We still do not know one thousandth of one percent of what nature has revealed to us. - Albert Einstein-

OXIDATION-REDUCTION AND ELECTROCHEMISTRY ELECTROCHEMISTRY IS THE STUDY OF THE INTERCHANGE OF CHEMICAL AND ELECTRIC ENERGY. IN STUDYING ELECTROCHEMISTRY, WE WILL BE CONCERNED WITH TWO PROCESSES THAT INVOLVE OXIDATION AND REDUCTION REACTIONS: 1)THE GENERATION OF AN ELECTRIC CURRENT FROM A SPONTANEOUS CHEMICAL REACTION 2)USE OF CURRENT TO PRODUCE CHEMICAL CHANGE IN DISCUSSING ELECTROCHEMISTRY REALIZE THAT AN ELECTRIC CURRENT IS SIMPLEY THE FLOW OF ELECTRONS FROM ONE POINT TO ANOTHER.

IN THE SIMPLE ELECTRIC CIRCUIT PICTURED BELOW, ELECTRONS ARE FLOWING THROUGH A WIRE FROM THE NEGATIVE TERMINAL OF A BATTERY BACK TO THE POSITIVE TERMINAL OF THE BATTERY. THE VALENCE ELECTRONS IN THE METAL WIRE ARE HELD RATHER LOOSELY, AND WITH THE HELP OF A VOLTAGE CAN BE MOVED THROUGH THE WIRE. THE CHEMICAL REACTIONS IN THE BATTERY ARE PROVIDING THE ENERGY FOR THIS. E = IR WHERE E = VOLTAGE I = CURRENT R = RESISTANCE

WE WILL BE CONCERNED FIRST WITH OXIDATION- REDUCTION REACTIONS – MOSTLY IN AQUEOUS SOLUTIONS. AN OXIDATION-REDUCTION OR REDOX REACTION IS ONE THAT INVOLVES THE LOSS OR GAIN OF ELECTRONS. A GOOD EXAMPLE IS: 2Ce +4 (aq) + Sn +2 (aq)  2Ce +3 (aq) + Sn +4 (aq) OXIDTION IS THE LOSS OF ELECTRONS. REDUCTION IS THE GAIN OF ELECTRONS.

WE CAN SPLIT THE ABOVE REACTION INTO TWO HALF REACTIONS. 2Ce +4 (aq) + 2e -  2Ce +3 (aq) reduction half reaction Sn +2 (aq)  Sn +4 (aq) + 2e - oxidation half reaction NOTE: THE SAME NUMBER OF ELECTRONS ARE LOST AS GAINED

HALF REACTION METHOD FOR BALANCING OXIDATION-REDUCTION REACTION EQUATIONS 1)WRITE SEPARATE EQUATIONS FOR THE OXIDATION AND REDUCTION HALF-REACTIONS 2)FOR EACH HALF-REACTION a. BALANCE ALL THE ELEMENTS EXCEPT HYDROGEN AND OXYGEN b. BALANCE OXYGEN USING H 2 O c. BALANCE HYDROGEN USING H + d. BALANCE THE CHARGE USING ELECTRONS 3) BALANCE THE TWO HALF REACTIONS SO THAT THE TWO HAVE THE SAME NUMBER OF ELECTRONS LOSS AS GAINED 4) ADD THE TWO HALF REACTIONS AND CANCEL IDENTICAL SPECIES 5) CHECK TO MAKE SURE THE EQUATION IS BALANCED

MnO Fe +2  Fe +3 + Mn +2 Oxidation: Fe +2  Fe +3 + e - Reduction: MnO 4 -  Mn +2 if we assume that each oxygen has an oxidation state of -2, the Mn in MnO 4 - would have an oxidation state of +7, so we have a gain of 5 electrons from MnO 4 - to Mn +2. MnO e -  Mn +2 Balancing oxygen MnO H + + 5e -  Mn H 2 O Note: net charges balance. 5Fe +2  5Fe e - ADDING THE TWO HALF-REACTIONS GIVES: MnO Fe H +  Mn Fe H 2 O

TREAT THESE AS PUZZLES – NOT BURDENS!

RULES FOR ASSIGNING OXIDATION NUMBERS The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O 2, O 3, P 4, S 8, and aluminum metal all have an oxidation number of 0. The oxidation number of monatomic ions is equal to the charge on the ion. The oxidation number of sodium in the Na + ion is +1, for example, and the oxidation number of chlorine in the Cl - ion is -1. The oxidation number of hydrogen is +1 when it is combined with a nonmetal. Hydrogen is therefore in the +1 oxidation state in CH 4, NH 3, H 2 O, and HCl. The oxidation number of hydrogen is -1 when it is combined with a metal. Hydrogen is therefore in the -1 oxidation state in LiH, NaH, CaH 2, and LiAlH 4.

The metals in Group IA form compounds (such as Li 3 N and Na 2 S) in which the metal atom is in the +1 oxidation state. The elements in Group IIA form compounds (such as Mg 3 N 2 and CaCO 3 ) in which the metal atom is in the +2 oxidation state. Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O 2, O 3, H 2 O 2, and the O 2 2- ion. The nonmetals in Group VIIA often form compounds (such as AlF 3, HCl, and ZnBr 2 ) in which the nonmetal is in the -1 oxidation state. The sum of the oxidation numbers of the atoms in a molecule is equal to the charge on the molecule. The most electronegative element in a compound has a negative oxidation number.

WHAT IS THE OXIDATION STATE OF THE METAL IN THE FOLLOWING COMPOUNDS? A)Cu 2 O B)Cr 2 O 3 C)MnO 2 D)Al 2 S 3

WHAT IS THE OXIDATION STATE OF THE NONMETALS IN THE FOLLOWING COMPOUNDS? A)NO B)NO 2 -1 C)SO 4 -2 D)PCl 5 E)NH 4 +1

In each of the following equations, indicate the element that has been oxidized and the one that has been reduced. You should also label the oxidation state of each before and after the process: 1)2 Na + FeCl 2  2 NaCl + Fe 2)2 C 2 H O 2  4 CO H 2 O 3)2 PbS + 3 O 2  2 SO PbO 4)2 H 2 + O 2  2 H 2 O 5)Cu + HNO 3  CuNO 3 + H 2 6)AgNO 3 + Cu  CuNO 3 + Ag

Balance the following redox equations: a)HNO 3 (aq) + H 3 AsO 3 (aq)  NO (g) + H 3 AsO 4 (aq) + H 2 O (l) b)Cu(s) + HNO 3 (aq)  Cu(NO 3 ) 2 (aq) + NO(g) + H 2 O(l) c)Cr 2 O 7 2− (aq) + HNO 2 (aq)  Cr 3+ (aq) + NO 3 − (aq) (acidic)