In a redox reaction, what do oxidation and reduction mean?
1 Answer
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
#"Mn"# -containing species increased, indicating oxidation via the first definition. - The oxidation state of
#"Mn"# increased from#+2# to#+4# , thereby indicating oxidation via the second definition.
Example (full redox reaction):
#"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
#"Cr"# decreased from#+6# to#+4# , indicating reduction of#"Cr"# , therefore showing that#"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
#"P"# -containing species increased, indicating reduction via the first definition. - The oxidation state of
#"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
#"H"_2"CrO"_4# , following both definitions in its half-reaction.
This was that half-reaction:
#"H"_2stackrel(color(red)(+6))("Cr")"O"_4 + 2"H"^(+) + 2e^(-) -> "H"stackrel(color(red)(+4))("Cr")"O"_3^(-) + "H"_3"O"^(+)#