Balancing equations using oxidation numbers C3H8O + CrO3 + H2SO4  Cr2(SO4)3 + C3H6O + H2O Balancing equations using oxidation numbers

Review: balancing chemical equations Balance the following chemical reaction: CuCl2 + Al  Cu + AlCl3 In the past (e.g. in grade 11) we balanced equations by “inspection”. Balancing equations relied on having equal numbers of atoms on each side of the equation We now also concerned with transfer of electrons or the balance of charge We can balance equations using oxidation #s. This relies on the idea that the number of electrons lost by an element must be equal to the number gained by a different element.

Using Oxidation Numbers total oxidation # CuCl2 + Al  Cu + AlCl3 +2 -2 +3 -3 +2 -1 +3 -1 Notice: Cu has gained 2e– (oxidation #  by 2) Notice: Al has lost 3e– (oxidation #  by 3) But, number of e– gained must equal e– lost Multiply Cu by 3, Al by 2: change is 6 for both change total oxidation # 3CuCl2 + 2Al  3Cu + 2AlCl3 +6 -6 +6 +6 -1 -2 +2 +3 -3

Steps to balancing equations Write the skeleton equation Assign oxidation numbers to all atoms and identify the atom/ion whose oxidation numbers change. Using the change in Oxidation numbers, write the number of electrons transferred per atom. Using the chemical formulas, determine the number of electrons transferred per reactant. Balance the electron transfer by adding coefficients to the reactants. Balance the O atoms by adding water molecules, balance H atoms by adding H+ If the solution is basic, add OH- to both sides to neutralize the H+, combine to for water and reduce.

ClO3-1 + I2 → Cl-1 + IO3-1 in an acidic solution Practice : ClO3-1 + I2 → Cl-1 + IO3-1 in an acidic solution

Practice : CH3OH + MnO4-1  CO3-2 + MnO4-2 In a basic solution

Homework: Page 613 #1-4