Oxidation state (1 Viewer)

[tennille]

Why is the oxidation state of SO42-, 6, and in H2S it is 2? I really dont get this stuff.

 

[Xayma]

The oxidation state of sulfur is VI because the net oxidation state is -II.

As oxygen is -II unless it is peroxide in which case it is -I.

You have

x+4(-II)=-II
x-VIII=-II
x=VI

Similarly for the second one, Hydrogen is +I etc.

Here are some rules you must know:

1. The oxidation state of a pure element eg O₂ is 0.
2. The oxidation state of an ion is equal to the net charge of that ion.
3. Oxygen always has an oxidation state of -II, except if it is in it's elemental form or a peroxide in which case it is -I.
4. Hydrogen has an oxidation state of -I in metal hydrides and an oxidation state of +I in non-metal hydrides.

They are only abstract concepts it really just allows the tracing of flow of electrons.

Oxidation is an increase in oxidation number.
Reduction is a decrease (or reduction) in oxidation number.

 

[tennille]

So what would be the oxidation state of CO32-, just as an example? Sorry about this.

 

[grimreaper]

So what would be the oxidation state of CO32-, just as an example? Sorry about this.

the oxidation state is -2, as xayma said the oxidation state of an ion is just its charge

 

[tennille]

From what i've read, the oxidation state of CO23- is +4. This is really confusing and I'm not sure how to use the algebra stuff to work it out.

 

[tennille]

CO32- = -2

X + 3(-2) = -2

X - 6 = -2

X = 4

That's how I've worked it out, but for other compounds, i'm not sure what to do.

 

[~*HSC 4 life*~]

CO32- = -2

X + 3(-2) = -2

X - 6 = -2

X = 4

That's how I've worked it out, but for other compounds, i'm not sure what to do.

that's right i think...

you should do the same with other compounds...like with sulfate:

S o42-

S (-2)4 = -2

S -8 = -2

S= 6

 

[Xayma]

So what would be the oxidation state of CO32-, just as an example? Sorry about this.

You need to be more specific. You are looking for the oxidation state of carbon which is IV.

 

[Kirsti]

From what i've read, the oxidation state of CO32- is +4. This is really confusing and I'm not sure how to use the algebra stuff to work it out.

The oxidation state of the entire ion is -3
The oxidation state of the Carbon within the ion is +4


C + 3*O = CO3
replace the O's with -2

C + (-2*3)=-2 (the oxidation number of the ion)
C + -6 = -2
C = 4

 

[Xayma]

The oxidation state of the entire ion is -3

I know its just a typo but -II.

Because it is an abstract quality as such generally you should use roman numerals to show oxidation state.

 

[~*HSC 4 life*~]

arabic numbers 4 life*~ hahahah

i've never used roman numerals :/