Example 17

Arsenic acid (H₃AsO₄) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH₃, a highly toxic and unstable gas) and Zn²⁺(aq). Balance the equation for this reaction using oxidation states:

Given: reactants and products in acidic solution

Asked for: balanced chemical equation using oxidation states

Strategy:

Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.

Solution:

1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.

2. Assign oxidation states using the procedure described in Section 3.5 "Classifying Chemical Reactions" and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in Zn²⁺(aq) is +2:

   $${\text{H}}_{\text{3}}\stackrel{\text{+5}}{\text{As}}{\text{O}}_{\text{4}}\text{(aq)}+\stackrel{\text{0}}{\text{Zn}}\text{(s)}\to \underset{\text{–3}}{\text{As}}{\text{H}}_{\text{3}}\text{(g)}+\stackrel{\text{+2}}{{\text{Zn}}^{\text{2+}}}\text{(aq)}$$

3. Write separate equations for oxidation and reduction. The arsenic atom in H₃AsO₄ is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:

   $${\text{Reduction: As}}^{\text{5+}}+{\text{8e}}^{-}\to \underset{\text{–3}}{{\text{As}}^{\text{3–}}}$$

   Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:

   $$\text{Oxidation: Zn}\to {\text{Zn}}^{\text{2+}}+{\text{2e}}^{-}$$

4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain

   Reduction (× 1): As⁵⁺ + 8e^– → As^3– Oxidation (× 4): 4Zn⁰ → 4Zn²⁺ + 8e^–

5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives

   Reduction: H₃AsO₄(aq) + 8e^– → AsH₃(g) Oxidation: 4Zn(s) → 4Zn²⁺(aq) + 8e^–

6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is

   $${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}\text{(aq) }+\text{ 4Zn(s)}+\cancel{{\text{8e}}^{–}}\to {\text{AsH}}_{\text{3}}\text{(g)}+{\text{4Zn}}^{\text{2}+}\text{(aq)}+\cancel{{\text{8e}}^{–}}$$

   which then yields

   H₃AsO₄(aq) + 4Zn(s) → AsH₃(g) + 4Zn²⁺(aq)

7. Balance the charge by adding H⁺or OH⁻ions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H⁺ ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H⁺ ions to the left side gives a charge of +8 on both sides of the equation:

   H₃AsO₄(aq) + 4Zn(s) + 8H⁺(aq) → AsH₃(g) + 4Zn²⁺(aq)

8. Balance the oxygen atoms by adding H₂O molecules to one side of the equation. There are 4 O atoms on the left side of the equation. Adding 4 H₂O molecules to the right side balances the O atoms:

   H₃AsO₄(aq) + 4Zn(s) + 8H⁺(aq) → AsH₃(g) + 4Zn²⁺(aq) + 4H₂O(l)

   Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms.

9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:

The balanced chemical equation for the reaction is therefore:

Exercise

Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states:

Answer: 3CuS(s) + 8H⁺(aq) + 8NO₃⁻(aq) → 3Cu²⁺(aq) + 8NO(g) + 3SO₄²⁻(aq) + 4H₂O(l)

Example 18

The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:

Balance this equation using oxidation states.

Given: reactants and products in a basic solution

Asked for: balanced chemical equation

Strategy:

Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.

Solution:

We will apply the same procedure used in Example 17 but in a more abbreviated form.

1. The equation for the reaction is given, so we can skip this step.

2. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH)₄]⁻. The oxidation state of hydrogen changes from +1 in H₂O to 0 in H₂. Aluminum is oxidized, while hydrogen is reduced:

   $$\stackrel{\text{0}}{\text{Al}}\text{(s)}+{\stackrel{\text{+1}}{\text{H}}}_{\text{2}}\text{O(aq)}\to {\text{[}\stackrel{\text{+3}}{\text{Al}}{\text{(OH)}}_{\text{4}}\text{]}}^{\text{–}}\text{(aq)}+\stackrel{\text{0}}{{\text{H}}_{\text{2}}}\text{(g)}$$

3.  

   $$\begin{array}{l}{\text{Reduction: Al}}^{\text{0}}\to {\text{Al}}^{\text{3+}}+{\text{3e}}^{-}\\ {\text{Oxidation: H}}^{\text{ + }}+{\text{e}}^{-}\to {\text{H}}^{\text{0}}{\text{ (in H}}_{\text{2}}\text{)}\end{array}$$

4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:

   Reduction: 3H⁺ + 3e^– → 3H⁰ (in H₂) Oxidation: Al⁰ → Al³⁺ + 3e^–

5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of H₂O contains two protons, in this case, 3H⁺ corresponds to 3/2H₂O. Similarly, each molecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H₂.

   $$\begin{array}{l}\text{Reduction: }\frac{3}{2}{\text{H}}_{\text{2}}\text{O}+{\text{3e}}^{-}\to \frac{3}{2}{\text{H}}_{\text{2}}\\ \text{Oxidation: Al}\to {\left[\text{Al}{\left(\text{OH}\right)}_{\text{4}}\right]}^{–}+{\text{3e}}^{–}\end{array}$$

6. Adding the equations and canceling the electrons gives

   $$\begin{array}{c}\text{Al}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\text{O}+\cancel{{\text{3e}}^{-}}\to {{\text{[Al(OH)}}_{\text{4}}\text{]}}^{-}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}+\cancel{{\text{3e}}^{-}}\\ \text{Al}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\text{O}\to {{\text{[Al(OH)}}_{\text{4}}\text{]}}^{-}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\end{array}$$

   To remove the fractional coefficients, multiply both sides of the equation by 2:

   2Al + 3H₂O → 2[Al(OH)₄]^– + 3H₂

7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two OH⁻ ions to the left side:

   2Al + 2OH^– + 3H₂O → 2[Al(OH)₄]^– + 3H₂

8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H₂O molecules to the left side:

   2Al + 2OH^– + 6H₂O → 2[Al(OH)₄]^– + 3H₂

9. Be sure the equation is balanced:

   Atoms: 2Al + 8O + 14H = 2Al + 8O + 14H Total charge: (2)(0) + (2)(–1) + (6)(0) = (2)(–1) + (3)(0) −2 = −2

   The balanced chemical equation is therefore

   2Al(s) + 2OH^–(aq) + 6H₂O(l) → 2[Al(OH)₄]^–(aq) + 3H₂(g)

   Thus 3 mol of H₂ gas are produced for every 2 mol of Al.

Exercise

The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.

Answer: 2MnO₄⁻(aq) + 3NO₂⁻(aq) + H₂O(l) → 2MnO₂(s) + 3NO₃⁻(aq) + 2OH⁻(aq)