REDOX REACTIONS Chapters 4, 19.1 & 19.2 M. Shozi CHEM110 / 2014
REDOX REACTIONS Reactions involve the transfer of electrons between reactants When a substance loses electrons, it undergoes oxidation: Ca(s) + 2H + (aq) Ca 2+ (aq) + H 2 (g) When a substance gains electrons, it undergoes reduction: 2Ca(s) + O 2 (g) 2CaO(s)
OXIDATION NUMBERS To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity using the following five sets of rules.
RULES FOR ASSIGNING OXIDATION NUMBERS 1. For an atom in its elemental form, the oxidation number is always zero. e.g. Na = 0 and Cl 2 = 0 2. For any monatomic ion the oxidation number equals the charge on the ion. e.g. Na + = +1 and Cl = -1
OXIDATION NUMBERS 3. Non metals usually have negative oxidation numbers: a) The oxidation number of oxygen is usually -2 except in the peroxide ion in which it has an oxidation number of 1. b) The oxidation number of hydrogen is +1 when bonded to a metal and -1 when bonded to a nonmetal. c) The oxidation number of fluorine is -1 in all compounds. The other halogens usually have an oxidation number of -1 unless combined with oxygen (oxyanions) where they have a positive oxidation number.
OXIDATION NUMBERS 4. The sum of the oxidation numbers of all atoms in a neutral compound is zero. Eg. Pb(NO 3 ) 2, KCl, H 2 SO 4 5. The sum of the oxidation numbers in a polyatomic ion atom equals the charge of the ion. Eg. NH 4+, NO 3, Cr 2 O 7 2
OXIDISING & REDUCING AGENTS OXIDISING AGENT Causes the oxidation of another substance Gains electrons, i.e. it is reduced Example: MnO 4 and Cr 2 O 2-7 REDUCING AGENT Causes the reduction of another substance Loses electrons, i.e. it is oxidised Example: NaH and CaH 2
Exercise 1 Identify the reducing and oxidising agents in the following reaction: 5SO 3 2- + 2MnO 4 + 6H + 5SO 4 2- + 2Mn 2+ + 3H 2 O
Exercise 1 solution
Exercise 1 solution
DISPROPORTIONATION REACTION A redox reaction where an element is simultaneously oxidised and reduced. Example: Decomposition of hydrogen peroxide +1-1 +1-2 0 H 2 O 2 2 H 2 O + O 2-1 reduced to -2-1 oxidised to 0
DISPLACEMENT REACTION A redox reaction where one element in molecular/atomic form reacts with either an acid or metal salt Mg(s) + 2 HCl(aq) MgCl 2 (s) + H 2 (g) 12
OXIDATION & REDUCTION HALF -REACTIONS Separate the overall REDOX reaction oxidation and reduction half reaction: 0 +2 Zn(s) + Cu 2+ +2 (aq) Zn 2+ 0 (aq) + Cu(s) Half Reactions REDOX reactions are linked by gain/loss of e - Oxidation Reaction: Zn(s) Zn 2+ (aq) + 2 e - Reduction Reaction: Cu 2+ (aq) + 2 e - Cu(s) Overall Reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)
Exercise 2 Show the oxidation and reduction that occur, and write the overall ionic equation for the reaction of Mn with lead(ii) nitrate solution to produce Pb(s) and Mn(NO 3 ) 2 (aq). Mn(s) + Pb(NO 3 ) 2 (aq) Pb(s) + Mn(NO 3 ) 2 (aq)
Exercise 2 - Solution
BALANCING REDOX REACTIONS HALF REACTION METHOD ACID MEDIUM Follow the steps below: 1. Write skeleton half reactions 2. Balance a) Any atoms other than H & O; b) O by adding H 2 O; & c) H by adding H + 3. Balance the charge by adding electrons 4. Multiply half reactions so that electrons cancel & add reactions together.
Exercise 3 Balance the following reaction in acidic medium: MnO 4 + C 2 O 4 2- Mn 2+ + CO 2
Exercise 3 Solution 1. Assign oxidation numbers to determine what is oxidised and what is reduced. +7 +3 +2 +4 MnO 4 + C 2 O 4 2- Mn 2+ + CO 2 Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidised.
Exercise 3 Solution 2. Divide the equation into two halfreactions: one for oxidation and one for reduction. Oxidation C 2 O 2 4 CO 2 Reduction MnO 4 Mn 2+
Exercise 3 Solution 3. Balance each half-reaction Oxidation a. To balance the C, we add a coefficient of 2. C 2 O 4 2 2CO 2 b. H 2 O not added since O is balanced. c. H + not added as there is no H in the equation. d. To balance the charge, we must add 2 electrons to the right-hand side. C 2 O 4 2 2CO 2 + 2e
Reduction Exercise 3 Solution a. The manganese is balanced MnO 4 Mn 2+ b. To balance the O, we must add 4 H 2 O to the righthand side. MnO 4 Mn 2+ + 4H 2 O c. To balance the H, we add 8 H + to the left-hand side. 8H + + MnO 4 Mn 2+ + 4H 2 O d. To balance the charge, we add 5 e to the left-hand side. 5e + 8H + + MnO 4 Mn 2+ + 4H 2 O
Exercise 3 Solution 4. Multiply the half-reactions by integers if necessary so that the numbers of electrons equal each other Oxidation 5 x C 2 O 4 2 2CO 2 + 2e gives 5C 2 O 4 2 10CO 2 + 10e Reduction 2 x 5e + 8H + + MnO 4 Mn 2+ + 4H 2 O gives 10e + 16H + + 2MnO 4 2Mn 2+ + 8H 2 O
Exercise 3 Solution 5. Add the two half-reactions and simplify by cancelling species that appear on both sides of the combined equation. 5C 2 O 2 4 10CO 2 + 10e + 10e + 16H + + 2MnO 4 2Mn 2+ + 8H 2 O 5C 2 O 4 2 + 10e + 16H + + 2MnO 4 2Mn 2+ + 8H 2 O + 10CO 2 + 10e and simplify to: 5C 2 O 4 2 + 16H + + 2MnO 4 2Mn 2+ + 8H 2 O + 10CO 2
BALANCING REDOX REACTIONS HALF REACTION METHOD BASIC MEDIUM METHOD 1 Follow the steps below: 1. Balance reaction exactly as you would for ACIDIC MEDIUM, i.e. steps 1-4 2. In your overall balanced acidic medium reaction, add the same number of OH to each side to neutralise the same number of H + in the equation and create H 2 O in its place 3. The resulting water can be cancelled as needed
BALANCING REDOX REACTIONS HALF REACTION METHOD BASIC MEDIUM METHOD 2 Follow the steps below: 1. Write skeleton half reactions 2. Balance a) Any atoms other than H & O; b) O by adding H 2 O; & c) H by adding H + d) H + by adding OH - to both sides of the equation and form H 2 O 3. Balance the charge by adding electrons 4. Multiply half reactions so that electrons cancel & add reactions together.
Exercise 4 Balance the following reaction in basic medium: MnO 4 + C 2 O 4 2- Mn 2+ + CO 2
Exercise 4 Solution (Method 1) 5C 2 O 2 4 + 16H + + 2MnO 4 2Mn + 2 + 8H 2 O + 10CO 2 5C 2 O 2 4 + 16H + + 16OH - + 2MnO 4 2Mn + 2 + 8H 2 O + 10CO 2 + 16OH - 5C 2 O 2 4 + 16H 2 O + 2MnO 4 8 2Mn + 2 + 8H 2 O + 10CO 2 + 16OH - and simplify to: 5C 2 O 4 2 + 8H 2 O + 2MnO 4 2Mn 2+ + 10CO 2 + 16OH -