General Chemistry/Redox Reactions/Oxidation and Reduction equations

Redox

Redox reactions are chemical reactions in which elements are oxidized and reduced.

Specifically, at the most basic level one element gets oxidized by losing, or donating, electrons to the oxidizing agent. In doing so, the oxidizing agent gets reduced by accepting the electrons lost, or donated, by the reducing agent (i.e. the element getting oxidized).

If it seems as though there are two separate things going on here, you are correct: redox reactions can be split into two half-reactions, one dealing with oxidation, the other, reduction.

Mnemonic

Oil Rig

Oxidation Is Loss. Reduction Is Gain

Alternatively:

LEO GER

Lose Electrons Oxidation. Gain Electrons Reduction

Example

${\hbox{Fe}}+{\hbox{Cu}}^{2+}\to {\hbox{Fe}}^{2+}+{\hbox{Cu}}$	This is the complete reaction. Iron is oxidized, thus it is the reducing agent. Copper is reduced, making it the oxidizing agent.
${\hbox{Fe}}\to {\hbox{Fe}}^{2+}+2e^{-}$	This is the oxidation half-reaction.
${\hbox{Cu}}^{2+}+2e^{-}\to {\hbox{Cu}}$	This is the reduction half-reaction.

When the two half-reactions are summed, the result is:

Helpful Hint!
If you cancel out the electrons on both sides, you get the original equation.

${\begin{aligned}{\hbox{Fe}}&\to {\hbox{Fe}}^{2+}+2e^{-}\\{\hbox{Cu}}^{2+}+2e^{-}&\to {\hbox{Cu}}\\\hline {\hbox{Fe}}+{\hbox{Cu}}^{2+}+2e^{-}&\to {\hbox{Cu}}+{\hbox{Fe}}^{2+}+2e^{-}\\\end{aligned}}$

Balancing Redox Equations

In a redox reaction, all electrons must cancel out. If you are adding two half-reactions with unequal numbers of electrons, then the equations must be multiplied by a common denominator. This process is similar to balancing regular equations, but now you are trying to balance the electrons between two half-reactions.

Example

${\begin{aligned}{\hbox{Fe}}^{2+}&\to {\hbox{Fe}}^{3+}+e^{-}\\{\hbox{H}}_{2}{\hbox{O}}_{2}+2e^{-}&\to 2{\hbox{OH}}^{-}\\\hline {\hbox{Fe}}^{2+}+{\hbox{H}}_{2}{\hbox{O}}_{2}+2e^{-}&\to 2{\hbox{OH}}^{-}+{\hbox{Fe}}^{3+}+e^{-}\\\end{aligned}}$

The electrons don't completely cancel out. There is one electron more on the left. However, if you double all terms in the first half-reaction, then add it to the second half-reaction, the electrons will cancel out completely. That means the half-reactions for this redox reaction are actually:

${\begin{aligned}2{\hbox{Fe}}^{2+}&\to 2{\hbox{Fe}}^{3+}+2e^{-}\\{\hbox{H}}_{2}{\hbox{O}}_{2}+2e^{-}&\to 2{\hbox{OH}}^{-}\\\hline 2{\hbox{Fe}}^{2+}+{\hbox{H}}_{2}{\hbox{O}}_{2}&\to 2{\hbox{OH}}^{-}+2{\hbox{Fe}}^{3+}\\\end{aligned}}$

Balancing Redox Equations in an Acidic or Basic Solution

If a reaction occurs in an acidic or basic environment, the redox equation is balanced as follows:

1. Write the oxidation and reduction half reactions, but with the whole compound, not just the element that is reduced/oxidized.
2. Balance both reactions for all elements except oxygen and hydrogen.
3. If the oxygen atoms are not balanced in either reaction, add water molecules to the side missing the oxygen.
4. If the hydrogen atoms are not balanced, add hydrogen ions until the hydrogen atoms are balanced.
5. Multiply the half reactions by the appropriate number (so that they have equal numbers of electrons).
6. Add the two equations to cancel out the electrons, as in the previous method, and the equation is balanced!

If the reaction occurs in a basic environment, proceed as if it is in an acid environment, but, after step 4, for each hydrogen ion added, add a hydroxide ion to both sides of the equation. Then, combine the hydroxide ions and hydrogen ions to form water. Then, cancel all the water molecules that appear on both sides.