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The metal cadmium tends to form Cd \(^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Short Answer

Expert verified
(a) The net ionic equations for the given reactions are: (i) \( Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s) \) (ii) \( Cd(s) + Ni^{2+}(aq) \rightarrow Cd^{2+}(aq) + Ni(s) \) (b) Cadmium lies between zinc and nickel in the activity series, where zinc is more reactive and nickel is less reactive. (c) To locate the position of cadmium more precisely, we can test its reactivity with other metal salts and perform electrochemical cell experiments with different metals in the activity series. By comparing the outcomes, we can narrow down cadmium's position accurately.

Step by step solution

01

Identify relevant ions in each reaction.

To write net ionic equations, we will first identify the relevant ions involved in the given reactions: (i) The reaction between a strip of zinc metal and aqueous CdCl2. (ii) The reaction between a strip of cadmium metal and aqueous Ni(NO3)2.
02

Write net ionic equations for each reaction.

(i) In the reaction between zinc metal and aqueous CdCl2, zinc replaces cadmium, and cadmium metal is deposited on the zinc strip. So, the net ionic equation is: \[ Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s) \] (ii) In the reaction between cadmium metal and aqueous Ni(NO3)2, cadmium displaces nickel, depositing nickel metal on the cadmium strip. The net ionic equation is: \[ Cd(s) + Ni^{2+}(aq) \rightarrow Cd^{2+}(aq) + Ni(s) \]
03

Determine the position of cadmium in the activity series.

From the given observations, we can conclude the following: - Zinc is more reactive than cadmium since it can replace cadmium in the reaction with CdCl2. - Cadmium is more reactive than nickel since it can replace nickel in the reaction with Ni(NO3)2. Therefore, in the activity series, the position of cadmium lies between zinc and nickel, with zinc being more reactive and nickel being less reactive.
04

Suggest experiments to locate the position of cadmium more precisely in the activity series.

To locate the position of cadmium more precisely in the activity series, we can perform the following experiments: 1. Test the reactivity of cadmium with other metal salts (e.g., solution of copper sulfate or magnesium chloride) in a similar displacement reaction. This will help determine the reactivity of cadmium relative to these metals. 2. Conduct a series of electrochemical cells using cadmium as an electrode, and couple it with other metals in the activity series. By comparing the cell voltage, we can determine which metals are more reactive than cadmium and which are less reactive. The relative positions of metals can be inferred by comparing their redox potentials, also called standard electrode potentials. Based on the outcomes of these experiments, we can further narrow down the position of cadmium in the activity series.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Net Ionic Equations
Understanding the net ionic equations is essential when examining chemical reactions in solutions. Net ionic equations simplify reactions to their core components, showing only the species that undergo a change. They exclude spectator ions that do not participate in the reaction.

For example, when zinc metal reacts with cadmium chloride in solution, the net ionic equation is \[ Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s) \] This represents the actual chemical change occurring: zinc metal becomes zinc ions while cadmium ions are reduced to cadmium metal.

In a similar reaction, when cadmium metal is placed into a nickel nitrate solution, the net ionic equation is \[ Cd(s) + Ni^{2+}(aq) \rightarrow Cd^{2+}(aq) + Ni(s) \] showing that cadmium displaces nickel. Here, cadmium undergoes oxidation while nickel ions are reduced. These ionic equations provide a clearer understanding of the electrochemical behavior of the involved metals.
Reactivity of Metals
The reactivity of metals is an intriguing subject that refers to a metal's tendency to lose electrons and form positive ions, known as cations. The activity series is a chart that ranks metals from most reactive to least reactive.

In our examples, the cadmium's position in the activity series is inferred from its reactions with zinc and nickel. As zinc can deposit cadmium from its ionic solution, and cadmium can deposit nickel from its ionic solution, it suggests that zinc is more reactive than cadmium, while cadmium is more reactive than nickel.

To further explore this, additional experiments can be done with cadmium and other metal salts to precisely determine its reactivity. These experiments are not just a means of comparison but can also illustrate the practical applications of metal reactivity, such as in corrosion prevention and metallurgical processes.
Electrochemical Cells
Electrochemical cells harness the chemical energy of redox reactions to produce electrical energy. Such cells consist of two electrodes, an anode, and a cathode, submerged in electrolyte solutions connected by a salt bridge.

By using cadmium as an electrode in an electrochemical cell paired with other metals, we can assess its reactivity. If we form a cell with cadmium and zinc, and the cell shows a positive voltage, this indicates that zinc is more reactive and is the anode where oxidation occurs. Conversely, a cell with cadmium and copper showing a negative voltage would indicate that cadmium acts as an anode.

Such experiments are not only crucial for academic understanding, but also play a significant role in the development of batteries and corrosion prevention methods in the industry.
Standard Electrode Potentials
The standard electrode potentials are a measure of the individual potentials of electrodes compared to a standard hydrogen electrode under specific conditions. They are central to understanding the driving force behind redox reactions in electrochemical cells.

For example, the standard potential can predict whether a metal will tend to oxidize or reduce. Metals with a more negative standard electrode potential than cadmium will be more likely to lose electrons, signaling a higher reactivity, whereas those with a more positive potential will be less reactive. To place cadmium accurately within the activity series, experimental measurements of cell voltages (which are related to electrode potentials) can be compared.

This understanding is pivotal not only academically but also in real-world applications such as the designing of galvanic cells in batteries, and comprehending the electrochemical behavior of metals in various environments.

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Most popular questions from this chapter

(a) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) aqueous solution of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) A certain volume of a \(0.50 \mathrm{M}\) solution contains \(4.5 \mathrm{~g}\) of a salt. What mass of the salt is present in the same volume of a \(2.50 \mathrm{M}\) solution?

Write the balanced molecular and net ionic equations for each of the following neutralization reactions: (a) Aqueous acetic acid is neutralized by aqueous barium hydroxide. (b) Solid chromium(III) hydroxide reacts with nitrous acid. (c) Aqueous nitric acid and aqueous ammonia react.

Determine the oxidation number for the indicated element (b) \(\mathrm{Al}\) in each of the following compounds: (a) \(\mathrm{Co}\) in \(\mathrm{LiCoO}_{2}\), in \(\mathrm{NaAlH}_{4}\), (c) \(\mathrm{C}\) in \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol), \((\mathrm{d}) \mathrm{N}\) in \(\mathrm{GaN},\) (e) Cl in \(\mathrm{HClO}_{2},\) (f) \(\mathrm{Cr}\) in \(\mathrm{BaCrO}_{4}\)

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with \(\mathrm{HCl}\) as follows: $$\mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)$$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

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