Introduction/Background
In 1791 Luigi Galvani noticed electrical activity in the nerves of the frogs that he was dissecting. He thought that electricity was of animal origin and could be found only in living tissues. A few years later, in 1800 Alessandro Volta discovered that electricity could be produced through inorganic means. In fact, by using small sheets of copper and zinc and cloth spacers soaked in an acid solution, he built a battery.
Key Concepts
Example 1: What is the oxidation state of Na in the compound NaCl?
Answer 1: The oxidation rules state that in compounds: Group 1A metals are assigned an oxidation number of +1. Therefore the oxidation state of Na will be +1. Also, since there is no net charge on NaCl (the compound is neutral and not ionic), the oxidation state of Cl must be -1. |
Example 2: What is the oxidation state of C in the compound CO2?
Answer: click to reveal |
A redox reaction is a type of reaction that involves both the loss of electrons (termed oxidation) and the gain of electrons (termed reduction). In order to understand redox reactions, we must first understand oxidation states. The idea of oxidation states began with whether or not a metal was attached to an oxygen. Atoms that are unattached to oxygens have an oxidation state of zero. To see the full set of rules, refer to your lab manual. Now let's do a couple practice problems.
Now we need to be able to balance redox reactions. In order to do this, we must first separate the redox reaction into its two half reactions. These half reactions consist of an oxidation reaction and a reduction reaction. Both of these half reactions must be balanced in charge as well as mass. Let's look at a couple examples.
Example 3: Separate the redox reaction, Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s). Then balance each half reaction.
Let's first separate the two half reactions. In order to do this, we must first look to see what is being oxidized and what is being reduced. We do this by assigning oxidation numbers to everything. The oxidation states for each species in the above reaction are as follows: Cu: 0, Ag+: +1, Cu2+: +2, Ag: 0. From this we see that Cu is being oxidized (went from 0 to +2), and Ag+ is being reduced (went from +1 to 0). Now we need to separate them into half reactions and balance.
The half reactions are:
Cu(s) → Cu2+(aq) oxidation
Ag+(aq) → Ag(s) reduction
The rules for balancing half reactions are:
- First, balance all atoms except H and O.
- Second, balance O atoms by adding as many molecules of H2O as needed.
- Third, balance H atoms by adding as many H+ ions as needed.
- Lastly, balance the charge by adding as many e- as needed.
Let's balance the oxidation half reaction first.
Cu(s) → Cu2+(aq) (The number of Cu's are balanced, and there are no oxygens or hydrogens to balance, so all we need to balance now is the number of electrons). The total charge on the left side of the equation is zero, and the total charge on the right side of the equation is +2. Therefore, we need to add two electrons to the right side of the equation. The final half reaction equation looks like this:
Cu(s) → Cu2+(aq) + 2e-
Now you try balancing the reduction half reaction.
Answer: click to reveal |
There is still another step we have to do and that is combining the two half reactions to make one full, balanced redox reaction. First, we have to balance the number of electrons in both half reactions. This means that the number of electrons in one half reaction must equal the number of electrons in the other half reaction. In the above half reactions, the oxidation reaction has 2 electrons, and the reduction reaction only has 1 electron. Therefore, we must multiply the entire reduction half reaction by 2. The final half reactions look like this:
Cu(s) → Cu2+(aq) + 2e-
2Ag+(aq) + 2e- → 2Ag(s)
Finally, we just add the two half reactions together to get the balanced redox reaction. The electrons cancel each other out, and we end up with:
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Here is a summary of the rules for balancing redox reactions.
- Separate an oxidation-reduction equation into two half reactions, one for oxidation and one for reduction.
- Balance the atoms and the electric charge in each half reaction. Electrons appear on the left in the reduction reaction and on the right in the oxidation reaction.
- Adjust the coefficients in the half reactions so that the same number of electrons appears in each half reaction. This may require multiplying one or both half reactions by an appropriate factor.
- Add together the two adjusted half reactions to obtain an overall oxidation/reduction reaction. |
Example 4: Now you try balancing this redox reaction:
Zn(s) + Ag+(aq) → Zn2+(aq) + Ag(s)
Answer: Zn(s) + 2Ag+(aq) → Zn2+(aq) + Ag(s)
Glossary
Related Materials
Redox (http://v.d.singleton.home.att.net/genchem/electrch.htm)
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