Reduction-Oxidation Reactions PDF

59 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions ACTIVITY 2 REDUCTION-OXIDATION REACTIONS The term oxidation was originally applied to reactions in which substances are c...

59 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions ACTIVITY 2 REDUCTION-OXIDATION REACTIONS The term oxidation was originally applied to reactions in which substances are combined with oxygen. On the other hand, reduction was defined as the removal of oxygen from an oxygen-containing compound. Gradually, the definition of the terms broadened. Today, oxidation and reduction are defined on the basis of change in oxidation numbers. Rules in Determining Oxidation Number Oxidation Number is a positive or negative number (or zero) assigned to an atom in a compound according to arbitrary rules that take into account bond polarity. The following are the rules used to assign oxidation numbers: 1. Any uncombined atom or any atom in a molecule of an element (Free State) is assigned an oxidation number of zero. (example: Fe, S, I2, Cl2, O2) 2. The oxidation number of a monatomic ion is the same as the charge of the ion. a. In their compounds, Group IA metals (Li, Na, K, Rb and Cs) always have oxidation numbers of +1 while Group IIA elements (Be, Mg, Ca, Sr and Ba) always have oxidation numbers of +2. b. The oxidation number of fluorine, the most electronegative element, is -1 in all fluorine-containing compounds. c. In most oxygen-containing compounds, the oxidation number of oxygen is -2. There are, however, a few exceptions: i. In peroxides, the oxygen atom has an oxidation number of -1. The two O atoms of the peroxide ion, O22-, are equivalent. Each must be assigned an oxidation number of -1 so that the sum equals the charge on the ion. ii. In the superoxide ion, O2-, each oxygen atom has an oxidation number of -½. iii. In OF2, oxygen has an oxidation number of +2 (see Rule 5). d. The oxidation number of hydrogen is +1 in all its compounds except for metallic hydrides (CaH2 and NaH are examples) in which hydrogen has a -1 oxidation state. For the exclusive use of UST Engineering students 60 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions 3. The sum of the oxidation numbers of the atoms that constitute a polyatomic ion equals the charge on the ion. Ex. NH41+ : 1(N) + 4(H) = 1(-3) + 4(+1) = +1 4. The sum of the oxidation numbers of the atoms in a compound is zero, since compounds are electrically neutral. Ex. HCl: 1(H) + 1(Cl) = 1(+1) + 1(-1) = 0 HClO3: 1(H) + 1(Cl) + 3(O) = 1(+1) + 1(+5) + 3(-2) = 0 5. In a combination of two non-metals (either a molecule or a polyatomic ion) the oxidation number of the more electronegative element is negative and equal to the charge on the common monatomic ion of that element. In PCl3, for example, the oxidation number of Cl is -1 and that of P is +3. In CS2, the oxidation number of S is -2, and that of C is +4. Oxidation is any chemical change in which a substance loses electrons and thus, increases in oxidation state. Example: Zn → Zn2+ + 2e- Reduction is a chemical change in which a substance gains electrons and thus, decreases in oxidation state. Example: S + 2e- → S2- Oxidation and reduction reactions occur simultaneously. The substance in the reaction that gives up electrons is referred to as the reducing agent (RA) or reductant, while the substance that takes up those electrons is referred to as the oxidizing agent (OA) or the oxidant. Example: Zn + S → ZnS (RA) (OA) oxidation reduction For the exclusive use of UST Engineering students 61 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions Take note that the total increase in oxidation number must equal the total decrease in oxidation number. Equations for oxidation-reduction reactions, usually called redox reactions, are normally more difficult to balance than equations for non-redox reactions. Methods of Balancing Redox Reactions: Two methods are commonly used to balance oxidation-reduction equations, the oxidation-number method and the ion-electron method. I. Oxidation-Number Method or Valence-Change Method There are three steps in the oxidation-number method for balancing oxidation- reduction equations: 1. The oxidation numbers of the atoms in the equation are determined in order to identify atoms undergoing oxidation or reduction. 2. Coefficients are adjusted so that the total increase in oxidation state for all atoms involved in oxidation and the total decrease in oxidation state for all atoms involved in the reduction will be equal. 3. Balancing is completed by inspection. The final, balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left. Water may be added if necessary. The oxidation-number method can also be used to balance net ionic equations, in which only ions and molecules that take part in the reaction are shown. An ionic equation must indicate charge balance as well as mass balance. For the exclusive use of UST Engineering students 62 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions Example 1: Consider the reaction in a blast furnace during the processing of iron ore to metallic iron: Fe2O3(s) + CO (g) → Fe(s) + CO2 (g) Step 1 Assign the oxidation numbers to each atom in the equation +3 -2 +2- 2 0 +4 -2 Fe2O3 + CO(g) → Fe(s) + CO2(g). Step 2 Determine the atoms that are oxidized and reduced. ü Fe atoms: Fe (+3) to Fe (0) there is a DECREASE in oxidation state ü C atoms: C (+2) to C (+4) there is an INCREASE in oxidation state Hence, Fe underwent REDUCTION while C underwent OXIDATION Step 3 Use a line to connect the atoms that are undergoing a change in oxidation number. On that line write the oxidation-number change. +2 +3 -2 +2- 2 0 +4 -2 Fe2O3 + CO(g) → Fe(s) + CO2(g) Note: -3 ü The number of electrons lost does not equal the number of electrons gained. ü The increase in oxidation number of one atom must be made equal to the decrease in oxidation number of the other. Step 4 Coefficients are adjusted so that the total increase in oxidation state for all atoms involved in oxidation and the total decrease in oxidation state for all atoms involved in the reduction will be equal. +2 x 3 = 6 +3 -2 +2- 2 0 +4 -2 Fe2O3 + CO(g) → Fe(s) + CO2(g) -3 x 2 = - 6 Continued next page For the exclusive use of UST Engineering students 63 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions ü The least common multiple of 2 and 3 is 6. therefore, the oxidation- number increase should be multiplied by 3, while the oxidation-number decrease should be multiplied by 2. ü The coefficient is also applied to the formulas in the equation. Therefore, 3 is placed in front of the CO and in front of the CO2 and 2 is placed in front of the Fe on the right side of the equation. However, Fe2O3 does not require a coefficient because the subscript of 2 after the Fe indicates that there are already two iron atoms. Step 5 Balancing is completed by inspection. The final, balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left. Fe2O3 + 3 CO(g) → 2Fe(s) + 3CO2(g) Example 2: Balance the following reaction: FeCl2 + SnCl4 → SnCl2 + FeCl3 Step 1 Assign the oxidation numbers to each atom in the equation +2 -1 +4 -1 +2 -1 +3 -1 FeCl2 + SnCl4 → SnCl2 + FeCl3. Step 2 Determine the atoms that are oxidized and reduced. ü Fe atoms: Fe (+2) to Fe (+3) there is an INCREASE in oxidation state ü Sn atoms: Sn (+4) to Sn (+2) there is a DECREASE in oxidation state Hence, Fe underwent OXIDATION while Sn underwent REDUCTION Step 3 Use a line to connect the atoms that are undergoing a change in oxidation number. On that line write the oxidation-number change. -2 +2 -1 +4 -1 +2 -1 +3 -1 FeCl2 + SnCl4 → SnCl2 + FeCl3 +1 For the exclusive use of UST Engineering students 64 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions ü The number of electrons lost does not equal the number of electrons gained. ü The increase in oxidation number of one atom must be made equal to the decrease in oxidation number of the other. Step 4 Coefficients are adjusted so that the total increase in oxidation state for all atoms involved in oxidation and the total decrease in oxidation state for all atoms involved in the reduction will be equal. -2 x 1 = -2 +2 -1 +4 -1 +2 -1 +3 -1 FeCl2 + SnCl4 → SnCl2 + FeCl3 +1 x 2 = +2 ü The least common multiple of 1 and 2 is 2. therefore, the oxidation- number increase should be multiplied by 2, while the oxidation-number decrease should be multiplied by 1. ü The coefficient is also applied to the formulas in the equation. Therefore, 2 is placed in front of the Fe2Cl2 and in front of the FeCl3 and 1 is placed in front of the SnCl4 and SnCl2. However, 1 may not appear as stoichiometric coefficient. Step 5 Balancing is completed by inspection. The final, balanced equation should be checked to ensure that there are as many atoms of each element on the right as there are on the left. 2FeCl2 + 1SnCl4 → 1SnCl2 + 2FeCl3 For the exclusive use of UST Engineering students 65 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions Very often REDOX reactions are dependent on the acidity or basicity of a reaction. When this occurs, we need to balance the numbers of O and H atoms that appear in H+, OH- and H2O species in the reaction. The additional steps in balancing the REDOX reactions are: Acidic Solution ü Balance O by adding H2O, then balance H by adding H+ Basic Solution ü For each O, add two OH- to side needing O and one H2O to other side for each H+, add one H2O to side needing H+ and one OH- to other side These additional steps are illustrated in the following examples. Example 3: Balance the following REDOX reaction in ACIDIC solution Fe2+ + MnO4- → Mn2+ + Fe3+ Step Assign +2 +7 -2 +2 +3 1 Oxidation Fe2+ + MnO4- → Mn2+ + Fe3+ Numbers ü Fe atoms: Fe (+2) to Fe (+3) there is an INCREASE in Determine the oxidation state Step atoms that are ü Mn atoms: Mn (+7) to Mn (+2) there is a DECREASE in 2 oxidized and oxidation state reduced. Hence, Fe underwent OXIDATION while Mn underwent REDUCTION +1 Use a line to connect the +2 +7 -2 +2 +3 Step atoms that are Fe2+ + MnO4- → Mn2+ + Fe3+ 3 undergoing a change in -5 oxidation number +1 x 5 = +5 Step Find common 4 factor, Adjust +2 +7 -2 +2 +3 stoichiometric Fe2+ + MnO4- → Mn2+ + Fe3+ coefficients -5 x 1 = -5 For the exclusive use of UST Engineering students 66 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions Step Acidic Balance 5 of O atoms 5 Fe2+ + 1MnO4- → 1Mn2+ + 5Fe3+ 4H2O Step Acidic Balance 6 of H atoms 5 Fe2+ +1 MnO4- + 8 H+ → 1Mn2+ + 5 Fe3+ 4H2O Example 4: Balance the following REDOX reaction in BASIC solution Fe2+ + MnO4- → MnO2 + Fe3+ Step Assign +2 +7 -2 +4 -2 +3 1 Oxidation Fe2+ + MnO4- → MnO2 + Fe3+ Numbers ü Fe atoms: Fe (+2) to Fe (+3) there is an INCREASE in Determine the oxidation state Step atoms that are ü Mn atoms: Mn (+7) to Mn (+4) there is a DECREASE in 2 oxidized and oxidation state reduced. Hence, Fe underwent OXIDATION while Mn underwent REDUCTION +1 Use a line to connect the +2 +7 -2 +4 -2 +3 Step atoms that are Fe2+ + MnO4- → MnO2 + Fe3+ 3 undergoing a change in -3 oxidation number +1 x 3 = +3 Find common factor, Adjust +2 +7 -2 +2 +3 Step stoichiometric Fe2+ + MnO4- → MnO2 + Fe3+ 4 coefficients Step Basic Balance of 5 O atoms 3 Fe2+ + 1 MnO4- → 1 MnO2 + 3 Fe 3+ + 4 OH - For the exclusive use of UST Engineering students 67 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions Step Basic Balance of 3 Fe2+ + 1 MnO4- + 2 H2O → 1 MnO2 + 3 Fe 3+ + 4 OH - 6 H atoms II. Ion-Electron Method or Method of Half-Reactions This method of balancing redox equations employs partial equations representing half-reactions. One partial equation describes the oxidation while the other describes the reduction. In balancing each half-reaction, it is necessary to find the same number of each atom on each side of the equation, adding water and H+ or OH- where needed. The balanced equation is obtained by putting electrons on either side of the partial equation. A complete redox equation cannot have any excess electrons on either side. Two slightly different procedures are employed to balance equations by the ion- electron method. One is used for reactions that take place in acid solution, the other for reactions that occur in alkaline solution. Example 5: Consider the following reaction in ACIDIC solution: H2C2O4 + MnO4¯ → CO2 + Mn2+ (unbalanced) Step Separate half- Oxidation: H2C2O4 → 2 CO2 + 2 H+ + 2e¯ reactions 1 Reduction: 5 e¯ + 8 H+ + MnO4¯ → Mn2+ + 4 H2O Balance each half- reaction Note: Balance O by adding H2O and Balance H by adding H+ (atom and charge) Step Equalize electrons Oxidation, multiply by a factor of 5: 2 5 H2C2O4 → 10 CO2 + 10 H+ + 10e¯ Reduction, multiply by a factor of 10 10 e¯ + 16 H+ + 2 MnO4¯ →2 Mn2+ + 8H2O Note: Multiply by some integer to make electrons (lost) equal to electrons (gained) For the exclusive use of UST Engineering students 68 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions Step Combine half- 5 H2C2O4 → 10 CO2 + 10 H+ + 10e¯ reactions (electrons 10 e¯ + 16 H+ + 2 MnO4¯ →2 Mn2+ + 8H2O 3 and some H+ ions get cancelled) 5 H2C2O4 + 6 H+ + 2 MnO4¯ → 10 CO2 + 2 Mn2+ + 8 H2O Note: Add half equations and cancel substances on both sides. Check atom balance and charge balance on both sides of the equation. Example 6: Consider the following reaction in BASIC solution: Cr2O72¯ + I2 → Cr3+ + IO3¯ Step Separate half- Oxidation: I2 → IO3¯ reactions Reduction: Cr2O72¯ → Cr3+ 1 Note: ü You can determine the oxidation and reduction half reactions by accounting for the changes in the oxidation states of Iodine and Chromium atoms Step Balance each Oxidation: 6H2O + I2 → 2IO3¯ + 12H+ + 10e¯ half-reaction Reduction: 6e¯ + 14H+ + Cr2O72¯ → 2Cr3+ + 7H2O 2 first in acidic Note: condition ü Balance O by adding H2O and Balance H by adding H+ (atom and charge) Step Equalize Oxidation: 18H2O + 3I2 → 6IO3¯ + 36H+ + 30e¯ electrons Reduction: 30e¯ + 70 H+ + 5Cr2O72¯ → 10Cr3+ + 35H2O 2 Note: ü Multiply by some integer to make electrons (lost) equal to electrons (gained) Step Combine half- 18 H2O + 3 I2 → 6 IO3¯ + 36 H+ + 30 e¯ reactions 30 e¯ + 70 H+ + 5 Cr2O72¯ → 10 Cr3+ + 35 H2O 3 (electrons and 34 H+ + 5 Cr2O72¯ + 3 I2 → 6 IO3¯ + 10 Cr3+ + 17 H2O some H+ ions Note: get cancelled) For the exclusive use of UST Engineering students 69 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions ü Add half equations and cancel the same substances on both sides. Check atom balance and charge balance on both sides of the equation. Step Only in basic Add 34 OH¯ to each side and eliminate duplicates: solution: 34 OH¯+34 H+ + 5 Cr2O72¯ + 3 I2 → 6IO3¯+10 Cr3+ + 17 H2O + 34 OH¯ 4 add OH− and cancel H2O 34 H2O + 5 Cr2O72¯ + 3 I2 → 6IO3¯+10 Cr3+ + 17 H2O + 34 OH¯ Finally, the balanced REDOX reaction is: 17 H2O + 5 Cr2O72¯ + 3 I2 → 6 IO3¯ + 10 Cr3+ + 34 OH ¯ For the exclusive use of UST Engineering students 70 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions PRACTICE 1 Activity 2. 1 OXIDATION NUMBERS DIRECTION: State the oxidation number of the underlined element. Number Substance 1 U2Cl10 2 BiO+ 3 Na6V10O28 4 K2SnO3 5 Ta6O18 6 N2H4 7 NH2OH 8 S2O5Cl2 9 Mg3UO6 10 Na2S4O6 For the exclusive use of UST Engineering students 71 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions PRACTICE 2 Activity 2.2 REDUCING AGENT AND OXIDIZING AGENT DIRECTION: For each of the following reactions: identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent. Numbe REACTIONS r 1 Zn + Cl2 → ZnCl2 2 2ReCl5 + SbCl3 → 2ReCl4 + SbCl5 3 Mg + CuCl2 → MgCl2 + Cu 4 2NO + O2 → 2NO2 5 WO3 + 3H2 → W + 3H2O For the exclusive use of UST Engineering students 72 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions PRACTICE 3 Activity 2.3 BALANCING MOLECULAR EQUATIONS DIRECTION: Balance the following molecular equations: Numbe Molecular Equations r 1 FeCl3 + H2S → FeCl2 + S + HCl 2 HNO3 + I2 → NO2 + HlO3 + H2O 3 Bi(OH)3 + K2SnO2 → Bi + K2SnO3 + H2O 4 I2O5 + CO → I2 + CO2 5 HNO3 + Fe → Fe(NO3)3 + NO + H2O 6 Na2Cr2O7 + FeCl2 + HCl → CrCl3 + FeCl3 + NaCl + H2O 7 NiS + HCl + HNO3 → NiCl2 + NO + S + H2O 8 Bi(OH)3 + Na2SnO2 → Na2SnO3 + Bi + H2O 9 MnS + HCl + HNO3 → MnCl2 + NO + S + H2O 10 HNO3 + Hl → NO + I2 + H2O For the exclusive use of UST Engineering students 73 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions PRACTICE 4 Activity 2.4 BALANCING IONIC EQUATIONS DIRECTION: Balance the following ionic equations Numbe Ionic Equations r 1 Sn2+ + IO3- → Sn4+ + I- (acid) 2 AsO2- + MnO4- → AsO3- + Mn2+ (acid) 3 Cl2 → ClO3- + Cl- (base) 4 MnO4- + ClO2- → MnO2 + ClO4- (base) 5 S2- + NO3- → S + NO (acid) For the exclusive use of UST Engineering students 74 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions PRACTICE 5 Activity 2.5 BALANCING HALF REACTIONS DIRECTION: Write a complete, balanced half-reaction for each of the following. Number Half Reactions 1 MnO4- → Mn2+ (acid) 2 SnO22- → SnO32- (base) 3 Cr2O72- → Cr3+ (acid) 4 ClO- → Cl- (acid) 5 NO3- → NH4+ (acid) For the exclusive use of UST Engineering students 75 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions DATA SHEET Activity 2. 1 OXIDATION NUMBERS NAME: ____________________________ INSTRUCTOR: _________________________ _____________________________ ______________________________ SECTION: _______________ GROUP NO: _________________DATE: _______________ Number Substance Oxidation Number 1 U2Cl10 2 BiO+ 3 Na6V10O28 4 K2SnO3 5 Ta6O18 6 N2H4 7 NH2OH 8 S2O5Cl2 9 Mg3UO6 10 Na2S4O6 For the exclusive use of UST Engineering students  76 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions DATA SHEET Activity 2.2 REDUCING AGENT AND OXIDIZING AGENT NAME: ____________________________ INSTRUCTOR: _________________________ _____________________________ ______________________________ SECTION: _______________ GROUP NO: _________________DATE: _______________ Numbe REACTION Substanc Substanc RA OA r e e Oxidized Reduced 1 Zn + Cl2 → ZnCl2 2 2ReCl5 + SbCl3 → 2ReCl4 + SbCl5 3 Mg + CuCl2 → MgCl2 + Cu 4 2NO + O2 → 2NO2 5 WO3 + 3H2 → W + 3H2O For the exclusive use of UST Engineering students 77 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions DATA SHEET Activity 2.3 BALANCING MOLECULAR EQUATIONS NAME: ____________________________ INSTRUCTOR: _________________________ _____________________________ ______________________________ SECTION: _______________ GROUP NO: _________________DATE: _______________ Numbe Balanced Molecular Equation r 1 2 3 4 5 6 7 8 9 10 For the exclusive use of UST Engineering students 78 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions DATA SHEET Activity 2.4 BALANCING IONIC EQUATIONS NAME: ____________________________ INSTRUCTOR: _________________________ _____________________________ ______________________________ SECTION: _______________ GROUP NO: _________________DATE: _______________ Numbe Balanced Ionic Equation r 1 2 3 4 5 For the exclusive use of UST Engineering students 79 CHEMISTRY APPLICATIONS IN ENGINEERING LABORATORY (ENG 202) Activity 2: Reduction-Oxidation Reactions DATA SHEET Activity 2.5 BALANCING HALF REACTIONS NAME: ____________________________ INSTRUCTOR: _________________________ _____________________________ ______________________________ SECTION: _______________ GROUP NO: _________________DATE: _______________ Number Balanced Half Reaction 1 2 3 4 5 For the exclusive use of UST Engineering students