Problem 24 - 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) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(\mathrm{g})\) (acidic solution) (d) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (acidic solution) (e) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(\) aq \() \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}\) (aq) (basic solution)

Step-by-Step Solution

Answer

(a) Reduction: \(\mathrm{Mo}^{3+}(a q) + 3\mathrm{e}^- \longrightarrow \mathrm{Mo}(s)\) (b) Oxidation: \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) + 4\mathrm{H}^+(aq) + 2\mathrm{e}^-\) (c) Reduction: \(\mathrm{NO}_{3}^{-}(a q) + 4\mathrm{H}^+(aq) + 3\mathrm{e}^- \longrightarrow \mathrm{NO}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) Reduction (acidic): \(\mathrm{O}_{2}(\mathrm{~g}) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (e) Reduction (basic): \(\mathrm{O}_{2}(\mathrm{~g}) + 4\mathrm{e}^- + 2\mathrm{H}_2\mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+ 4\mathrm{OH}^-(aq)\) (f) Oxidation: \(\mathrm{Mn}^{2+}(aq) + 2\mathrm{OH}^-(aq) \longrightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{e}^-\) (g) Oxidation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 6\mathrm{e}^- +3\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{OH}^-(aq)\)

Step 1: Identify the oxidation states

We can see that Mo changes from an oxidation state of +3 to an oxidation state of 0.

Step 2: Balance atoms and charge

To balance the equation, we need to add 3 electrons to the right side of the equation: \(\mathrm{Mo}^{3+}(a q) + 3\mathrm{e}^- \longrightarrow \mathrm{Mo}(s)\) Since the oxidation state of Mo is being reduced, this half-reaction represents a reduction process. (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution)

Step 1: Identify the oxidation states

In this case, S changes from an oxidation state of +4 in H2SO3 to +6 in SO4^2-.

Step 2: Balance atoms and charge

Balance the oxygen atoms by adding 2 H2O molecules to the left side and balance the hydrogen atoms by adding 4 H+ ions to the right side. To balance the charges, add 2 electrons to the right side of the equation: \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) + 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) + 4\mathrm{H}^+(aq) + 2\mathrm{e}^-\) Since the oxidation state of S is being increased, this half-reaction represents an oxidation process. (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(\mathrm{g})\) (acidic solution)

Step 1: Identify the oxidation states

In this case, N changes from an oxidation state of +5 in NO3^- to +2 in NO.

Step 2: Balance atoms and charge

Balance the oxygen atoms by adding 2 H2O molecules to the right side and balance the hydrogen atoms by adding 4 H+ ions to the left side. To balance the charges, add 3 electrons to the left side of the equation: \(\mathrm{NO}_{3}^{-}(a q) + 4\mathrm{H}^+(aq) + 3\mathrm{e}^- \longrightarrow \mathrm{NO}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(l)\) Since the oxidation state of N is being reduced, this half-reaction represents a reduction process. (d) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (acidic solution)

Step 1: Identify the oxidation states

In this case, O changes from an oxidation state of 0 in O2 to -2 in H2O.

Step 2: Balance atoms and charge

Balance the oxygen atoms by having 1 O2 molecule and 2 H2O molecules in the equation. To balance the hydrogen atoms, add 4 H+ ions to the left side of the equation. To balance the charges, add 4 electrons to the left side of the equation: \(\mathrm{O}_{2}(\mathrm{~g}) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) Since the oxidation state of O is being reduced, this half-reaction represents a reduction process. (e) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution)

Step 1: Balance the half-reaction as if it were in acidic solution

As seen in part (d), we balanced this half-reaction for acidic solution as: \(\mathrm{O}_{2}(\mathrm{~g}) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

Step 2: Adjust for basic solution

In order to convert the half-reaction from acidic to basic solution, add 4 OH- ions to both sides of the equation to cancel 4 H+ ions. The resulting equation is: \(\mathrm{O}_{2}(\mathrm{~g}) + 4\mathrm{e}^- + 2\mathrm{H}_2\mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+ 4\mathrm{OH}^-(aq)\) Since the oxidation state of O is being reduced, this half-reaction also represents a reduction process. (f) \(\mathrm{Mn}^{2+}(\) aq \() \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution)

Step 1: Identify the oxidation states

In this case, Mn changes from an oxidation state of +2 in Mn^2+ to +4 in MnO2.

Step 2: Balance atoms and charge

Balance the oxygen atoms by adding 2 OH- ions to the left side of the equation. To balance the charges, add 2 electrons to the right side of the equation: \(\mathrm{Mn}^{2+}(aq) + 2\mathrm{OH}^-(aq) \longrightarrow \mathrm{MnO}_{2}(s) + 2\mathrm{e}^-\) Since the oxidation state of Mn is being increased, this half-reaction represents an oxidation process. (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution)

Step 1: Identify the oxidation states

In this case, Cr changes from an oxidation state of +3 in Cr(OH)3 to +6 in CrO4^2-.

Step 2: Balance atoms and charge

Balance the oxygen atoms by adding 3 OH- ions and 3 H2O molecules to the equation. To balance the charges, add 6 electrons to the left side of the equation: \(\mathrm{Cr}(\mathrm{OH})_{3}(s) + 6\mathrm{e}^- +3\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) + 6\mathrm{OH}^-(aq)\) Since the oxidation state of Cr is being increased, this half-reaction represents an oxidation process.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that involves making sure each type of atom has the same number on both sides of the equation. This conservation reflects the law of conservation of mass. To start, write the unbalanced equation and identify the elements involved. Then, look at the number of atoms for each element on both sides of the equation. Use coefficients, which are numbers placed in front of compounds, to balance these levels without altering the chemical formulas themselves. Always begin by balancing elements that appear in only one reactant and one product. Then, proceed to balance elements that appear in multiple reactants or products. Finally, double-check to confirm both mass and charge are balanced, especially in complex redox reactions.

Oxidation States

Oxidation states, also known as oxidation numbers, are essential for tracking electron transfer in redox reactions. They indicate the degree of oxidation or reduction of an element in a compound. To determine oxidation states, there are several guidelines that can help:

• The oxidation state of an element in its natural form is zero. For example, O₂ or N₂ are zero.
• The oxidation state of a monoatomic ion is equal to its charge. For example, Na⁺ is +1.
• In compounds, hydrogen is usually +1, oxygen is typically -2, and halogens have negative oxidation states unless bonded to oxygen or another halogen.
• The sum of oxidation states in a neutral molecule is zero, and in an ion is equal to the ion's charge.

Understanding these rules can simplify the process of identifying which elements are oxidized and which are reduced.

Acidic and Basic Solutions

The medium of a reaction, whether acidic or basic, affects how we balance redox reactions. In acidic solutions, we use hydrogen ions (H⁺) to help balance the equation. Balance the oxygen atoms by adding water ( H₂O) molecules and the hydrogen by adding H⁺ ions as needed. Once atoms are balanced, add electrons to balance the charge on both sides.
In basic solutions, the process is slightly different. We first solve for acidic conditions and then convert to basic by adding hydroxide ions (OH⁻). These OH⁻ ions will balance the extra H⁺ ions by forming water, which must be counted in the final balance. This dual approach, handling acidic and then basic balancing, ensures both mass and charge conservation. Adjustments between acidic and basic conditions show the flexibility and adaptability in redox balancing.

Half-Reactions

Half-reactions are an essential concept for understanding redox reactions. They break down the full redox reaction into two separate parts: the oxidation part and the reduction part. This separation simplifies the balancing of redox reactions by focusing on one part at a time. Each half-reaction involves either the gain or loss of electrons, indicated by adding electrons to one side of the equation. In oxidation half-reactions, electrons are products, while in reduction half-reactions, electrons are reactants. This distinction helps clarify which chemical species is losing electrons and which is gaining, highlighting the electron transfer process. By balancing each half-reaction independently for mass and charge, and then combining them in a way that cancels out the electrons involved, a balanced redox equation is achieved. This step-by-step method not only facilitates understanding but also ensures accurate balancing in complex redox systems.