Oxidation and Reduction

[naved_s9994]

Hey everyone,
Can anyone pleasee, express their knowledge on how to calculate the Oxidation state before and after the process:

EQ:    2Na + FeCl2 --> 2NaCl + Fe 

Also how to write the half equation.. Ive read about KOHES, but it makes no sense.

Thanks!

[d0minicz]

Na = Initially 0    ------> (x)+ (-1) = 0    , x=1 so it is now +1

Fe= Initially (x) + (-2) = 0        -------> It is now 0 as it is in a single element
                 x = +2      

As for half equations

Remove spectator ions

         
then balance the charges



=

[naved_s9994]

Thanks !

However you can you elaborate, how did you decipher that 0 --> (x) + (-1) = 0  for Na?
and similarly for Fe

[d0minicz]

Its 0 initially because its just Na
in the textbook it should say, single elements have an oxidation number of 0
after the reaction it becomes NaCl right
if you look in ya book, theres a table for ion charges
Cl has a -1 charge
and the NaCl molecule is neutral yeah (total = 0)
we want to find X (which is oxidation number of Na in NaCl)
so (X) + (-1) = 0
so X = +1 , therefore Na+ in NaCl has an oxidation number of +1

same thing for Fe.

[naved_s9994]

THANK YOU
so much !

its all locked into place...thnx !

[naved_s9994]

Also, btw.....
when i asked for Ox state, how did you know you had to
only calculate the Fe and Na??

[Collin Li]

You can instantly tell from the charges that the chlorine atoms exist as chloride ions (-), and hence their oxidation number is -1, and doesn't change.

[naved_s9994]

No but my equation was  ( 2Na + FeCl2 --> 2NaCl + Fe  )
Now, I want to calculate the Ox state for this.

From the respnse, he directly calculated the ox state for Na and Fe.
What i am asking is why not the Cl?    - It has 2 on both sides?

[Collin Li]

Oh, he removed spectator ions, which is a similar idea (removing ions that don't change in the process).

You don't need to consider spectator ions because they have not changed their charge, and hence, not their oxidation state. I guess you're implicitly calculating the oxidation state for the chlorine atom by doing this, by recognising it's (-1) and doesn't change.

And chloride having a charge of -1 is assumed knowledge.

[naved_s9994]

Ahhh Thank-you !

[naved_s9994]

Eg.2

Hence:  2C2H2 + 5 O2  ----> 4 CO2 + 2 H20

I.e Oxygen is reduced (0 to -2)?
     Carbon is oxidised..i think?...by how much though?

[Collin Li]

Left-side:

2 O.N(C) + 2(+1) = 0
O.N(C) = -1

Right-side:

O.N(C) + 2(-2) = 0
O.N(C) = 4

Oxidised by 5 electrons per carbon.

[naved_s9994]

Can you explain the O.N(C)

[naved_s9994]

Also Right hand..how is it obtained...because O2 has -2 ...correct
but im lost from there

[Collin Li]

O.N.(C) = oxidation number of carbon

each O is -2, so O2 is 2*(-2) = -4