[Kartiky14]

N₂H₄ loses 10 moles of electron to form Y.Y has every atom of nitrogen in N₂H₄ and the oxidation state of hydrogen remains same.Find out the oxidation state of nitrogen in Y.

[fledarmus]

How would you start?

[Kartiky14]

By the equation
2x + 4 =10 ?

[confusedstud]

I'm also pretty curious about this question. Since N2H4 loses 10 moles electrons and the number of nitrogen atoms remains the same in the product so nitrogen is oxidised and something else is reduced. So originally Nitrogen has an oxidation state of +2 and since there are 2 atoms at the end, so it can be +2+7? But could it be 0 as well? I'm confused...

[confusedstud]

N2H4 is oxidised as 10 moles of electrons is given out and the oxidation of H remains the same. so each nitrogen increases by 5 so the oxidation is +7? But is it possible that it just forms a compound that is also N2 so its oxidation state is 0...

[UG]

But is it possible that it just forms a compound that is also N2 so its oxidation state is 0...

But then nitrogen would have to gain electrons.

[confusedstud]

What do you mean?I understand that the nitrogen is oxidised but why can't it just be oxidised 0? Or is that only possible if the question stated that it loses only 4 moles if electrons instead then N2 is formed?

So we are ably to find out the new oxidation state with these information:
1) original oxidation state
2) number of that atom in the original and final product
3) number of electrons given out or taken in

So we are asked to find eg the number of electrons given out then we need 1) and 2)  to solve it? Eg. X2Cl reacts to form XCl so original oxidation state is +0.5 and final is +1 so it gives out 1 electron?

[UG]

Ok, lets get our terminology right here. Oxidation is the loss of electrons, so after an atom is oxidised, the oxidation number of the atom must increase/become more positive. Refering to the original post, the original oxidation state of N is +2 and after 10 moles of electrons are lost, the oxidation state of N becomes +7 as you have already said. You cannot 'oxidise' N from +2 to 0 because that would be a reduction reaction. If the question had stated that N₂H₄ gained 4 moles of electrons, then you would get N₂ as this would be a reduction reaction. N is reduced from +2 to 0.

[confusedstud]

Oh wait, I think it should be -2 to +3 instead so sorry about that. Because originally its -2 and afterwards it loses 5 electrons to +3.

In a similar question XCl2 is oxidised to X2Cl. How many electrons are given out/taken in? (I made this up)

So if I look at it this way: 2XCl2+n electrons-->X2Cl
2(X)+4(-1)=0 so 2X is +4 originally
2(X')+(-1)=0 so 2X' is +1 at the end. So 2 moles of XCl2 gives out 3 electrons?
But if I look at it in another way:XCl2+n electrons-->2X2Cl
X+(2)(-1)=0 so X is +2 originally
4(X')+2(-1)=0 so X is +0.5 so 1.5 moles of electrons is givrn out by 1 mole of XCl2 which is the same thing.
why would I http through same result? Since I'm balancing them in different waYs? Thanks!

[Borek]

X2Cl

Assuming Cl is -1 (which it often is) X would be -½. Quite unlikely. That means Cl is most likely not -1 - which in turn means we don't know oxidation number of X in X₂Cl.

None of your reaction "equations" is balanced, no wonder you get different results.