Problem 101 - Chemistry The Science in Context

Question

Balance the following half-reactions by adding the appropriate number of electrons. Identify the oxidation half-reactions and the reduction half- reactions. a. \(\operatorname{Br}_{2}(\ell) \rightarrow 2 \operatorname{Br}^{-}(a q)\) b. \(\mathrm{Pb}(s)+2 \mathrm{Cl}^{-}(a q) \rightarrow \mathrm{PbCl}_{2}(s)\) c. \(\mathrm{O}_{3}(g)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\) d. \(2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}^{+}(a q) \rightarrow \mathrm{HS}_{2} \mathrm{O}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

a. \(Br_{2}(\ell) \rightarrow 2 Br^{-}(aq)\) b. \(Pb(s) + 2 Cl^{-}(aq) \rightarrow PbCl_{2}(s)\) c. \(O_{3}(g) + 2 H^{+}(aq) \rightarrow O_{2}(g) + H_{2} O(\ell)\) d. \(2 H_{2}SO_{3}(aq) + H^{+}(aq) \rightarrow HS_{2}O_{4}^{-}(aq) + 2 H_{2}O(\ell)\) Answer: a. The reduction half-reaction is \(Br_{2}(\ell) \rightarrow 2 Br^{-}(aq) + 2e^{-}\). b. The oxidation half-reaction is \(Pb(s) \rightarrow PbCl_{2}(s) + 2e^{-}\). c. The reduction half-reaction is \(3O_{3}(g)+6H^{+}(aq) \rightarrow 3O_{2}(g)+2H_{2}O(\ell) + 2e^{-}\). d. The oxidation half-reaction is \(2H_{2}SO_{3}(aq)+H^{+}(aq) \rightarrow HS_{2}O_{4}^{-}(aq)+2H_{2}O(\ell) + 4e^{-}\).

Step 1: Determine the oxidation state of each element

Bromine has an oxidation state of 0 in Br2 and -1 in 2Br-.

Step 2: Identify the change in oxidation state

The oxidation state of bromine changes from 0 to -1, which means a decrease of 1 electron charge.

Step 3: Add the appropriate number of electrons

Since there are 2 bromine atoms, we need to add 2 electrons to balance the charges: \(Br_{2}(\ell) \rightarrow 2 Br^{-}(aq) + 2e^{-}\).

Step 4: Determine if the half-reaction is an oxidation or reduction process

The process is a reduction half-reaction as the oxidation state of bromine decreases. b. \(\mathrm{Pb}(s)+2 \mathrm{Cl}^{-}(a q) \rightarrow \mathrm{PbCl}_{2}(s)\)

Step 1: Determine the oxidation state of each element

Lead has an oxidation state of 0 in Pb and +2 in PbCl2, while chlorine has an oxidation state of -1 in 2 Cl- and -1 in PbCl2.

Step 2: Identify the change in oxidation state

The oxidation state of lead changes from 0 to +2, which means an increase of 2 electron charges.

Step 3: Add the appropriate number of electrons

We need to add 2 electrons to balance the charges: \(Pb(s) \rightarrow PbCl_{2}(s) + 2e^{-}\).

Step 4: Determine if the half-reaction is an oxidation or reduction process

The process is an oxidation half-reaction as the oxidation state of lead increases. c. \(\mathrm{O}_{3}(g)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Step 1: Determine the oxidation state of each element

Oxygen has an oxidation state of 0 in O2, +2 in H2O, and +2/3 in O3. Hydrogen has an oxidation state of +1 in H+ and H2O.

Step 2: Identify the change in oxidation state

The oxidation state of oxygen in O3 changes from +2/3 to 0, which means a decrease of 2/3 electron charge.

Step 3: Add the appropriate number of electrons

We need to multiply the half-reaction by 3 to remove fractions and add 2 electrons to balance the charges: \(3O_{3}(g)+6H^{+}(aq) \rightarrow 3O_{2}(g)+2H_{2}O(\ell) + 2e^{-}\).

Step 4: Determine if the half-reaction is an oxidation or reduction process

The process is a reduction half-reaction as the oxidation state of oxygen in O3 decreases. d. \(2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}^{+}(a q) \rightarrow \mathrm{HS}_{2} \mathrm{O}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step 1: Determine the oxidation state of each element

Sulfur has an oxidation state of +4 in H2SO3 and +6 in HS2O4-. Oxygen has an oxidation state of -2 in all compounds. Hydrogen has an oxidation state of +1 in all compounds.

Step 2: Identify the change in oxidation state

The oxidation state of sulfur changes from +4 to +6, which means an increase of 2 electron charges.

Step 3: Add the appropriate number of electrons

Since there are 2 sulfur atoms, we need to add 4 electrons to balance the charges: \(2H_{2}SO_{3}(aq)+H^{+}(aq) \rightarrow HS_{2}O_{4}^{-}(aq)+2H_{2}O(\ell) + 4e^{-}\).

Step 4: Determine if the half-reaction is an oxidation or reduction process

The process is an oxidation half-reaction as the oxidation state of sulfur increases.

Key Concepts

Oxidation State

Oxidation states are a way to keep track of electrons in chemical reactions, especially redox reactions. The oxidation state, often referred to as the oxidation number, signifies the degree of oxidation of an atom. It's an imaginary charge that an atom would have if the compound were purely ionic. For example, consider bromine in the molecule \(\operatorname{Br}_2(\ell)\). The bromine atoms each have an oxidation state of 0 because it exists as a diatomic molecule consisting of the same element. When \(\operatorname{Br}_2(\ell)\) is reduced to \(2 \operatorname{Br}^-(aq)\), the oxidation state changes to -1, reflecting a gain in electrons. Recognizing changes in oxidation states helps identify which elements are oxidized and which are reduced in a redox process.

Reduction Half-Reaction

A reduction half-reaction involves the gain of electrons by a molecule, atom, or ion. In these reactions, the oxidation state of the reactant decreases. This is evident in reaction \(a\), where \(\operatorname{Br}_2(\ell)\) becomes \(2 \operatorname{Br}^-(aq)\). The bromine atoms in \(\operatorname{Br}_2\) each accept an electron, resulting in the reduction of their oxidation state from 0 to -1. Writing a half-reaction shows this transfer effectively:

• The reaction is: \(\operatorname{Br}_2(\ell) \rightarrow 2 \operatorname{Br}^-(aq) + 2e^-\)
• The 2 electrons are added on the products side to balance the equation.
• This process decreases the bromine's oxidation state, indicating it is a reduction reaction.

Reduction half-reactions are essential for tracking electron flow in redox reactions.

Oxidation Half-Reaction

In contrast to reduction, an oxidation half-reaction describes the loss of electrons. It involves an increase in the oxidation state of the reactant. Let's look at example \(b\), where lead, with an initial oxidation state of 0 in \(\mathrm{Pb}(s)\), becomes \(\mathrm{PbCl}_2(s)\) with an oxidation state of +2. Here, lead loses two electrons:

• The oxidation process is written as: \(\mathrm{Pb}(s) \rightarrow \mathrm{PbCl}_2(s) + 2e^-\)
• To signify this loss, electrons are shown on the product side.
• The oxidation state increase from 0 to +2 confirms this is oxidation.

Recognizing and writing oxidation half-reactions is crucial for balancing redox equations.

Balancing Equations

Balancing equations is a fundamental skill in chemistry to ensure both mass and charge are conserved in chemical reactions. For redox reactions, you not only balance atoms but also the charge. Let's elaborate on part \(c\), which involves ozone \(\mathrm{O}_3(g)\):

• First, examine the atoms: 3 oxygens on the reactant side should match 3 oxygens on the product side.
• Check the charges, ensuring the charge is the same on both sides after adding electrons.
• Since the oxidation states involve fractions, multiply by a whole number to eliminate fractional coefficients, here by 3 to get \(3\mathrm{O}_3(g)\) on the reactant side, eventually yielding \(3\mathrm{O}_2(g)\) and \(2e^-\) on the product side.
• Balancing is completed when both atoms and charges are equal on both sides of the equation.

With practice, balancing complex redox reactions becomes more intuitive.