# In a redox reaction, what do oxidation and reduction mean?

Jul 9, 2016

TWO WAYS TO DEFINE OXIDATION

The intuitive definition of oxidation is:

The increase in the number of oxygens, or the decrease in the number of hydrogens on a species.

An alternative definition is:

When an element's oxidation state becomes more positive, or less negative.

Example: (half-reaction)

stackrel(color(red)(+2))("Mn"^(2+)) + 2"H"_2"O" -> stackrel(color(red)(+4))("Mn")"O"_2 + 4"H"^(+) + 2e^(-)

• The number of oxygens in the $\text{Mn}$-containing species increased, indicating oxidation via the first definition.
• The oxidation state of $\text{Mn}$ increased from $+ 2$ to $+ 4$, thereby indicating oxidation via the second definition.

Example (full redox reaction):

${\text{H"_3"C"("HC"-"OH")"CH"_3 + "H"_2"CrO"_4 -> "H"_3"C"("C"="O")"CH"_3 + "HCrO"_3^(-) + "H"_3"O}}^{+}$

• The number of hydrogens in the alcohol decreased, indicating oxidation via the first definition on its half-reactoin.
• The oxidation state of $\text{Cr}$ decreased from $+ 6$ to $+ 4$, indicating reduction of $\text{Cr}$, therefore showing that ${\text{H"_2"CrO}}_{4}$ is an oxidizing agent, causing the alcohol to oxidize into the ketone, in accordance with the second definition.

TWO WAYS TO DEFINE REDUCTION

Reduction is simply the opposite:

The decrease in the number of oxygens, or the increase in the number of hydrogens on a species.

An alternative definition is:

When an element's oxidation state becomes more negative, or less positive.

Example (half-reaction):

stackrel(color(red)(0))("P") + 3"H"^(+) + 3e^(-) -> stackrel(color(red)(-3))("P")"H"_3

• The number of hydrogens in the $\text{P}$-containing species increased, indicating reduction via the first definition.
• The oxidation state of $\text{P}$ decreased from $0$ to $- 3$, thereby indicating reduction via the second definition.

Example (half-reaction):

• See the chromium-based part of the reaction above. Chromium was reduced to a less positive oxidation state, and one oxygen was lost from ${\text{H"_2"CrO}}_{4}$, following both definitions in its half-reaction.

This was that half-reaction:

${\text{H"_2stackrel(color(red)(+6))("Cr")"O"_4 + 2"H"^(+) + 2e^(-) -> "H"stackrel(color(red)(+4))("Cr")"O"_3^(-) + "H"_3"O}}^{+}$