REDOX REACTIONS

OXIDATION NUMBER  The oxidation state of an element in an elemental state is zero. (O 2, Fe, He)  The oxidation state of an element in a monoatomic ion is the charge of that ion. (oxidation number of iron in Fe +3 is +3)  The oxidation number of Group 1 elements in all their compounds is +1.  The oxidation number of Group 2 elements in all their compounds is +2.

OXIDATION NUMBER  The oxidation number for fluorine in a compound is -1.  The oxidation number of H is almost always +1.  The oxidation of O is ordinarily -2.  The sum of oxidation numbers in a molecule is zero. H 2 SO 4  In a polyatomic ion, the sum of the oxidation numbers is the charge of the ion. Cr 2 O 7 -2

DEFINITIONS Oxidation: An increase in oxidation number Reduction: A decrease in oxidation number Reducing agent: Whatever is oxidized Oxidizing agent: Whatever is reduced Zn (s) + 2H + (aq)  Zn +2 (aq) + H 2(g) Zn is oxidized (ox # 0  +2) Reducing agent H + is reduced (ox # +1  0) Oxidizing agent

BALANCING HALF-EQUATIONS 1)Balance the atoms of the atoms of the element being oxidized or reduced. 1)Balance the oxidation number by adding electrons. 1)Balance charge by adding H + in acidic solution, OH - in basic solution. 1)Balance hydrogen by adding H 2 O molecules. 1)Check to make sure oxygen is balanced.

BALANCING EQUATIONS IN ACIDIC SOLUTION

MnO 4 - (aq)  Mn +2 (aq) MnO 4 -  Mn +2 MnO e -  Mn +2 MnO e - + 8H +  Mn +2 MnO e - + 8H +  Mn H 2 O

BALANCING EQUATIONS IN BASIC SOLUTION

Cr(OH) 3(s)  CrO 4 -2 (aq) Cr(OH) 3  CrO 4 -2 Cr(OH) 3  CrO e - Cr(OH) 3 + 5OH -  CrO e - Cr(OH) 3 + 5OH -  CrO e - + 4H 2 O

BALANCING REDOX EQUATIONS

1)Split the equation into two half-equations. 1)Balance one of the half equations with respect to atoms and charge. 1)Balance the other half equation. 1)Combine the two half equations in such a way as to eliminate electrons.

Balance the following redox equation in acidic solution. Fe +3 (aq) + MnO 4 - (aq)  Fe +2 (aq) + Mn +2 (aq) Oxidation: Fe +3  Fe +2 Reduction: MnO 4 -  Mn +2 Balanced half equations Fe +2 -  Fe +3 + e - MnO e - + 8H +  Mn H 2 O To eliminate e -, multiply the oxidation half-equation by 5 and add to the reduction half equation. 5Fe +2 + MnO e - + 8H +  5Fe +3 + Mn H 2 O + 5e - 5Fe +2 + MnO H +  5Fe +3 + Mn H 2 O

Balance the following redox equation in basic solution. Cl 2(g) + Cr(OH) 3(s)  Cl - (aq) + CrO 4 -2 (aq) Cl 2 + 2e -  2Cl - Cr(OH) 3  CrO e - + 4H 2 O Multiply reduction equation by 3 and the oxidation equation by 2. 3Cl 2 + 6e - + 2Cr(OH) 3  6Cl - + 2CrO e - + 8H 2 O 3Cl 2 + 2Cr(OH) 3  6Cl - + 2CrO H 2 O

STOICHIOMETRY

What volume of of M KMnO 4 solution is required to react completely with mL of M Fe(NO 3 ) 2 ? 5Fe +2 + MnO H +  5Fe +3 + Mn H 2 O Moles Fe +2 = L x moles Fe(NO 3 ) 2 x 1 mole Fe +2 = 1 L 1 mole Fe(NO 3 ) x moles Fe +2 Moles MnO 4 - = 6.88 x moles Fe +2 x 1 mole MnO 4 - = moles MnO moles Fe +2 Moles KMnO 4 = moles MnO 4 - = moles KMnO 4 V = moles/M moles KMnO 4 /0.684 M KMnO 4 = 2.02 x L = 2.02 mL

REDOX REACTION PROBLEMS Page , 66, 68 70, 72