Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Jefgreen109 on March 18, 2011, 07:25:26 PM
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In each of the following two redox reactions, I need to name the following:
- Oxidized Element
- Reduced Element
- Reducing Agent
- Oxidizing Agent
N2H4 (aq) + 2O2 (g) --> N2 (g) + 2H2O (g)
3Cl2 (g) + NaI (aq) + 3H2O (l) --> 6HCl (aq) + NaIO3
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For the first reaction, I know that water and oxygen gas both have zero as their oxidation state. However, my worksheet has N(sub 2) H(super 4), which makes no sense. Also, I'm guessing that Nitrogen gas (on the product side) would be negative 6? I'm confused.
For the second reaction, I know that both chlorine and water have oxidation states of zero. Na is a +1 and I is a -1. On the product side H is +1 and Cl is -1, but I have no idea how to find the oxidation state of NaIO(sub3), or Sodium Iodate.
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Water doesn't have oxidation state. Atoms do, molecules don't.
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^ That's what I said. And I meant the atoms in Sodium Iodate...
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H2O has a net charge of zero, so...:
O = -2
H = +1*(2) = +2
Thus, H2O = 0
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For NaIO3
Na = +1, but I don't understand how the oxidation numbers for O3 and I can possibly add up together, and with Na, to equal zero.
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Anyone?
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H2O has a net charge of zero, so...:
(...)
Thus, H2O = 0
Yes, water molecule has a charge of zero, so it has a charge of zero. You are mixing charge with an oxidation state. As I wrote earlier, one is property of an atom, other is property of a molecule. It may happen that single atom has to be treated as molecule (for example Na+).
For NaIO3
Na = +1, but I don't understand how the oxidation numbers for O3 and I can possibly add up together, and with Na, to equal zero.
OK for Na. What is oxidation number of oxygen?
http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method