Oxidation numbers (1 Viewer)

[mitch_f1]

Hey all,
I am having a tiny bit of trouble with oxidation numbers, and was wondering if someone could help me out please.

I Understand the basic ones (MN4 2- for instance has an OxN of +6) thanks to the introduction of a simple formula following a quick google search (http://www.chemguide.co.uk/inorganic/redox/oxidnstates.html). But the larger compounds have me stumped.

Here is one I need help with please: The question is: " For the following, assign oxidation numbers to the underlined atom."

(the 'N' is underlined) [Cu(NH3)4] 2+
So I am just stuck as to how to work it out. I have tried it with this new mothod i've found (above) but it doesn't work

There is possibly (almost certainly) an easier way to do this, so by all means introduce it to me.

ANY help apprecieated
Thanks all
Mitch

P.S. Is it possible to have this compound?: [Fe(CN)6] 3-

 

[Riviet]

Hi mitch. ^^

The reason why Cu(NH₃)₄ ²⁺ has a valence number of 2+ is because the sum of the valence number of Cu (2+) and (NH₃)₄ (0) is +2. Therefore I think you only need to look at (NH₃)₄ for finding the O.N of N. Now according to the rules, H has a O.N of +1 in compounds. There are three of these in the compound so N+3=0, since the sum of the separate valencies need to add up to 0. therefore the oxidation number of N is -3. That should be right.

Is it possible to have this compound?: [Fe(CN)6] 3-

I'm not sure, but I would question the (CN)₆ bit as that is not very common.

I hope that helped.

Riv.

 

[mitch_f1]

ok, cool thanks heaps. Makes sence.

 

[mitch_f1]

One more question, might as well keep it to the same topic:

This time it's to do with redox reactions. The question is in a question booklet our teacher gave us to revise with. It says to write a ox and red 1/2 equations. This is quite simple really, but it's the answers that are given that have me perplexed. Here is the question:

MnO4^- + Pb2+ >>> MnO2 + Pb4+

here is my answers:
Mn3+ +e- >>> Mn2+ (reduction)
Pb2+ >>> Pb4+ +2e- (oxidation)

I left out the oxygen cause it's a spectator ion (doesn't change its charge) But the answer states it as this:
3e- + MnO4^- + 4H+ >>> MnO2 + 2H2O


Where did the Hydrogen come from,a nd why is the O and the H included?

thanks, and sorry if i'm being a nuisance
Mitchell
Happy new years all BTW

 

[insert-username]

MnO4^- + Pb2+ >>> MnO2 + Pb4+

Do you mean MnO^4- + Pb²⁺ >>> MnO₂ + Pb⁴⁺?

3e- + MnO4^- + 4H+ >>> MnO2 + 2H2O

Are you sure that's the right answer? That doesn't really make sense to me either since there's no hydrogen to begin with AND it's not written as reduction/oxidation half reactions...


I_F

 

[mitch_f1]

Ok, how do you do those superscript numbers in this?

And yes, I am 100% certain that is the answer, they have reduction equation, then they list it. They have done it before as well.....in the beginning of the booklet they did the exact same; added a hydrogen, and spectator ions.

 

[xvelidras]

Though it may be true in the reduction of MnO4- to MnO2, the oxygen's oxidation number does not change, it is NOT a spectator ion, since MnO4- and MnO2 are individual compounds that will not dissociate in aqueous soln (unlike NaCl for e.g. in which components may be treated separately) they must be treated as a whole, not the Mn and O parts separately. so your basic unbalanced reduction half equation is

MnO4^- --> MnO2

However, this is not balanced, mass wise, nor charge wise. To balance the excess oxygen on the left side, we add water.

MnO4^- --> MnO2 + 2H2O

Yet, with the addition of water, there is now an excess of hydrogen on the right hand side. we balance this by adding H+

MnO4^- + 4H+ --> MnO2 + 2H2O

Hence, we have completed balancing the mass discrepancies, now for charge

MnO4^- + 4H+ + 3e- --> MnO2 + 2H2O

Hence,

But the answer states it as this:
3e- + MnO4^- + 4H+ >>> MnO2 + 2H2O

 

[Riviet]

Wow, that was very well explained xvelidras, I understood all of what you said. Well done.

 

[mitch_f1]

So wait, in the redox reaction, are you meant to ignore the H and H2O???

xvelidras that was an awesome explanation, so easy to understnad and all. Well done, thanks HEAPS.

Mitch

 

[Riviet]

So wait, in the redox reaction, are you meant to ignore the H and H2O???

If you are trying to work out whether oxidation or reduction has occured, the H and H₂0 can be neglected since they are there for the main purpose of balancing the number of H's and O's on the LHS and RHS of the equation.

 

[xvelidras]

So wait, in the redox reaction, are you meant to ignore the H and H2O???

No, you're not. the H2O and H+ are required for the reaction to proceed.

EXT: There is an alternative soln:

MnO4^- + 4H+ + 3e- --> MnO2 + 2H2O

is only applicable in acidic solns. We may convert this equation for basic solns by adding OH- to both sides

MnO4^- + 4H+ + 4OH- + 3e- --> MnO2 + 2H2O + 4OH-

simplifying:
MnO4^- + 2H2O + 3e- --> MnO2 + 4OH-

MnO4^- + Pb2+ >>> MnO2 + Pb4+

Your teacher gave out the question as above so as not to provide the answer in the question, not because the H+ and H2O aren't required in the redox reaction =P

 

[angelxtearz]

?

What about this q? (Name the oxidant + reductant, identify the species that has oxidised and reduced)

1) I^2O^5 + 3CO --> I^2 +3C0^2

the ^ represents a subscript

I got :
oxidised: C
red": I
oxidant- O
red"- C

i dunno, say a element moves from the ON of -1 to +5 how would you kno whether that element has oxidised or reduced. or is it (non metals gain, metals lose)

thankyooz

 

[pLuvia]

If it goes up electrons then it is Reduced, if it loses electrons it Oxidised Hence the acronym OILRIG - Oxidation is loss Reduction is gain of electrons

So in that case it would be Oxidised

 

[Riviet]

oxidised: C Correct!
red": I Correct!
oxidant- O Incorrect
red"- C Correct!

Explanation for incorrect answer: Oxygen does not change oxidation number, since its ON is -2 in each product/reactant in the equation. Therefore, it is a SPECTATOR, which doesn't change ON, hence not part of the redox reaction occuring. From your first answer, you said carbon is oxidised, hence it is the reductant, because a reductant causes something else to be reduced, therefore it is oxidised itself.
The correct answer is Iodine, because from your second answer, you said that Iodine is reduced, hence it is the oxidant, because a oxidant causes something else to be oxidised, therefore it is reduced itself.

i dunno, say a element moves from the ON of -1 to +5 how would you kno whether that element has oxidised or reduced. or is it (non metals gain, metals lose)

You will never get this wrong if you think of it like this:

--------->----------->------------>---------->--------->OXIDATION

-NEGATIVE -6|-5|-4|-3|-2|-1|0|+1|+2|+3|+4|+5|+6 POSITIVE+

REDUCTION<-----------<-------------<-------------<--------------

* If the ON changes from one number to another on its RIGHT(->), then oxidation has occured.

* If the ON changes from one number to another on its LEFT(<--), then reduction has occured.

In your example, if the ON of something changes from -1 to +5, then you look at the number line and see that the number has moved from -1 to the right, where +5 is, therefore oxidation has occured.

I hope that helps.

 

[mitch_f1]

ok, I have another one, please.

Copper metal is added to 20mL of 2M nitric acid and heated. Write the oxidation and reduction half equation, and the redox equation.

Ok, I know to ignore the numbers, because they have nothing to do with what the question asks. I have got as far as writing out the redox reaction, as far as I think it is:

Cu + 2HNO3 --> Cu (NO3)2 +H2
but, the answers state it as this:

3Cu + 2NO3- +8H+ ---> 3CU2+ + 2NO + 4H2O

I just don't understand how the NO3 looses the 4 O atoms, I thought that it would be one of those substances that don’t separate?

Thanks
Mitch

 

[Riviet]

but, the answers state it as this:

3Cu + 2NO3- +8H+ ---> 3CU2+ + 2NO + 4H2O

I just don't understand how the NO3 looses the 4 O atoms, I thought that it would be one of those substances that don’t separate?

The 4 oxygens in 4H₂0 on the right side balances the oxygens.

 

[Dreamerish*~]

The 4 oxygens in 4H₂0 on the right side balances the oxygens.

It balances the oxygen atoms, but I don't get it either.

How can the products - Cu(NO₃)₂ and H₂ - be written as 3CU²⁺ + 2NO + 4H₂O?

Copper ions is okay, because copper nitrate is not a precipitate in solution, but nitrogen oxide and water? Those aren't even products of acid + metal reactions. And as far as I'm concerned, nitrogen oxides don't come out of nitric acid as gases.

 

[Riviet]

That's what I was thinking, I reckon the answer is wrong, what Mitch had before should be right.

 

[mitch_f1]

I'll check, see what the teacher says tmrw. But first of all. Can someone pelase explain how the NO3 can go to NO ??? i mean like I understand that the 4 infront of water balances it, but why does the oxygen leave the NO3 in the first place?

Thanks
Mitch

 

[YBK]

That's what I was thinking, I reckon the answer is wrong, what Mitch had before should be right.

I'm pretty sure the answer is right actually. His answer doesn't really make much sense to me.


If you check the table of standard potentials on pg 69(lol - that's our chem teacher's favourite joke ) of chemistry contexts you'd see the reduction process that nitric acid undergoes. You don't need to know why Oxygen leaves the NO3, that's just the reaction. It's in the table of standard reduction potentials.

I got the same answer when I did it. Remember 2mol of nitric acid are heated, which is why you multiply the coefficents of the compounds by 2.

The only thing that I was unsure of was Cu+ or Cu2+. Cause the question only specifies that copper metal is added, and copper metal can undergo oxidation to form either Cu+ or Cu2+. I luckily guessed Cu2+, but does anyone know how come it's that and not Cu+ ?