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Chapter 18: 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
For the electrolysis of: a. molten KF: Cathode (reduction): \(K^+ + e^- \rightarrow K\) Anode (oxidation): \(2 F^- \rightarrow F_2 + 2 e^-\) b. molten \(CuCl_2\): Cathode (reduction): \(Cu^{2+} + 2 e^- \rightarrow Cu\) Anode (oxidation): \(2 Cl^- \rightarrow Cl_2 + 2 e^-\) c. molten \(MgI_2\): Cathode (reduction): \(Mg^{2+} + 2 e^- \rightarrow Mg\) Anode (oxidation): \(2 I^- \rightarrow I_2 + 2 e^-\)

Step by step solution

01

Reaction of molten KF

For the electrolysis of molten KF, we have the following ions: K⁺ and F⁻. 1. Reduction at Cathode (gain of electrons): K⁺ + e⁻ → K 2. Oxidation at Anode (loss of electrons): 2 F⁻ → F₂ + 2 e⁻
02

Reaction of molten CuCl₂

For the electrolysis of molten CuCl₂, we have the following ions: Cu²⁺ and 2 Cl⁻. 1. Reduction at Cathode (gain of electrons): Cu²⁺ + 2 e⁻ → Cu 2. Oxidation at Anode (loss of electrons): 2 Cl⁻ → Cl₂ + 2 e⁻
03

Reaction of molten MgI₂

For the electrolysis of molten MgI₂, we have the following ions: Mg²⁺ and 2 I⁻. 1. Reduction at Cathode (gain of electrons): Mg²⁺ + 2 e⁻ → Mg 2. Oxidation at Anode (loss of electrons): 2 I⁻ → I₂ + 2 e⁻

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Most popular questions from this chapter

The Ostwald process for the commercial production of nitric acid involves the following three steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Which reactions in the Ostwald process are oxidationreduction reactions? b. Identify each oxidizing agent and reducing agent.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) 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 \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.957 \mathrm{~V}\) \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.775 \mathrm{~V}\) a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations that are not already balanced. Standard reduction potentials are found in Table \(18.1 .\) a. \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}^{-}(a q)\) b. \(\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)\)

An electrochemical cell is set up using the following unbalanced reaction: $$ \mathrm{M}^{a+}(a q)+\mathrm{N}(s) \longrightarrow \mathrm{N}^{2+}(a q)+\mathrm{M}(s) $$ The standard reduction potentials are: $$ \begin{array}{ll} \mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=0.400 \mathrm{~V} \\ \mathrm{~N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} & \mathscr{E}^{\circ}=0.240 \mathrm{~V} \end{array} $$ The cell contains \(0.10 \mathrm{M} \mathrm{N}^{2+}\) and produces a voltage of \(0.180 \mathrm{~V}\). If the concentration of \(\mathrm{M}^{a+}\) is such that the value of the reaction quotient \(Q\) is \(9.32 \times 10^{-3}\), calculate \(\left[\mathrm{M}^{a+}\right]\). Calculate \(w_{\max }\) for this electrochemical cell.

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).
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