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Chapter 20: Problem 73

A solution containing both \(\mathrm{Ag}^{+}\) and \(\mathrm{Cu}^{2+}\) ions is subjected to electrolysis. (a) Which metal should plate out first? (b) Plating out is finished after a current of \(0.75 \mathrm{A}\) is passed through the solution for 2.50 hours. If the total mass of metal is \(3.50 \mathrm{g},\) what is the mass percent of silver in the product?

Short Answer

Expert verified
The metal that plates out first is silver. The mass percent of silver in the product is 100%.

Step by step solution

01

Determine the metal that plates out first

The metal with a higher reduction potential (or less negative) will be reduced first. Looking at the standard reduction potentials, Ag+ has a higher reduction potential (+0.80V) compared to Cu2+ (+0.34V), and thus Ag+ will plate out first.
02

Calculate the total charge passed

The total charge passed in the electrolysis process can be calculated using the formula Q=It, where I is the current and t is the time. Here, \(I=0.75 \mathrm{A}\) and \(t=2.5 \mathrm{hr} = 9000 \mathrm{s}\). Substituting these into the formula gives \(Q=0.75 \mathrm{A} \times 9000 \mathrm{s} = 6750 \mathrm{C}\)
03

Calculate the moles of electrons

Using Faraday's law, we can calculate the moles of electrons transferred. According to this law, one mole of electrons carries a charge of \(1 \mathrm{F}=96485 \mathrm{C}\). Therefore, the moles of electrons transferred are \( \frac{6750 \mathrm{C}}{96485 \mathrm{C/mol}} \approx 0.07 \mathrm{mol}\) of electrons.
04

Calculate the moles and mass of Ag and Cu plated

From the stoichiometry of the half reactions, the reduction of Ag+ to Ag requires 1 mole of electrons for 1 mole of Ag, while the reduction of Cu2+ to Cu requires 2 moles of electrons for 1 mole of Cu. However, as Ag plates out first, all electrons will reduce Ag+ until it is completely plated out. Hence, approximately 0.07 moles of Ag are plated out. Using the molar mass of silver (107.87 g/mol), the mass of Ag plated out is \(0.07 \mathrm{mol} \times 107.87 \mathrm{g/mol} \approx 7.55 \mathrm{g}\). However, as the total mass of metal is given to be 3.50 g, this means that all electrons could not have reduced Ag and thus, all 3.50 g is silver.
05

Calculate the mass percent of silver

The mass percent of silver can be calculated as \( \frac{mass \, of \, Ag}{total \, mass} \times 100\). Thus, \( \frac{3.50 \, g}{3.50 \, g} \times 100 = 100\%\)

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Most popular questions from this chapter

A test for completeness of electrodeposition of \(\mathrm{Cu}\) from a solution of \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is to add \(\mathrm{NH}_{3}(\mathrm{aq}) .\) A blue color signifies the formation of the complex ion \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\left(K_{\mathrm{f}}=1.1 \times 10^{13}\right) .\) Let \(250.0 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{CuSO}_{4}(\text { aq })\) be electrolyzed with a \(3.512 \mathrm{A}\) current for 1368 s. At this time, add a sufficient quantity of \(\mathrm{NH}_{3}(\text { aq })\) to complex any remaining \(\mathrm{Cu}^{2+}\) and to maintain a free \(\left[\mathrm{NH}_{3}\right]=0.10 \mathrm{M} .\) If \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is detectable at concentrations as low as \(1 \times 10^{-5} \mathrm{M}\) should the blue color appear?

You must estimate \(E^{\circ}\) for the half-reaction \(\operatorname{In}^{3+}(\mathrm{aq})+\) \(3 \mathrm{e}^{-} \longrightarrow \operatorname{In}(\mathrm{s}) .\) You have no electrical equipment, but you do have all of the metals listed in Table 20.1 and aqueous solutions of their ions, as well as \(\operatorname{In}(\mathrm{s})\) and \(\operatorname{In}^{3+}(\text { aq })\). Describe the experiments you would perform and the accuracy you would expect in your result.

Natural gas transmission pipes are sometimes protected against corrosion by the maintenance of a small potential difference between the pipe and an inert electrode buried in the ground. Describe how the method works.

A common reference electrode consists of a silver wire coated with \(\mathrm{AgCl}(\mathrm{s})\) and immersed in \(1 \mathrm{M} \mathrm{KCl}\) $$\mathrm{AgCl}(\mathrm{s})+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}^{-}(1 \mathrm{M}) E^{\circ}=0.2223 \mathrm{V}$$ (a) What is \(E_{\text {cell }}^{\circ}\) when this electrode is a cathode in combination with a standard zinc electrode as an anode? (b) Cite several reasons why this electrode should be easier to use than a standard hydrogen electrode. (c) By comparing the potential of this silver-silver chloride electrode with that of the silver-silver ion electrode, determine \(K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\).

Explain the important distinctions between each pair of terms: (a) half- reaction and overall cell reaction; (b) voltaic cell and electrolytic cell; (c) primary battery and secondary battery; (d) \(E_{\text {cell }}\) and \(E_{\text {cell }}^{\circ}\).
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