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Chapter 18: Problem 97

An unknown metal \(\mathrm{M}\) is electrolyzed. It took \(74.1 \mathrm{~s}\) for a current of \(2.00 \mathrm{~A}\) to plate out \(0.107 \mathrm{~g}\) of the metal from a solution containing \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{3} .\) Identify the metal.

Short Answer

Expert verified
The charge passed through the electrolyte is calculated using \(Q = I \times t\), where Q is the charge, I is the current, and t is the time. The charge is found to be \(148.2 C\). Using Faraday's law, \(Q = nF\), the moles of electrons transferred are found to be approximately \(0.001536 \ \text{mol}\). Since the metal ion is trivalent, moles of the unknown metal are calculated as \(0.000512 \ \text{mol}\). The molar mass of the metal is calculated using \(\frac{\text{Mass of M}}{\text{Moles of M}} = 209 \ \frac{\text{g}}{\text{mol}}\). The metal is identified as Thallium (Tl) based on its molar mass from the periodic table.

Step by step solution

01

1. Calculate the charge passed through the electrolyte

To calculate the charge, we'll use the given current of \(2.00 A\) and the given time of \(74.1 s\). We can find the charge (Q) using the formula: \[Q = I \times t\] where Q is the charge, I is the current, and t is the time. Using the given values, we have: \[Q = (2.00 A) (74.1 s) = 148.2 C\] So, the charge passed through the electrolyte is \(148.2 C\).
02

2. Determine the moles of electrons transferred

To find the moles of electrons transferred from the unknown metal ions in the electrolyte to the metal, we will use Faraday's law: \[Q = nF\] where Q is the charge, n is the number of mole of electrons, and F is the Faraday constant, which is approximately 96485 C/mol. We can rearrange the equation to solve for n: \[n = \frac{Q}{F}\] Now, we substitute the given values: \[n = \frac{148.2 C}{96485 C/mol} \approx 0.001536 \ \text{mol}\] So, \(0.001536 \ \text{mol}\) of electrons were transferred.
03

3. Calculate the molar mass of the metal and identify it

Since we are given that the unknown metal comes from a compound with the formula \(\mathrm{M(NO_{3})_{3}}\), it means that the metal ion is trivalent, which means that 3 moles of electrons are needed for each mole of the metal to form the metal from the ion. So, we can find the moles of the unknown metal M formed from the given electrons as: Moles of M = \(\frac{\text{Moles of electrons transferred}}{3} = \frac{0.001536 \ \text{mol}}{3} = 0.000512 \ \text{mol}\) Now, we will calculate the molar mass by dividing the mass of the metal plated out (0.107 g) by the moles of the metal: Molar mass of M = \(\frac{\text{Mass of M}}{\text{Moles of M}} = \frac{0.107 \ \text{g}}{0.000512 \ \text{mol}} = 209 \ \frac{\text{g}}{\text{mol}}\) Now, since the molar mass of M is approximately 209 g/mol, we can identify the metal as Thallium (Tl) based on its molar mass from the periodic table.

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Most popular questions from this chapter

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

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ} &=0.34 \mathrm{~V} \\ \mathrm{~V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ} &=-1.20 \mathrm{~V} \end{aligned} $$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M\), and the vanadium compartment contains a vanadium electrode and \(\mathrm{V}^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{~L}\) of solution) was titrated with \(0.0800 M \mathrm{H}_{2} \mathrm{EDTA}^{2-}\), resulting in the reaction $$ \begin{array}{r} \mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \\ K=? \end{array} $$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{~mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{~V}\). The solution was buffered at a pH of \(10.00\). a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K\), for the titration reaction. c. Calculate \(\mathscr{B}_{\text {cell }}\) at the halfway point in the titration.

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume \(\mathrm{Al}\) is oxidized.)

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{8}^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{C}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\)
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