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Chapter 18: Problem 106

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 \mathrm{M}\) KF solution b. \(1.0 \mathrm{M} \mathrm{CuCl}_{2}\) solution c. \(1.0 \mathrm{M} \mathrm{MgI}_{2}\) solution

Short Answer

Expert verified
For the electrolysis of different solutions, the following reactions occur at the cathode and anode: a) 1.0 M KF solution: Cathode: \( K^{+} + e^{-} \rightarrow K \) Anode: \( 2F^{-} \rightarrow F_{2} + 2e^{-} \) b) 1.0 M CuCl鈧� solution: Cathode: \( 2H^{+} + 2e^{-} \rightarrow H_{2} \) Anode: \( 2Cl^{-} \rightarrow Cl_{2} + 2e^{-} \) c) 1.0 M MgI鈧� solution: Cathode: \( Mg^{2+} + 2e^{-} \rightarrow Mg \) Anode: \( 2I^{-} \rightarrow I_{2} + 2e^{-} \)

Step by step solution

01

Solution a: 1.0 M KF

Step 1: Identify ions in solution In the 1.0 M KF solution, we have the following ions: K+ and F-. Step 2: Determine half-reactions at the cathode (reduction) and anode (oxidation) Since potassium (K) is higher on the activity series of metals than hydrogen (H), reduction of K+ ions to K will take place at the cathode: Cathode reaction: \( K^{+} + e^{-} \rightarrow K \) At the anode, oxidation of F- ions to form F2 gas will occur: Anode reaction: \( 2F^{-} \rightarrow F_{2} + 2e^{-} \)
02

Solution b: 1.0 M CuCl2

Step 1: Identify ions in solution In the 1.0 M CuCl2 solution, we have the following ions: Cu2+ and 2Cl-. Step 2: Determine half-reactions at the cathode (reduction) and anode (oxidation) Copper (Cu) is lower on the activity series of metals than hydrogen (H), so reduction of H+ ions to H2 gas will take place at the cathode instead of Cu2+ ions to Cu: Cathode reaction: \( 2H^{+} + 2e^{-} \rightarrow H_{2} \) At the anode, oxidation of Cl- ions to form Cl2 gas will occur: Anode reaction: \( 2Cl^{-} \rightarrow Cl_{2} + 2e^{-} \)
03

Solution c: 1.0 M MgI2

Step 1: Identify ions in solution In the 1.0 M MgI2 solution, we have the following ions: Mg2+ and 2I-. Step 2: Determine half-reactions at the cathode (reduction) and anode (oxidation) Magnesium (Mg) is higher on the activity series of metals than hydrogen (H), so reduction of Mg2+ ions to Mg will take place at the cathode: Cathode reaction: \( Mg^{2+} + 2e^{-} \rightarrow Mg \) At the anode, oxidation of I- ions to form I2 molecules will happen (instead of O2 gas since I- is more easily oxidized than water): Anode reaction: \( 2I^{-} \rightarrow I_{2} + 2e^{-} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cathode and Anode Reactions
Electrolysis involves two key half-reactions at the electrodes: reduction at the cathode and oxidation at the anode. During electrolysis of a substance like potassium fluoride (KF), positive potassium ions (K+) are attracted to the cathode, where they gain electrons in a reduction process:
Cathode reaction: \( K^{+} + e^{-} \rightarrow K \).
At the anode, fluoride ions (F-) give up electrons through oxidation, producing fluorine gas:
Anode reaction: \( 2F^{-} \rightarrow F_{2} + 2e^{-} \).
In solutions where metal ions are less reactive than hydrogen ions, like copper(II) chloride (CuCl2), hydrogen will be reduced at the cathode instead of the metal ion.
Activity Series of Metals
The activity series of metals is a chart that ranks metals by their reactivity in displacement reactions and, by extension, in electrolysis. More active metals, such as potassium (K) and magnesium (Mg), are more likely to be reduced at the cathode compared to hydrogen (H), which is intermediate in the series. Conversely, less active metals, like copper (Cu), will not reduce in the presence of H+ ions, leading to the reduction of H+ rather than the metal ion itself.
Half-Reactions in Electrolysis
Understanding half-reactions is crucial for predicting the outcomes of electrolysis. A half-reaction is either the reduction or oxidation process that occurs at an electrode. Reduction half-reactions always occur at the cathode, as electrons are gained by the chemical species. Oxidation half-reactions happen at the anode, where electrons are lost. For example, the reduction of magnesium ions to magnesium metal (Mg2+ to Mg) and the oxidation of iodide ions to iodine (I- to I2) are half-reactions in the electrolysis of magnesium iodide (MgI2).
Standard Electrode Potentials
Standard electrode potentials are essential for predicting the direction of half-reactions in electrolysis. They are a measure of a half-cell鈥檚 potential under standard conditions, with higher potentials indicating greater tendency to be reduced. In a cell, the difference in standard electrode potential between two half-cells will drive the flow of electrons from the anode to the cathode. For instance, because Cu has a higher standard electrode potential than H2, copper ions in CuCl2 solution will be less likely to reduce than hydrogen ions from water present in the solution.

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Most popular questions from this chapter

A copper penny can be dissolved in nitric acid but not in hydrochloric acid. Using reduction potentials from the book, show why this is so. What are the products of the reaction? Newer pennies contain a mixture of zinc and copper. What happens to the zinc in the penny when the coin is placed in nitric acid? Hydrochloric acid? Support your explanations with data from the book, and include balanced equations for all reactions.

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of \(2.50 \mathrm{~A}\) in \(15.0 \mathrm{~min} ?\)

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) b. \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\)

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

Consider the following half-reactions: $$\begin{array}{cc}\mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 \mathrm{M}\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)
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