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Chapter 18: Problem 27

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 during the electrolysis of a molten salt, you need to know the chemical formula of the salt, the amount of electric charge passed through the cell, Faraday's constant, and the atomic mass of the metal ion. Determine the metal ion and its charge from the salt's formula, calculate moles of electrons using the charge and Faraday's constant, determine moles of metal deposited by dividing moles of electrons by the stoichiometric coefficient (equal to the charge of metal ion), and finally, calculate the mass of the metal by multiplying moles of metal ions by the molar mass of the metal.

Step by step solution

01

Determine the metal ion and its charge in the molten salt

First, you need to identify the metal ion present in the molten salt, as well as its oxidation state (charge). Look at the chemical formula of the salt and identify the positively charged metal ion.
02

Calculate the moles of metal ions reduced during the electrolysis

Next, use the amount of electric charge (Q) passed through the electrolytic cell in coulombs, and the Faraday's constant (F) to find the moles of electrons exchanged during the electrolysis. The relationship between the charge and the moles of electrons is given by: Moles of electrons = \(\frac{Q}{F}\)
03

Determine the moles of metal deposited

To calculate the moles of metal deposited, divide the moles of electrons obtained in step 2 by the stoichiometric coefficient of the electrons in the balanced half-reaction of the metal ion being reduced. The stoichiometric coefficient is equal to the charge on the metal ion.
04

Calculate the mass of metal plated out

Finally, multiply the moles of metal ions determined in step 3 by the molar mass (atomic mass) of the metal to find the mass of the metal plated out during the electrolysis: Mass of metal = moles of metal 脳 molar mass of metal

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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?

The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing \(\mathrm{Mn}^{2+}\) ions. These ions are then oxidized to the deeply colored \(\mathrm{MnO}_{4}^{-}\) ions by periodate ion \(\left(\mathrm{IO}_{4}^{-}\right)\) in acid solution. a. Complete and balance an equation describing each of the above reactions. b. Calculate \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each reaction.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) 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 \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

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 \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)
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