Problem 23 - Chemistry: The Central Science in SI Units, Global Edition

Question

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (acidic solution) (b) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (d) \(\mathrm{S}(s)(a q) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (acidic solution) (e) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{OH}^{-}(a q)\) (basic solution)

Step-by-Step Solution

Answer

(d) \(\underline{Oxidation:} \quad S(s) + 2H_2O(l) \longrightarrow H_2S(aq) + 4H^+(aq) + 2e^-\) (e) \(\underline{Reduction:} \quad NO_3^-(aq) + 2H_2O(l) + e^- \longrightarrow NO_2^-(aq) + 4OH^-(aq) + H_2O(l)\) (f) \(\underline{Oxidation:} \quad H_2O_2(aq) \longrightarrow OH^-(aq) + H_2O(l) + e^-\)

Step1: Balance the Chromium atoms

We have 2 Cr atoms on the left side and 1 Cr atom on the right side. To balance them, multiply the right side by 2: \[Cr_2O_7^{2-}(aq) \longrightarrow 2Cr^{3+}(aq)\]

Step 2: Balance Hydrogen atoms

There are no hydrogen atoms in this half-reaction.

Step 3: Balance Oxygen atoms

There are 7 oxygen atoms on the left and none on the right. Add 7 H2O molecules on the right to balance the oxygen atoms: \[Cr_2O_7^{2-}(aq) \longrightarrow 2Cr^{3+}(aq) + 7H_2O(l)\]

Step 4: Balance charges

The total charge on the left side is -2, and the total charge on the right side is +6. To balance the charges, add 6 electrons (e-) to the right side: \[\underline{Reduction:} \quad Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \longrightarrow 2Cr^{3+}(aq) + 7H_2O(l)\] (b) \(Mn^{2+}(aq) \longrightarrow MnO_4^{-}(aq)\) (acidic solution)

Step1: Balance the Manganese atoms

Manganese atoms are already balanced (1 Mn atom on each side).

Step 2: Balance Hydrogen atoms

There are no hydrogen atoms in this half-reaction.

Step 3: Balance Oxygen atoms

There are 4 oxygen atoms on the right side and none on the left. Add 4 H2O molecules on the left to balance the oxygen atoms: \[Mn^{2+}(aq) + 4H_2O(l) \longrightarrow MnO_4^{-}(aq)\]

Step 4: Balance charges

The total charge on the left side is +2, and the total charge on the right side is -1. To balance the charges, add 3 electrons (e-) to the right side: \[\underline{Oxidation:} \quad Mn^{2+}(aq) + 4H_2O(l) \longrightarrow MnO_4^{-}(aq) + 8H^+(aq) + 5e^-\] (c) \(I_2(s) \longrightarrow IO_3^{-}(aq)\) (acidic solution)

Step1: Balance the Iodine atoms

We have 2 I atoms on the left side and 1 I atom on the right side. To balance them, multiply the right side by 2: \[I_2(s) \longrightarrow 2IO_3^{-}(aq)\]

Step 2: Balance Hydrogen atoms

There are no hydrogen atoms in this half-reaction.

Step 3: Balance Oxygen atoms

There are 6 oxygen atoms on the right side and none on the left. Add 6 H2O molecules on the left to balance the oxygen atoms: \[I_2(s) + 6H_2O(l) \longrightarrow 2IO_3^{-}(aq)\]

Step 4: Balance charges

The total charge on the left side is 0, and the total charge on the right side is -2. To balance the charges, add 2 electrons (e-) to the left side: \[\underline{Reduction:} \quad I_2(s) + 6H_2O(l) + 2e^- \longrightarrow 2IO_3^{-}(aq) +12H^+(aq)\] Continue with the remaining half-reactions in a similar manner.

Key Concepts

Oxidation and Reduction

Redox reactions consist of two key processes: oxidation and reduction. Oxidation refers to the loss of electrons from an atom or molecule, while reduction involves the gain of electrons. These complementary processes occur simultaneously in a redox reaction. It is important to remember this mnemonic: "OIL RIG"—Oxidation Is Loss, Reduction Is Gain, which helps in recalling that oxidation involves losing electrons, whereas reduction pertains to gaining them.
For example, consider the reduction of dichromate ions \[ ext{Cr}_2 ext{O}_7^{2-} + 14 ext{H}^+ + 6e^- ightarrow 2 ext{Cr}^{3+} + 7 ext{H}_2 ext{O} \] In this reaction, the Cr ions are reduced from a +6 to a +3 oxidation state, signifying a gain of electrons. Conversely, in the oxidation of manganese: \[ ext{Mn}^{2+} + 4 ext{H}_2 ext{O} ightarrow ext{MnO}_4^- + 8 ext{H}^+ + 5e^- \] manganese ions lose electrons, increasing their oxidation state. Recognizing oxidation and reduction helps in balancing redox reactions effectively.

Balancing Chemical Equations

Balancing chemical equations is an essential skill for solving redox reactions. It ensures that the quantities of atoms for each element are balanced on both sides of a reaction. To balance an equation, follow these general steps: identify and label all reactants and products, balance the number of atoms for each element, and finally ensure that the charges are balanced between both sides.

• **Balancing Atoms:** Start by balancing atoms that appear in only one reactant and one product. Continue with the others. For example, in balancing \( ext{Cr}_2 ext{O}_7^{2-} \rightarrow 2 ext{Cr}^{3+} \), adjust the number of chromium atoms by writing a coefficient of 2 for \( ext{Cr}^{3+} \).
• **Balancing Charges:** Once the atoms are balanced, look at the overall charges. Add electrons to one or both sides to balance the charge. In acidic solutions, you might need to add \( ext{H}^+ \) ions, while in basic solutions, use \( ext{OH}^- \) ions.

Balancing chemical reactions requires practice, but with these steps, it becomes manageable and straightforward.

Half-Reactions in Acidic and Basic Solutions

Half-reactions are a way to simplify and understand redox reactions more clearly by splitting them into two parts: the oxidation and the reduction. This separation helps focus on the changes happening to each species independently. Half-reactions can occur in either acidic or basic solutions, impacting how they are balanced.
In **acidic solutions**, the addition of hydrogen ions \( ext{H}^+ \) is common. Oxygen atoms may be balanced by adding water molecules \( ext{H}_2 ext{O} \). For example, \[ ext{I}_2(s) + 6 ext{H}_2 ext{O}(l) + 2e^- ightarrow 2 ext{IO}_3^-(aq) +12 ext{H}^+(aq) \] shows iodine being oxidized with added water and hydrogen ions.
In **basic solutions**, balancing often requires adding hydroxide ions \( ext{OH}^- \) instead of \( ext{H}^+ \). Equations balanced in basic conditions usually involve converting \( ext{H}^+ \) ions into \( ext{H}_2 ext{O} \) by adding \( ext{OH}^- \) to both sides.
This method demands careful attention to both atom counts and charge balance. With practice, choosing the correct strategy for balancing half-reactions in different pH environments becomes intuitive.