Problem 104 - Chemistry The Science in Context

Question

Balance the following net ionic reactions, and identify which elements are oxidized and which are reduced: a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{S}(s)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(s)\) c. \(\mathrm{Mn}^{2+}(a q)+\mathrm{BiO}_{3}^{-}(a q) \rightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Bi}^{3+}(a q)\)

Step-by-Step Solution

Answer

Question: In each of the balanced net ionic reactions, identify the oxidizing and reducing agents. a. 2 MnO4- + 5 S2- → 2 MnO2 + 5 S b. 5 IO3- + 6 I- → 3 I2 c. 2 Mn2+ + 5 BiO3- → 2 MnO4- + 5 Bi3+ + 6 H2O Answer: a. The oxidizing agent is S2-, and the reducing agent is MnO4-. b. The oxidizing agent is I-, and the reducing agent is IO3-. c. The oxidizing agent is Mn2+, and the reducing agent is BiO3-.

Step 1: Determine the oxidation states

In reactants: Mn has +7, O has -2, S has -2 In products: Mn has +4, O has -2, S has 0

Step 2: Identify the elements that are being oxidized and reduced

Mn is reduced (from +7 to +4), S is oxidized (from -2 to 0)

Step 3: Balance the equation

2 \(\mathrm{MnO}_{4}^{-}(a q)+5\mathrm{S}^{2-}(a q) \rightarrow 2\mathrm{MnO}_{2}(s)+5\mathrm{S}(s)\)

Step 4: Check the balanced equation

Atoms are balanced for each element on both sides.

Step 5: Identify the oxidation and reduction agents

MnO4- is the reducing agent, S2- is the oxidizing agent. b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(s)\)

Step 1: Determine the oxidation states

In reactants: I in IO3- has +5, O has -2, I in I- has -1 In products: I in I2 has 0

Step 2: Identify the elements that are being oxidized and reduced

I (in IO3-) is reduced (from +5 to 0), I (in I-) is oxidized (from -1 to 0)

Step 3: Balance the equation

5 \(\mathrm{IO}_{3}^{-}(a q)+6\mathrm{I}^{-}(a q) \rightarrow 3\mathrm{I}_{2}(s)\)

Step 4: Check the balanced equation

Atoms are balanced for each element on both sides.

Step 5: Identify the oxidation and reduction agents

IO3- is the reducing agent, I- is the oxidizing agent. c. \(\mathrm{Mn}^{2+}(a q)+\mathrm{BiO}_{3}^{-}(a q) \rightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Bi}^{3+}(a q)\)

Step 1: Determine the oxidation states

In reactants: Mn has +2, Bi has +3, O has -2 In products: Mn has +7, O has -2, Bi has +3

Step 2: Identify the elements that are being oxidized and reduced

Mn is oxidized (from +2 to +7), Bi is reduced (from +5 to +3)

Step 3: Balance the equation

2 \(\mathrm{Mn}^{2+}(a q)+5\mathrm{BiO}_{3}^{-}(a q) \rightarrow 2\mathrm{MnO}_{4}^{-}(a q)+5\mathrm{Bi}^{3+}(a q)+6 \mathrm{H}_{2}\mathrm{O}(l)\)

Step 4: Check the balanced equation

Atoms are balanced for each element on both sides.

Step 5: Identify the oxidation and reduction agents

Mn2+ is the oxidizing agent, BiO3- is the reducing agent.

Key Concepts

Oxidation States

Understanding oxidation states is crucial when working with net ionic equations. An oxidation state, also known as an oxidation number, is a figure that represents the total number of electrons an atom uses to bond with other atoms. This number helps determine which element in a chemical reaction is oxidized and which is reduced.

To figure out oxidation states, consider the charge of each atom if the compound were entirely ionic. For example, in the compound \( \text{MnO}_4^- \), the manganese (Mn) has an oxidation state of +7, while each oxygen (O) is -2. The sum of oxidation states should equal the total charge of the compound.

• The oxidation state changes when an atom gains or loses electrons during a reaction.
• A positive change in oxidation state indicates oxidation.
• A negative change in oxidation state signifies reduction.

This knowledge is foundational when balancing redox reactions because it helps identify the flow of electrons.

Balancing Equations

Balancing equations is an essential skill in chemistry to ensure mass and charge conservation in a reaction. A balanced equation means that there are equal numbers of each type of atom on both sides of the equation. This is especially important in net ionic equations, which show only the particles directly involved in the reaction process.

When balancing, start by writing down the unbalanced equation with the correct formulas for each species. Then, follow these steps:

• Identify the elements that get oxidized and reduced by comparing their oxidation states.
• Use these changes to help balance the overall number of electrons lost in oxidation with those gained in reduction.
• Adjust coefficients to ensure that all atoms have equivalent numbers on both the reactant and product sides.
• Keep in mind the conservation of charge; the overall charge on both sides must be the same.

Practicing with examples helps reinforce these steps and ensures a deeper understanding of balancing net ionic equations.

Oxidizing and Reducing Agents

In redox reactions, it is crucial to know about oxidizing and reducing agents. These agents drive the transfer of electrons during the reaction, leading to changes in oxidation states.

- The oxidizing agent is the substance that brings about oxidation by gaining electrons. By gaining electrons, the oxidizing agent undergoes reduction.
- The reducing agent, on the other hand, causes reduction by losing electrons. Therefore, the reducing agent is oxidized in the process.

• In our example, \( \text{MnO}_4^- \) acts as an oxidizing agent because it gains electrons (Mn goes from +7 to +4).
• \( \text{S}^{2-} \) is the reducing agent since it loses electrons (S goes from -2 to 0).

Understanding these roles gives insight into the electron exchange and helps in predicting the products of a redox reaction.