[sameeralord]

Hello guys ,

Now I have moved to redox equations and I have bit of trouble with it. I'm reading my teacher's notes at the moment. Here is the question

Pottassium manganese (VI) oxide (K2Mn04) reacts with concentrated hydrocholoric acid to form chlorine gas and manganese (II) ions

A. Skeleton equation

MnO4 ^2-  + HCL ------- Cl₂ + Mn ²⁺

Ok I know that Pottassium is a spector ion here

Now the teacher has written half equations

Mn04 ^2-  --------------  Mn²⁺

HCL---------Cl₂

I know how to balance and do the rest from here. My simple question is why is that we take HCl-------Cl₂  rather than Cl------ Cl₂.  Any help would be apppreciate thanks

[Borek]

I would go for Cl^- -> Cl₂. H⁺ takes place in the reaction as well, but the redox part is just about chloride/chlorine.

[sameeralord]

I would go for Cl^- -> Cl₂. H⁺ takes place in the reaction as well, but the redox part is just about chloride/chlorine.

Thank you for the help Borek . So is it wrong having HCl----- Cl₂ or it doesn't make  a differenece. Why is that in this reaction we don't get any product with H⁺ in it. H⁺ can't be a  specator ion. If there was a H in one of the products I could have worked out if H was oxidised or reduced but there is no H product.

EDIT: I just realized with the Mn04 2-  --------------  Mn2+
why is that we take the whole compound Mn04 instead of just the Mn in it.

[Borek]

H⁺ will be neither reduced nor oxidized, but it will be consumed (MnO₄^- contains a lot of oxygen that ends in water molecules, you need hydrogen for that).

IMHO putting HCl in the reaction equation is wrong, as it is 100% dissociated.

However, if you decide to write reaction in molecular form (HCl + KMnO₄ -> MnCl₂ + Cl₂ + KCl + H₂O) HCl is OK.

[Borek]

EDIT: I just realized with the Mn04 2-  --------------  Mn2+
why is that we take the whole compound Mn04 instead of just the Mn in it.

Because you don't have separate Mn and O in solution, but MnO₄^- ions.

[sameeralord]

Really nice answer Borek . After combining the half equations and writing the redox equation I get H20 as a product. So yeah it hasn't gone anywhere ( I was scared that matter can be neither created or destroyed theory would be violoated in this question ) .

Thanks for writing the formula in molecular it further made it clear K+ is the specator ion. I only have one slight question

EDIT: I just realized with the Mn04 2-  --------------  Mn2+
why is that we take the whole compound Mn04 instead of just the Mn in it.

Because you don't have separate Mn and O in solution, but MnO₄^- ions.

With HCL we have separate ions H+ and Cl- yeah. So we can write Cl-----Cl2. But I think you could also write HCl-----------Cl2. I don't think the teacher made a mistake.

[Borek]

With HCL we have separate ions H+ and Cl- yeah. So we can write Cl-----Cl2. But I think you could also write HCl-----------Cl2. I don't think the teacher made a mistake.

In a way that's a matter of style. In both cases it will finally boil down to the same answer. However, if you put ions into one half equation - because everything is 100% dissociated - you should also put everything dissociated as ions into the second equation. It is not a mistake, rather lack of consistency.

Note that this lack of consistency confused you - and that's the best proof that it was not the best approach

[sameeralord]

You are right . It does give you the same answer in both ways. One last question and this topic would also be concluded . Thanks a lot for help so far

Copper metal doesn not dissolve in hydrochloric acid, but can be dissolved by reaction with concentrated nitric acid, HNO₃. The products are Cu²⁺ , nitrogen dioxide, NO₂, and water. Give balanced ionic equation for

b) The reduction process

The half equation is NO₃ ^- = NO₂

Why is that only  NO₃ ^-  is taken and not HNO3
. Is it because H is not oxidised or reduced and NO3 is reduced. The reason why we can break it like this is because it is aqeuous and if it was ion like MnO4- before we can't break right. Am I right?

[Rabn]

I like that this thread has lots of smileys...makes it seem like some solid understanding is happening. I'm a little confused by the way you worded your question sameeralord.  I think I know what you mean though.  Half reactions are written so that the only species that are being oxidized or reduced are in the quation.  The half reaction you wrote is incomplete.  The complete half-reaction is:

NO₃^- + 2 H⁺ + e^- <-> NO₂ + H₂O

You really should get in the habit of writing out the complete half reactions now so that later on you don't get tripped up. You are definitely getting it though.  Good work.

[sameeralord]

I like that this thread has lots of smileys...makes it seem like some solid understanding is happening. I'm a little confused by the way you worded your question sameeralord.  I think I know what you mean though.  Half reactions are written so that the only species that are being oxidized or reduced are in the quation.  The half reaction you wrote is incomplete.  The complete half-reaction is:

NO₃^- + 2 H⁺ + e^- <-> NO₂ + H₂O

You really should get in the habit of writing out the complete half reactions now so that later on you don't get tripped up. You are definitely getting it though.  Good work.

Thanks a lot Rabin for your help . Sorry I didn't make it clear before. I understand how to balance the equation. My problem is I don't know how to pick which ones are the half equations. Balancing I have no problems.

Why is that we take

NO3---------- NO2

why can't we take

HNO3--------- NO2

How do I know I have to eliminate the H. Thanks for your help anway Rabn. Rereading my post before even I can't understand what I just said

[Astrokel]

Why is that we take

NO3---------- NO2

why can't we take

HNO3--------- NO2

How do I know I have to eliminate the H.

Hey!

As stated by Rabn, you only write ions that are being reduced or oxidized in half equations. H+ does participate in the reaction BUT THEY ARE NOT REDUCED OR OXIDIZED. So i'd say if H+ is being reduced or oxidized, then you include it in the half eq, otherwise no.

Kelvin 

[Borek]

As stated by Rabn, you only write ions that are being reduced or oxidized in half equations. H+ does participate in the reaction BUT THEY ARE NOT REDUCED OR OXIDIZED. So i'd say if H+ is being reduced or oxidized, then you include it in the half eq, otherwise no.

That's not correct. Take a look at Rabn half reaction:

NO₃^- + 2 H⁺ + e^- <-> NO₂ + H₂O

It contains H⁺, even if it is not being reduced nor oxidized.

Define what do you mean by half reaction.

[Rabn]

Where you are running into problems is that you aren't writing the complete half reaction as I did, which you can find in any reduction table as you know.  As for why you don't use HNO₃...it is because in reality it is a strong acid so it will completely dissociate.  Therefore the reaction will look like this before redox occurs:

(H+) + NO3- + (reducing agent) ...note the reducing agent may dissociate in aqueous solution as well

So when you think about it this way, the H+ is more of a spectator ion and is not written in the redox half reaction. It may be the case that it forms water but it also may diffuse away.   

[Astrokel]

As stated by Rabn, you only write ions that are being reduced or oxidized in half equations. H+ does participate in the reaction BUT THEY ARE NOT REDUCED OR OXIDIZED. So i'd say if H+ is being reduced or oxidized, then you include it in the half eq, otherwise no.

That's not correct. Take a look at Rabn half reaction:

NO₃^- + 2 H⁺ + e^- <-> NO₂ + H₂O

It contains H⁺, even if it is not being reduced nor oxidized.

Ok i got what u mean, guess i phrased my last sentence incorrectly. well im trying to say that its the NO3- thats being reduced and not the H+, so u take NO3- --> NO2 instead of HNO3 in half eq.

thanks borek for pointing out my mistake,

Kelvin

[sameeralord]

Thank you Rabn, Borek and Kelvin for your help . Ok I think I get it now. So as Borek has pointed before using HNO3 would get you the same answer and using that would be lack of consistency. That makes sense because my first question was teacher's notes and the next one was text book question.
So one last question. So is it only if the reactant is in aqueous form I can break it and use one thing for the half reaction.What I mean is HNO3 is aqueuous and if only nitogern oxidised/reduced then I can use NO3 for my half reaction. But if the reactant was CO2 (gas) and only carbon oxidised I can't use O only for my half equation I have to use whole CO2. Am I right. If I am another topic conlcuded . Thanks heaps guys!!    Great forum you got here