Question

How would I get the oxidation numbers and balance this redox reaction? `"H"_2"O"_2 + 2"C""H"_3"-""S""-""H" → 2"H"_2"O" + "CH"_3"-S-S-CH"_3`

Answer 1

WARNING! Long answer! Here's how I do it.

Explanation:

You get the oxidation numbers from the Lewis structures.

The rules for counting the valence electrons are:

1. Electrons shared between identical atoms are shared equally.
2. Electrons shared between nonidentical atoms belong entirely to the more electronegative atom.
3. Lone pairs belong entirely to the atom that has them.

You count the electrons using these rules. Let's call this number E.

Then you subtract from the number of valence electrons in an isolated atom V.

The formal charge `"FC = V-E"`.

`bb("H"_2"O"_2)`

The Lewis structure is

I have marked the allocation of electrons by vertical red lines.

Each `"H"` atom loses its shared electron to the more electronegative `"O"` atom, and the `"O-O"` shared pair is shared equally (1 electron to each atom).

For H, `"FC" = "1 - 0" = "+1"`

For O, `"FC" = "6 - 7" = "-1"`

`bb("CH"_3"SH""`

The structure is

Here the formal charges are

For H, `"FC" = "1 - 0" = "+1"`

For C, `"FC" = "4 - 6" = "-2"`

For S, `"FC" = "6 - 8" = "-2"`

`bb("H"_2"O")`

You can do this one.

The formal charges are `"H = +1"` and `"O = -2"`.

`bb("CH"_3"S-SCH"_3)`

The formal charges are

For `"H: FC = +1"`

For `"C: FC = -2"`

For `"S: FC = 6 - 7 = -1"`.

Now, we put these oxidation numbers above the atoms in the equation.

`stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-1")("O")_2 + 2stackrelcolor(blue)("-2")("C")stackrelcolor(blue)("+1")("H")_3"-"stackrelcolor(blue)("-2")("S")"-"stackrelcolor(blue)("+1")("H") → 2stackrelcolor(blue)("+1")("H")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)("-2")("C")stackrelcolor(blue)("+1")("H")_3"-"stackrelcolor(blue)("-1")("S")"-"stackrelcolor(blue)("-1")("S")"-"stackrelcolor(blue)("-2")("C")stackrelcolor(blue)("+1")("H")_3`

The changes in oxidation number are

`"O: -1 → -2 = -1"` (reduction)
`"S: -2 → -1 = +1"` (oxidation)

`"O"` is reduced, so `"H"_2"O"_2` is the oxidizing agent.

`"S"` is oxidized, so `"CH"_3"SH"` is the reducing agent.