# What is the first step to balance a redox equation using the half-reaction method?

Jul 26, 2017

Assign $\text{oxidation numbers.....}$

#### Explanation:

Redox reactions are conceived to occur on the basis of electron transfer. A species that is oxidized is conceived to LOSE electrons, and a species that is reduced is conceived to gain electrons. I write $\text{conceived}$ because we introduce the electron as a virtual particle. Nevertheless, its use allows us to represent the transfer of charge, which is a fundamental property that is certainly CONSERVED in every chemical reaction.

As a practical example we could consider the oxidation of alcohols to ketones, and carboxylic acid, which is normally accomplished in the organic lab by the use of strong oxidizing agents such as $M n {O}_{4}^{-}$ or $C {r}_{2} {O}_{7}^{2 -}$. So we will try it for the oxidation of $\text{2-propanol}$ to give $\text{acetone}$.

$\text{Oxidation half equation:}$
${H}_{3} C - \stackrel{0}{C} H \left(O H\right) C {H}_{3} \rightarrow {H}_{3} C \stackrel{+ I I}{C} \left(= O\right) C {H}_{3} + 2 {H}^{+} + 2 {e}^{-}$

Please observe that mass and charge are balanced here. Are they? And what do I mean when I say this?

$\text{Reduction half equation:}$
$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$

Chromium is reduced from $C r \left(V I +\right)$ to $C r \left(I I I +\right)$........

And we add the half-equations in such a way that the electrons are eliminated.....we take three of the former equation, and one of the latter to give......

$C {r}_{2} {O}_{7}^{2 -} + 8 {H}^{+} + 3 {\text{H"_3"CCH(OH)CH"_3rarr2Cr^(3+) +3"H"_3"CC(=O)CH}}_{3} + 7 {H}_{2} O$

We would see the reaction give a macroscopic colour change from orange due to $\text{dichromate}$ to green due to $C {r}^{3 +}$.

Please observe that mass and charge are balanced here. Are they? And what do I mean when I say this?