Transcript Chapter 6:
Chapter 6: Oxidation-Reduction Reactions • Reactions that involve the transfer of electrons are called oxidation-reduction or redox reactions • Oxidation is the loss of electrons by a reactant • Reduction is the gain of electrons by a reactant • Oxidation and reduction always occur together • The total number of electrons lost by one substance is the same as the total number of electrons gained by the other • For a redox reaction to occur, something must accept the electrons that are lost by another substance • The substance that accepts the electrons is called the oxidizing agent • The substance that lost the electrons is called the reduction agent • Note that the oxidizing agent is reduced and the reducing agent is oxidized • For example: 2 Na + Cl2 2 NaCl – Na is the reducing agent because it lost electrons and was oxidized – Cl2 is the oxidizing agent because it gained electrons and was reduced • Oxidation numbers provide a way to keep track of electron transfers : 1) The oxidation number of any free element is zero. 2) The oxidation number of any simple, monoatomic ion is equal to the charge on the ion. 3) The sum of all oxidation numbers of the atoms in a molecule or polyatomic ion must equal the charge on the particle. 4) In its compounds, fluorine has an oxidation number of –1. 5) In its compounds, hydrogen has an oxidation number of +1. 6) In its compounds, oxygen has an oxidation number of –2. • If there is a conflict between two rules apply the rule with the lower number and ignore the conflicting rule • In binary ionic compounds with metals, the nonmetals have oxidation numbers equal to the charges on their anions Example: What is the oxidation number of Fe in Fe2O3? ANALYSIS: This binary compound is ionic. Apply rule 3 and 6 Fe: 2x O: 3(-2) = -6 0 = 2x + (-6) or x = +3 = ox. number of Fe • Note that fractional values of oxidation numbers are allowed • In terms of oxidation numbers: – Oxidation is an increase in oxidation number – Reduction is a decrease in oxidation number • This provides a simple way to follow redox reactions • Many redox reactions take place in aqueous solution • A procedure called the ion-electron method provides a way to balance these equations • The oxidation and reduction are divided into equations called half-reactions • The half-reactions are balanced separately, then combined into the fully balanced net ionic equation • Both mass and charge must be balanced • Charge is balanced by adding electrons to the side of the equation that is more positive or less negative Example: Balance the following skeleton equation Al(s) Cu 2 (aq) Al3 (aq) Cu( s) ANALYSIS: T hisis a redox reaction. SOLUT ION: Oxidation: Al(s) Al3 3eReduction: Cu 2 (aq) 2e- Cu( s) T heleast commonfactoris 6, combining 2 Al(s) 3Cu2 (aq) 2Al3 (aq) 3Cu(s) • Many reactions occur in either acidic or basic solutions The Ion-Electron Method in Acidic Solution: 1) 2) 3) 4) 5) 6) Divide the equation into two half-reactions. Balance atoms other than H and O. Balance O by adding water. Balance H by adding hydrogen ion. Balance net charge by adding electrons. Make electron gain and loss equal: add halfreactions. 7) Cancel anything that’s the same on both sides of the equation. • The simplest way to balance reactions in basic solution is to first balance them as if they were in acidic solution, then “convert” to basic solution: Additional Steps for Basic Solutions 8) Add to both sides of theequation the same number of OH- ions as thereare H . 9) CombineH and OH- to form H 2 O. 10) Cancelany H 2 O that you can. Example: Balance the following in basic solution: MnO-4 C 2 O 24- MnO2 CO 32ANALYSIS: Balanceas if in acidic solution hen t " convert". SOLUT ION: C 2 O 24- 2H2 O 2CO32- 4H 2eMnO-4 4H 3e- MnO2 2H2 O Net ionic: 3C2 O 24- 2H2 O 2MnO-4 6CO32 4H 2MnO2 Add OH3C2 O 24- 2H2 O 4OH- 2MnO-4 6CO32 4(H OH ) 2MnO2 Form H 2 O 3C2 O 24- 2H2 O 4OH- 2MnO-4 6CO32 4H2 O 2MnO2 Simplify 3C2 O 24- 4OH- 2MnO-4 6CO32 2H2 O 2MnO2 • Metals more active than hydrogen (H2) dissolve in oxidizing acids Nonoxidizing acids : HCl(aq), cold dilute H 2SO 4 (aq), H 3 P O4 (aq), and most organicacids. Ox. Acid HNO3 Reduct ion Reaction (conc.)NO3- 2H e - NO2 ( g ) H 2 O (dilute) NO3- 4H 3e- NO(g ) 4H2 O (verydilute, with strongreducing agent ) NO3- 10H 8e- NH4 3H2 O H 2SO 4 (hot,conc.)SO 24- 4H 3e- SO 2 ( g ) 2H2 O (hot conc.,with strongreducing agent ) SO -4 10H 8e- H 2S( g ) 4H2 O • Some examples: Concentrated HNO3 : Cu( s ) 4HNO3 (aq) Cu(NO3 ) 2 (aq) 2NO2 ( g ) 2H2 O Dilute HNO3 : 3Cu(s ) 8HNO3 (aq) 3Cu(NO3 ) 2 (aq) 2NO(g ) 4H2 O Hot concentrated H 2SO 4 : Cu( s ) 2H2SO 4 (aq) CuSO 4 (aq) SO 2 ( g ) 2H2 O • More active metals will displace a less active metal from its compound • This often occurs in solution and is called a single replacement reaction Zinc is a more active metal than copper. Copper ions (blue) collide with zinc metal (gray) picking up electrons. Copper ions become copper atoms (red-brown) and stick to the zinc surface. Zinc ions (yellow) replace the copper ions in solution. • An activity series arranges metals according to their ease of oxidation • They can be used to predict reactions Activity Series for Some Metals and Hydrogen Element OxidationProduct Least active Gold Au3 Silver Ag Copper Cu 2 Hydrogen H Sodium Na Most active Cesium Cs (See Table 6.2 for a more extensive list.) • A given element will be displaced from its compounds by any element below it in the table • Oxygen reacts with many substances • The products depends, in part, on how much oxygen is available • Combustion of hydrocarbons O2 plentiful: CH 4 2O2 CO 2 2H2O O2 limited: 2CH4 3O2 2CO 4H2O O2 verylimited: CH 4 O2 C 2H2O • Organic compounds containing O also produce carbon dioxide and water C2H5OH 3O2 2CO2 3H2O • Organic compounds containing S produce sulfur dioxide 2C2H5SH 9O2 4CO2 6H2O 2SO2 • Many metals corrode or tarnish when exposed to oxygen Corrosionof iron : 4Fe 3O2 2Fe2O3 T arnishingof silver : 4Ag O2 2Ag2O • Most nonmetals react with oxygen directly O2 plentiful: C O2 CO 2 O2 limited: 2C O2 2CO • Redox reactions are more complicated than most metathesis reactions • In general, it is not possible to balance a redox reaction by inspection • This is especially true when acid or bases are involved in the reaction • Once balanced, they can be used for stoichiometric calculations • Redox titrations are common because they often involve dramatic color changes • Mole-to-mole ratios are usually involved Example: A 0.3000 g sample of tin ore was dissolved in acid solution converting all the tin to tin(II). In a titration, 8.08 mL of 0.0500 M KMnO4 was required to oxidize the tin(II) to tin(IV). What was the percentage tin in the original sample? ANALYSIS: This is a redox titration in acidic solution. SOLUTION: • Form skeleton equation and use the ion-electron method to produce a balanced equation 3Sn 2 2MnO-4 8H 3Sn 4 2MnO2 4H2O • Use the balanced equation to define equivalence relations and determine the mass of Sn in the original sample Sn 2 1.000 mol Sn 0.0719g Sn 11mol mol Sn 118.7 g Sn 0.00808L MnO sol 0.0500 mol MnO 4 4 - 1.00 L MnO -4 sol - 2 mol MnO -4 3 mol Sn 2 • Convert to percentage g % Sn 0.0719 0.3000 g 100% 24.0% Sn in ore