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Chapter 21: Problem 120

Commercial electrolysis is performed on both molten \(\mathrm{NaCl}\) and aqueous \(\mathrm{NaCl}\) solutions. Identify the anode product, cathode product, species reduced, and species oxidized for the electrolysis of (a) molten \(\mathrm{NaCl}\) and (b) an aqueous solution of \(\mathrm{NaCl}\).

Short Answer

Expert verified
For molten NaCl: Cathode product Na, Anode product Cl2, Reduced species Na+, Oxidized species Cl-. For aqueous NaCl: Cathode product H2, Anode product Cl2, Reduced species H+, Oxidized species Cl-.

Step by step solution

01

Title - Identify Products for Molten NaCl

For molten \(\text{NaCl}\), the substance is in a liquid state and only contains \(\text{Na}^+\) and \(\text{Cl}^-\). No water is involved. At the cathode, the \(\text{Na}^+\) ions will be reduced, and at the anode, the \(\text{Cl}^-\) ions will be oxidized. Thus, the cathode product is sodium metal (Na), and the anode product is chlorine gas (Cl₂).
02

Title - Identify Oxidation and Reduction for Molten NaCl

In the molten \(\text{NaCl}\) electrolysis, the \(\text{Na}^+\) ions gain electrons (reduction) to form sodium metal, and the \(\text{Cl}^-\) ions lose electrons (oxidation) to form chlorine gas. Therefore, the species reduced is \(\text{Na}^+\) and the species oxidized is \(\text{Cl}^-\).
03

Title - Identify Products for Aqueous NaCl Solution

In an aqueous \(\text{NaCl}\) solution, both water and \(\text{NaCl}\) are present. At the cathode, hydrogen ions \(\text{H}^+\) (from water) are more easily reduced than \(\text{Na}^+\), resulting in the production of hydrogen gas \(\text{H}_2\). At the anode, \(\text{Cl}^-\) ions are oxidized to produce chlorine gas \(\text{Cl}_2\). Thus, the cathode product is hydrogen gas (H₂), and the anode product is chlorine gas (Cl₂).
04

Title - Identify Oxidation and Reduction for Aqueous NaCl Solution

In the aqueous \(\text{NaCl}\) electrolysis, hydrogen ions \(\text{H}^+\) are reduced to form hydrogen gas \(\text{H}_2\), and chloride ions \(\text{Cl}^-\) are oxidized to form chlorine gas \(\text{Cl}_2\). Therefore, the species reduced is \(\text{H}^+\), and the species oxidized is \(\text{Cl}^-\).

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Key Concepts

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

Molten NaCl Electrolysis
Molten NaCl (sodium chloride) electrolysis occurs when NaCl is in a liquid state, without any water involved. This process separates sodium and chlorine ions present in the salt. At the cathode (negative electrode), sodium ions (\(\text{Na}^+\)) gain electrons, which is a reduction process. This forms sodium metal (Na).
At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) lose electrons, an oxidation process, resulting in chlorine gas (\(\text{Cl}_2\)).
Therefore, the products of molten NaCl electrolysis are:
  • Sodium metal (Na) at the cathode
  • Chlorine gas (\(\text{Cl}_2\)) at the anode
Aqueous NaCl Electrolysis
In aqueous NaCl (saltwater) electrolysis, both water and NaCl are present. Here, things work a bit differently due to the involvement of water. At the cathode, hydrogen ions (\(\text{H}^+\)) from water are more easily reduced than sodium ions (\(\text{Na}^+\)). This means that hydrogen gas (\(\text{H}_2\)) is produced instead of sodium.
At the anode, chloride ions (\(\text{Cl}^-\)) still get oxidized to produce chlorine gas (\(\text{Cl}_2\)).
The final products are:
  • At the cathode, hydrogen gas (\(\text{H}_2\))
  • At the anode, chlorine gas (\(\text{Cl}_2\))
Oxidation and Reduction Processes
In electrolysis, reduction and oxidation occur simultaneously at different electrodes. For molten NaCl:
  • Reduction (gain of electrons) occurs at the cathode: \(\text{Na}^+ + e^- \rightarrow \text{Na}\)
  • Oxidation (loss of electrons) occurs at the anode: \(\text{2Cl}^- \rightarrow \text{Cl}_2 + 2e^-\)
In an aqueous NaCl solution:
  • Reduction at the cathode involves hydrogen ions from water: \(\text{2H}^+ + 2e^- \rightarrow \text{H}_2\)
  • Oxidation at the anode still involves chloride ions: \(\text{2Cl}^- \rightarrow \text{Cl}_2 + 2e^-\)
Electrolysis Products
The products of electrolysis depend on the substances involved and the conditions of the process. For molten NaCl, the products are sodium metal and chlorine gas. In aqueous NaCl, water changes the outcome, leading to different products - hydrogen gas at the cathode and chlorine gas at the anode.
It's essential to consider the environment (molten vs. aqueous) because this impacts which ions get reduced or oxidized, altering the final products.
Chemical Reactions in Electrolysis
The chemical reactions occurring during electrolysis are crucial to understand the process. For molten NaCl:
  • At the cathode: \(\text{Na}^+ + e^- \rightarrow \text{Na}\)
  • At the anode: \(\text{2Cl}^- \rightarrow \text{Cl}_2 + 2e^-\)
For aqueous NaCl:
  • At the cathode: \(\text{2H}^+ + 2e^- \rightarrow \text{H}_2\)
  • At the anode: \(\text{2Cl}^- \rightarrow \text{Cl}_2 + 2e^-\)
Understanding these reactions helps us know why different products form under different conditions, allowing better control and use of electrolysis in various applications.

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Most popular questions from this chapter

A voltaic cell consists of an \(\mathrm{Mn} / \mathrm{Mn}^{2+}\) half-cell and a \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. Calculate \(\left[\mathrm{Pb}^{2+}\right]\) when \(\left[\mathrm{Mn}^{2+}\right]\) is \(1.4 M\) and \(E_{\text {cell }}\) is \(0.44 \mathrm{~V}\).

Bubbles of \(\mathrm{H}_{2}\) form when metal \(\mathrm{D}\) is placed in hot \(\mathrm{H}_{2} \mathrm{O}\). No reaction occurs when \(\mathrm{D}\) is placed in a solution of a salt of metal E, but D is discolored and coated immediately when placed in a solution of a salt of metal \(\mathrm{F}\). What happens if \(\mathrm{E}\) is placed in a solution of a salt of metal F? Rank metals \(D, \mathrm{E},\) and \(\mathrm{F}\) in order of increasing reducing strength.

The \(\mathrm{MnO}_{2}\) used in alkaline batteries can be produced by an electrochemical process of which one half-reaction is $$\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}$$ If a current of \(25.0 \mathrm{~A}\) is used, how many hours are needed to produce \(1.00 \mathrm{~kg}\) of \(\mathrm{MnO}_{2} ?\) At which electrode is the \(\mathrm{MnO}_{2}\) formed?

21.115 A thin circular-disk earring \(4.00 \mathrm{~cm}\) in diameter is plated with a coating of gold \(0.25 \mathrm{~mm}\) thick from an \(\mathrm{Au}^{3+}\) bath. (a) How many days does it take to deposit the gold on one side of one earring if the current is \(0.013 \mathrm{~A}\left(d\right.\) of gold \(\left.=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\) (b) How many days does it take to deposit the gold on both sides of the pair of earrings? (c) If the price of gold is \(\$ 1595\) per troy ounce \((31.10 \mathrm{~g}),\) what is the total cost of the gold plating?

Aqua regia, a mixture of concentrated \(\mathrm{HNO}_{3}\) and \(\mathrm{HCl}\) was developed by alchemists as a means to "dissolve" gold. The process is a redox reaction with this simplified skeleton reaction: $$\mathrm{Au}(s)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AuCl}_{4}^{-}(a q)+\mathrm{NO}_{2}(g)$$ (a) Balance the reaction by the half-reaction method. (b) What are the oxidizing and reducing agents? (c) What is the function of \(\mathrm{HCl}\) in aqua regia?
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