4.2 Classifying chemical reactions  (Page 8/34)

Balancing redox reactions via the half-reaction method

Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:

1. Write the two half-reactions representing the redox process.

2. Balance all elements except oxygen and hydrogen.

3. Balance oxygen atoms by adding H ₂ O molecules.

4. Balance hydrogen atoms by adding H ⁺ ions.

5. Balance charge The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced. by adding electrons.

6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.

7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.

8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:

1. Add OH ⁻ ions to both sides of the equation in numbers equal to the number of H ⁺ ions.
2. On the side of the equation containing both H ⁺ and OH ⁻ ions, combine these ions to yield water molecules.
3. Simplify the equation by removing any redundant water molecules.

9. Finally, check to see that both the number of atoms and the total charges The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced. are balanced.

Balancing redox reactions in acidic solution

Solution

1. Write the two half-reactions .

   Each half-reaction will contain one reactant and one product with one element in common.

   ${\text{Fe}}^{\text{2+}}⟶{\text{Fe}}^{\text{3+}}$
   ${\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}⟶{\text{Cr}}^{\text{3+}}$

2. Balance all elements except oxygen and hydrogen . The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.

   ${\text{Fe}}^{\text{2+}}⟶{\text{Fe}}^{\text{3+}}$
   ${\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}⟶2{\text{Cr}}^{\text{3+}}$

3. Balance oxygen atoms by adding H ₂ O molecules . The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.

   $\begin{array}{c}{\text{Fe}}^{\text{2+}}⟶{\text{Fe}}^{\text{3+}}\\ \\ \\ \\ {\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}⟶2{\text{Cr}}^{\text{3+}}+7{\text{H}}_2\text{O}\end{array}$

4. Balance hydrogen atoms by adding H ⁺ ions . The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.

   ${\text{Fe}}^{\text{2+}}⟶{\text{Fe}}^{\text{3+}}$
   ${\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}+14{\text{H}}^{+}+⟶2{\text{Cr}}^{\text{3+}}+7{\text{H}}_2\text{O}$

5. Balance charge by adding electrons . The iron half-reaction shows a total charge of 2+ on the left side (1 Fe ²⁺ ion) and 3+ on the right side (1 Fe ³⁺ ion). Adding one electron to the right side bring that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved.

   The chromium half-reaction shows a total charge of (1 $\times$ 2−) + (14 $\times$ 1+) = 12+ on the left side ${(\text{1 Cr}}_2{\text{O}}_7{}^{\text{2−}}$ ion and 14 H ⁺ ions). The total charge on the right side is (2 $\times$ 3+) = 6 + (2 Cr ³⁺ ions). Adding six electrons to the left side will bring that side’s total charge to (12+ + 6−) = 6+, and charge balance is achieved.

   ${\text{Fe}}^{\text{2+}}⟶{\text{Fe}}^{\text{3+}}+{\text{e}}^{-}$
   ${\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}+14{\text{H}}^{+}+6{\text{e}}^{-}⟶2{\text{Cr}}^{\text{3+}}+7{\text{H}}_2\text{O}$

6. Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction . To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6.

   ${\text{6Fe}}^{\text{2+}}⟶6{\text{Fe}}^{\text{3+}}+6{\text{e}}^{-}$
   ${\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}+6{\text{e}}^{-}+14{\text{H}}^{+}⟶2{\text{Cr}}^{\text{3+}}+7{\text{H}}_2\text{O}$

7. Add the balanced half-reactions and cancel species that appear on both sides of the equation .

   $6{\text{Fe}}^{\text{2+}}+{\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}+6{\text{e}}^{-}+14{\text{H}}^{+}⟶6{\text{Fe}}^{\text{3+}}+6{\text{e}}^{-}+2{\text{Cr}}^{\text{3+}}+7{\text{H}}_2\text{O}$

   Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:

   $6{\text{Fe}}^{\text{2+}}+{\text{Cr}}_2{\text{O}}_7{}^{\text{2−}}+14{\text{H}}^{+}⟶6{\text{Fe}}^{\text{3+}}+2{\text{Cr}}^{\text{3+}}+7{\text{H}}_2\text{O}$