1. A redox reaction is one in which the reactants’ oxidation numbers change. What are the oxidation numbers of the metals in the reaction below? 2 FeCl 3 + SnCl 2  2 FeCl 2 + SnCl 4 +3+2 +4 The iron’s charge has become more negative, so it is reduced. The tin’s charge has become more positive, so it is oxidized. reduced oxidized 0 +2 +4 +6 +8 -2 -6 -4 0 +2 +4 +6 +8 -2 -6 -4 Fe Sn

2. Which reactant is oxidized and which is reduced in the following? 2) Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag 0+1+20 reduced oxidized 1) Cl 2 + SnCl 2  SnCl 4 0 reduced oxidized +2+4 3) 2KClO 3  2KCl + 3O 2 -20 +5 reduced oxidized Synthesis Single- replacement Decomposition Note that a redox reaction can also be another type of reaction.

3. Terminology/Memory Aids: OIL RIGO xidation I s L oss of an electron R eduction I s G ain of an electron LEO GERL ose an E lectron: O xidation G ain an E lectron: R eduction The Oxidizing Agent is reduced by the reaction. The Reducing Agent is oxidized by the reaction.

4. 0 Balancing Redox Reactions: Half-Reaction Method Balance the following: Cr 2 O 7 -2 + I -1  Cr +3 + I 2 Step 1: Break the reaction up into its half reactions Cr 2 O 7 -2  Cr +3 I -1  I 2 22 Step 2: Determine the oxidation numbers 2x + 7(-2) = -2 x = +6 +6 Calculate how many electrons are involved in each half reaction and balance the element changing oxidation# 6 e + + 2 e

5. 6 e + Cr 2 O 7 -2  2 Cr +3 2 I -1  I 2 + 2 e Step 3: Multiply the half reactions so that they both have the same number of electrons 3 6 e + Cr 2 O 7 -2  2 Cr +3 6 I -1  3 I 2 + 6 e Step 4: Balance oxygens with water molecules 6 e + Cr 2 O 7 -2  2 Cr +3 + 7 H 2 O 6 e + 14H + + Cr 2 O 7 -2  2 Cr +3 + 7 H 2 O Step 5: Balance hydrogens with H +

6. 6 e + 14H + + Cr 2 O 7 -2  2 Cr +3 + 7 H 2 O 6 I -1  3 I 2 + 6 e Step 6: Add the two balanced half reactions + 6 e + 14H + + Cr 2 O 7 -2 + 6 I -1  2 Cr +3 + 7 H 2 O + 3 I 2 + 6 e and cancel like terms Final balanced redox reaction: 14H + + Cr 2 O 7 -2 + 6 I -1  2 Cr +3 + 7 H 2 O + 3 I 2

7. What if the reaction isn’t occurring in an acidic solution? (ie. There’s no H+ for balancing the hydrogens) Balance the reaction as though it were in acid: 14H + + Cr 2 O 7 -2 + 6 I -1  2 Cr +3 + 7 H 2 O + 3 I 2 Add enough OH -1 to both sides to cancel out the H + 14 OH -1 + + 14 OH -1 14H 2 O + Cr 2 O 7 -2 + 6 I -1  2 Cr +3 + 7 H 2 O + 3 I 2 + 14 OH -1 Cancel out the excess water molecules 7 7 H 2 O + Cr 2 O 7 -2 + 6 I -1  2 Cr +3 + 3 I 2 + 14 OH -1