Aim # 21: How do we balance oxidation-reduction equations? H.W. # 21 Study pp. 170 – 177 (sec. 4.9 – 4.10) pp. 833-839 (sec. 18.1) Ans. ques. p. 877 # 15-17 p. 878 # 29a, 30d, 31b, 32a

Oxidation refers to any chemical change in which there is I Oxidation – a loss, or an apparent loss, of electrons. e.g. 2Ca(s) + O2(g) → 2CaO(s) Ca ox. State: 0 (-2e-) +2 Oxidation refers to any chemical change in which there is an increase in oxidation state (number). The atom (or ion) that increases in oxidation number is said to be oxidized. II Reduction – a gain, or an apparent gain, of electrons. e.g. 2Mg(s) + O2(g) → 2MgO(s) O ox. State: 0 (+2e-) -2 Reduction refers to any chemical change in which there is a decrease in oxidation state (number). The atom (or ion) that decreases in oxidation number is said to be reduced.

IV Types of redox reactions A. Oxidation of metals by oxygen III Redox Reaction – a reaction in which oxidation and reduction occur simultaneously. One cannot occur without the other. The substance being oxidized acts as a reducing agent, while the substance being reduced acts as an oxidizing agent. IV Types of redox reactions A. Oxidation of metals by oxygen 2Ca(s) + O2(g) → 2CaO(s) 4Al(s) + 3O2(g) → 2Al2O3(s) B. Oxidation of metals by acids Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)

Net Ionic Equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) C. Oxidation of metals by salts Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s) Net Ionic Equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Ca(s) + SnCl2(aq) → CaCl2(aq) + Sn(s) Net Ionic Equation: Ca(s) + Sn2+(aq) → Ca2+(aq) + Sn(s) V Oxidation States (numbers) – the oxidation number of an atom is the charge, which an atom has, or appears to have, when electrons are counted according to certain rules. VI Rules for determining oxidation numbers SEE HANDOUT SHEET

VII Balancing redox equations for reactions in acidic solution. Problem: Balance the following equation for the reaction of potassium dichromate, K2Cr2O7, with ethyl alcohol, C2H5OH, in acidic solution. H+(aq) + Cr2O72-(aq) + C2H5OH(ℓ) → Cr3+(aq) + CO2(g) + H2O(ℓ) Step 1: Determine what is being oxidized and what is being reduced. Write separate “half-reactions” for both. +1 +6 -2 -2 +1 -2 +1 +3 +4 -2 +1 -2 ox: C2H5OH ℓ) → CO2(g) red: Cr2O72- (aq) → Cr3+(aq)

C2H5OH(ℓ) + 3H2O(ℓ) → 2CO2(g) + 12H+(aq) Step 2: For each half-reaction, balance a) all elements except H and O b) O by adding H2O c) H by adding H+ d) the charge by adding electrons (e-) to the side with greater overall positive charge ox: C2H5OH(ℓ) → 2CO2(g) C2H5OH(ℓ) + 3H2O(ℓ) → 2CO2(g) C2H5OH(ℓ) + 3H2O(ℓ) → 2CO2(g) + 12H+(aq) C2H5OH(ℓ) + 3H2O(ℓ) → 2CO2(g) + 12H+(aq) + 12e- red: Cr2O72-(aq) → 2Cr3+(aq) Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(ℓ) Cr2O72-(aq) + 14H+(aq) → 2Cr2+(aq) + 7H2O(aq) (-2) + 14(+1) = +12 2(+3) = +6 ••• add 6e- Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(ℓ)

2Cr2O72-(aq) + 28H+(aq) +12e- → 4Cr3+(aq) + 14H2O(ℓ) Step 3: Multiply one or both half- reactions by an integer to equalize the electrons lost and gained. 2Cr2O72-(aq) + 28H+(aq) +12e- → 4Cr3+(aq) + 14H2O(ℓ) Step 4: Add the half reactions and cancel identical species. C2H5OH(ℓ) + 3H2O(ℓ) → 2CO2(g) + 12H+(aq) + 12e- 2Cr2O72-(aq) + 28H+(aq) + 12e- → 4Cr3+(aq) + 14H2O(ℓ) 2Cr2O72-(aq) C2H5OH(ℓ) + 16H+(aq) → 4Cr3+(aq) + 2CO2(g) + 11H2O(ℓ)

Step 5: check that all elements and charges are balanced VIII Balancing redox equations for reactions in basic solution Balance the half-reactions as if they occurred in acidic solution. Then “neutralize” the H+ ions by adding equal numbers of OH- ions to both sides of the equation. Problem: Balance the following redox equation using the half-reaction method. NO2-(aq) + Al(s) → NH3(g) + AlO2-(aq) (basic solution)

+3 -2 0 -3 +1 +3 -2 Step 1: NO2-(aq) + Al(s) → NH3(g) + AlO2-(aq) ox: Al(s) → AlO2-(aq) red: NO2-(aq) → NH3(g) Step 2: ox: Al(s) + 2H2O(ℓ) → AlO2-(aq) Al(s) + 2H2O(ℓ) → AlO2-(aq) + 4H+(aq) Al(s) + 2H2O(ℓ) → AlO2-(aq) + 4H+(aq) + 3e- red: NO2-(aq) → NH3(g) + 2H2O(ℓ) NO2-(aq) + 7H+(aq) → NH3(g) + 2H2O(ℓ) NO2-(aq) + 7H+(aq) + 6e- → NH3(g) + 2H2O(ℓ)

2Al(s) + 8OH-(aq) → 2AlO2-(aq) + 4H2O(ℓ) + 6e- Step 3: “Neutralize” the half-reactions by adding OH- to both sides of each half-reaction. ox: Al(s) + 2H2O(ℓ) + 4OH-(aq) → AlO2-(aq) + 4H+(aq) + 3e- + 4OH-(aq) Al(s) + 2H2O(ℓ) + 4OH-(aq) → AlO2-(aq) + 4H2O(ℓ) + 3e- red: NO2-(aq) + 7H+(aq) + 6e- + 7OH-(aq) → NH3(g) + 2H2O(ℓ) + 7OH-(aq) NO2-(aq) + 7H2O(ℓ) + 6e- → NH3(g) +2H2O(ℓ) + 7OH-(aq) Step 4: Equalize the electron loss and gain, add half-reactions, and simplify. 2Al(s) + 8OH-(aq) → 2AlO2-(aq) + 4H2O(ℓ) + 6e- NO2-(aq) + 5H2O(ℓ) + 6e- → NH3(g) + 7OH-(aq) 2Al(s) + NO2-(aq) + H2O(ℓ) + OH-(aq) → 2AlO2-(aq) + NH3(g) Step 5: Check for balance in atoms and charge.

Balance the following equations: Practice Problems Balance the following equations: 1. Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s) 2. H3AsO4(aq) + Zn(s) → AsH3(g) + Zn2+(aq) (acidic solution) 3. MnO4-(aq) + S2-(aq) → MnS(s) + S(s) (basic solution) 4. Al(s) + MnO4-(aq) → MnO2(s) + Al(OH)4-(aq) (basic solutio)