Question

How can I do redox reactions step by step?

Answer 1

A rather open-ended question...

Explanation:

And the first step is assignment of oxidation state...for elemental ions, i.e. `Na^+`, or `K^+` or `Cr^(3+)`, the oxidation state of the metal is the CHARGE on the ion, so that we gots `Na(+I), K(+I), Cr(+III)` respectively...

For compound ions, we have to divide up the anion/cation according to predetermined rules …

Now `"permanganate"`, `MnO_4^(-)`, i.e. `Mn(+VII)` is a brightly coloured anion, that commonly undergoes reduction to give COLOURLESS `Mn^(2+)`..and since `+VII-II=V`... WE CONCEIVE that manganese undergoes a FIVE ELECTRON REDUCTION....

`underbrace(MnO_4^(-))_"deep purple"+8H^+ +5e^(-) rarr Mn^(2+)+4H_2O` `(i)`

And note for this half equation both mass and charge are balanced, as is absolutely required. Now the electrons are conceived to COME from somewhere...to cause a corresponding oxidation...let us assume that it OXIDIZES a primary alcohol as shown...i.e. here we gots the oxidation `C(-I)rarrC(+III)`, a four electron oxidation....

`Rstackrel(-I)CH_2OH+H_2O(l)rarrRstackrel(+III)CO_2H+4H^+ +4e^(-)` `(ii)`

We add `(i)` and `(ii)` in such a way that the electrons, virtual particles of convenience, are RETIRED....and thus `4xx(i)+5xx(ii)` gives....

`4MnO_4^(-)+32H^+ +5RCH_2OH+5H_2O(l) +20e^(-) rarr 4Mn^(2+)+16H_2O+5RCO_2H+20H^+ +20e^(-)`

To give after cancellation...

`4MnO_4^(-) +5RCH_2OH +12H^+rarr 4Mn^(2+)+5RCO_2H+11H_2O`

The which is balanced with respect to mass and charge as is absolutely required...

Confused yet? But all I have done is assign oxidation numbers according to predetermined rules...and based electron transfer on the basis of the change in oxidation number....