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Chapter 17: Problem 110

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten KF b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)

Short Answer

Expert verified
a. In molten KF, potassium (K) metal is produced at the cathode, and fluorine (F₂) gas is produced at the anode. b. During the electrolysis of molten CuCl₂, copper (Cu) metal is produced at the cathode, and chlorine (Cl₂) gas is produced at the anode. c. During the electrolysis of molten MgI₂, magnesium (Mg) metal is produced at the cathode, and iodine (I₂) gas is produced at the anode.

Step by step solution

01

a. Molten KF

In molten KF, there are potassium (K⁺) and fluoride (F⁻) ions. At the cathode (reduction): K⁺ + e⁻ -> K At the anode (oxidation): 2F⁻ -> F₂ + 2e⁻ Therefore, during the electrolysis of molten KF, potassium metal is produced at the cathode, and fluorine gas is produced at the anode.
02

b. Molten CuCl₂

In molten CuCl₂, there are copper (Cu²⁺) and chloride (Cl⁻) ions. At the cathode (reduction): Cu²⁺ + 2e⁻ -> Cu At the anode (oxidation): 2Cl⁻ -> Cl₂ + 2e⁻ During the electrolysis of molten CuCl₂, copper metal is produced at the cathode, and chlorine gas is produced at the anode.
03

c. Molten MgI₂

In molten MgI₂, there are magnesium (Mg²⁺) and iodide (I⁻) ions. At the cathode (reduction): Mg²⁺ + 2e⁻ -> Mg At the anode (oxidation): 2I⁻ -> I₂ + 2e⁻ During the electrolysis of molten MgI₂, magnesium metal is produced at the cathode, and iodine gas is produced at the anode.

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Most popular questions from this chapter

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table \(17-1\) a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block. Why?

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.
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