(a) Use the following reactions to prepare an activity series for the halogens: $$ \begin{aligned} \mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) & \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\ \mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) & \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q) \end{aligned} $$ (b) Relate the positions of the halogens in the periodic table with their locations in this activity series. (c) Predict whether a reaction occurs when the following reagents are mixed: \(\mathrm{Cl}_{2}(a q)\) and \(\mathrm{KI}(a q) ; \operatorname{Br}_{2}(a q)\) and \(\operatorname{LiCl}(a q)\)

Short Answer

Expert verified
In summary, an activity series for the given halogens is: Cl₂ > Br₂ > I₂. The reactivity decreases as we move down Group 17 in the periodic table. A reaction will occur between Cl₂ and KI, forming KCl and I₂. However, no reaction will occur between Br₂ and LiCl as Br₂ is less reactive than Cl₂.

Step by step solution

01

Analyze the given reactions

The reactions are as follows: \( \begin{aligned} \mathrm{Br}_{2}(a q)+2 \mathrm{NaI}(a q) & \longrightarrow 2 \mathrm{NaBr}(a q)+\mathrm{I}_{2}(a q) \\ \mathrm{Cl}_{2}(a q)+2 \mathrm{NaBr}(a q) & \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{Br}_{2}(a q) \end{aligned} \) In reaction 1, bromine (Br₂) displaces iodine (I₂) from its salt, sodium iodide (NaI), resulting in the formation of sodium bromide (NaBr) and iodine (I₂). In reaction 2, chlorine (Cl₂) displaces bromine (Br₂) from its salt, sodium bromide (NaBr), resulting in the formation of sodium chloride (NaCl) and bromine (Br₂).
02

Prepare the halogen activity series

Based on the given reactions, we can arrange the halogens in decreasing order of their reactivity as: 1. Chlorine (Cl₂) 2. Bromine (Br₂) 3. Iodine (I₂) This activity series tells us that chlorine is the most reactive among the given halogens, followed by bromine, and then iodine.
03

Relate halogen positions in the periodic table and the activity series

In the periodic table, all halogens belong to Group 17 (or Group VIIA). Their positions in the periodic table are as follows: 1. Chlorine (Cl) - Period 3 2. Bromine (Br) - Period 4 3. Iodine (I) - Period 5 When comparing their locations in the periodic table with their positions in the activity series, we can notice that the reactivity of halogens decreases as we move down the group. This is due to the decrease in the electron affinity and the increase in atomic size, which make it more difficult for the halogens to attract an additional electron to complete their octets.
04

Predict the outcome of reactions based on the activity series

We have two reactions to predict: 1. \( \mathrm{Cl}_{2}(a q) + \mathrm{KI}(a q)\) 2. \( \mathrm{Br}_{2}(a q) + \mathrm{LiCl}(a q)\) (a) Since chlorine (Cl₂) is more reactive than iodine (I₂), it will displace iodine from potassium iodide (KI) to form potassium chloride (KCl) and iodine (I₂). Thus, a reaction will occur, and the products will be KCl and I₂. (b) Bromine (Br₂) is less reactive than chlorine (Cl₂), so it cannot displace chlorine from lithium chloride (LiCl). Therefore, no reaction will occur in this case.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(3.455-\mathrm{g}\) sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If \(0.2815 \mathrm{~g}\) of barium sulfate was obtained, what was the mass percentage of barium in the sample?

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) \(\mathrm{H}_{2} \mathrm{SO}_{3}\), (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethanol), (c) \(\mathrm{NH}_{3}\), (d) \(\mathrm{KClO}_{3}\), (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\)

Explain the following observations: (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\) ions, and yet its aqueous solutions are basic; (b) HF is called a weak acid, and yet it is very reactive; (c) although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) ions than \(\mathrm{SO}_{4}^{2-}\) ions.

Indicate the concentration of each ion or molecule present in the following solutions: (a) \(0.25 \mathrm{M} \mathrm{NaNO}_{3}\), (b) \(1.3 \times 10^{-2} \mathrm{M}\) \(\mathrm{MgSO}_{4}\), (c) \(0.0150 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), (d) a mixture of \(45.0 \mathrm{~mL}\) of \(0.272 \mathrm{M} \mathrm{NaCl}\) and \(65.0 \mathrm{~mL}\) of \(0.0247 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) Assume that the volumes are additive.

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free