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Chapter 4

by: Biljana Martic
Biljana Martic
SIM-UB (University at Buffalo-The State University of New York)
GPA 3.7
Gen Chem 1
Keister, J B

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About this Document

There is a complete description of how to do Redox Reactions and assign oxidation numbers. There is also descriptions of other reactions.
Gen Chem 1
Keister, J B
Class Notes
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This 4 page Class Notes was uploaded by Biljana Martic on Friday October 2, 2015. The Class Notes belongs to CHE 107LLR at SIM-UB (University at Buffalo-The State University of New York) taught by Keister, J B in Summer 2015. Since its upload, it has received 80 views. For similar materials see Gen Chem 1 in Chemistry at SIM-UB (University at Buffalo-The State University of New York).

Similar to CHE 107LLR at SIM-UB (University at Buffalo-The State University of New York)


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Date Created: 10/02/15
Chapter 4 Reactions in Solutions Precipitation Reactions a solid precipitate is created Usually happens during double displacement reactions when the anions and cations of two ionic compounds change partners 2 NaOH aq CUSO4aq gt Na2504aq CUOH7 s NaOH lt ionic because Na is a Group 1 element and therefore always has a 1 charge and OH hydroxide always has a 1 charge CuSO4 lt ionic because Cu is a metal and has a positive charge and 04 sulfate always has a charge of 2 Hydroxide OH39 and sulfate SO42 are anions because they have negative charges 0 Sodium Na and copper Cu are cations because they have positive charges Think of the t in cations as a sign Acidbase Neutralization an acid reacts with a base to create products of salt and water HCI NaOH gt NaCl H20 HCI is an acid because it donates H et unknown element M HM gt H M39 NaOH is a base because it donates OH39 et unknown element M MOH gt M OH39 Hydrolysis a reversible reaction where water breaks down 1 molecule into 2 fairly straightforward CH3C02CH3 aq H20 I gt HOCH3 aq CH3COOH aq A dehydration reaction would have 2 molecules as the and then 1 molecule and water as the products Oxidationreduction Redox Reaction one or more electrons are transferred changing the oxidation number of a 39 or atom 3Im4AI gt3Mn2Alzo3 i 4 0 0 3 Oxidation is when the oxidation number gets bigger or more positive loses electrons Reduction is when the oxidation number gets smaller or more negative gains electrons Oxidation Number an indicator of how many or how little electrons the element has or if it is neutral 0 There are a few rules to consider when assigning oxidation numbers 1 An atom in its elemental state has an oxidation number of 0 Like in the example above when the Al and Mn were alone with no charge they had an oxidation number of 0 Even if they have a subscript like 02 they are 0 as long as there is no superscript charge 2 An atom in a monoatomic ion has an oxidation number identical to its charge Like 3 4 Al3 has an oxidation number of 3 and Ca2 has an oxidation number of 2 An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monoatomic ion A lot of atoms have a set charge that they rarely deviate from The elements in Group 1 usually have always have a 1 charge Metals which are on the left of the periodic table will be mostly be positive while nonmetals will mostly have a negative charge 0 When assigning oxidation numbers there are a few major things that can be used 0 Oxygen usually has an oxidation number of 2 so whenever you see oxygen in an equation assume it is a 2 unless it doesn t give you the result you need then it has likely changed 0 Hydrogen is usually 1 but it can be either 1 or 1 depending on what kind of atom it has bonded to If it has bonded to a metal which are usually positive it will have an oxidation number of 1 If it is bonded to a nonmetal which are usually negative it will have an oxidation number of 1 0 Lastly Group 7 the halogens mostly have a 1 oxidation number because they are nonmetals Unless they are bonded to an oxygen atom in which case they will change to 1 A polyatomic ion will have an oxidation number of 0 if it is neutral If it has a charge on it then the oxidation numbers must equal the charge Take the polyatomic ion H3PO4 First we look at the oxygen and assign it a charge of 2 2 x 4 8 charge The 2 is still the oxidation number for the oxygen 8 is for the 04 So if this were a question on a test your answer for oxygen would be 2 Second we look at the hydrogen Since it is bonded to nonmetals it will have a charge of 1 1 x 3 3 Lastly we assign the phosphorous Since we need to neutralize the charge get it to 0 the phosphorous will have to have a charge of 5 So 8 3 5 0 Now take the polyatomic C0321 We need the overall charge to equal 2 As usual we rst assign the oxygen a 2 charge 2 x 3 6 Since we need the polyatomic ion to equal 2 we can give the carbon a charge of 4 So 6 4 i Another about oxidation is when they say Reducing Agent or Oxidizing Agent Back to the Redox equation l l a v 3 Mn02 4 3 M 2 A203 4 0 0 3 Manganese Mn is being reduced gaining electrons but it is the Oxidizing Agent because it causes oxidation Mn takes the electrons away from the Al enabling the Al to be oxidized And vice versa the Al is being oxidized losing electrons it is the Reducing Agent because it gives its electrons away enabling Mn to take them and be reduced I know it s confusing so just try to remember that the names are switched The HalfReaction method is used to balance Redox Reactions The idea here is to split up the oxidation and the reduction equations Taking the equation Cus HNO3aq gt Cu2 aq NOg Assign oxidation 5 Cus Cu O HN03H1N5O6 Cu Cu 2 NOg N 2 0 2 Cu loses electrons so it becomes more positive it oxidizes 0 to 2 Oxidation HalfReaction Cu gt Cu2 1st thing to balance the oxygen with water so Cu gt Cu2 there aren t any oxygens 2nCI balance out the hydrogens so Cu gt Cu2 there was no water so no H to balance out 3rCI add electrons to balance charges Cu gt Cu2 2 e39 the electrons make the right side neutral N gains electrons so it becomes more negative it reduces 5 to 2 Reduction HalfReaction 1st balance out the oxygens HNO3 gt NO 2 H20 there are now three oxygens on both sides 2nCI balance out the hydrogens HNO3 3 H gt NO 2 H20 there are four hydrogens on both sides 3rCI add electrons HNO3 3 H 3 e39 gt NO 2 H20 the negatives and the positives cancel out Next the electron transfer must be balanced out There need to be equal amount of electrons in both of the half reactions There are 2 e39 in the oxidation halfreaction and 3 e39 in the reduction halfreaction The lowest common denominator for 2 and 3 is 6 2HNO33H2e39 gtNO2H202HNO36H6e39 gt2NO4H20 3Cu gtCu22e393Cu gt3Cu26e39 Now there is the same amount of electrons in both half reactions The next step is to put them back together 2HNO36H6 e 3Cu gt2NO4H203Cu26 e39 The electrons cancel out and now the equation is balanced 2HNO36H3CUgt2NO4H203CU2 Solubility Rules Soluble Compounds 0 Group 1A Li Na K Rb Cs always soluble no exceptions Ammonium lon NH4 always soluble no exception Nitrate NO339 always soluble no exception Perchlorate CIO439 always soluble no exception Acetate CH3C0239 always soluble no exception Halides Cl39 Br39 I39 soluble unless combined with halides of Ag Hg22 Pb2 Sulfate SO4239 soluble unless combined with sulfates of Sr Ba Hg22 Pb2 lnsoluble Compounds Carbonate CO3239 insoluble unless combined with carbonates of group 1A and NH4 Sul de SZ39 unless combined with sul des of group 1A cations NH4 Ca Sr Ba2 Phosphate PO4339 unless combined with phosphates of group 1A cations NH4 Hydroxide OH39 unless combined with group 1A cations NH4 Ca Sr Ba2


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