Problem 19 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) \(\mathrm{Sn}^{2+}(aq) \rightarrow \mathrm{Sn}^{4+}(aq) + 2e^-\), Oxidation (b) \(4\mathrm{H}^+(aq) + \mathrm{TiO}_{2}(s) + 2e^- \rightarrow \mathrm{Ti}^{2+}(aq) + 2\mathrm{H}_2\mathrm{O}(l)\), Reduction (c) \(5e^- + \mathrm{ClO}_{3}^{-}(aq) + 6\mathrm{H}^+(aq) \rightarrow \mathrm{Cl}^{-}(aq) + 2\mathrm{H}_2\mathrm{O}(l)\), Reduction (d) \(6e^- + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(aq)\), Reduction (f) \(\mathrm{SO}_{3}^{2-}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq) + 2e^-\), Oxidation

Step 1: Balance the atoms

Sn is already balanced here.

Step 2: Oxidation or reduction?

Since Sn loses electrons, it is an oxidation half-reaction.

Step 3: Balancing the charge

Since Sn loses two electrons, balance the charges by adding two electrons: \(\mathrm{Sn}^{2+}(aq) \rightarrow \mathrm{Sn}^{4+}(aq) + 2e^-\) (b) \(\mathrm{TiO}_{2}(s) \rightarrow \mathrm{Ti}^{2+}(aq)\) (acidic solution)

Step 1: Balance the atoms

Ti and O are already balanced here.

Step 2: Oxidation or reduction?

Since Ti gains electrons, it is a reduction half-reaction.

Step 3: Balance hydrogen and charge

In an acidic solution, we balance hydrogen using H+. So we need 4 H+ ions on the left to balance the two O atoms: \(4\mathrm{H}^+(aq) + \mathrm{TiO}_{2}(s) \rightarrow \mathrm{Ti}^{2+}(aq)\) Now balance the charges by adding 2e- on the left: \(4\mathrm{H}^+(aq) + \mathrm{TiO}_{2}(s) + 2e^- \rightarrow \mathrm{Ti}^{2+}(aq) + 2\mathrm{H}_2\mathrm{O}(l)\) (c) \(\mathrm{ClO}_{3}^{-}(aq) \rightarrow \mathrm{Cl}^{-}(aq)\) (acidic solution)

Step 1: Balance the atoms

Cl is already balanced here.

Step 2: Balance oxygen atoms

Add 2 water molecules on the right side to balance the three O atoms on the left: \(\mathrm{ClO}_{3}^{-}(aq) \rightarrow \mathrm{Cl}^{-}(aq) + 2\mathrm{H}_2\mathrm{O}(l)\)

Step 3: Balance hydrogen atoms

Add 4 H+ ions on the left side to balance the 4 hydrogen atoms in the 2 water molecules: \(\mathrm{ClO}_{3}^{-}(aq) + 6\mathrm{H}^+(aq) \rightarrow \mathrm{Cl}^{-}(aq) + 2\mathrm{H}_2\mathrm{O}(l)\)

Step 4: Oxidation or reduction?

Since Cl gains electrons, it is a reduction half-reaction.

Step 5: Balance charge

Balance charges by adding 5 e- on the left: \(5e^- + \mathrm{ClO}_{3}^{-}(aq) + 6\mathrm{H}^+(aq) \rightarrow \mathrm{Cl}^{-}(aq) + 2\mathrm{H}_2\mathrm{O}(l)\) (d) \(\mathrm{N}_{2}(g) \rightarrow \mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(aq)\) (acidic solution)

Step 1: Balance atoms

Add 2 NH4+ on the right side to balance the nitrogen atoms: \(\mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(aq)\)

Step 2: Oxidation or reduction?

Since N gains electrons, it is a reduction half-reaction.

Step 3: Balance charge

Add 6 e- to balance the charge: \(6e^- + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(aq)\) (e) is not a valid half-reaction as there are no redox processes happening. (f) \(\mathrm{SO}_{3}^{2-}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq)\) (basic solution)

Step 1: Balance atoms

S and O are already balanced.

Step 2: Oxidation or reduction?

Since S loses electrons, it is an oxidation half-reaction.

Step 3: Balance charge

Add 2 e- on the right side to balance the charges: \(\mathrm{SO}_{3}^{2-}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq) + 2e^-\) The reaction is complete as it is a redox half-reaction in a basic solution. (g) is not a valid half-reaction as there are no redox processes happening.

Key Concepts

Oxidation

Oxidation is a chemical process where an element loses electrons. This change results in an increase in the oxidation state of the element. A classic example of oxidation at work can be seen in the half-reaction involving tin (Sn) ions as such: - In the transformation from Sn²⁺ to Sn⁴⁺, tin is oxidized. - Here, Sn loses two electrons, evident by the equation: \[ \text{Sn}^{2+}(aq) \rightarrow \text{Sn}^{4+}(aq) + 2e^- \]This loss of electrons is marked by a change from a lower to a higher positive charge. In oxidation reactions, it's essential to balance both the atoms and the charge to maintain equilibrium. Remember: When electrons are lost, oxidation occurs.
This is easy to remember with the mnemonic: **LEO** (Loss of Electrons is Oxidation).

Reduction

Reduction is the counter process to oxidation and involves the gain of electrons. This decreases the oxidation state of the element involved. In the half-reaction for titanium (Ti - Ti²⁺) reduction is clearly observed:- The process changes titanium dioxide (TiO₂) to a Ti²⁺ ion.- This requires the addition of electrons, shown in the equation: \[ 4\mathrm{H}^+(aq) + \text{TiO}_{2}(s) + 2e^- \rightarrow \text{Ti}^{2+}(aq) + 2\text{H}_2\text{O}(l) \] The titanium here gains two electrons, demonstrating a reduction reaction. You will often see this in acidic solutions where hydrogen ions (H⁺) and electrons contribute to balancing the reaction. Always remember: **GER** or "Gain of Electrons is Reduction."

Half-Reaction Balancing

Balancing half-reactions involves several steps to ensure that the number of atoms and charges are the same on both sides of the reaction. Each half-reaction should keep the principles of conservation of mass and charge.- **Start by balancing atoms other than **O** and **H**. - **Next, balance the oxygen atoms by adding water molecules (H₂O) where needed.** - **Balance the hydrogen atoms with H⁺ ions.** - **Finally, balance the charges with electrons.**For instance, in balancing the reaction for **ClO₃^-** to **Cl^-**: - Three water molecules help balance oxygen: \( \text{ClO}_3^-(aq) \rightarrow \text{Cl}^-(aq) + 3\mathrm{H}_2\mathrm{O}(l) \) - Next, six H⁺ are added to balance the hydrogens. - Finally, add five electrons on the left to account for charge: \( 5e^- + \text{ClO}_3^-(aq) + 6\text{H}^+(aq) \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l) \).By following these steps, the half-reaction is complete and balanced.

Acidic and Basic Solutions

In redox chemistry, the nature of the solution—acidic or basic—affects how reactions are balanced, especially regarding atoms and charges.- **In acidic solutions:** Use H⁺ ions to balance hydrogen atoms. - When balancing a reduction half-reaction in an acidic solution, as seen with **N₂** transforming to **NH₄⁺**, add electrons and H⁺ ions: \( 6e^- + \mathrm{N}_{2}(g) \rightarrow 2\mathrm{NH}_{4}^{+}(aq) \).- **In basic solutions:** Instead, add OH^- ions to balance hydrogen after completing the acidic balancing steps. - For instance, the transformation from **SO₃^2-** to **SO₄^2-**: \( \mathrm{SO}_{3}^{2-}(aq) \rightarrow \mathrm{SO}_{4}^{2-}(aq) + 2e^- \).These distinctions ensure that reactions respect the different pH conditions, which influence how electrons and ions behave in the solution. Understanding the solution type helps predict the necessary ions needed for balancing.