Chapter 21- Electrochemistry Reduction-Oxidation or REDOX chemistry

21-1: Introduction to Electrochemistry Background: In Chemistry 11 we saw many Single Replacement reactions, such as: Background: In Chemistry 11 we saw many Single Replacement reactions, such as: Mg (s) + 2HCl (aq) ssd MgCl 2(aq) + H 2(g) Mg (s) + 2HCl (aq) ssd MgCl 2(aq) + H 2(g) In fact, this is a two part “REDOX” reaction: In fact, this is a two part “REDOX” reaction:

1) Mg (s) ssd Mg 2+ (aq) + 2 e — Mg (s) has LOST two Electrons during an OXIDATION reaction (* LEO = loss of electrons is oxidation) 2) 2H + (aq) + 2 e — ssd H 2 (g) The H + ions have GAINED two Electrons during a REDUCTION reaction (*GER = gain of electrons is reduction) Note: The Cl — ions in the original reaction did not participate and are SPECTATOR ions

Background…. Cont’d Oxidation and Reduction reactions occur simultaneously… every electron lost in the oxidation is gained in the reduction. Oxidation and Reduction reactions occur simultaneously… every electron lost in the oxidation is gained in the reduction. As each of these reactions is one-half of the overall reaction, we call them… HALF- REACTIONS! As each of these reactions is one-half of the overall reaction, we call them… HALF- REACTIONS!

Back to our example: Back to our example: 1) Mg (s) sd Mg 2+ (aq) + 2 e — OXIDATION 2) 2H + (aq) + 2 e — sd H 2 (g) REDUCTION Mg (s) + 2H + (aq) sd Mg 2+ (aq) + H 2 (g) These two reactions can be combined to yield the overall NET reaction. Mg gets oxidized, but CAUSES the reduction of H +, and Mg is called the REDUCING AGENT H + gets reduced, but CAUSES the oxidation of Mg and H + is called the OXIDIZING AGENT

The Reducing Agent gets oxidized and the Oxidizing Agent gets reduced

2+

Another Example 3CuCl 2(aq) +2Al (s) ssd 2AlCl 3(aq) + 3Cu (s) 3CuCl 2(aq) +2Al (s) ssd 2AlCl 3(aq) + 3Cu (s) Two half reactions: Two half reactions: 1) Cu 2+ (aq) + 2e — ssd Cu (s) 1) Cu 2+ (aq) + 2e — ssd Cu (s) Cu 2+ ion GAINS 2 electrons and is Cu 2+ ion GAINS 2 electrons and is REDUCED REDUCED 2) Al (s) ssd Al 3+ (aq) + 3e — 2) Al (s) ssd Al 3+ (aq) + 3e — Aluminum atom LOSES 3 electrons and is Aluminum atom LOSES 3 electrons and is OXIDIZED OXIDIZED

Cu 2+ ion is the OXIDIZING AGENT and gets REDUCED Cu 2+ ion is the OXIDIZING AGENT and gets REDUCED Al atom is the REDUCING AGENT and gets OXIDIZED Al atom is the REDUCING AGENT and gets OXIDIZED

REMEMBER: LEO the Lion says GER REMEMBER: LEO the Lion says GER -- LEO= Loss of Electrons is Oxidation -- LEO= Loss of Electrons is Oxidation -- GER= Gain of Electrons is Reduction -- GER= Gain of Electrons is Reduction

21-2 Oxidation Numbers Oxidation and reduction can be explained in terms of the loss and gain of electrons. Oxidation numbers are used to help keep track of electrons in chemical reactions Oxidation and reduction can be explained in terms of the loss and gain of electrons. Oxidation numbers are used to help keep track of electrons in chemical reactions OXIDATION NUMBER = the theoretical charge an atom would have in a molecule if every molecule was IONIC. This is as if the electrons were completely transferred. OXIDATION NUMBER = the theoretical charge an atom would have in a molecule if every molecule was IONIC. This is as if the electrons were completely transferred.

For example, in the compound H 2 O: For example, in the compound H 2 O: O has an electronegativity of 3.16 O has an electronegativity of 3.16 H has an electronegativity of 2.1 H has an electronegativity of 2.1 Since O is more electronegative than H, then we assume that the O removes the electron from each H and the oxidation numbers are: H = +1 O = -2 Since O is more electronegative than H, then we assume that the O removes the electron from each H and the oxidation numbers are: H = +1 O = -2

Redox reactions can be defined in terms of oxidation number: Redox reactions can be defined in terms of oxidation number: An element is said to be oxidized if its oxidation number increases in a reaction. An element is said to be oxidized if its oxidation number increases in a reaction. An element is said to be reduced if its oxidation number decreases in a reaction. An element is said to be reduced if its oxidation number decreases in a reaction.

Rules for Assigning Oxidation #’s Rule #1. The oxidation number of an atom of any element in its elemental state (uncombined form) is ZERO, no matter how complex the molecule. Rule #1. The oxidation number of an atom of any element in its elemental state (uncombined form) is ZERO, no matter how complex the molecule. H 2 = 0, F 2 = 0. Be = 0. Li = 0, Na = 0, O 2 = 0, P 4 = 0, S 8 = 0 H 2 = 0, F 2 = 0. Be = 0. Li = 0, Na = 0, O 2 = 0, P 4 = 0, S 8 = 0

RULE #2. For an ion composed of only one atom, the oxidation number is equal to the charge on the ion. RULE #2. For an ion composed of only one atom, the oxidation number is equal to the charge on the ion. K + = +1, Mg 2+ = +2, Al 3+ = +3, F - = -1, O 2- = -2 K + = +1, Mg 2+ = +2, Al 3+ = +3, F - = -1, O 2- = -2 All alkali metal ions (Li +, Na +, K +, Rb +, Cs +, Fr + ) have an oxidation number of +1. All alkali metal ions (Li +, Na +, K +, Rb +, Cs +, Fr + ) have an oxidation number of +1. All alkaline earth metal ions (Be +2, Mg +2, Ca +2, Sr +2, Ba +2, Ra +2 ) have an oxidation number of +2. All alkaline earth metal ions (Be +2, Mg +2, Ca +2, Sr +2, Ba +2, Ra +2 ) have an oxidation number of +2.

Rule #3: The oxidation number of oxygen in most compounds is –2 (e.g. H 2 O, CaO) Rule #3: The oxidation number of oxygen in most compounds is –2 (e.g. H 2 O, CaO) There are three exceptions to this rule: There are three exceptions to this rule: In OF 2 O=+2 since F is more electronegative than oxygen In OF 2 O=+2 since F is more electronegative than oxygen In H 2 O 2, or HO 2 -, or O 2 2- (the peroxide family) O= -1 In H 2 O 2, or HO 2 -, or O 2 2- (the peroxide family) O= -1 O 2 : the oxidation number is 0 (see rule #1) O 2 : the oxidation number is 0 (see rule #1)

Rule # 4. Fluorine has an oxidation number of –1 in all of its compounds. Rule # 4. Fluorine has an oxidation number of –1 in all of its compounds. Rule #5. The oxidation number of hydrogen is mostly +1, except when it is bonded to elements less electronegative than itself or if it is in its elemental form. Rule #5. The oxidation number of hydrogen is mostly +1, except when it is bonded to elements less electronegative than itself or if it is in its elemental form. NaHNa = +1, H = -1 NaHNa = +1, H = -1 CH 4 C = -4, H = +1 CH 4 C = -4, H = +1 H 2 H = 0 H 2 H = 0

Rule #6. In a NEUTRAL molecule, the sum of the oxidation numbers of all the atoms must be ZERO. In a POLYATOMIC ION, the sum of the oxidation numbers of all the atoms in the ion must be equal to the NET CHARGE of the ion. Rule #6. In a NEUTRAL molecule, the sum of the oxidation numbers of all the atoms must be ZERO. In a POLYATOMIC ION, the sum of the oxidation numbers of all the atoms in the ion must be equal to the NET CHARGE of the ion.

Examples Assign Oxidation Numbers to each element in the following: Assign Oxidation Numbers to each element in the following: a) NH 4 + a) NH 4 + b) Rb 2 O b) Rb 2 O c) H 2 Cr 2 O 7 c) H 2 Cr 2 O 7

Homework Optional: Read HEBDEN WORKBOOK Section 5-1 "Introduction". Do Exercises p.192 #1-2. Optional: Read HEBDEN WORKBOOK Section 5-1 "Introduction". Do Exercises p.192 # Optional: Read HEBDEN WORKBOOK Section 5-2 "Oxidation Numbers". Do Exercise p.194 # Optional: Read HEBDEN WORKBOOK Section 5-2 "Oxidation Numbers". Do Exercise p.194 # Do HANDOUT "Section 211/212 Provincial Exam Questions". 4. Do HANDOUT "Section 211/212 Provincial Exam Questions".