Problem 111 - Chemistry The Science in Context

Question

Chromium is more toxic and more soluble in natural waters as \(\mathrm{HCrO}_{4}^{-}\) than as chromium(III) ion. In the presence of \(\mathrm{H}_{2} \mathrm{S}\), the following reaction takes place in neutral solution: $$ \mathrm{HCrO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{4}^{2-} $$ a. Assign oxidation numbers to the reactants and products. b. Balance the equation. c. How many electrons are transferred for each atom of chromium that reacts?

Step-by-Step Solution

Answer

Answer: Each chromium atom transfers 3 electrons during the reaction.

Step 1: Assigning oxidation numbers

In order to assign oxidation numbers to the reactants and products, we need to follow a set of rules that allow us to determine the number of electrons that each atom loses or gains in the process. For HCrO\(_4$$^{-}\): - The oxidation number of hydrogen (H) is always +1. - The oxidation number of oxygen (O) is always -2. - The overall molecule must have a charge of -1, so the oxidation number of chromium (Cr) must be +6. For H\(_2\)S: - The oxidation number of hydrogen (H) is still +1. - The oxidation number of sulfur (S) in this molecule must be determined, since overall charge of the molecule must be 0. Therefore, the oxidation number of sulfur (S) should be -2. For Cr\(_2\)O\(_3\): - The oxidation number of oxygen (O) is still -2. - Now we must determine the oxidation number of chromium (Cr) in this molecule since there are two chromium atoms in the molecule. Since overall charge must be 0, the oxidation number of chromium (Cr) should be +3. For SO\(_4$$^{2-}\): - The oxidation number of oxygen (O) is still -2. - The sulfur (S) oxidation number should be determined. The overall molecule must have a charge of -2, so the oxidation number of sulfur (S) should be +6.

Step 2: Balancing the equation

To balance the equation, we will follow the steps below: 1. Balance the non-hydrogen and non-oxygen atoms (Cr and S). 2. Balance the hydrogen atoms (H). 3. Balance the oxygen atoms (O). 4. Ensure the charges are balanced. $$2\ \mathrm{HCrO}_{4}^{-} + 3\ \mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3} + 3\ \mathrm{SO}_{4}^{2-} + 6\ \mathrm{H}^{+}$$ The balanced equation also takes into account the transfer of 6 positive charges from the 6 hydrogen ions that were created in the reaction.

Step 3: Calculating electrons transferred for chromium

To determine the number of electrons transferred for each chromium atom during the reaction, we will compare the oxidation numbers of the chromium in the reactants and products. In the reactants, the oxidation number of chromium in HCrO\(_4$$^{-}\) is +6. In the products, the oxidation number of chromium in Cr\(_2\)O\(_3\) is +3. Considering one chromium atom, the change in oxidation number is (+6 - (+3)) = +3, which means that each chromium atom loses 3 electrons in the reaction.

Key Concepts

Oxidation Numbers

Understanding and assigning oxidation numbers is key to analyzing redox reactions. Oxidation numbers indicate the electron status of an atom in a compound. They help determine how electrons are distributed and if any atoms are oxidized or reduced during a reaction. Assigning oxidation numbers can be done by following simple rules:

• Elemental state: Elements in their natural state have an oxidation number of 0. For example, O₂ or H₂.
• Monatomic ions: The oxidation number is equal to the ion’s charge. For instance, Na⁺ has an oxidation number of +1.
• Hydrogen and oxygen: Usually, hydrogen is +1 and oxygen is -2.

In compounds like \( ext{HCrO}_{4}^{-}\), you calculate chromium's oxidation number by following the algebraic sum of oxidation numbers equal to the compound's charge. Hydrogen and oxygen have known numbers (+1 and -2, respectively), letting us solve for chromium as +6. This basic understanding aids in predicting the electron flow of the chemical reaction.

Chemical Equation Balancing

Balancing chemical equations ensures compliance with the law of conservation of mass. It reflects that matter cannot be created or destroyed during a chemical reaction. Each element must have the same number of atoms in both reactants and products' side. Here's how you can balance the given reaction:

• Step 1: Focus on non-hydrogen and non-oxygen elements first, such as chromium (Cr) and sulfur (S). Match them on both sides.
• Step 2: Balance hydrogen atoms. In the given example equation, hydrogen in \( ext{H}_{2} ext{S}\) needs to be balanced with hydrogen on the product side.
• Step 3: Balance oxygen atoms next. If oxygen can't match initially, adjust the coefficients of compounds with oxygen on one side.
• Step 4: Ensure that charges on both sides of the equation are balanced, especially in ionic reactions.

Proper balancing helps in reflecting the correct stoichiometry of the reaction, which is critical for real-life chemical interactions.

Electron Transfer

Electron transfer is the crux of redox reactions. These reactions involve the movement of electrons from one element to another, changing their oxidation states. This transfer defines oxidation, the loss of electrons, and reduction, the gain of electrons. In the exercise, the oxidation number of chromium changes from +6 in \( ext{HCrO}_{4}^{-}\) to +3 in \( ext{Cr}_{2} ext{O}_{3}\), indicating a reduction in chromium. Chromium has gained electrons here. Similarly, since sulfur's oxidation number changes from -2 in \( ext{H}_{2} ext{S}\) to +6 in \( ext{SO}_{4}^{2-}\), it demonstrates oxidation, as it loses electrons.

• To find the number of electrons transferred, examine the difference between the initial and final oxidation states. Each change indicates how many electrons have moved.
• In our specific reaction, for each atom of chromium, there is a loss of 3 electrons, underscoring the reduction phase of chromium.

This careful accounting reflects the overall changes occurring during the redox process and helps identify which substances act as oxidizing or reducing agents in the chemical equation.