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# Slightly confused by oxidation states/rules....

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Why bother with a post grad course - waste of time? 17-10-2016
1. And now for the products......

NO ( N = +2; O = -2)

H3AsO4 =

(H = (3x1 =3)
(O4=4x-2=-8)
(As=+5)

H2O = H2;
O = -2

Is this correct?

Therefore, As is oxidised, and N is reduced.
2. (Original post by apronedsamurai)
And now for the products......

NO ( N = +2; O = -2)

H3AsO4 =

(H = (3x1 =3)
(O4=4x-2=-8)
(As=+5)

H2O = H2;
O = -2

Is this correct?

Therefore, As is oxidised, and N is reduced.
yeah N goes from +5 to +2 so it's reduced
As goes from +3 to +5 so it's oxidised
3. (Original post by z33)
yeah N goes from +5 to +2 so it's reduced
As goes from +3 to +5 so it's oxidised
So...net changes

As (+2) x3
N (-3) x2

Therefore ratio of Oxidation to Reduction: 3:2?
4. (Original post by apronedsamurai)
So...net changes

As (+2) x3
N (-3) x2

Therefore ratio of Oxidation to Reduction: 3:2?
u wot m8? why do you need the ratio of oxidation to reduction never worked with it before xD
5. (Original post by z33)
u wot m8? why do you need the ratio of oxidation to reduction never worked with it before xD
Apparently you plug it into the original equation to determine coeffecients to balance the equation :s
6. (Original post by apronedsamurai)
Apparently you plug it into the original equation to determine coeffecients to balance the equation :s
there's an equation?
http://preparatorychemistry.com/Bish...cing_Redox.htm i found that <---
never used this before tbh

we use half equations and stuff to balance
like you have the e- on either side and them times them so they both have the same amount of e- on either side and then you cancel the e- and the equation is balanced!

like that ^
7. (Original post by Serine Soul)
Yeah, it's right.

Could you work out the oxidation states of each atom in SO42-?
Hey, I don't know whether your question has been answered or not but im gonna answer it anyways If you're familiar with the rules of oxidation numbers, we know that the overall oxidation state of a compound is the ionic charge which in this case is 2-. Also, we know that one Oxygen atom usually has a 2- oxidation state and in this case we have 4 oxygen atoms so its -2 x 4 = -8. How we get that -8 to increase to -2? We have the sulphur and one atom of the element to get -8 to -2 we add 6. Therefore the oxidation state of sulfur will be +6 and the oxidation state of oxygen (one oxygen not all 4) will be -2.

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