Reduction: 2 H + (aq) + 2 e H2(g)

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Balancing Redox Reactions Homework Answers page 617 # 1-3, 5, 7-9 Mrs. Giovannone 1. (a) Solution: Step 1: Write the unbalanced equation for the reaction. Mg(s) + HCl(aq) H2(g) + MgCl2(aq) Step 2: Write the unbalanced net ionic equation, without the spectator ion, Cl (aq). Mg(s) + H + (aq) H2(g) + Mg 2+ (aq) Step 3: Assign oxidation numbers to all entities. Step 4: Write unbalanced equations for oxidation and reduction half-reactions. Step 5: Balance each half-reaction equation. Mg(s) Mg 2+ (aq) 2 H + (aq) H2(g) Step 6: Use electrons to balance the charge in each half-reaction equation. Mg(s) Mg 2+ (aq) + 2 e 2 H + (aq) + 2 e H2(g) Statement: The balanced half-reaction equations for this reaction are Oxidation: Mg(s) Mg 2+ (aq) + 2 e Reduction: 2 H + (aq) + 2 e H2(g) (b) Add the two half-reactions to obtain the balanced equation. Include the spectator ion, Cl (aq). The balanced equation for the reaction is Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq) (c) The number of electrons transferred in the balanced equation is 2. 2. (a) Solution: Step 1: Write the unbalanced equation with its ionic compounds dissociated. Ag + (aq) + NO3 (aq) + Cu(s) Cu 2+ (aq) + 2 NO3 (aq) + Ag(s) Step 2: Write the unbalanced net ionic equation, without the spectator ion, NO3 Ag + (aq) + Cu(s) Cu 2+ (aq) + Ag(s) Step 3: Assign oxidation numbers to all entities.

Step 4: Write unbalanced equations for oxidation and reduction half-reactions. Step 5: The half-reaction equations are each balanced. Use electrons to balance the charge in each halfreaction equation. Cu(s) Cu 2+ (aq) + 2 e Ag + (aq) + e Ag(s) Statement: The balanced half-reaction equations for this reaction are Oxidation: Cu(s) Cu 2+ (aq) + 2 e Reduction: Ag + (aq) + e Ag(s) (b) Solution: Step 1: Assign oxidation numbers to all entities. Step 2: Write unbalanced equations for oxidation and reduction half-reactions. Step 3: Balance each half-reaction equation. Use H2O(l) to balance oxygen and H (aq) to balance hydrogen. 2 Cr 3+ (aq) + 7 H2O(l) Cr2O7 2 (aq) + 14 H + (aq) Cl2(g) 2 Cl (aq) Step 4: Use electrons to balance the charge in each half-reaction equation. 2 Cr 3+ (aq) + 7 H2O(l) Cr2O7 2 (aq) + 14 H + (aq) + 6 e Cl2(g) + 2 e 2 Cl (aq) Statement: The balanced half-reaction equations are: Oxidation: 2 Cr 3+ (aq) + 7 H2O(l) Cr2O7 2 (aq) + 14 H + (aq) + 6 e Reduction: Cl2(g) + 2 e 2 Cl (aq) (c) Solution: Step 1: Write the unbalanced equation with its ionic compounds dissociated. 2 H + (aq) + SO4 2 (aq) + Ca(s) Ca 2+ (aq) + SO4 2 (aq) + H2(g) Step 2: Write the unbalanced net ionic equation, without the spectator ion, SO4-2 2 H + (aq) + Ca(s) Ca 2+ (aq) + H2(g)

Step 3: Assign oxidation numbers to all entities. Step 4: Write unbalanced equations for oxidation and reduction half-reactions. Step 5: The half-reaction equations are each balanced. Use electrons to balance the charge in each halfreaction equation. Ca(s) Ca 2+ (aq) + 2 e 2 H + (aq) + 2 e H2(g) Statement: The balanced half-reaction equations are Oxidation: Ca(s) Ca 2+ (aq) + 2 e Reduction: 2 H + (aq) + 2 e H2(g) 3. (a) Solution: Step 1: Assign oxidation numbers to all entities. The oxidation number of chlorine changes from +5 to 1, so chlorine gains 6 electrons. The oxidation number of iodine changes from 0 to +5, so iodine loses 5 electrons. Step 2: To balance electrons, adjust the coefficients by multiplying electrons gained by 5 and electrons lost by 6. Since each iodine molecule has 2 atoms, the coefficient for I2(aq) would be 3. 5 ClO3 (aq) + 3 I2(aq) 5 Cl (aq) + 6 IO3 (aq) Step 3: Balance the rest of the equation by inspection. Balance oxygen by adding water. 5 ClO3 (aq) + 3 I2(aq) + 3 H2O(l) 5 Cl (aq) + 6 IO3 (aq) Balance hydrogen by adding hydrogen ions. 5 ClO3 (aq) + 3 I2(aq) + 3 H2O(l) 5 Cl (aq) + 6 IO3 (aq) + 6 H + (aq) Statement: The balanced equation is: 5 ClO3 (aq) + 3 I2(aq) + 3 H2O (l) 5 Cl (aq) + 6 IO3 (aq) + 6 H + (aq) (b) Solution: Step 1: Assign oxidation numbers to all entities. The oxidation number of chromium changes from +6 to +3, so each chromium atom gains 3 electrons.

Therefore, each Cr2O7 2 (aq) gains 6 electrons. The oxidation number of chlorine changes from 1 to 0, so each chlorine atom loses 1 electron. Step 2: To balance electrons, adjust the coefficients by multiplying electrons lost by 6. Since each chlorine molecule has 2 atoms, the coefficient for Cl2(aq) is 3. Cr2O7 2 (aq) + 6 Cl (aq) Cr 3+ (aq) + 3 Cl2(aq) Step 3: Balance the rest of the equation by inspection. Balance chromium. Cr2O7 2 (aq) + 6 Cl (aq) 2 Cr 3+ (aq) + 3 Cl2(aq) Balance oxygen by adding water. Cr2O7 2 (aq) + 6 Cl (aq) 2 Cr 3+ (aq) + 3 Cl2(aq) + 7 H2O(l) Balance hydrogen by adding hydrogen ions. Cr2O7 2 (aq) + 6 Cl (aq) + 14 H + (aq) 2 Cr 3+ (aq) + 3 Cl2(aq) + 7 H2O(l) Statement: The balanced equation is Cr2O7 2 (aq) + 6 Cl (aq) + 14 H + (aq) 2 Cr 3+ (aq) + 3 Cl2(aq) + 7 H2O(l) 5. (a) Determine the oxidation numbers for each element in the equation and identify the elements for which the oxidation numbers change. The oxidation number of manganese changes from +7 to +2, so 5 electrons are transferred for each MnO4 (aq) ion. (b) The oxidation number of carbon changes from +3 to +2, and there are two carbon atoms in each C2O4 2 (aq) ion, so 2 electrons are transferred for each C2O4 2 (aq) ion. (c) Solution: Step 1: Manganese gains 5 electrons, and each carbon atom loses 1 electron. To balance electrons, adjust the coefficients by multiplying electrons gained by 2 and electrons lost by 5. 2 MnO4 (aq) + 5 C2O4 2 (aq) 2 Mn 2+ (aq) + 10 CO2(aq) Step 2: Balance the remaining entities by inspection. Balance oxygen by adding water. 2 MnO4 (aq) + 5 C2O4 2 (aq) 2 Mn 2+ (aq) + 10 CO2(aq) + 8 H2O(l) Balance hydrogen by adding hydrogen ions. 2 MnO4 (aq) + 5 C2O4 2 (aq) + 16 H + (aq) 2 Mn 2+ (aq) + 10 CO2(aq) + 8 H2O(l) Statement: The balanced equation is 2 MnO4 (aq) + 5 C2O4 2 (aq) + 16 H + (aq) 2 Mn 2+ (aq) + 10 CO2(aq) + 8 H2O(l)

7. (a) Solution: Cr2O7 2 (aq) + C2H5OH(aq) Cr 3+ (aq) + CO2(g) Step 1: Write unbalanced equations for oxidation and reduction half-reactions. The reduction half-reaction equation is: Cr2O7 2 (aq) Cr 3+ (aq) The oxidation half-reaction equation is C2H5OH(aq) CO2(g) Step 2: Balance each half-reaction equation. For the reduction half-reaction, balance chromium. Cr2O7 2 (aq) 2 Cr 3+ (aq) Balance oxygen by adding water. Cr2O7 2 (aq) 2 Cr 3+ (aq) + 7 H2O(l) Balance hydrogen by adding hydrogen ions. Cr2O7 2 (aq) + 14 H + (aq) 2 Cr 3+ (aq) + 7 H2O(l) For the oxidation half-reaction, balance carbon. C2H5OH(aq) 2 CO2(g) Balance oxygen by adding water. C2H5OH(aq) + 3 H2O(l) 2 CO2(g) Balance hydrogen by adding hydrogen ions. C2H5OH(aq) + 3 H2O(l) 2 CO2(g) + 12 H + (aq) Step 3: Use electrons to balance the charge in each half-reaction equation. Cr2O7 2 (aq) + 14 H + (aq) + 6 e 2 Cr 3+ (aq) + 7 H2O(l) C2H5OH(aq) + 3 H2O(l) 2 CO2(g) + 12 H + (aq) + 12 e Step 4: Equalize the electron transfer in the two half-reaction equations. Multiply the reduction halfreaction equation by 2. 2 Cr2O7 2 (aq) + 28 H + (aq) + 12 e 4 Cr 3+ (aq) + 14 H2O(l) C2H5OH(aq) + 3 H2O(l) 2 CO2(g) + 12 H + (aq) + 12 e Step 5: Add the two half-reaction equations. 2 Cr2O7 2 (aq) + C2H5OH(aq) + 16 H + (aq) 4 Cr 3+ (aq) + 2 CO2(g) + 11 H2O(l) Statement: The balanced equation is 2 Cr2O7 2 (aq) + C2H5OH(aq) + 16 H + (aq) 4 Cr 3+ (aq) + 2 CO2(g) + 11 H2O(l) 8. (a) Solution: Step 1: Write unbalanced equations for oxidation and reduction half-reactions. The reduction half-reaction equation is MnO4 (aq) Mn 2+ (aq) The oxidation half-reaction equation is H2O2(aq) O2(g) Step 2: Balance each half-reaction equation. For the reduction half-reaction, balance oxygen by adding water. MnO4 (aq) Mn 2+ (aq) + 4 H2O(l) Balance hydrogen by adding hydrogen ions. MnO4 (aq) + 8 H + (aq) Mn 2+ (aq) + 4 H2O(l) Add 5 electrons to the reactant side of the equation to balance the charge. MnO4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H2O(l)

For the oxidation half-reaction, balance hydrogen by adding hydrogen ions. H2O2(aq) O2(g) + 2 H + (aq) Add 2 electrons to the product side of the equation to balance the charge. H2O2(aq) O2(g) + 2 H + (aq) + 2 e Statement: The balanced half-reaction equations Oxidation: H2O2(aq) O2(g) + 2 H + (aq) + 2 e Reduction: MnO4 (aq) + 8 H + (aq) + 5 e Mn 2+ (aq) + 4 H2O(l) (b) Solution: Step 1: Equalize the electron transfer in the two half-reaction equations. Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5. 2 MnO4 (aq) + 16 H + (aq) + 10 e 2 Mn 2+ (aq) + 8 H2O(l) 5 H2O2(aq) 5 O2(g) + 10 H + (aq) + 10 e Step 2: Add the two-half-reaction equations. 2 MnO4 (aq) + 5 H2O2(aq) + 6 H + (aq) 2 Mn 2+ (aq) + 5 O2(g) + 8 H2O(l) Statement: The balanced equation for the redox reaction in an acidic solution is the following: 2 MnO4 (aq) + 5 H2O2(aq) + 6 H + (aq) 2 Mn 2+ (aq) + 5 O2(g) + 8 H2O(l) 9. Solution: Step 1: Write an unbalanced equation for the reaction. SO3 2 (aq) + NO3 (aq) SO4 2 (aq) + NO2(g) Step 2: Assign oxidation numbers to all entities. The oxidation number of sulfur changes from +4 to +6, so sulfur loses 2 electrons. The oxidation number of nitrogen changes from +5 to +4, so nitrogen gains 1 electron. Step 3: To balance electrons, adjust the coefficients by multiplying electrons gained by 2. SO3 2 (aq) + 2 NO3 (aq) SO4 2 (aq) + 2 NO2(g)

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