Lesson 1

 Earlier in the year we learned that if a copper strip was placed in silver nitrate, the silver would be displaced by the copper and solid silver crystals would form on the copper. The reaction is classified as a single displacement reaction.

 Cu (s) + 2 Ag NO 3­(aq)  2 Ag (s) ­ + Cu(NO 3 ) 2(aq)

 If we look at the transfer of electrons in this reaction we will see that Cu looses e - and Ag gains e - Cu (s) + 2 Ag + (aq)  2 Ag (s) + Cu 2+ (aq)  Cu becomes Cu 2+, a loss of 2 e -  Each Ag + becomes Ag, a gain of 2 e

Oxidation - A process in which chemical entities lose e -  Cu was oxidized to Cu 2+

Reduction – A process in which chemical entities gain e -  Each Ag + was reduced to Ag

 reaction in which one reactant is oxidized and the other is reduced is called a redox reaction.  Therefore, in any redox reaction, the number of e - lost must equal the number of e - gained. It must balance. 

 To remember which is oxidation and which is reaction you can use of these memory aids.  LEO says GER  Lose Electrons Oxidation  Gain Electrons Reduction  OR  OIL RIG  Oxidation Is Loss – Reduction Is Gain

 A substance that removes electrons from another substance is known as a oxidizer or a Oxidizing agent   A substance that gives electrons to another substance is known as an reducer or a reducing agent.

Step 1) Identify repeating entities  Write the total ionic equation  Eliminate ions common on both sides Step 2) Label charges  Uncombined have a charge of 0 Step 3) Identify loss and gain of electrons

Zn (s) + CuSO 4(aq)  ZnSO 4(aq) + Cu (s) Step 1) Zn + Cu 2+ (aq) + SO 4 2- (aq)  Zn 2+ (aq) + SO 4 2- (aq) + Cu (s) Step 2)  Zn 0 (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu 0 (s)

Step 3)  Zn becomes Zn 2+, loss of 2 e - (oxidation)  Cu 2+ becomes Cu, gain of 2e - (reduction)

 Practice page 376  Questions page 377 # 2-5, 7

Not all redox reactions involve metals  2H 2(g)­ + O 2(g)  2H 2 O (g) This reaction is similar to the reaction of sodium with oxygen  2Na (s) + Cl 2(g)  2NaCl (s)

 The main difference is that there are NO ionic charges.  Earlier in the year we learned that polar covalent bonds have partial charges δ - and δ +.  The more electronegative an element is the more it will pull the electrons of the other element towards it. This apparent charge is called an oxidation number, which is the charge it would have if it gained or lost electrons.

 The sum of all of the oxidation numbers in a molecule must be zero because the molecule does not have an overall charge.  If it was a polyatomic ion it would have a charge but the oxidation numbers can still be calculated.

 2H 2(g)­ + O 2(g)  2H 2 O (g)  2Na (s) + Cl 2(g)  2NaCl (s)  In the reaction seen above, O is more electronegative than H. Since oxygen gained 2 e - to from O 2- in the ionic compound. We will assume it has a oxidation # of -2.

 To find the oxidation #of the other element, solve for the unknown. H 2 O  2( H) + (O) = 0  2(H) + (-2) = 0  2H = 2  H = 1  Therefore, H has an oxidation number of +1.

 Note:  The oxidation number of an atom can change if the atom is involved in a redox reaction.

I) The oxidation number of a atom in an uncombined element is always Zero.  Example: Na, K, O 2, H 2, Cl 2, S 8, Li,

II) The oxidation number of a simple ion is the charge of the ion.  Example: Ca 2+ = + 2  Na + = + 1  Cl - = - 1

III) The oxidation number of Hydrogen in most compounds is +1.  Example: H­ 2 O, H 2 SO 4, NH 3, H = +1  Except in metal hydrides: NaH, H = -1

IV) The oxidation number of oxygen in most compounds is -2  Example : MgO, HNO 3, O = -2  Except in peroxides: H 2 O 2, O = -1

V) The oxidation number of group 1 elements is +1  Example: Na, Li, = + 1  The oxidation number of group 2 elements is +2  Example: Mg, Ca = + 2

VI) The sum of oxidation numbers in a compound must = 0  Example: H 2 O = 2 (+1) + (-2) = 0

 The sum of oxidation numbers in a polyatomic ion must equal the charge of the ion.  Example: OH - = (-2) + (+1) = -1

H 2 SO 4 : 2(H) + (S) + 4(O) = 0 2 (+1) + (S) +4(-2) = (S) -8 = 0 S - 6 = 0 S = 6

 NO 3 - : (N) + 3(O) = -1 (N) + 3(-2) = -1 N – 6 = -1 N = 5

 Not all reactions are redox  By using oxidation numbers, you can tell if a reaction is redox. A redox reaction will have a change in oxidation numbers where a reaction that is not redox will not.  Example: AgNO 3 + HCl  HNO 3 + AgCl Ag + + NO H + + Cl -  H + + NO Ag + + Cl -

Questions  Page 383# 3

 The Activity series is an arrangement of metals in order of their tendency to react (become oxidized).  The most reactive metals (most easily oxidized) are at the top of the list.  The least reactive are at the bottom.

 This allows us to predict whether a single displacement reaction will occur not. For it to occur, the metal must be higher on the list than the metal in the compound that it is trying to displace.

 The halogens are also arranged in order of most reactive to least reactive.  Unlike the metals, the most reactive are the easiest to reduce. Fluorine is the strongest oxidizing agent because of its very high electronegativity.

MetalsDecreasing ActivityHalogens Lithium Fluorine Potassium Chlorine Calcium Bromine Sodium Iodine Aluminum Zinc Chromium Iron Nickel Tin Lead Hydrogen * Copper Mercury Silver Platinum Gold

 The Activity series is used to predict the products of single displacement reactions.  A + BC  AC + B  In general, an element that is higher on the activity series will displace an element that is lower. The lower element is, thus left as a pure metal.

 Zn (s) + AgNO 3(aq)  Zn(NO 3 ) 2(aq) + Ag  Is this a redox reaction?  Zn (s) + 2Ag + (aq) + 2NO 3 - (aq)  Zn 2+ 2(NO 3 ) - (aq) + Ag

 Zn 0 (s) + Ag + (aq)  Zn 2+ + Ag 0  Zn becomes Zn 2+, loss of 2 e - (oxidation)  Ag + becomes Ag, gain of 1e - for 2atoms (reduction)

 Practice page 388

 Activity Series Lab