What are oxidation numbers?

Dec 11, 2016

These are typically used to simplify redox equations....

Explanation:

Sodium dichromate, $N {a}_{2} C {r}_{2} {O}_{7}$ is a powerful and widely used oxidant for both organic and inorganic substrates. It is typically reduced to $C {r}^{3 +}$.

Now I can write the reduction half equation reasonably easily. You know the drill; the difference in oxidation number is accounted for by electrons, which we treat as virtual particles:

Chromium is reduced from $+ V I$ to $+ I I I$, and thus we introduce 6 electrons into the half equation.

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$ $i .$

Both charge is balanced, and mass is balanced (is it?), as indeed it must be if it reflects chemical reality. I could have included the counterions, i.e. $N {a}_{2} C {r}_{2} {O}_{7} \left(a q\right)$ or $H C l \left(a q\right)$, but the sodium and chloride ions are along for the ride, and I am not really concerned with them. Thus they can be removed from the net ionic equation, and simply ignored.

Dichromate is sometimes used to oxidize alcohols to carboxylic acids, where the carbon is oxidized from $C \left(- I\right)$ to $C \left(+ I I I\right)$, a four electron oxidation:

${H}_{3} C - C {H}_{2} O H + {H}_{2} O \rightarrow {H}_{3} C - C {O}_{2} H + 4 {H}^{+} + 4 {e}^{-}$ $i i .$

The overall redox equation typically removes the electrons, and so we take $2 \times \left(i\right) + 3 \times \left(i i\right) :$

$2 C {r}_{2} {O}_{7}^{2 -} + 16 {H}^{+} + 3 {H}_{3} C - C {H}_{2} O H$
$\rightarrow 3 {H}_{3} C - C {O}_{2} H + 4 C {r}^{3 +} + 11 {H}_{2} O$

Which I think is balanced with respect to mass and charge.

Anyway, is this what you want? I am willing to have another try if you refine your question.