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.

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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 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 Fe Sn

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

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.

0 Balancing Redox Reactions: Half-Reaction Method Balance the following: Cr 2 O 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 = Calculate how many electrons are involved in each half reaction and balance the element changing oxidation# 6 e e

6 e + Cr 2 O 7 -2  2 Cr +3 2 I -1  I 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 e Step 4: Balance oxygens with water molecules 6 e + Cr 2 O 7 -2  2 Cr H 2 O 6 e + 14H + + Cr 2 O 7 -2  2 Cr H 2 O Step 5: Balance hydrogens with H +

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

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 I -1  2 Cr H 2 O + 3 I 2 Add enough OH -1 to both sides to cancel out the H + 14 OH OH -1 14H 2 O + Cr 2 O I -1  2 Cr H 2 O + 3 I OH -1 Cancel out the excess water molecules 7 7 H 2 O + Cr 2 O I -1  2 Cr I OH -1