/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 127 Predict the anode, cathode, and ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Predict the anode, cathode, and overall cell reactions when an aqueous solution of each of the following salts is electrolyzed in a cell having inert electrodes: (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) KI

Short Answer

Expert verified
Ag2SO4: Anode - \( \text{O}_2 \); Cathode - \( \text{Ag} \). Ca(OH)2: Anode - \( \text{O}_2 \); Cathode - \( \text{H}_2 \). KI: Anode - \( \text{I}_2 \); Cathode - \( \text{H}_2 \).

Step by step solution

01

Understanding the Concept

In an electrolytic cell, oxidation occurs at the anode and reduction occurs at the cathode. The reactions depend on the ions present in the solution and their electrode potentials.
02

Identify Ions for Ag2SO4

Decompose \( \text{Ag}_2\text{SO}_4 \) in water to get \( \text{Ag}^+ \) and \( \text{SO}_4^{2-} \). The water can also dissociate into \( \text{H}^+ \) and \( \text{OH}^- \).
03

Ag2SO4 - Determine Anode Reaction

Potential anode reactions include oxidation of \( \text{SO}_4^{2-} \) or \( \text{H}_2\text{O} \). Water oxidizes more easily, giving \( \text{O}_2 \) gas and \( \text{H}^+ \). The half-reaction: \( 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \).
04

Ag2SO4 - Determine Cathode Reaction

At the cathode, \( \text{Ag}^+ \) ions are reduced to form silver metal: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \).
05

Ag2SO4 - Overall Cell Reaction

Combine half-reactions: \( 2\text{Ag}^+ + 2\text{H}_2\text{O} \rightarrow 2\text{Ag} + \text{O}_2 + 4\text{H}^+ \).
06

Identify Ions for Ca(OH)2

Decompose \( \text{Ca(OH)}_2 \) to get \( \text{Ca}^{2+} \) and \( \text{OH}^- \), with water dissociating further into \( \text{H}^+ \) and \( \text{OH}^- \).
07

Ca(OH)2 - Determine Anode Reaction

Anode reactions include oxidation of \( \text{OH}^- \) to form \( \text{O}_2 \) gas: \( 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^- \).
08

Ca(OH)2 - Determine Cathode Reaction

Cathode reactions can include water being reduced: \( 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \).
09

Ca(OH)2 - Overall Cell Reaction

Combine half-reactions: \( 2\text{H}_2\text{O} \rightarrow \text{H}_2 + \text{O}_2 \).
10

Identify Ions for KI

Dissociate \( \text{KI} \) into \( \text{K}^+ \) and \( \text{I}^- \), with water dissociating into \( \text{H}^+ \) and \( \text{OH}^- \).
11

KI - Determine Anode Reaction

Anode reaction involves oxidation of \( \text{I}^- \) ions: \( 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \).
12

KI - Determine Cathode Reaction

At the cathode, water is reduced: \( 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \).
13

KI - Overall Cell Reaction

Combine half-reactions: \( 2\text{I}^- + 2\text{H}_2\text{O} \rightarrow \text{I}_2 + \text{H}_2 + 2\text{OH}^- \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Electrolytic Cell
An electrolytic cell is a fascinating device used to induce a chemical reaction through the application of electrical energy. It’s quite the opposite of a galvanic cell, where spontaneous reactions occur to produce electrical energy. In an electrolytic cell, electrical energy is used to drive non-spontaneous chemical reactions. This setup typically includes:
  • An external power source supplying electrical energy.
  • Two electrodes: an anode (positive electrode) and a cathode (negative electrode).
  • An electrolyte solution containing ions that participate in the reactions.
The main processes happening at the electrodes are:
  • Oxidation occurs at the anode, where loss of electrons takes place.
  • Reduction occurs at the cathode, where gain of electrons occurs.
Understanding these components helps in predicting the products of electrolysis for different electrolytes.
Oxidation and Reduction
Oxidation and reduction are central processes occurring in any electrochemical reaction. Combined, these reactions are referred to as redox reactions. Oxidation involves the loss of electrons. Remember "OIL RIG"—Oxidation Is Loss (of electrons). In an electrolytic cell, this process happens at the anode. For example, when electrolyzing water, the reaction at the anode is:\[ 2 ext{H}_2 ext{O} ightarrow ext{O}_2 + 4 ext{H}^+ + 4e^- \]Reduction involves the gain of electrons. Again, think "OIL RIG"—Reduction Is Gain (of electrons). This takes place at the cathode, such as:\[ ext{Ag}^+ + e^- ightarrow ext{Ag} \]In every electrochemical process, oxidation and reduction are inseparable—one cannot happen without the other. Together they allow for the transfer of electrons from one species to another, leading to the formation of new substances.
Half-Reaction
A half-reaction is a part of the overall electrochemical reaction that shows either oxidation or reduction separately. Each half-reaction includes all the species involved along with the electrons transferred. This concept is crucial for analyzing and balancing the reactions in electrolytic cells.When considering half-reactions, each one can be independently balanced in terms of both mass and charge. For example, in the electrolytic decomposition of water, the half-reaction for oxidation at the anode is:\[ 2 ext{H}_2 ext{O} ightarrow ext{O}_2 + 4 ext{H}^+ + 4e^- \]At the cathode, where reduction occurs, the half-reaction might be:\[ 2 ext{H}_2 ext{O} + 2e^- ightarrow ext{H}_2 + 2 ext{OH}^- \]Half-reactions help us understand which species undergo oxidation and which undergo reduction. By combining these half-reactions, we get the complete reaction taking place in the cell.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The silver oxide-zinc battery used in watches delivers a voltage of \(1.60 \mathrm{~V}\). Calculate the free-energy change (in kilojoules) for the cell reaction $$ \mathrm{Zn}(s)+\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{ZnO}(s)+2 \mathrm{Ag}(s) $$

What is the reduction potential at \(25^{\circ} \mathrm{C}\) for the hydrogen electrode in each of the following solutions? The half-reaction is $$ 2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g, 1 \mathrm{~atm}) $$ (a) \(1.0 \mathrm{M} \mathrm{HCl}\) (b) A solution having \(\mathrm{pH} 4.00\) (c) Pure water (d) \(1.0 \mathrm{M} \mathrm{NaOH}\)

A steam-hydrocarbon reforming process is one method for producing hydrogen from fossil fuels for use in a fuel cell. In the first step, steam reacts with hydrocarbons, such as \(\mathrm{CH}_{4}\), at high temperatures in the presence of a catalyst, yielding \(\mathrm{H}_{2}\) and CO. In the second step, the reaction of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\), called the water- gas shift reaction, removes toxic carbon monoxide and produces more hydrogen. Step \(1: \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CH}_{4}(g) \stackrel{1100^{\circ} \mathrm{C} \mathrm{N}_{1} \text { cathmst }}{\longrightarrow} \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(g)\) Step 2: \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{400^{\circ} \mathrm{C}, \text { cualyst }}{\longrightarrow} \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (a) In Step 1 , which element is oxidized and which is reduced? (b) In Step 2, which element is oxidized and which is reduced? What is the oxidizing agent and reducing agent? (c) What are drawbacks of steam reforming in the production of hydrogen?

A galvanic cell is constructed from a \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half- cell (anode) and a \(\mathrm{Cl}_{2} / \mathrm{Cl}^{-}\) half-cell (cathode). (a) Sketch the cell, indicating the direction of electron and ion flow. (b) Write balanced equations for the electrode and overall cell reactions. (c) Give the shorthand notation for the cell.

Consider a galvanic cell that uses the following half-reactions: $$ \begin{aligned} &2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) \\ &\mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s) \end{aligned} $$ (a) What materials are used for the electrodes? Identify the anode and cathode, and indicate the direction of electron and ion flow. (b) Write a balanced equation for the cell reaction, and calculate the standard cell potential. (c) Calculate the cell potential at \(25{ }^{\circ} \mathrm{C}\) if the ion concentrations are \(0.10 \mathrm{M}\) and the partial pressure of \(\mathrm{H}_{2}\) is \(10.0 \mathrm{~atm}\). (d) Calculate \(\Delta G^{\circ}\) (in kilojoules) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). (e) Calculate the mass change (in grams) of the aluminum electrode after the cell has supplied a constant current of \(10.0 \mathrm{~A}\) for \(25.0 \mathrm{~min}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.