Question

Classify each half-reaction occurring in acidic aqueous solution as an oxidation or a reduction and balance the half-reaction. (a) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)\) (b) \(\mathrm{Pb}^{2+}(a q) \longrightarrow \mathrm{PbO}_{2}(s)\) (c) \(\mathrm{IO}_{3}^{-}(a q) \rightarrow \mathrm{I}_{2}(s)\) (d) \(\mathrm{SO}_{2}(g) \rightarrow \mathrm{SO}_{4}^{2-}(a q)\)

Short answer

(a) Reduction: 8H+ + MnO4- + 5e- -> Mn2+ + 4H2O. (b) Oxidation: Pb2+ + 2H2O -> PbO2 + 4H+ + 2e-. (c) Reduction: 2IO3- + 12H+ + 10e- -> 3I2 + 6H2O. (d) Oxidation: SO2 + 2H2O -> SO42- + 4H+ + 2e-.

Step-by-step solution

- Identifying the Change in Oxidation States for MnO4- to Mn2+

Determine the oxidation states of Mn in MnO4- and Mn2+. In MnO4-, the oxidation state of Mn is +7 (oxygen is typically -2). In Mn2+, the oxidation state of Mn is +2. Therefore, Mn is reduced as its oxidation state decreases from +7 to +2. Next, the half-reaction needs to be balanced.

- Balancing Atoms Other Than Oxygen and Hydrogen

Ensure that the Mn atoms are balanced on both sides. In this case, there is one Mn atom on both sides, so the Mn atoms are already balanced.

- Balancing Oxygen Atoms with Water

Balance oxygen atoms by adding water molecules (H2O) to the side that needs more oxygen. Since there are 4 oxygen atoms in MnO4- and none in Mn2+, add 4 H2O to the right side of the equation.

- Balancing Hydrogen Atoms with H+ Ions

Balance hydrogen by adding H+ to the side with less hydrogen. There are 8 hydrogen atoms in the 4 H2O molecules, so add 8 H+ to the left side of the equation.

- Balancing Charge with Electrons

Balance the charge by adding electrons. The left side has a charge of -1 (MnO4-) + 8(+1) (8H+) = +7, and the right side has a charge of +10 (5 Mn2+). Add 5 electrons to the left side to balance the charges.

- Identifying the Change in Oxidation States for Pb2+ to PbO2

Determine the oxidation states of Pb in Pb2+ and PbO2. In Pb2+, the oxidation state of Pb is +2. In PbO2, Pb is typically +4 (each oxygen is -2). Therefore, Pb is oxidized as its oxidation state increases from +2 to +4. Next, balance the half-reaction.

- Balancing Atoms Other Than Oxygen and Hydrogen for Pb2+ to PbO2

Ensure that the Pb atoms are balanced on both sides. In this case, there is one Pb atom on both sides, so they are already balanced.

- Balancing Oxygen Atoms with Water for Pb2+ to PbO2

Balance oxygen atoms by adding water molecules to the side with less oxygen. Since there are 2 oxygen atoms in PbO2 and none in Pb2+, add 2 H2O to the left side of the equation.

- Balancing Hydrogen Atoms with H+ Ions for Pb2+ to PbO2

Balance hydrogen by adding H+ to the side with less hydrogen. There are 4 hydrogen atoms in the 2 H2O molecules, so add 4 H+ to the right side of the equation.

- Balancing Charge with Electrons for Pb2+ to PbO2

Balance the charge by adding electrons. The left side has no charge (Pb2+) + 4(+1) (4H2O) = +4 and the right side has a charge of 0 (PbO2) + 4(+1) (4H+) = +4. Add 2 electrons to the left side to balance the charges and maintain the oxidation state change from +2 to +4.

- Identifying the Change in Oxidation States for IO3- to I2

Determine the oxidation states of I in IO3- and I2. In IO3-, I has an oxidation state of +5 (oxygen is -2). In I2, the oxidation state of I is 0. Therefore, I is reduced as its oxidation state decreases from +5 to 0. Next, balance the half-reaction.

- Balancing Iodine Atoms

Ensure that I atoms are balanced. Since there are 3 I atoms on the left and 2 on the right, multiply the IO3- by 2 and the I2 by 3 to balance the iodine atoms.

- Balancing Oxygen Atoms with Water for IO3- to I2

Balance oxygen atoms by adding water molecules. There are 6 oxygen atoms in 2 IO3- and none in 3 I2, so add 6 H2O to the right side of the equation.

- Balancing Hydrogen Atoms with H+ Ions for IO3- to I2

Balance hydrogen by adding H+ to the side with less hydrogen. There are 12 hydrogen atoms in the 6 H2O molecules, so add 12 H+ to the left side of the equation.

- Balancing Charge with Electrons for IO3- to I2

Balance the charge by adding electrons. The left side has a charge of -2 (2 IO3-) + 12(+1) (12H+) = +10, and the right side has 0 charge (3 I2). Add 10 electrons to the left side to balance the charges.

- Identifying the Change in Oxidation States for SO2 to SO42-

Determine the oxidation states of S in SO2 and SO42-. In SO2, the oxidation state of S is +4 (oxygen is -2). In SO42-, the oxidation state of S is +6. Therefore, S is oxidized as its oxidation state increases from +4 to +6. Next, balance the half-reaction.

- Balancing Oxygen Atoms with Water for SO2 to SO42-

Balance oxygen atoms by adding water molecules. There are 2 oxygen atoms in SO2 and 4 in SO42-, so add 2 H2O to the left side of the equation.

- Balancing Hydrogen Atoms with H+ Ions for SO2 to SO42-

Balance hydrogen by adding H+ to the side with less hydrogen. There are 4 hydrogen atoms in the 2 H2O molecules, so add 4 H+ to the right side of the equation.

- Balancing Charge with Electrons for SO2 to SO42-

Balance the charge by adding electrons. The left side has no charge (SO2) + 4(+1) (4H2O) = +4 and the right side has a charge of -2 (SO42-) + 4(+1) (4H+) = +2. Add 2 electrons to the left side to balance the charges and maintain the oxidation state change from +4 to +6.

Key concepts

Oxidation States

Understanding oxidation states is crucial when balancing redox reactions. Oxidation states, often referred to as oxidation numbers, are used to describe the degree of oxidation (loss of electrons) or reduction (gain of electrons) of an element in a chemical compound. For example, in the half-reaction \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}\), the oxidation state of manganese (Mn) changes from +7 to +2, indicating a reduction process since the oxidation state decreases.

To determine these states, we apply certain rules, considering the known oxidation states of certain elements such as oxygen (typically -2) and hydrogen (typically +1 when combined with nonmetals). The algebraic sum of the oxidation states of all atoms in a neutral molecule must be zero, and in an ion, it must equal the charge of the ion. This concept is the foundation for identifying and balancing the changes that occur during redox reactions.

Half-Reactions

Half-reactions are the two parts of a redox reaction: oxidation and reduction. They show the loss or gain of electrons separately. When balancing redox reactions in an acidic aqueous solution, it is essential to balance these half-reactions by ensuring that the number of atoms and the electrical charges are equal on both sides of the reaction.

For example, in the balance process for \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}\), after balancing manganese atoms, oxygen is balanced with water (H2O), hydrogen with hydrogen ions (H+), and the overall charge with electrons. Balancing half-reactions is an organized method that helps in systematically approaching the conservation of mass and charge.

Oxidation and Reduction

Oxidation and reduction are two halves of the same coin in redox reactions. Oxidation involves the loss of electrons, whereas reduction involves gaining electrons. Each half-reaction in a redox process involves either oxidation or reduction. In the reduction example \(\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}\), manganese is gaining electrons, decreasing its oxidation state from +7 to +2. On the other hand, for the oxidation case \(\mathrm{SO}_{2} \rightarrow \mathrm{SO}_{4}^{2-}\), sulfur is losing electrons, increasing its oxidation state from +4 to +6.

To remember which is which, you can use the mnemonic 'LEO the lion says GER': Loss of Electrons is Oxidation, Gain of Electrons is Reduction. In balancing these reactions, it is vital to link the electron transfer between the oxidation and reduction half-reactions to ensure that the electrons lost are equal to the electrons gained, keeping the entire system electrically neutral.