always check my calculations.

[Edited on 3-5-2014 by CHRIS25]

Mildronate - 3-5-2014 at 12:53

It doesn't matter is reaction possible to this way or to that way by Gibs potential, redox balancing method works also if reaction doesnt happen.

Mildronate - 3-5-2014 at 12:55

For solubility you look at solubility costants in handbook not in wiki, ok sometimes wiki is faster

CHRIS25 - 3-5-2014 at 13:25

Ok so I googled solubility constants, and great, there is a ton of data...only problem now is that I have to learn what on earth all those strange alien numbers mean, my brain is already twice the size as it was yesterday....

blogfast25 - 4-5-2014 at 04:45

For AgCl I find a value for Ksp = [Ag+] x [Cl-] = 1.77 x 10-10 (= 0.000000000177).

What does this mean? If in a solution the PRODUCT of [Ag+] and [Cl-] exceeds 1.77 x 10-10, the AgCl will start precipitating.

If you take some solid AgCl and allow it to dissolve in water, the maximum concentration is governed by the Ksp.

Since as AgCl(s) < === > Ag+(aq) + Cl-(aq), this also means that in such a solution [Ag+] = [Cl-], so we have:

Ksp = [Ag+] x [Cl-] = [Ag+]2 = 1.77 x 10-10

It follows that [Ag+] = SQRT (1.77 x 10-10

= 1.33 x 10-5 M or 0.0000133 M is the theoretical maximum solubility of AgCl in pure water. We call that 'insoluble' but there is some AgCl in solution, just not very much...

It also explains why when you add a bit of silver nitrate solution to a solution that contains some chloride you get a precipitation of AgCl, a classic test for chloride ions.

Brain + 10 % size, now reaching explosion limit.

[Edited on 4-5-2014 by blogfast25]