Red-ox Reactions

What are red-ox reactions?

Reactions that involve the transfer of electrons which can be divided into two reactions representing the reactants which are being oxidized and the reactants being reduced:

Complete Reaction:

$2 MnO_{4 \ (aq)}^{-} + 5 H_{2}C_{2}O_{4 \ (aq)} + 6 H^{+}_{(aq)} \rightarrow 10 CO_{2 \ (g)} + 2 Mn^{2+}_{(aq)} + 8 H_{2}O_{(l)}$

Reactants being oxidized:

$H_{2}C_{2}O_{4} \rightarrow CO_{2}$

Reactants being reduced:

$MnO_{4} \rightarrow Mn^{2+}$

What are the oxidizing and reducing agents in the above reaction?

$MnO_{4}$ is the oxidizing agent and $H_{2}C_{2}O_{4}$ is the reducing agent.

Think of oxidizing/reducing agents as the ÒmediatorsÓ that control oxidation and reduction. That is, the ``culprit'' which causes a reactant to be oxidized.

So, in a red-ox reaction where there is one molecule being oxidized and one molecule being reduced, the molecule undergoing oxidation (losing $e^{-}$) serves as the reducing agent (or $e^{-}$ donor) and the molecule undergoing reduction (gaining $e^{-}$) serves as the oxidizing agent (or $e^{-}$ remover).

What is a disproportionation reaction?

A reaction where a substance is being oxidized and reduced in the same reaction. This commonly occurs with H$_{2}$, which can dissociate oxidatively and lose electrons (H$_{2}$ to 2H$^{+}$) and also dissociate reductively and gain electrons (H$_{2}$ to 2H$^{-}$), all in the same reaction. The reverse is called comproportionation.