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Slightly confused by oxidation states/rules....

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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.

(edited 8 years ago)

Reply 21

z33

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 :biggrin:

Reply 22

User947387

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 :biggrin:

So...net changes

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

Therefore ratio of Oxidation to Reduction: 3:2?

Reply 23

z33

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 :s-smilie:

never worked with it before xD

Reply 24

User947387

Original post by z33

u wot m8? why do you need the ratio of oxidation to reduction :s-smilie:

never worked with it before xD

Apparently you plug it into the original equation to determine coeffecients to balance the equation :s

Reply 25

z33

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/Bishop_Balancing_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 ^

Reply 26

JAFF2400

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 :smile:

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.