Unit 2 Chem Help!

[suskieanna]

I really don't get this sort of questions. Please Help!

This question is forming half equations (oxidation and reduction half equations).

H₂S + SO₂ -> S + H₂O

- In the above question I am not sure how to pair up things to make oxidation and reduction half equations.

Na₃PO₄

Also can anyone show me how to get oxidation numbers for each atoms in this compound?

I am really struggling a lot Somebody plz help

[Seno72]

You need to know that for most of the time, the oxidation number of hydrogen atom is +1 (except when it is a metal hydride, where oxidation number is -1) and the oxidation number for Oxygen is -2 (except for peroxides where it is is -1 I believe).

A neutral compound or atom's oxidation number is always equal to zero.

So let's look at H2S. We know that hydrogen's oxidation number is +1. There are two hydrogens, so the overall oxidation number of hydorogen is 2 x 1 = 2. We don't know what Sulfur's oxidation number is. We can find it out by

(2 x 1) + (x) = 0 (where x is the Sulfur's oxidation number and 0 is there because H2S is a neutral compound).

 Therefore x=-2 (Sulfur's oxidation number is -2).

Let's look at SO2 (keep in mind that we need to calculate Sulfur's oxidation number here as well because it is bonded to a completely different atom).

Oxidation no. of oxygen is -2. There are two oxygen, so overall oxidation number for oxygen is -4

Therefore, (-4) + x = 0 (where X is again the oxidation number is Sulfur and 0 is there because SO2 is neutral)

Let's look at the product side.

Sulfur is by itself and so is neutral, therefore its oxidation no. is zero.

H2O:

We know the oxidation numbers of Hydrogen and Oxygen (+1 and -2 respectively).

/part 1

[Seno72]

Now in order for us to find a reduction and oxidation reaction, we have to look at the changes in oxidation number. There have been no changes in the Oxygen or Hydrogen oxidation numbers, only sulfur.

The change in oxidation number of Sulfur in H2S (reactant) to S (product) is from -2 to 0.

The change in oxidation number from Sulfur in SO2 (reactant) to S (product) is from + 4 to 0.

Remember that if there has been an increase in oxidation number from reactant to product, then oxidation has occurred. Therefore H2S has been oxidised (or is the reductant).

If there has been a decrease in oxidation number of an elementary from reactant to product, then reduction has occurred. Therefore SO2 has been reduced (or is the oxidant).

Now we know that H2S undergoes oxidation and SO2 undergoes reduction. The product for each reaction is S and (H2O using KOHES). You should try yourself and see if you can come up with proper oxidation and reduction reactions. Keep in mind that when you join these to reactions together to form an overall reaction, this won't be the same as the reaction you provided me because that reaction is not balanced.

Feel free to ask any questions if you don't get part 1 or 2.

Mod edit: merged double posts --Calebark