(reduction/oxidation) Redox Reactions (reduction/oxidation)

Redox Reactions Involve the exchange of electrons in a chemical reaction Electrons are lost by one substance and gained by another substance One substance goes through oxidation, while the other substance goes through reduction

Oxidation loss of electrons the gain of oxygen the loss of hydrogen an increase in oxidation number for the substance being oxidized

Reducing Agent experiences oxidation is an electron donor

Oxidizing Agent experiences reduction is an electron acceptor

Oxidation States Also known as oxidation numbers Allows understanding of what is oxidized and what is reduced Imaginary charges that atoms would have if shared electrons were divided equally in a covalent bond Or real charges that monatomic ions have in an ionic bond

Assigning Oxidation States Written directly above a symbol with the sign and then the number…unlike charges

Assigning Oxidation States The oxidation state of an atom of an element in its natural state is zero. Na(s) Cl2(g) Br2(l) C(s)

Assigning Oxidation States The oxidation state of a monatomic ion is equal to its charge. Na1+(aq) NaCl(g) Fe2+(aq) Al3+(aq) +1 +1 -1 +2 +3

Assigning Oxidation States Oxygen is assigned an oxidation number of -2 in compounds; an exception is found in the peroxide ion, O22-, where each oxygen is assigned an oxidation number of -1. Na2O Fe2O3 +1 -2 +3 -2

Assigning Oxidation States Hydrogen is assigned an oxidation number of +1 in its covalent compounds with nonmetals. In compounds with metals, the oxidation number of hydrogen is -1. H2O HCl +1 -2 +1 -1

Assigning Oxidation States The sum of the oxidation states in a compound must be zero, as a compound’s charge is zero. Na2O Fe2O3 H2O HCl +1 -2 +3 -2 +1 -2 +1 -1

Assigning Oxidation States The sum of the oxidation states in a polyatomic ion must be equal to that ion’s charge. OH1- SO32- CN1- ClO41- -2 +1 +4 -2 +2 -3 +7 -2

Assigning Oxidation States Non-integer oxidation states do exist and indicate the average division of electrons among the elements. Fe3O4 +8/3 -2

Assigning Oxidation States +2 -2 +4 -2 CO CO2 NO NO2 NO21- N2 +2 -2 +4 -2 +3 -2

Half-Reaction Method of Balancing Assign oxidation numbers to every atom in the reaction Identify what is oxidized and what is reduced. Anything whose oxidation number does not change is a spectator.

Half-Reaction Method of Balancing 1. Write a ½ reaction for the reduction reaction without the spectators. 2. If an atom being reduced is part of a solid, a polyatomic ion or part of a molecular compound, you will keep that ion or cpd together when you bring it down for the ½ rxn.

Half-Reaction Method of Balancing 3. Balance the ½ rxn by first balancing the non-H, non-O atoms. 4. Then balance the H’s and O’s using the following guide: a. Use H1+ and H2O if the rxn occurs in an acidic medium. b. Use OH1- and H2O if the rxn occurs in a basic medium.

Half-Reaction Method of Balancing 5. Balance the ½ rxn electrically (charge-wise) by adding electrons (e-) to the left side since a reduction rxn goes GER

Half-Reaction Method of Balancing Repeat steps 1-5, but LEO in step 5

Half-Reaction Method of Balancing 6. Normalize the electrons in each ½ rxn by finding the LCM of the two numbers of electrons and distributing the multipliers through each entire ½ rxn.

Half-Reaction Method of Balancing 7. Add the two ½ rxns and cancel any like atoms, ions, or cpds that appear on both sides. 8. Add the spectators back into the rxn and balance the rest by inspection.

½ Rxn Method of Balancing PbO(s) + CO(g)  Pb(s) + CO2(g) +2 -2 +2 -2 +4 -2

½ Rxn Method of Balancing PbO(s)  Pb(s) + 2H1+ + 2e- + H2O

½ Rxn Method of Balancing CO(g)  CO2(g) + H2O + 2H1+ + 2e-

½ Rxn Method of Balancing PbO(s)  Pb(s) CO(g)  CO2(g) 1 + 2H1+ + 2e- + H2O 1 + H2O + 2H1+ + 2e-

½ Rxn Method of Balancing PbO(s)  Pb(s) CO(g)  CO2(g) 1 + 2H1+ + 2e- + H2O 1 + H2O + 2H1+ + 2e- PbO(s) + 2H++ 2e- + CO(g) + H2O  Pb(s) + H2O + CO2 + 2H++ 2e-

½ Rxn Method of Balancing PbO(s) + CO(g)  Pb(s) + CO2(g) Ta-da!

½ Rxn Method of Balancing CeCl4(aq) + SnCl2(aq)  CeCl3(aq) + SnCl4(aq) +4 -1 +2 -1 +3 -1 +4 -1

½ Rxn Method of Balancing Ce4+  Ce3+ + 1e-

½ Rxn Method of Balancing Sn2+  Sn4+ + 2e-

½ Rxn Method of Balancing Ce4+  Ce3+ Sn2+  Sn4+ + 2e- 2 + 1e- 1

½ Rxn Method of Balancing Ce4+  Ce3+ Sn2+  Sn4+ + 2e- 2 + 1e- 1 2Ce4+ + Sn2++ 2e-  2Ce3+ + Sn4+ + 2e-

½ Rxn Method of Balancing 2CeCl4 + SnCl2  2CeCl3 + SnCl4 Whoohoo!

½ Rxn Method of Balancing HCl + FeCl2 + KNO3  FeCl3 + NO + H2O + KCl