Question

Balancing RedOx Reactions in Acidic or Basic Aqueous Solution Read: Section 4.11, Zumdahl, "Chemical Principles" 8th ed. Compare ⇒ A worked example using the oxidation number method. Compose the balanced reaction equation for the following reaction occuring in acidic aqueous medium. Reduce all coefficients to the lowest possible integers. Hg(l) + NO3−(aq) → Hg22+(aq) + NO(g) Select the coefficients below, (appropriately including or excluding H2O(l) and H+(aq)). NO3−(aq) + Hg(l) + H+(aq) + H2O(l) → NO(g) + Hg22+(aq) + H+(aq) + H2O(l) 1pts Tries 0/5 Now for something perhaps more difficult. Compose the balanced reaction equation for the following reaction occuring in acidic aqueous medium. Reduce all coefficients to the lowest possible integers. As2S3(s) + NO3−(aq) → H3AsO4(aq) + NO(g) + S8(s) Select the coefficients below, (appropriately including or excluding H2O(l) and H+(aq)). As2S3(s) + NO3−(aq) + H+(aq) + H2O(l) → H3AsO4(aq) + NO(g) + S8(s) + H+(aq) + H2O(l)

Answer 1

Ans 1:

Here ‘Hg’ is undergoing oxidation and ‘N’ is undergoing reduction.

We can divide the reaction into two half cells to balance.

1. Oxidation Half cell:

a). Balancing ‘Hg’ atoms

2Hg --> Hg₂²⁺

b). Balancing electrons

2Hg --> Hg₂²⁺ + 2 e

2. Reduction Half cell:

a). Balancing ‘N’ atoms

1NO₃^- --> 1NO

b). Balancing electrons

1NO₃^- + 3e --> 1NO

c). Balancing ‘O’ using water

1NO₃^- + 3e --> 1NO + 2H₂O

d). Balancing ‘H’ using ‘H⁺’

1NO₃^- + 4 H⁺ +3e --> 1NO + 2H₂O

Oxidation Half cell: 2Hg -->  Hg₂²⁺ + 2 e -----‘A’

Reduction Half cell: 1NO₃^- + 4 H⁺ +3e -->  1NO + 2H₂O ---------‘B’

Multiply equation ‘A’ by 3 and equation ‘B’ by 2 to balance the electron count in two half cells.

Oxidation Half cell: 6Hg -->  3Hg₂²⁺ + 6 e

Reduction Half cell: 2NO₃^- + 8 H⁺ +6e --> 2NO + 4H₂O

Combining and simplifying two half-cell reactions we will have,

6Hg(l) + 2NO₃^- (aq) + 8 H^{+ ­­}(aq) --> 3Hg₂²⁺ (aq) + 2NO (g) + 4H₂O (l)

Ans 2:

Here ‘As’ and ‘S’ are undergoing oxidation and ‘N’ is undergoing reduction.

We can divide the reaction into two half cells to balance.

1. Oxidation Half cell:

a). Balancing ‘As’ and ‘S’ atoms

8 As₂S₃ --> 16 H₃AsO₄ + 3 S₈

b). Balancing electrons

8 As₂S₃ --> 16 H₃AsO₄ + 3 S₈ + 80 e

c) Balancing ‘O’ using water

8 As₂S₃ + 64 H₂O --> 16 H₃AsO₄ + 3 S₈ + 80 e

d) Balancing ‘H’ using ‘H⁺’

8 As₂S₃ + 64 H₂O --> 16 H₃AsO₄ + 3 S₈ + 80 e + 80 H⁺

2. Reduction Half cell:

a). Balancing ‘N’ atoms

1NO₃^- --> 1NO

b). Balancing electrons

1NO₃^- + 3e --> 1NO

c). Balancing ‘O’ using water

1NO₃^- + 3e --> 1NO + 2H₂O

d). Balancing ‘H’ using ‘H⁺’

1NO₃^- + 4 H⁺ +3e --> 1NO + 2H₂O

Oxidation Half cell: 8 As₂S₃ + 64 H₂O -->16 H₃AsO₄ + 3 S₈ + 80 e + 80 H⁺ ----------‘A’

Reduction Half cell: 1NO₃^- + 4 H⁺ +3e --> 1NO + 2H₂O ---------‘B’

Multiply equation ‘A’ by 3 and equation ‘B’ by 80 to balance the electron count in two half cells.

Oxidation Half cell: 24 As₂S₃ + 192 H₂O --> 48 H₃AsO₄ + 9 S₈ + 240e + 240 H⁺

Reduction Half cell: 80 NO₃^- + 320 H⁺ + 240e --> 80 NO + 160 H₂O

Combining two half-cell reactions we will have,

24 As₂S₃ + 192 H₂O + 80 NO₃^- + 320 H⁺ + 240e ----> 48 H₃AsO₄ + 9 S₈ + 240e + 240 H⁺ + 80 NO + 160 H₂O

On simplification we get,

24 As₂S₃ + 32 H₂O + 80 NO₃^- + 80 H⁺ ----> 48 H₃AsO₄ + 9 S₈ + 80 NO