Balancing+of+Equations+(including+redox)

A chemical equation is the description of a chemical reaction. By using the symbols of the elements in the reaction, you can write an equation. The elements on the left side of the equation are called the reactants, which are the chemicals that react. The elements on the right side of the equation are the products, which are the results of the reaction. An arrow is placed between the reactants and products to show what direction the reaction is occuring in.  **How to Balance Basic Equations ** Now, if you look at the chemical equation above, it is not balanced. The subscripts on the elements are the number of atoms in one molecule of that element. If you have a compound, like aluminum oxide (above product), the atom numbers get switched, so that aluminum's subscript is 2 and oxygen's subscript is 3. When balancing a chemical equation, you have to have the same number of each individual element on both the reactant and product sides. In order to do this you place a coefficient in front of the element. That coefficient then multiplies with the subscripts that are attached to each element after it. This will give you a balanced chemical equation. Example: 4Al + 3O 2 à 2Al 2 O 3 // **Tip - A way to help you balance basic chemical equations is to list each of the elements (on both sides) underneath the reactant and product sides of the equation, with the number of each elements next to its symbol. Make sure you line up the elements on both sides, so that you can see when you have the same number of each on both sides. This will help you know when the equation is balanced.** // Example: Al + O 2 à Al 2 O 3 Al 1 Al 2 O 2 O 3 4Al + 3O 2 à 2Al 2 O 3 Al 4 Al 4 O 6 O 6 Another Example: C 2 H 6 + O 2 à CO 2 + H 2 O    C 2 C 1 H 6 H 2 O 2 O 3 2C 2 H 6 + 7O 2 à 4CO 2 + 6H 2 O   C 4 C 4 H 12 H 12 O 14 O 14 What's a Redox Equation? A redox equation describes what happens in a reaction that involves the processes of oxidation and reduction. Oxidation and reduction reaction cannot occur without each other. Oxidation happens when one substance loses electrons. Reduction happens when another substance gains electrons. Balancing Redox Equations Step 1: Assign oxidation numbers to all the atoms in the reaction. This requires rules in a hierarchy: Example: KMnO 4
 * What is a Chemical Equation? **
 * The oxidation numbers of all atoms add up to the charge on the molecule or ion
 * If it is a compound, the charge must equal zero
 * if it is a polyatomic ion, the atom charges must equal the charge on the polyatomic ion
 * Alkali metals (group 1) are always +1
 * Alkaline-earth metals (group 2) are always +2
 * Metals in IIIA are +3
 * Hydrogen is +1
 * except when in a metal hydride where it is -1
 * Fluorine is always -1
 * Oxygen is always -2
 * except in peroxide ions where it is -1
 * Halogens are -1
 * Nonmetals in group VIA are -2
 * An element in its most stable state has an oxidation number of 0 (example O2)
 * A monatomic ion has the same oxidation number as its charge (example Zn²+). Therefore, Zn2+ has an oxidation number of +2.
 * The sum of the oxidation numbers of the atoms in a compound must equal zero; if it's a polyatomic ion then it must equal the charge

Step 2: Separate the reaction into half - reactions, make sure atoms being oxidized or reduced are balanced Example: Sn2+ + 2Fe 3+ --> Sn4+ + 2Fe 2+ Sn2+ --> Sn 4+ 2Fe 3+ --> 2Fe 2+ Step 3: Add the correct number of electrons to the equations Example: Sn2+ --> Sn4+ + 2e- 2Fe3+ + 1e- --> 2Fe 2+ Step 4: Make sure there is an equal number of electrons in both reactions Example: Sn2+ --> Sn4+ + 2e- 2(2Fe3+ + 1e- --> 2Fe2+) 4Fe3+ + 2e- --> 4Fe 2+ Step 5: ACIDS ONLY. Add H+ ions to the side of the equation that's more negative, to balance the charges. Example: Cr 2 O 7 2- + 6e- + 14H+ -->2Cr 3+ Step 5: BASES ONLY. Add OH- ions to the side of the equation that's more positive to balance the charges. Add twice the amount of OH- ions as you need. Example: Cr 2 O 7 2- + 6e- --> 2Cr 3+ + 14OH- Step 6: Add water to the side of the equation that needs either more H's or more O's in order to balance the equation correctly. In bases, you can see why you added twice the amount of OH- ions; the H2O's balance the extra O's on the right side of the reaction. Example: Cr 2 O 7 2- + 6e- + 14H+ --> 2Cr3+ + 7H 2 O (in acids) Cr 2 O 7 2- + 6e- + 7H 2 O --> 2Cr 3+ + 14OH- (in bases) Step 7: Put the two halves back together Example: Cr 2 O 7 2- + 6e- + 14H+ --> 2Cr 3+ + 7H 2 O 6Cl- --> 3Cl 2 + 6e- Cr 2 O 7 2- + 14H+ + 6Cl- --> 2Cr3+ + 7H 2 O + 3Cl 2 Another Example: MnO 4 - + SO 2 --> Mn2+ + SO 4 2- (MnO 4 - + 5e- --> Mn2+)2 (SO 2 - 2e- --> SO 4 2-)5 2MnO 4 - + 10e- --> 2Mn2+ 5SO 2 - 10e- --> 5SO 4 2- 2MnO 4 + 10e- --> 2Mn2+ + 8H 2 O 10H 2 O + 5SO 2 - 10e- --> 5SO 4 2- 2MnO 4 + 16H+ + 10e- --> 2Mn2+ + 8H 2 O 10H 2 O + 5SO 2 - 10e- --> 5SO 4 2- + 20H+ 2MnO 4 - + 2H 2 O + 5SO 2 --> 2Mn2+ + 5SO 4 2- + 4H+
 * K has an oxidation number of +1
 * O has an oxidation number - 2
 * Mn has an oxidation number of +7
 * Ignore anything in the half reactions that isn't being oxidized or reduced
 * Ignore any charges
 * Think about balancing the oxidation numbers
 * If the electrons aren't equal then you have to multiply each half reaction by the other equations electrons to make them equal
 * If the electrons are correctly balanced in each half reaction, then they should be able to cancel each other out
 * The number of H2O's in each reaction should be able to be canceled out in one of the equations


 * Tip - Take it step by step and just think about what you're doing before you do it. It's a long process, but just take each step one at a time.

Work Cited http://richardbowles.tripod.com/chemistry/balance.htm#part0