Oxidation states

How is it possible to work out the oxidation states of the following?

MNO4-, MN042- and MN02?.

Also how can we work out whether something is an oxidising agent or a reducing agent in one of these reactions. Could someone demonstrate with an example?

Reply 1

Eau

15

covalentbond

How is it possible to work out the oxidation states of the following?

MNO4-, MN042- and MN02?.

Oxygen ions usually have an oxidation number of 2-, because of its electronegative nature.

[MnO4]- Now if oxygen atoms are 2-, and there are 4 oxygens, that means charge due to oxygen atoms is 4 * 2- = 8-. The overall charge of the ion is 1-, so the oxidation number of Mn must be 7+.

Try the rest.

Reply 2

Eau

15

covalentbond

Also how can we work out whether something is an oxidising agent or a reducing agent in one of these reactions. Could someone demonstrate with an example?

Something is an oxidising agent if it oxidises (oxidation number increase) something else; hence, the oxidising agent itself is reduced. Something is a reducing agent if it reduces (oxidation number decrease) something else.

E.g.

KMnO4 + 3CH3CH2OH (ethanol) + 2H+ ----> Mn2+ + 3CH3CHO (ethanal) + 4H2O

Oxidation of Mn: 7+ to 2+ (reduced)

Oxidation of C: 3- to 0 (approximately - so oxidised)

Therefore the [MnO4]- is the oxidising agent.

Reply 3

covalentbond

OP

Mauve

Oxygen ions usually have an oxidation number of 2-, because of its electronegative nature.

[MnO4]- Now if oxygen atoms are 2-, and there are 4 oxygens, that means charge due to oxygen atoms is 4 * 2- = 8-. The overall charge of the ion is 1-, so the oxidation number of Mn must be 7+.

Try the rest.

thanks I get how to work it but I was actually stuck on mno42-, could you plesae show me how you would work it out for MNO42- because I cant seem to put it in right into my calculator to equal the overall charge on the ion.

Reply 4

Eau

15

covalentbond

thanks I get how to work it but I was actually stuck on mno42-, could you plesae show me how you would work it out for MNO42- because I cant seem to put it in right into my calculator to equal the overall charge on the ion.

Hmmm? Don't use a calculator for these simple calculations! Seriously.

[MnO4]2-

Charge due to 4 oxygen atoms = 8-

Overall charge = 2-

So charge due to Mn = -2 - (-8) = 6+

Reply 5

sufiankane

2

quick and easy rule i found helpful, is that the sum of the oxidation states HAS TO equal the charge. So [MnO4]2-, the sum of all the oxidation states has to equal -2. Now oxygen most of the time has an oxidation state of -2 so.... Mn + (-2*4) = -2. Rearrange with simple maths to get +6 stated by mauve. Hope my methood helps.

Reply 6

dream123

2

Okay, so I am a little raw on oxidation states myself.

When you have a compund, is it meant to add up to 0? Except for ions which represent the charge. Im confused.

So, is this correct for H2O....
O = -2 *-2 = 4
H = +1
is that it? ^^^^^

and what about...
HCL
Cl = -1
H= +1
So, is the oxidation number...0 ?

hellp plz

Compounds add up to zero
ions add up to the charge on the ion

..... it really is that simple

Reply 8

dream123

2

charco

Compounds add up to zero
ions add up to the charge on the ion

..... it really is that simple

FANTASTIC! Thats cleared... I thought that was the case but ppl just go on and on and on...and it just confuses me. So, yes...thanks for clearing that up.

Oxidation states

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