Half equation question

[dekoyl]

When we do redox half equations, are we meant to omit the spectator ions? And do we only omit the spectator ions at the end, when it's been balanced?
For example, I have to give reduction/oxidation half equations of the following:


I know S is oxidised and O is reduced. But when I try to write the reduction half equation it doesn't work out with the answers.

[NE2000]

First identify the key elements, in this case we are looking at the oxidation of S
Then balance oxygens, hydrogens and charge


Same thing again

I'm not sure what you mean by spectator ions in this case. Essentially the half equations show the electron transfer that has occurred and that is their function. As the other substances are present in liquid/solid form there are no spectator ions in that case.

If you put the equations back together, you get the same and on both sides, which you can then omit as long as they are balanced. Is that what your question was?

[ed_saifa]

To balance redox half equations

1. Balance the atoms


2. Balance the with


3. Balance the with


4. Balance the charges with e^-

[dekoyl]

If you put the equations back together, you get the same and on both sides, which you can then omit as long as they are balanced. Is that what your question was?

Yeah, kinda. I thought the Pb was a spectator ion (as it doesn't change state) but I think I got the theory confused.  (Consequence for not learning chemistry 1/2 properly.)

Thanks NE2000 and ed_saifa ^^

I suck at chemistry =[

[shinny]

From my understanding, a spectator ion isn't simply something that doesn't change state; it's something that doesn't change at all. In this case, although the state doesn't change, the Pb certainly does. The 'no state change' rule would only apply for aqueous stuff because the ions are dissociated, and even if a displacement occurs, the behaviour of the ions doesn't change as they are still just all separately dissolved inside the solution. Secondly, well, there's ION in the name so the most obvious thing is that spectator ions must be well..ions?

EDIT: From what I see, the reason why this occurs is because only aqueous ions are the ones capable of changing form without really changing at all since they just dissociate anyway. Molecules in any other state must change to be included in the overall equation anyway, otherwise they wouldn't have reacted and they'd just cancel out.

[dekoyl]

From my understanding, a spectator ion isn't simply something that doesn't change state; it's something that doesn't change at all. In this case, although the state doesn't change, the Pb certainly does. The 'no state change' rule would only apply for aqueous stuff because the ions are dissociated, and even if a displacement occurs, the behaviour of the ions doesn't change as they are still just all separately dissolved inside the solution. Secondly, well, there's ION in the name so the most obvious thing is that spectator ions must be well..ions?

Ah okay thanks for clearing that up. Like I said I got stuff confused Was carrying my limited knowledge from doing displacement/precipitation reactions to redox equations

[NE2000]

EDIT: From what I see, the reason why this occurs is because only aqueous ions are the ones capable of changing form without really changing at all since they just dissociate anyway. Molecules in any other state must change to be included in the overall equation anyway, otherwise they wouldn't have reacted and they'd just cancel out.

I just look at it from a collision theory perspective. If the ions are dissociated in both the left and right side of the equations, they aren't doing any collisions during the reaction. If a solid is involved, then even if the same atoms are present in a solid molecule on the other side, the atoms of that solid will be participating in a collision that results in the reaction.

[shinny]

If a solid is involved, then even if the same atoms are present in a solid molecule on the other side, the atoms of that solid will be participating in a collision that results in the reaction.

I'm probly just misunderstanding you, but if the solid doesn't change, then how is it even reacting? Just because other stuff collides with it doesn't mean its reacting. Something would have to change wouldn't it?

[NE2000]

If a solid is involved, then even if the same atoms are present in a solid molecule on the other side, the atoms of that solid will be participating in a collision that results in the reaction.

I'm probly just misunderstanding you, but if the solid doesn't change, then how is it even reacting? Just because other stuff collides with it doesn't mean its reacting. Something would have to change wouldn't it?

What I meant is if is in the reactants and is in the products then don't omit the . Is that right?

If were in both reactants and products and were not to change in the course of the reaction then I think you would omit it in a net balanced equation. Not sure though because for a lot of photosynthesis reactions I've read in bio they've got in the reactants and in the products because all 12 water molecules were used up in the course of the complex reaction and 6 water molecules were produced in a different stage of the reaction.

[shinny]

Oh, yeh that's what I was trying to say. The basis of this is that and are totally different. That's what I meant by a change. What isn't change is when say i.e. react to become other ions (e.g. and or something), but in their new aqueous form, still technically remain as

[NE2000]

Oh, yeh that's what I was trying to say. The basis of this is that and are totally different. That's what I meant by a change. What isn't change is when say i.e. react to become other ions (e.g. and or something), but in their new aqueous form, still technically remain as

OK all good then

But the reaction you picked was interesting. If you had a reaction:


Then what of the ionic equation? Would such a reaction even occur (or would the acid rather react with water or something)? Do the spectator ions actually react at all or are they just there 'spectating' as the reaction occurs, because in the above case (if a reaction were to ever occur like that) then the solution before and after the reaction would not be different at all I would guess. I once got this type of equation in a test this semester and I got marked right for saying 'all are spectator ions' but it still bugs me.

[shinny]

From my understanding, I think that reaction can occur, and I've seen it come up a few times where something will simply have no ionic equation. Like in the above case, this seems to happen when everything's aqueous, and you basically just have a displacement of ions occurring. I'm not too great at chem overall, so I'm actually quite lost as to how both sides of that equation differ if they can equally be written as the sum of the individual ions anyway. I'm pretty sure it's something to do with the bonding between the particles, but yeh I'm not too sure about the net effect.

[dekoyl]

I got another two..
Write the oxidation and reduction half equations.

1) Acidified potassium permanganate solution changes from deep purple color to a colorless Mn(2+) solution with the addition of iron(II) sulfate solution. Fe(3+) forms (rusty brown color) in the solution.

Reduction Alright I think...
Oxidation What the? I probably made a mistake somewhere that I can't spot =\

2)H2S gas is bubbled through a solution of iron(III) nitrate, a pale yellow precipitate forms.

I don't know how to write the half equations for this at all. I don't think you can, either? =\

[Wizard]

1) In the first question, your oxidation and reduction reactions are slightly wrong

MnO₄^- + 8H⁺ + 5e^- ---> Mn²⁺ + 4H₂O   (REDUCTION)

Fe²⁺ ---> Fe³⁺ + e^-                                            (OXIDATION)

Sorry about having no states, I cant be bothered Thats the general idea though

[Collin Li]

Yeah, you should get rid of the spectator ion.

Start with: , and work from there.

Similarly: and work from there.

To do the second question, first identify the oxidation states of all the species, to figure out precisely which species is oxidising and which one is reducing. When you do this, you will find out that the oxidation states have not changed for any of the species. This is a trick question. (Normally, you'd then write the half equations after figuring out exactly what oxidises and what reduces)

However, are you sure the question didn't ask for an "ionic equation"?