Question

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. $$ \text { (a)} \mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Sn}^{4+}(a q) \text {(acidic solution)} \\ \text {(b)} \mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{2+}(a q) \text {(acidic solution)} \\ \text {(c)} \mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) \text {(acidic solution)} \\ \text {(d)} \mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{4}^{+}(a q) \text {(acidic solution)} \\ \text {(e)} \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{O}_{2}(g) \text {(acidic solution)} \\ \text {(f)} \operatorname{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text {(acidic solution)} \\\\(\mathrm{g}) \mathrm{N}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) \text {(acidic solution)} $$

Short answer

Following the steps for all the half-reactions, we get: (a) \( \mathrm{Sn}^{2+}(a q) - 2e^{-} \longrightarrow \mathrm{Sn}^{4+}(a q) \) (Oxidation) (b) \( \mathrm{TiO}_{2}(s) + 2e^{-} + 4\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \) (Reduction) (c) \( \mathrm{ClO}_{3}^{-}(a q) + 6e^{-} + 6\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cl}^{-}(a q) + 3\mathrm{H}_{2}\mathrm{O}(l) \) (Reduction) (d) \( \mathrm{N}_{2}(g) + 6\mathrm{H}^{+}(a q) + 6e^{-} \longrightarrow 2\mathrm{NH}_{4}^{+}(a q) \) (Reduction) (e) \( 2\mathrm{OH}^{-}(a q) - 4e^{-} \longrightarrow \mathrm{O}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \) (Oxidation) (f) \( \mathrm{SO}_{3}^{2-}(a q) + 2\mathrm{H}^{+}(a q) - 2e^{-} \longrightarrow \mathrm{SO}_{4}^{2-}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \) (Oxidation) (g) \( \mathrm{N}_{2}(g) + 6\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NH}_{3}(g) \) (Reduction)

Step-by-step solution

Determine the oxidation number of Sn

In this half-reaction, the initial oxidation state of Sn is +2, and the final oxidation state is +4.

Determine the number of electrons gained or lost

To go from an oxidation number of +2 to +4, two electrons are lost: \[ \mathrm{Sn}^{2+}(a q) - 2e^{-} \longrightarrow \mathrm{Sn}^{4+}(a q) \]

Balance the half-reaction

The half-reaction is already balanced, as there are equal numbers of charges on both sides of the equation.

Identify the type of half-reaction

Since the oxidation number of Sn increased and electrons were lost, this half-reaction is an oxidation process. ##Half-reaction (b)##

Determine the oxidation number of Ti

In TiO2, the oxidation number of Ti is +4, while in Ti^2+ the oxidation number of Ti is +2.

Determine the number of electrons gained or lost

To go from an oxidation number of +4 to +2, two electrons are gained: \[ \mathrm{TiO}_{2}(s) + 2e^{-} \longrightarrow \mathrm{Ti}^{2+}(a q) \]

Balance the half-reaction

Add four H+ ions and two H2O molecules to balance the half-reaction: \[ \mathrm{TiO}_{2}(s) + 2e^{-} + 4\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ti}^{2+}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) \]

Identify the type of half-reaction

Since the oxidation number of Ti decreased and electrons were gained, this half-reaction is a reduction process. Follow the same steps for the remaining half-reactions.

Key concepts

Oxidation and Reduction

Oxidation and reduction are two key concepts in the study of chemical reactions, particularly in the context of redox (reduction-oxidation) reactions. These processes involve the transfer of electrons between atoms, ions, or molecules.

Oxidation

Oxidation refers to the loss of electrons by a substance. When a substance undergoes oxidation, its oxidation number increases. This is seen in our exercise where Sn^{2+} loses two electrons to become Sn^{4+}, thus undergoing oxidation. A helpful mnemonic to remember oxidation is 'LEO' which stands for 'Loss of Electrons is Oxidation'.

Reduction

Reduction is the gain of electrons by a substance, causing a decrease in oxidation number. The mnemonic here is 'GER', meaning 'Gain of Electrons is Reduction'. In the exercise,TiO_{2} gains two electrons, resulting in Ti^{2+}, showcasing a typical reduction.

In a redox reaction, one species is oxidized (loses electrons), while another is reduced (gains electrons). It's essential to remember that oxidation and reduction always occur together; for every electron lost by one substance, another must gain it. This interdependence is the core of redox reactions and highlights the importance of simultaneously considering both half-reactions.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is upheld. This principle states that matter cannot be created or destroyed in a chemical reaction, thus the quantity of each element must remain constant throughout the reaction.

When balancing half-reactions, we make sure that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction. In our exercise, for instance, Sn^{2+} turning into Sn^{4+} involves a loss of two electrons. This balance is crucial for accurately representing the stoichiometry of the redox process.

Consideration of charge is also important. For reactions in acidic solutions, H^+ ions and water molecules are added to balance both the charge and the hydrogen and oxygen atoms. In the case of TiO_{2} to Ti^{2+}, the addition of 4H^+ and 2H_2O is required to balance the equation. Remember to confirm that both sides of the equation have the same number of each type of atom and the same overall charge.

Determining Oxidation Numbers

Determining oxidation numbers (also known as oxidation states) is a crucial step in the study of redox reactions. An oxidation number is a theoretical charge on an atom if it were in its most stable ionic form. This number can either be an actual charge for monatomic ions or, more commonly, a hypothetical charge assigned to individual atoms within a molecule or ion based on certain rules.

Some general rules for assigning oxidation numbers include:

• Oxygen usually has an oxidation number of -2, except in peroxides where it is -1.
• Hydrogen is typically given +1 when paired with nonmetals, and -1 with metals.
• The oxidation number of a pure element is always zero.
• For a molecule or ion, the sum of the oxidation numbers must equal the total charge of the molecule or ion.

In our exercise, tin (Sn) in Sn^{2+} has an oxidation number of +2, but it becomes +4 in Sn^{4+}. These numbers help in understanding the redox processes taking place, identifying whether a half-reaction involves oxidation or reduction, and ultimately are invaluable for balancing the chemical equation.