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Chapter 18: Problem 107

What reactions take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{NiBr}_{2} \quad\) b. molten \(\mathrm{AlF}_{3} \quad\) c. molten \(\mathrm{MgI}_{2}\)

Short Answer

Expert verified
For the electrolysis of molten compounds: a. \(NiBr_2\): Cathode (Reduction): \(Ni^{2+} + 2e^- \rightarrow Ni\) Anode (Oxidation): \(2Br^- \rightarrow Br_2 + 2e^-\) b. \(AlF_3\): Cathode (Reduction): \(Al^{3+} + 3e^- \rightarrow Al\) Anode (Oxidation): \(2F^- \rightarrow F_2 + 2e^-\) c. \(MgI_2\): Cathode (Reduction): \(Mg^{2+} + 2e^- \rightarrow Mg\) Anode (Oxidation): \(2I^- \rightarrow I_2 + 2e^-\)

Step by step solution

01

NiBr2

First, let's identify the ions in the compound: Cation: Ni2+ Anion: Br- Now, let's analyze the reactions that take place at the cathode and the anode during the electrolysis of molten NiBr2: Cathode (Reduction): The cation (Ni2+) will gain 2 electrons and be reduced to form metallic nickel. Ni2+ + 2e- → Ni Anode (Oxidation): The anion (Br-) will lose 1 electron and be oxidized to form bromine gas. Since there are two bromide ions for every nickel ion, this reaction occurs twice. 2Br- → Br2 + 2e-
02

AlF3

First, let's identify the ions in the compound: Cation: Al3+ Anion: F- Now, let's analyze the reactions that take place at the cathode and the anode during the electrolysis of molten AlF3: Cathode (Reduction): The cation (Al3+) will gain 3 electrons and be reduced to form metallic aluminum. Al3+ + 3e- → Al Anode (Oxidation): The anion (F-) will lose 1 electron and be oxidized to form fluorine gas. Since there are three fluoride ions for every aluminum ion, this reaction occurs 1.5 times (3/2). 2F- → F2 + 2e-
03

MgI2

First, let's identify the ions in the compound: Cation: Mg2+ Anion: I- Now, let's analyze the reactions that take place at the cathode and the anode during the electrolysis of molten MgI2: Cathode (Reduction): The cation (Mg2+) will gain 2 electrons and be reduced to form metallic magnesium. Mg2+ + 2e- → Mg Anode (Oxidation): The anion (I-) will lose 1 electron and be oxidized to form iodine gas. Since there are two iodide ions for every magnesium ion, this reaction occurs twice. 2I- → I2 + 2e-

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Most popular questions from this chapter

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111\()\) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) . The \(\mathscr{E}^{\circ}\) value for the following half-reaction is 0.446 \(\mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{E}_{\text { cell } \text { and }} \Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at $25^{\circ} \mathrm{C} )\( in which \)\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} M,$ what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is 0.504 \(\mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$.

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 consists of a silver metal electrode immersed in a solution with [Ag \(^{+} ]=1.0 M\) separated by a porous disk from a copper metal electrode. If the copper electrode is placed in a solution of 5.0$M \mathrm{NH}_{3}\( that is also 0.010 \)\mathrm{M}$ in \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\) $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

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?
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