Question

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{~S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{~S}(s)+\) \(2 \mathrm{MnO}_{2}(s)+8 \mathrm{OH}^{-}(a q)\) (b) \(4 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+2 \mathrm{OH}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\) \(5 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{O}_{2}(g)\) (c) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Short answer

In the given equations, the following elements undergo changes in oxidation number: (a) Mn: from +7 to +4 (change of 3) and S: from -2 to 0 (change of 2). (b) H in \( \mathrm{H}_{2}\mathrm{O}_{2} \): from +1 to 0 (change of 1) and Cl in \( \mathrm{Cl}_{2}\mathrm{O}_{7} \): from +7 to +3 (change of 4). (c) O in \( \mathrm{H}_{2}\mathrm{O}_{2} \): from -1 to 0 (change of 1).

Step-by-step solution

Equation (a)

1. Assign oxidation numbers to the atoms in the reactants and products: - Mn in \(\mathrm{MnO}_{4}^{-}\): +7 - S in \(\mathrm{S}^{2-}\): -2 - O in \(\mathrm{H}_{2}\mathrm{O}\): -2 - H in \(\mathrm{H}_{2}\mathrm{O}\): +1 - S in \(\mathrm{S}\): 0 - Mn in \(\mathrm{MnO}_{2}\): +4 - O in \(\mathrm{OH}^{-}\): -2 - H in \(\mathrm{OH}^{-}\): +1 2. Identify the elements that undergo changes in oxidation numbers: - Mn: from +7 to +4 - S: from -2 to 0 3. Indicate the magnitude of the change in each case: - Mn: 3 - S: 2

Equation (b)

1. Assign oxidation numbers to the atoms in the reactants and products: - O in \(\mathrm{H}_{2}\mathrm{O}_{2}\): -1 - H in \(\mathrm{H}_{2}\mathrm{O}_{2}\): +1 - Cl in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\): +7 - O in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\): -2 - O in \(\mathrm{OH}^{-}\): -2 - H in \(\mathrm{OH}^{-}\): +1 - Cl in \(\mathrm{ClO}_{2}^{-}\): +3 - O in \(\mathrm{ClO}_{2}^{-}\): -2 - O in \(\mathrm{H}_{2}\mathrm{O}\): -2 - H in \(\mathrm{H}_{2}\mathrm{O}\): +1 - O in \(\mathrm{O}_{2}\): 0 2. Identify the elements that undergo changes in oxidation numbers: - H in \(\mathrm{H}_{2}\mathrm{O}_{2}\): from +1 to 0 - Cl in \(\mathrm{Cl}_{2}\mathrm{O}_{7}\): from +7 to +3 3. Indicate the magnitude of the change in each case: - H: 1 - Cl: 4

Equation (c)

1. Assign oxidation numbers to the atoms in the reactants and products: - Ba in \(\mathrm{Ba}^{2+}\): +2 - O in \(\mathrm{OH}^{-}\): -2 - H in \(\mathrm{OH}^{-}\): +1 - O in \(\mathrm{H}_{2}\mathrm{O}_{2}\): -1 - H in \(\mathrm{H}_{2}\mathrm{O}_{2}\): +1 - Cl in \(\mathrm{ClO}_{2}\): +4 - O in \(\mathrm{ClO}_{2}\): -2 - Ba in \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}\): +2 - Cl in \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}\): +4 - O in \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}\): -2 - O in \(\mathrm{H}_{2}\mathrm{O}\): -2 - H in \(\mathrm{H}_{2}\mathrm{O}\): +1 - O in \(\mathrm{O}_{2}\): 0 2. Identify the elements that undergo changes in oxidation numbers: - O in \(\mathrm{H}_{2}\mathrm{O}_{2}\): from -1 to 0 3. Indicate the magnitude of the change in each case: - O: 1

Key concepts

Oxidation Numbers

An oxidation number is a helpful tool in chemistry to keep track of electron transfer in oxidation-reduction (redox) reactions. It is assigned to each atom in a substance to show its degree of oxidation or reduction.
Oxidation numbers are like a book-keeping system for charge. They help identify which elements are oxidized and which are reduced during a reaction. Here are some rules to keep in mind:

• The oxidation number of an element in its natural state (e.g., O₂, H₂) is zero.
• For monoatomic ions, the oxidation number equals the ion's charge (e.g., Na⁺ has +1).
• Fluorine always has an oxidation number of -1 in compounds.
• Hydrogen usually has an oxidation number of +1, except when it is bonded to metals in binary compounds, where it is -1.
• Oxygen typically has an oxidation number of -2, except in peroxides where it is -1.
• The sum of the oxidation numbers in a neutral compound is zero, while in a polyatomic ion, it equals the ion's charge.

By determining the changes in oxidation numbers in a reaction, you can identify which atoms have lost or gained electrons. This is crucial in recognizing and balancing redox reactions.

Balancing Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation, adhering to the Law of Conservation of Mass.
In the context of redox reactions, balancing requires a bit more attention because it involves both an exchange of electrons and atoms. Here’s a step-by-step approach for balancing redox equations:

• Assign oxidation numbers to all elements in the equation to identify the changes in the oxidation state.
• Determine the number of electrons lost or gained by each atom that undergoes a change in oxidation number.
• Balance the electrons transferred between the oxidized and reduced elements. This may involve multiplying the changes to have an equal electron exchange.
• Next, balance all other atoms in the equation, usually starting with H and O. You might add water, H⁺, or OH^- to balance hydrogen or oxygen, especially in acidic or basic solutions.
• Recheck all elements and charges to confirm they are balanced.

Balancing redox reactions guarantees mass and charge conservation across the reaction, upholding the fundamental chemical laws.

Redox Chemistry

Redox chemistry is the study of oxidation-reduction reactions, a type of chemical reaction where electrons are transferred between two substances.
Redox reactions comprise two main processes: oxidation and reduction.

• Oxidation: This involves the loss of electrons from a substance. The substance that loses electrons is called the reducing agent because it reduces the other reactant.
• Reduction: This involves the gain of electrons by a substance. The substance that gains electrons is termed the oxidizing agent because it oxidizes the other reactant.

In a redox reaction, one element is oxidized, losing electrons, and another element is reduced, gaining those electrons. Identifying the oxidizing and reducing agents is crucial in analyzing these reactions.
A deeper understanding of redox chemistry is essential in fields like biochemistry, metallurgy, and environmental chemistry, where these reactions play a significant role. Ultimately, mastering redox reactions involves recognizing changes in oxidation states and understanding electron transfer between species.