Coefficients while balancing ionic equations

Cromptu

Hello. I'm having some trouble balancing ionic equations..

Are we supposed to consider the coefficients of the reactants/products?

I came across a contradiction in the examples given in my book :

1) Na + H(+1) → Na(+1) + H2
So in order to figure out the oxidation/reduction part, we will find out the oxidation number. The oxidation number of Na would be 0 as it is in elemental form and since in the RHS it has an oxidation number of +1, it is being oxidized. In case of H, we will balance the element first. So, 2H(+1) → H2
My question is, in this case, what will be the oxidation number of H in the LHS? +1 or +2?
According to the book it is +2 so that means we would be considering the coefficients.

2) I(-1) + Br2 → I2 + Br(-1)
So again we would figure out what is being oxidized and what is being reduced. And here, since the oxidation number of I is being increased from -1 to 0, it is being oxidized. And in case of Br, we would balance the element first. So, Br2 → 2Br
Again, what would be the oxidation number of Br? -1 or -2? Do we consider the coefficient?

Please explain the answer..Thankyou

Borek

Oxidation number changes when the element loses or gains the electrons. So, when H⁺ (ON equal +1) gets reduced to hydrogen H₂ (ON equal 0) half reaction equation is

2H⁺ + 2e^- -> H₂

as you need two electrons to balance the charge.

It is +1, but you need two electrons to reduce two H⁺.

Cromptu

So we do not consider the coefficients but we do add the number of electrons according to the charge that needs to be balanced. But suppose that we have to figure out whether a substance is getting oxidized or reduced, then will consider the coefficient?
For example, Cl2 + HC2O4(-1) ---> 2CO3(-2) +2Cl(-1)
So here the oxidation number of Cl2 on the left hand side is 0. What will it be on the right hand side? -1 or -2? Whatever it will be, it is getting reduced.
Now in the case of carbon, it is 6 ( we do consider the subscript right? ) and if we balance it on the left hand side, it has a coefficient 2. So what will be the oxidation number of carbon here? 2 or 4? Whatever it will be, it would be reduction. So what is getting oxidized here?

Borek

No. We assign ON to each atom separately, we use coefficients for balancing.

-1

Oxidation number is a property of a single atom. In oxalic acid you have two identical carbon atoms, so it is 2*(oxidation number of carbon)+4*(-2)+1*(+1)=-1, so each carbon atom has ON +3. On the right you have CO₃^2-, so the ON of carbon is +4. That means each carbon atom got from +3 to +4, so it is an oxidation.