Problem 21 - General Chemistry: Principles and Modern Applications

Question

Write plausible half-equations to represent each of the following in acidic solution. (a) \(\mathrm{VO}^{2+}(\mathrm{aq})\) as an oxidizing agent (b) \(\mathrm{Cr}^{2+}(\text { aq })\) as a reducing agent

Step-by-Step Solution

Answer

(a) \(\mathrm{VO}^{2+} + 2H^+ + 3e^- \rightarrow V^{2+} + H_2O\) (b) \(\mathrm{Cr}^{2+} \rightarrow Cr^{3+} + e^-\)

Step 1: Identify oxidation state changes for a)

First, identify the starting oxidation state of the vanadium in the \(\mathrm{VO}^{2+}\) compound which is +5. Considering the \(\mathrm{VO}^{2+}\) acts as an oxidizing agent, it will receive electrons and be reduced, so the oxidation number should decrease.

Step 2: Write and balance the half-equation for a)

Create a half-equation by assigning a lower oxidation state to vanadium, commonly +2 or +3. For our task, we will say it’s +2 as \(\mathrm{V^{2+}}\). The equation becomes: \(\mathrm{VO}^{2+} + 3e^- \rightarrow V^{2+} + O^{2-}\). The oxygen has an oxidation number of -2, so to balance the charges, add 2H+ on the left side: \(\mathrm{VO}^{2+} + 2H^+ + 3e^- \rightarrow V^{2+} + H_2O\). Now the equation is balanced in terms of matter and charges.

Step 3: Identify oxidation state changes for b)

The starting oxidation state of the chromium in the \(\mathrm{Cr}^{2+}\) compound is +2. Since \(\mathrm{Cr}^{2+}\) is acting as a reducing agent, it will give away electrons and be oxidized, so the oxidation number should increase.

Step 4: Write and balance the half-equation for b)

Create a half-equation by assigning a higher oxidation state to chromium, commonly +3 as in \(\mathrm{Cr}^{3+}\). The equation becomes: \(\mathrm{Cr}^{2+} \rightarrow Cr^{3+} + e^-\). The equation is already balanced in terms of matter and charges.

Key Concepts

Oxidation State

The concept of oxidation state, or oxidation number, is a fundamental part of understanding redox reactions in chemistry. It is a hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. Oxidation states help chemists keep track of electron movement in redox reactions.

For instance, in the vanadium compound \( \mathrm{VO}^{2+} \) mentioned in the exercise, vanadium has an oxidation state of +5. This means if all the bonds were ionic, vanadium would carry a charge of +5. Determining oxidation states involves some basic rules, like the fact that the sum of oxidation states in a neutral compound is zero, and in a polyatomic ion, the sum equals the charge of the ion.

Calculating oxidation state is crucial for writing half-equations, as it shows whether an atom is being oxidized (increase in oxidation state) or reduced (decrease in oxidation state). Through the use of these states, one can balance half-reactions by ensuring that the total charge before and after the reaction remains equal.

Reducing Agent

A reducing agent is a substance that loses electrons in a chemical reaction and, as a consequence, reduces another substance. It undergoes oxidation itself. The key aspect of a reducing agent is its ability to donate electrons.

In the given example, \( \mathrm{Cr}^{2+} \) acts as a reducing agent by losing an electron to form \( \mathrm{Cr}^{3+} \). When a reducing agent loses electrons, its oxidation state increases. It's important for students to identify the reducing agent in a redox reaction because it helps in determining the direction of electron flow. The reducing agent's role is crucial for balancing the electrons lost and gained in the complete redox reaction.

Oxidizing Agent

An oxidizing agent gains electrons and is itself reduced in a chemical reaction. In other words, it causes another substance to lose electrons. For a reactant to be considered an oxidizing agent, it must take on electrons and, consequently, its oxidation state decreases.

In the example provided, \( \mathrm{VO}^{2+} \) is the oxidizing agent. It accepts electrons, and as a result, the oxidation state of vanadium decreases from +5 to +2 when it is reduced to \( \mathrm{V}^{2+} \). Understanding the role of the oxidizing agent is critical in writing half-equations because it highlights which species is gaining electrons, and it also dictates the substances required to balance the half-equation for both mass and charge.

Acidic Solution

An acidic solution has properties that alter the structure of half-equations. It donates protons (\( H^+ \) ions) and usually has a pH less than 7. In redox reactions occurring in acidic solutions, \( H^+ \) ions can be added or removed to balance the charge and hydrogen atoms.

For example, in balancing the vanadium half-equation, \( 2H^+ \) were added to the left side to counterbalance the charge introduced by the addition of three electrons. Similarly, water \( H_2O \) molecules are often included in these half-equations to balance the number of oxygen atoms. The proper treatment of acid or basic solutions is vital, as it ensures the half-equations accurately reflect the reaction conditions, which is essential for the overall balance of the redox reaction.