Oxidation Numbers for H3AsO4 and H2S Equation

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Specter

Homework Statement


Identify the element oxidized and the element reduced, in this chemical equation:
H3AsO4+H2S ---> H3AsO3+S+H2O

Homework Equations


none

The Attempt at a Solution


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So I have learned the rules for oxidation numbers:

1. The oxidation number of an element is always 0 regardless of its subscript. This is true as long as they are not part of compound.
2. The oxidation number of a simple ion is the charge of the ion.
3.The oxidation number of oxygen in compounds is -2.
4. The oxidation number of hydrogen in compounds is +1.
5.Group 1 and group 2 metals always have a +1, and +2 value, respectively, in compounds.
6. All other oxidation numbers are assigned so that the net sum of the oxidation numbers equals the net charge of the molecule or complex ion.

So with what I know I can figure out the oxidation numbers for the oxygens and hydrogens in the chemical equation but how do I figure out the numbers for arsenic and sulfur?
 
So let's start with H2S. It has no net charge, so you can apply rules 6 and 4.
 
DrDu said:
So let's start with H2S. It has no net charge, so you can apply rules 6 and 4.
H2 would be +1 but I'm not sure what S would be. Would it be -2 so that it equals +1 overall?
 
Specter said:
H2 would be +1 but I'm not sure what S would be. Would it be -2 so that it equals +1 overall?

No, H2 would be not +1. You have two H at +1 each, so it is +2 total for hydrogen. If so, what must be the charge on the sulfur atom so that the molecule is neutral?
 
Borek said:
No, H2 would be not +1. You have two H at +1 each, so it is +2 total for hydrogen. If so, what must be the charge on the sulfur atom so that the molecule is neutral?
Oh I understand. Thats why S is -2.
 
Specter said:
Oh I understand. Thats why S is -2.

That's it.
 

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