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Coefficients while balancing ionic equations

  1. May 2, 2012 #1
    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
  2. jcsd
  3. May 3, 2012 #2


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    Staff: Mentor

    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+.
  4. May 3, 2012 #3
    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?
  5. May 3, 2012 #4


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    Staff: Mentor

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


    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 CO32-, so the ON of carbon is +4. That means each carbon atom got from +3 to +4, so it is an oxidation.
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