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Chapter 17: Problem 157

Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g chromium metal was deposited from a chromium(III) nitrate solution. In the second cell, 3.15 g osmium was deposited from a solution made of \(\mathrm{Os}^{n+}\) and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing \(\mathrm{X}^{2+}\) ions caused deposition of \(2.11 \mathrm{g}\) metallic \(\mathrm{X}\). What is the electron configuration of X?

Short Answer

Expert verified
The name of the salt in the second cell is osmium(IV) nitrate with the formula \(\mathrm{Os(NO}_{3}\mathrm{)_4}\). The electron configuration of element X, which we identified as copper (Cu), is [Ar] 3d10 4s1.

Step by step solution

01

Calculate the moles of chromium deposited from the first cell

Using the mass of chromium deposited and its molar mass, we can find the moles of chromium deposited: Moles of \(\mathrm{Cr} = \frac{Mass\ of\ \mathrm{Cr}}{Molar\ mass\ of\ \mathrm{Cr}} = \frac{1.15\mathrm{g}}{51.996\mathrm{g/mol}} = 0.02211\ mol\)
02

Calculate the charge passed through the first cell using Faraday's Law

Faraday's law states that charge = n * F, where n is the moles of electrons and F is Faraday's constant (96,485 C/mol). Charge passed = moles of electrons * F. For three moles of electrons (as \(Cr^{3+}\) gets reduced to \(Cr\)): Charge passed = 0.02211 mol * 3 * 96,485 C/mol = 6,389 C
03

Calculate moles of osmium deposited in the second cell

Using the mass of osmium deposited and its molar mass, find moles of osmium deposited: Moles of \(\mathrm{Os} = \frac{Mass\ of\ \mathrm{Os}}{Molar\ mass\ of\ \mathrm{Os}} = \frac{3.15\mathrm{g}}{190.23\mathrm{g/mol}} = 0.01656\ mol\)
04

Determine the oxidation number of osmium using Faraday's Law

Using the charge passed through the cell (6,389 C) and Faraday's Law, we can find the moles of electrons: Moles of electrons = \frac{6,389\mathrm{C}}{96,485\mathrm{C/mol}} = 0.06622\ mol Oxidation number (n) = \(\frac{\mathrm{Moles\ of\ electrons}}{\mathrm{Moles\ of\ Os}} = \frac{0.06622\ mol}{0.01656\ mol} = 4\)
05

Name the salt in the second cell

Since the osmium has an oxidation number of 4 and is combined with nitrate ions, the name of the salt is osmium(IV) nitrate with the formula \(\mathrm{Os(NO}_{3}\mathrm{)_4}\).
06

Calculate moles of X deposited in the third cell

Since element X has a 2+ charge, two moles of electrons are needed to deposit one mole of X. Using Faraday's law and the charge passed, we find the moles of X: Moles of X = \(\frac{1}{2} \times \frac{6,389\mathrm{C}}{96,485\mathrm{C/mol}} = 0.03311\ mol\)
07

Calculate the molar mass of element X

Using the mass of X deposited and the moles of X, find the molar mass of X: Molar mass of X = \(\frac{2.11\mathrm{g}}{0.03311\ mol} = 63.7\mathrm{g/mol}\)
08

Identify element X and determine its electron configuration

The molar mass of X (63.7 g/mol) corresponds to copper (Cu). The electron configuration of Cu is [Ar] 3d10 4s1.

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Most popular questions from this chapter

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is $$\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \quad \mathscr{E}^{\circ}=1.10 \mathrm{V} $$ For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

Consider only the species (at standard conditions) $$ \mathrm{Br}^{-}, \quad \mathrm{Br}_{2}, \quad \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd} $$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s) ?\)

Consider the following half-reactions: \(\begin{aligned} \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & & \mathscr{E}^{\circ}=1.188 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{E}^{\circ}=0.755 \mathrm{V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} & \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=0.96 \mathrm{V} \end{aligned}\) 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.

Consider the following half-reactions: $$\begin{array}{ll} \operatorname{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \operatorname{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 \(M\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?
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