Question

Question #20e6a

Answer 1

Here's what I got.

Explanation:

Start with what you know for sure, which is that oxygen has an oxidation number of `color(blue)(-2)` in almost all the compounds it forms.

Remember that the sum of the oxidation numbers of each atom that is part of a compound must be equal to the net charge of said compound.

For water, you will have

`"H"_ 2 stackrel(color(blue)(-2))("O")`

Since you have `2` hydrogen atoms, you will need an oxidation state of `color(blue)(+1)` for each of them to get

`2 xx (+1) + 1 xx (-2) = 0`

Therefore, you will have

`stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-2))("O")`

Do the same for sulfur dioxide. You will have

`"S" stackrel(color(blue)(-2))("O")_2`

Since you have `2` oxygen atoms here, you will need an oxidation state of `color(blue)(+4)` for the sulfur atom to get

`1 xx (+4) + 2 xx (-2) = 0`

Therefore, you will have

`stackrel(color(blue)(+4))("S")stackrel(color(blue)(-2))("O")_2`

Finally, move on to the permanganate ion, `"MnO"_4^(-)`. You will have

`"Mn" stackrel(color(blue)(-2))("O") _4""^(-)`

This time, the ion sum of the oxidation numbers must be equal to `-1` since that is the net charge of the ion. You will need an oxidation state of `color(blue)(+7)` for the manganese atom to get

`1 xx (+7) + 4 xx (-2) = -1`

Therefore, you will have

`stackrel(color(blue)(+7))("Mn") stackrel(color(blue)(-2))("O") _4""^(-)`