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Chapter 18: Problem 101

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 \mathrm{M} \mathrm{Cd}^{2+}, 1.0 \mathrm{MAg}^{+}, 1.0 \mathrm{M} \mathrm{Au}^{3+}\), and \(1.0 \mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

Short Answer

Expert verified
The metals will be plated out in the order of their reduction potentials: Ag from \(Ag^{+}\), Au from \(Au^{3+}\), Ni from \(Ni^{2+}\), and Cd from \(Cd^{2+}\). The order of plating will be Silver (Ag), Gold (Au), Nickel (Ni), and Cadmium (Cd) as the voltage is gradually increased.

Step by step solution

01

Determine the reduction potentials of the given metal ions

First, let's find the reduction potentials of the given metal ions. We can consult a table of standard reduction potentials or textbook, where the data can be found: For simplicity: \(E^{0}\) = Standard Reduction Potential \(E^{0}(Cd^{2+}/Cd) = -0.402 \, V\) \(E^{0}(Ag^{+}/Ag) = 0.799 \, V\) \(E^{0}(Au^{3+}/Au) = 1.498 \, V\) \(E^{0}(Ni^{2+}/Ni) = -0.257 \, V\)
02

Arrange the metal ions according to their reduction potentials

Now let's arrange the metal ions in ascending order according to their reduction potentials: 1. \(Cd^{2+}/Cd\) : \(-0.402 \, V\) 2. \(Ni^{2+}/Ni\) : \(-0.257 \, V\) 3. \(Ag^{+}/Ag\) : \(0.799 \, V\) 4. \(Au^{3+}/Au\) : \(1.498 \, V\)
03

Predict the order of plating based on the reduction potentials

In the cathode compartment of an electrolytic cell, the metal ion with the least negative (or most positive) reduction potential will plate out first as the voltage is gradually increased. So, based on our ranking from Step 2, the order in which the metals will plate out is as follows: 1. Silver (Ag) from \(Ag^{+}\) 2. Gold (Au) from \(Au^{3+}\) 3. Nickel (Ni) from \(Ni^{2+}\) 4. Cadmium (Cd) from \(Cd^{2+}\) Thus, as the voltage is gradually increased, the metals will be plated out in the order: Ag, Au, Ni, and Cd.

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Most popular questions from this chapter

The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{C}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\text {ref }}=0.250 \mathrm{~V}\) and that \(\mathscr{C}_{\text {tme\pi }}=0.480 \mathrm{~V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}(\pm 0.001 \mathrm{~V})\) ? b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

Sketch a galvanic cell, and explain how it works. Look at Figs. \(18.1\) and \(18.2 .\) Explain what is occurring in each container and why the cell in Fig. \(18.2\) "works" but the one in Fig. \(18.1\) does not.

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{MAg}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\). a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

Consider the following half-reactions: $$\begin{aligned}\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & \mathscr{E}^{\circ}=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{C}^{\circ}=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{C}^{\circ}=0.96 \mathrm{~V}\end{aligned}$$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h} ?\)
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