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Chapter 18: Problem 31

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

Short Answer

Expert verified
To calculate the mass of metal plated out in the electrolytic cell, follow these steps: 1. Determine the chemical formula and molar mass of the molten salt of the metal. 2. Calculate the moles of electrons involved in the electrolysis process using the formula: Moles of electrons = Charge (in coulombs) / Faraday constant (96,485 C/mol) 3. Write the balanced half-reaction equation for the electrolysis of the molten salt. 4. Determine the moles of metal involved using the stoichiometry of the balanced half-reaction: Moles of metal = Moles of electrons * (Stoichiometric ratio of metal ions to electrons) 5. Calculate the mass of the plated metal by multiplying the moles of metal by its molar mass: Mass of plated metal = Moles of metal * Molar mass of metal (in g/mol)

Step by step solution

01

Identify the molten salt

Determine the chemical formula and molar mass of the molten salt of the metal. The molten salt will be provided in the problem.
02

Calculate the moles of electrons

The Faraday constant (F) relates the charge (in coulombs) and the number of moles of electrons in an electrochemical process. F = 96,485 C/mol. Determine how many moles of electrons are involved in the electrolysis process by dividing the amount of charge passed through the system (in coulombs) by the Faraday constant. Moles of electrons = Charge (in coulombs) / Faraday constant (96,485 C/mol)
03

Determine the balanced half-reaction

Write the balanced half-reaction equation for the electrolysis of the molten salt. This equation will help you understand the stoichiometry of the metal and electrons in the electrochemical process.
04

Calculate moles of metal involved

Using the half-reaction equation and the stoichiometry, determine the number of moles of metal involved in the electrolysis process. Pay attention to the ratio between the moles of electrons and the moles of metal ions. Moles of metal = Moles of electrons * (Stoichiometric ratio of metal ions to electrons)
05

Calculate mass of the plated metal

Now that you have the moles of the metal involved in the electrolysis process and the molar mass of the molten salt, you can calculate the mass of the plated metal. Multiply the number of moles of metal by its molar mass to find the mass (in grams). Mass of plated metal = Moles of metal * Molar mass of metal (in g/mol)

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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 \(\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \qquad \mathscr{E}^{\circ}=1.10 \mathrm{V}\) For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

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Balance the following oxidation鈥搑eduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)\) c. \(\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)\)

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}} & {\mathscr{E}^{\circ}=-0.76 \mathrm{V}} \\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {\mathscr{E}^{\circ}=-0.44 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text { cell }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text { cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 M\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} .\)

A galvanic cell is based on the following half-reactions: $$\begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{aligned}$$ In this cell, the silver compartment contains a silver electrode and excess AgCl(s) \(\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) . a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) . b. Assuming 1.0 \(\mathrm{L}\) of 2.0\(M \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of 0.52 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3} )\) . $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K=1.0 \times 10^{13}$$
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