ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: M_BONG on November 17, 2013, 09:15:14 pm
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I am really struggling with the redox concept.
Can anyone explain why the following equations are redox equations?
H2S (g) +Cl2 (g) -> 2HCl(g) + S (s)
Zn(s) + 2HCl(aq) -> ZnCl2 (aq) + H2 (g)
How do you actually determine if an equation is redox? Is it through oxidation number?
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A change in oxidation number is probably the most efficient way of determining if a chemical reaction can be classified as a redox reaction. If there is a change in oxidation number for any of the species present, then it's a redox reaction. Do you know the rules for oxidation numbers? :)
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what i would do is write the oxidation number of each other elements above them. So in the first equation: a +1 on H2, -2 on S etc. etc, for the whole equation. Now you'd want to compare the same species on both sides and see if their oxidation number differ. For instance in the first equation the Cl2(g) has a zero oxidation number, and on the other side (as part of HCl) has an oxidation number of -1. Therefore we can say that Cl2(g) has been reduced, a term that is used for redox reactions.
Hope that cleared some things up.
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Alright, let's revise over the rules of oxidation numbers as we do this (this is taken directly from Heinemann Chemistry 2):
1. Free elements have an oxidation number of zero.
Therefore the oxidation number of chlorine gas and sulfur is zero.
2. In ionic compounds composed of simple ions, the oxidation number is equal to the charge on the ion.
3. Some elements have oxidation numbers that are regarded as fixed, except in a few exceptional circumstances.
- Oxygen usually takes −2 in compounds. In peroxides such as H2O2 and BaO2 it has −1.
- Hydrogen takes +1 in compounds, except in metal hydrides such as NaH and CaH2 where it has −1.
- Electronegative elements such as F, Cl and O take numbers equal to the charges on their simple ions (−1, −1 and −2, respectively) when part of a compound, provided that they are the most electronegative element present in the compound.
Therefore the oxidation of hydrogen is +1 on both the left and right hand side. However, chlorine takes on an oxidation number of -1 on the right hand side of the equation - hence it has been reduced.
4. The sum of the oxidation numbers in a neutral compound is zero.
Therefore the oxidation number of sulfur is -2 on the left hand side - it has hence been oxidised.
5. In a polyatomic ion the sum of the oxidation numbers is equal to the charge on the ion.
6. The most electronegative element in a compound has the negative oxidation number.
I hope this helps. :)
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You're welcome! There were times where I forgot about oxidation numbers in multiple choice and tried to work out if a chemical reaction was a redox reaction by writing out half equations... =_= As you can see, this is a far more effective method. Good luck for Year 12. :)