Problem 142 - Chemistry The Science in Context

Question

Balance this net ionic reaction and answer the questions that follow: $$ \mathrm{BrO}_{3}^{-}(a q)+\mathrm{Br}^{-}(a q) \rightarrow \mathrm{Br}_{2}(a q) $$ a. Is \(\mathrm{BrO}_{3}^{-}(a q)\) reduced? b. What is the product of \(\mathrm{BrO}_{3}^{-}(a q)\) reduction in this reaction? c. Is \(\mathrm{Br}^{-}(a q)\) oxidized? d. What is the product of \(\mathrm{Br}^{-}(a q)\) oxidation in this reaction?

Step-by-Step Solution

Answer

Is \(\mathrm{Br}^{-}(a q)\) oxidized, and what is the product of its oxidation in the given reaction? Answer: \(\mathrm{BrO}_{3}^{-}(a q)\) is reduced, and its product of reduction is \(\mathrm{Br}_{2}(a q)\). \(\mathrm{Br}^{-}(a q)\) is oxidized, and its product of oxidation is also \(\mathrm{Br}_{2}(a q)\).

Step 1: Balance the atoms

Start by balancing the atoms in the reaction except for oxygen and hydrogen. In this case, the bromine atoms must be balanced on the left-hand side and the right-hand side of the equation. $$ \mathrm{BrO}_{3}^{-}(a q)+5 \mathrm{Br}^{-}(a q) \rightarrow 3 \mathrm{Br}_{2}(a q) $$

Step 2: Balance the oxygens (if present)

Balance the oxygen atoms in the reaction. In this case, the oxygen atoms are already balanced on both sides, so there is no need to make any changes.

Step 3: Balance the hydrogens (if present)

Balance the hydrogen atoms in the reaction. In this case, there are no hydrogen atoms in the reaction, so there is no need to make any changes. Now, the balanced reaction is: $$ \mathrm{BrO}_{3}^{-}(a q)+5 \mathrm{Br}^{-}(a q) \rightarrow 3 \mathrm{Br}_{2}(a q) $$ Now, we have to answer the four given questions: a. Is \(\mathrm{BrO}_{3}^{-}(a q)\) reduced? b. What is the product of \(\mathrm{BrO}_{3}^{-}(a q)\) reduction in this reaction? c. Is \(\mathrm{Br}^{-}(a q)\) oxidized? d. What is the product of \(\mathrm{Br}^{-}(a q)\) oxidation in this reaction?

Step 4: Identify the substances undergoing reduction and oxidation

Reduction is the process of gaining electrons, while oxidation is the process of losing electrons. We can identify the oxidation states of the substances involved in the reaction to find out which is being reduced or oxidized. For the balanced reaction: $$ \mathrm{BrO}_{3}^{-}(a q)+5 \mathrm{Br}^{-}(a q) \rightarrow 3 \mathrm{Br}_{2}(a q) $$ The oxidation numbers of each atom in the equation are: Br in \(\mathrm{BrO}_{3}^{-}(a q)\): +5 (since O is usually -2) Br in \(\mathrm{Br}^{-}(a q)\): -1 (since it carries a single negative charge) Br in \(\mathrm{Br}_{2}(a q)\): 0 (since it is in its elemental form) a. Is \(\mathrm{BrO}_{3}^{-}(a q)\) reduced? Since the oxidation number of Br in \(\mathrm{BrO}_{3}^{-}(a q)\) changes from +5 to 0 after the reaction, it gains electrons and is thus reduced. b. What is the product of \(\mathrm{BrO}_{3}^{-}(a q)\) reduction in this reaction? The reduced product of \(\mathrm{BrO}_{3}^{-}(a q)\) is \(\mathrm{Br}_{2}(a q)\). c. Is \(\mathrm{Br}^{-}(a q)\) oxidized? Since the oxidation number of Br in \(\mathrm{Br}^{-}(a q)\) changes from -1 to 0 after the reaction, it loses an electron and is thus oxidized. d. What is the product of \(\mathrm{Br}^{-}(a q)\) oxidation in this reaction? The oxidized product of \(\mathrm{Br}^{-}(a q)\) is also \(\mathrm{Br}_{2}(a q)\).

Key Concepts

Balancing Chemical Equations

The process of balancing chemical equations involves ensuring that the number of each type of atom is the same on both sides of the equation. This is essential because of the law of conservation of mass, which dictates that matter cannot be created or destroyed in a chemical reaction. In the given reaction, \( \mathrm{BrO}_{3}^{-}(aq) + \mathrm{Br}^{-}(aq) \rightarrow \mathrm{Br}_{2}(aq) \), the first step is to balance the bromine atoms. Initially, the reaction has 1 bromine atom in \( \mathrm{BrO}_{3}^{-}(aq) \) and 1 bromine atom in \( \mathrm{Br}^{-}(aq) \) on the left side, while the product \( \mathrm{Br}_2(aq) \) on the right side contains 2 bromine atoms. By adjusting the coefficients, the balanced equation becomes \( \mathrm{BrO}_{3}^{-}(aq) + 5 \mathrm{Br}^{-}(aq) \rightarrow 3 \mathrm{Br}_{2}(aq) \), ensuring both bromine and oxygen atoms are balanced. Breaking the task into smaller steps and adjusting only one element's coefficient at a time ensures a smooth balancing process.

Oxidation States

Oxidation states, or oxidation numbers, are hypothetical charges assigned to atoms to keep track of electron transfer in chemical reactions. Understanding oxidation states helps identify which atoms are oxidized and reduced. In the reaction \( \mathrm{BrO}_{3}^{-}(aq) + 5 \mathrm{Br}^{-}(aq) \rightarrow 3 \mathrm{Br}_{2}(aq) \), analyzing oxidation states helps reveal the changes in electron configuration. The oxidation state of bromine in \( \mathrm{BrO}_{3}^{-}(aq) \) is +5 because oxygen, usually -2, results in a total charge of -1 for the oxyanion. \( \mathrm{Br}^{-}(aq) \) has a -1 oxidation state, reflecting its negative charge. On the product side, \( \mathrm{Br}_2(aq) \) is in the elemental form with an oxidation state of 0. Observing that Br in \( \mathrm{BrO}_{3}^{-}(aq) \) shifts from +5 to 0 signifies a reduction, while Br in \( \mathrm{Br}^{-}(aq) \) moves from -1 to 0, indicating oxidation.

Electron Transfer

Electron transfer is the fundamental aspect of redox (reduction-oxidation) reactions, where electrons are moved from one atom or molecule to another. In our Br reaction, this transfer is evidenced by changing oxidation states. Let's break down what happens:

• Reduction happens when \( \mathrm{BrO}_{3}^{-}(aq) \) undergoes a drop in oxidation state from +5 to 0, meaning it gains electrons.
• Oxidation occurs when \( \mathrm{Br}^{-}(aq) \) shifts from -1 to 0, signifying it loses electrons.

This exchange balances out because the same number of electrons lost by bromine ions \( (\mathrm{Br}^{-}) \) is gained by bromate ions \( (\mathrm{BrO}_{3}^{-}) \). Understanding electron transfer and the involved oxidation states lets us accurately assess redox processes. Such comprehension aids in predicting the changes an element undergoes and can be applied to various chemical reactions beyond this specific example.