Problem 29 - Chemistry An Atoms First Approach

Question

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Step-by-Step Solution

Answer

The balanced equations for the reactions are: a. \(2\mathrm{I}^{-}(a q) + \mathrm{ClO}^{-}(a q) + 2\mathrm{H}^{+}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q) + \mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s) + 12\mathrm{H}^{+}(a q) + 2\mathrm{OH}^{-}(a q) + 2\mathrm{NO}_{3}^{-}(a q) \rightarrow 2\mathrm{H}_{3} \mathrm{AsO}_{4}(a q) + 2\mathrm{NO}(g) + 4\mathrm{H}_{2}\mathrm{O}(l)\) c. \(10\mathrm{Br}^{-}(a q) + 2\mathrm{MnO}_{4}^{-}(a q) + 16\mathrm{H}^{+}(a q) \rightarrow 5\mathrm{Br}_{2}(l) + 2\mathrm{Mn}^{2+}(a q) + 8\mathrm{H}_{2}\mathrm{O}(l)\) d. \(3\mathrm{CH}_{3} \mathrm{OH}(a q) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q) \rightarrow 3\mathrm{CH}_{2} \mathrm{O}(a q) + 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\)

Step 1: a. Balancing I- + ClO- → I3- + Cl-

Step 1: Split the reaction into two half-reactions Oxidation half-reaction: \(\mathrm{I}^{-} \rightarrow \mathrm{I}_{3}^{-}\) Reduction half-reaction: \(\mathrm{ClO}^{-} \rightarrow \mathrm{Cl}^{-}\) Step 2: Balance atoms and charges in each half-reaction Oxidation: Add 2 I- to the left side and 2e- to the right side Balanced oxidation half-reaction: \(2\mathrm{I}^{-} \rightarrow \mathrm{I}_{3}^{-} + 2e^{-}\) Reduction: Add H2O to the left side and 2H+ + 2e- to the right side Balanced reduction half-reaction: \(\mathrm{ClO}^{-} + 2\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{H}_{2}\mathrm{O}\) Step 3: Combine the balanced half-reactions \(2\mathrm{I}^{-} + \mathrm{ClO}^{-} + 2\mathrm{H}^{+} \rightarrow \mathrm{I}_{3}^{-} + \mathrm{Cl}^{-} + \mathrm{H}_{2}\mathrm{O}\) The balanced equation for reaction a is: \(2\mathrm{I}^{-}(a q) + \mathrm{ClO}^{-}(a q) + 2\mathrm{H}^{+}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q) + \mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\)

Step 2: b. Balancing As2O3 + NO3- → H3AsO4 + NO

Step 1: Split the reaction into two half-reactions Oxidation half-reaction: \(\mathrm{As}_{2} \mathrm{O}_{3} \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}\) Reduction half-reaction: \(\mathrm{NO}_{3}^{-} \rightarrow \mathrm{NO}\) Step 2: Balance atoms and charges in each half-reaction Oxidation: Add 6H+ to the right side and 2 OH- to the left side Balanced oxidation half-reaction: \(\mathrm{As}_{2} \mathrm{O}_{3} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{H}_{3} \mathrm{AsO}_{4} + 4e^{-}\) Reduction: Add 2H2O to the right side and 6H+ + 6e- to the left side Balanced reduction half-reaction: \(2(\mathrm{NO}_{3}^{-} + 3\mathrm{H}^{+}) \rightarrow 2\mathrm{NO} + 6\mathrm{H}_{2}\mathrm{O}\) Step 3: Combine the balanced half-reactions (multiply reduction half-reaction by 2) \(\mathrm{As}_{2} \mathrm{O}_{3} + 2\mathrm{OH}^{-} + 2(\mathrm{NO}_{3}^{-} + 3\mathrm{H}^{+}) \rightarrow 2\mathrm{H}_{3} \mathrm{AsO}_{4} + 2\mathrm{NO} + 6\mathrm{H}_{2}\mathrm{O}\) The balanced equation for reaction b is: \(\mathrm{As}_{2} \mathrm{O}_{3}(s) + 12\mathrm{H}^{+}(a q) + 2\mathrm{OH}^{-}(a q) + 2\mathrm{NO}_{3}^{-}(a q) \rightarrow 2\mathrm{H}_{3} \mathrm{AsO}_{4}(a q) + 2\mathrm{NO}(g) + 4\mathrm{H}_{2}\mathrm{O}(l)\)

Step 3: c. Balancing Br- + MnO4- → Br2 + Mn2+

Step 1: Split the reaction into two half-reactions Oxidation half-reaction: \(\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2}\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) Step 2: Balance atoms and charges in each half-reaction Oxidation: Add 2 Br- to the left side and 2e- to the right side Balanced oxidation half-reaction: \(2\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2} + 2e^{-}\) Reduction: Add 4H2O to the left side and 8H+ + 5e- to the right side Balanced reduction half-reaction: \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\) Step 3: Combine the balanced half-reactions (multiply oxidation half-reaction by 5, reduction half-reaction by 2) \(10\mathrm{Br}^{-} + 2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} \rightarrow 5\mathrm{Br}_{2} + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}\mathrm{O}\) The balanced equation for reaction c is: \(10\mathrm{Br}^{-}(a q) + 2\mathrm{MnO}_{4}^{-}(a q) + 16\mathrm{H}^{+}(a q) \rightarrow 5\mathrm{Br}_{2}(l) + 2\mathrm{Mn}^{2+}(a q) + 8\mathrm{H}_{2}\mathrm{O}(l)\)

Step 4: d. Balancing CH3OH + Cr2O72- → CH2O + Cr3+

Step 1: Split the reaction into two half-reactions Oxidation half-reaction: \(\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{2} \mathrm{O}\) Reduction half-reaction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+}\) Step 2: Balance atoms and charges in each half-reaction Oxidation: Add H+ to the right side and 2e- to the left side Balanced oxidation half-reaction: \(\mathrm{CH}_{3} \mathrm{OH} + 2e^{-} \rightarrow \mathrm{CH}_{2} \mathrm{O} + \mathrm{H}^{+}\) Reduction: Add 7H2O to the left side and 14H+ + 6e- to the right side Balanced reduction half-reaction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\) Step 3: Combine the balanced half-reactions (multiply oxidation half-reaction by 3, reduction half-reaction by 1) \(3\mathrm{CH}_{3} \mathrm{OH} + 6e^{-} + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \rightarrow 3(\mathrm{CH}_{2} \mathrm{O} + \mathrm{H}^{+}) + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\) The balanced equation for reaction d is: \(3\mathrm{CH}_{3} \mathrm{OH}(a q) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^{+}(a q) \rightarrow 3\mathrm{CH}_{2} \mathrm{O}(a q) + 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_{2}\mathrm{O}(l)\)