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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鈦� and F鈦�. 1. Reduction at Cathode (gain of electrons): K鈦� + e鈦� 鈫� K 2. Oxidation at Anode (loss of electrons): 2 F鈦� 鈫� F鈧� + 2 e鈦�
02

Reaction of molten CuCl鈧�

For the electrolysis of molten CuCl鈧�, we have the following ions: Cu虏鈦� and 2 Cl鈦�. 1. Reduction at Cathode (gain of electrons): Cu虏鈦� + 2 e鈦� 鈫� Cu 2. Oxidation at Anode (loss of electrons): 2 Cl鈦� 鈫� Cl鈧� + 2 e鈦�
03

Reaction of molten MgI鈧�

For the electrolysis of molten MgI鈧�, we have the following ions: Mg虏鈦� and 2 I鈦�. 1. Reduction at Cathode (gain of electrons): Mg虏鈦� + 2 e鈦� 鈫� Mg 2. Oxidation at Anode (loss of electrons): 2 I鈦� 鈫� I鈧� + 2 e鈦�

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Key Concepts

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

Cathode Reaction
In electrolysis, the cathode reaction is where reduction occurs. This means that electrons are gained by ions present in the solution or molten compound. At the cathode, cations (positively charged ions) are attracted because the cathode is negatively charged. When these ions reach the cathode, they gain electrons, effectively reducing their charge to become neutral atoms.
For example, during the electrolysis of molten KF, the potassium ions ( K^+) move to the cathode:
  • They gain electrons (reduction) and form potassium: K^+ + e^- 鈫� K
This process is consistent across different compounds in molten form. For molten CuCl鈧�, copper ions ( Cu^{2+}) reduce to copper metal, and for molten MgI鈧�, magnesium ions ( Mg^{2+}) reduce to magnesium metal.
Anode Reaction
At the anode, oxidation occurs. This is the process where ions lose electrons. The anode is positively charged, so it attracts anions, which are negatively charged ions.
When anions reach the anode, they undergo oxidation. This means they lose electrons and convert into neutral atoms or molecules.
Take the example of molten KF electrolysis:
  • The fluoride ions ( F^-) lose electrons forming fluorine gas: 2 F^- 鈫� F鈧� + 2 e鈦�
Similarly, in molten CuCl鈧�, chloride ions ( Cl^-) lose electrons to form chlorine gas, and in molten MgI鈧�, iodide ions ( I^-) undergo oxidation to form iodine gas. The anode reaction is crucial in releasing non-metal elements from their ionic state during electrolysis.
Oxidation
Oxidation is a core concept in electrolysis where substances lose electrons. This process most commonly occurs at the anode, where negatively charged ions (anions) are drawn. An easy way to remember this is through the phrase "OIL RIG," which stands for "Oxidation Is Losing (electrons)."
Oxidation results in the formation of elemental substances from their ionic counterparts.
Examples from the exercises include:
  • Molten KF: F^- ions lose electrons and form F鈧� gas at the anode.
  • Molten CuCl鈧�: Cl^- ions lose electrons to produce Cl鈧� gas.
  • Molten MgI鈧�: I^- ions lose electrons, becoming I鈧� gas.
Understanding oxidation helps explain how atoms are transformed and separated during electrolysis, enabling the retrieval of elemental forms from compounds.
Reduction
Reduction is the half-reaction that takes place at the cathode in electrolysis. It involves the gain of electrons by ions, converting them into neutral atoms or simpler molecules. This concept aligns with the "OIL RIG" mnemonic: "Reduction Is Gaining (electrons)."
The reduction process is essential for producing pure elements from ionic compounds.
  • In molten KF electrolysis, K^+ ions gain electrons to form solid potassium.
  • For molten CuCl鈧�, Cu^{2+} ions receive electrons, resulting in the formation of solid copper.
  • In the process of molten MgI鈧�, Mg^{2+} ions are reduced to create metallic magnesium.
Reduction effectively reverses the ionization of elements, making it possible to extract metals in their elemental state. Understanding reduction helps illuminate why electrolysis is a useful method for refining metals and other substances.

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Most popular questions from this chapter

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{l} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3} . \mathrm{Ox}-\) ide ions can move through this solid at high temperatures (about \(\left.800^{\circ} \mathrm{C}\right) . \Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is \(-380 \mathrm{~kJ}\). Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is \(\begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow & \mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned}\) Balance this equation.

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

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 \(1.0 \mathrm{~atm}\).
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