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Chapter 18: Problem 127

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

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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.

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Most popular questions from this chapter

What is the function of a salt bridge in a galvanic cell?

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 $$ \mathrm{Zn}(s)+\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{ZnO}(s)+2 \mathrm{Ag}(s) $$

Write balanced net ionic equations for the following reactions in acidic solution: (a) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{O}_{2}(g)+\mathrm{Cr}^{3+}(a q)\) (c) \(\mathrm{Sn}^{2+}(a q)+\mathrm{IO}_{4}^{-}(a q) \rightarrow \mathrm{Sn}^{4+}(a q)+\mathrm{I}^{-}(a q)\) (d) \(\mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{PbCl}_{2}(s)+\mathrm{O}_{2}(g)\)

Write balanced net ionic equations for the following reactions in basic solution: (a) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+\mathrm{I}^{-}(a q)\) (b) \(\mathrm{Mn}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{NH}_{3}(a q)+\mathrm{Zn}(\mathrm{OH})_{4}^{2-}(a q)\) (d) \(\mathrm{Bi}(\mathrm{OH})_{3}(s)+\mathrm{Sn}(\mathrm{OH})_{3}^{-}(a q) \rightarrow \mathrm{Bi}(s)+\mathrm{Sn}(\mathrm{OH})_{6}^{2-}(a q)\)

A galvanic cell has a silver electrode in contact with \(0.050 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) and a copper electrode in contact with \(1.0 \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) (a) Write a balanced equation for the cell reaction, and calculate the cell potential at \(25^{\circ} \mathrm{C}\). (b) Excess \(\mathrm{NaBr}(a q)\) is added to the \(\mathrm{AgNO}_{3}\) solution to precipitate AgBr. What is the cell potential at \(25^{\circ} \mathrm{C}\) after the precipitation of \(\mathrm{AgBr}\) if the concentration of excess \(\mathrm{Br}^{-}\) is \(1.0 \mathrm{M}\) ? Write a balanced equation for the cell reaction under these conditions. \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgBr}\) at \(25^{\circ} \mathrm{C}\) is \(5.4 \times 10^{-13}\).) (c) Use the result in part (b) to calculate the standard reduction potential \(E^{\circ}\) for the half-reaction $$ \mathrm{AgBr}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) $$
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