Oxidation+and+Reduction

//Definition of oxidation and reduction in terms of oxidation number: //

**Oxidation** is the **loss** of electrons: Fe 2+  (aq) ---> Fe 3+  (aq) + e - **Reduction** is the **gain** of electrons: 2H +  (aq) + 2e -  ---> H 2  (g) An easy way to remember this is to use the mnemonic ** OILRIG **
 * O **xidation ** I **s ** L **osing [electrons] ** R **eduction ** I **s ** G **<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">aining [electrons]

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Oxidation number/ Oxidation state: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The oxidation state of an atom is the charge that the atom would have if all of its covalent bonds (in the molecule) were broken such that the more electronegative element kept the electrons in the bond.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Eg: A molecule of water is H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O. We want to find the oxidation number of hydrogen (H) and oxygen (O). <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">To do so, we assume that the O-H bonds are broken. If the bond is broken O would keep the shared electrons since it is more electronegative than H. Thus you get H + <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> and O 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> showing that the oxidation number of H is +1 and the oxidation number of O is -2. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">You can use a series of simple rules to help you find the oxidation number of an atom in a molecule.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">//**(1)**// The total oxidation state of an atom in a species is equal to the electrical charge that it carries. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">- The total oxidation state of elements and neutral molecules is zero. Eg. In a molecule such as HCl, if you add up the oxidation state of H to the oxidation state of Cl you get zero. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">- The total oxidation state of ions is equal to their charge. Eg. SO 4 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> ion. If you add up the oxidation states of the S and the 4 O atoms you should get -2 (that is the charge on the ion) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">//**(2)**// Some elements almost always have the same oxidation state in their compounds. For example fluorine’s oxidation state is -1 and oxygen’s is -2 except when it’s in the from of H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> where its Oxidation state is +1. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Usually elements like // Carbon, Nitrogen, Phosphorus, Sulfur // <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">and // transition metals //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> have varying oxidation states.

//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Take a few examples: // <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">In H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> the oxidation state of each H atom is 0, since both H atoms have the same electronegativity and thus each atom keeps one of the shared electrons in the pair. It can also be seen that 0+0=0 so the total oxidation state of the atoms is equal to the charge on the molecule. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The oxidation state of sulphur in H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">SO 4 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is +6. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We already know that the oxidation state of H is +1 and the oxidation state of O is -2 and we know if we add up the oxidation states of all the atoms they should add up to zero since the charge on the molecule H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">SO 4 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is zero. Let’s assume that the Oxidation state of S is **X**. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">So we have: 2x(+1) + 4x(-2) + **X**= 0. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Rearranging this gives: **X**=8-2=+6 so the oxidation state of sulphur is +6.

//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Oxidation and Reduction can also be defined in terms of the Oxidation number. // <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">**Oxidation** is an **increase** in the //oxidation number//. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">**Reduction** is a **decrease** in the //oxidation number//.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If in a reaction the oxidation number of an atom doesn’t change then it hasn’t undergone Oxidation/reduction. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Eg1: If you convert N 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> to NH 3 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">, has N undergone oxidation or reduction?? <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">To know that, you have to find the oxidation number of N in each molecule. In N 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">, the oxidation number is 0 while in NH 3 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> it’s -3 (since 3x(+1) + -3 = 0). Thus, the oxidation number has decreased so N had undergone reduction.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Eg2: If you convert NO 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> to NO 3 - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">, has N undergone oxidation or reduction?? In NO 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> the oxidation number is +4 while in NO 3 - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> it’s +5. The oxidation number has increased so N has undergone oxidation.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Some elements have more than one oxidation states. Thus to be able to know the oxidation state of the atom of that element in a particular compound, the oxidation state is shown using roman numerals. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">For example FeCl 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> where the oxidation state of Fe is +2, its name is written as Iron (II) chloride, so that anyone reading the name can know that the oxidation state of iron is +2 and not +3 in this compound. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">FeCl 3 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is written as Iron (III) chloride. The oxidation state of iron is Fe 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">.

By the end of this lesson you should be able to: **<span style="color: #800080; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif;">NEXT UP- REDOX EQUATIONS **
 * Define oxidation and reduction in terms of gain and loss of electrons
 * Deduce the oxidation number of an element in a compound
 * State the oxidation state of an atom in a compound using roman numerals
 * Use oxidation numbers to deduce whether an element undergoes oxidation or reduction in a reaction
 * media type="googleplusone" key="" width="360" height="18" ||
 * media type="facebooklike" key="http%3A%2F%2Fibchem4u.wikispaces.com%2FOxidation%20and%20Reduction" width="360" height="74" ||