Electrochemistry

Galvanic Cells

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Learning Objectives

By the end of this section, you will be able to:

  • Use cell notation to describe galvanic cells
  • Describe the basic components of galvanic cells

Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.

Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate ([link]). As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. The reaction may be split into its two half-reactions. Half-reactions separate the oxidation from the reduction, so each can be considered individually.

\(\begin{array}{}\\ \\ \underset{¯}{\begin{array}{l}\text{oxidation:}\phantom{\rule{5.7em}{0ex}}\text{Cu}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}\left(aq\right)+{\text{2e}}^{\text{−}}\\ \text{reduction:}\phantom{\rule{0.2em}{0ex}}\text{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({\text{Ag}}^{\text{+}}\left(aq\right)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ag}\left(s\right)\right)\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}2{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Ag}\left(s\right)\end{array}}\\ \text{overall:}\phantom{\rule{1.5em}{0ex}}2{\text{Ag}}^{\text{+}}\left(aq\right)+\text{Cu}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2Ag}\left(s\right)+{\text{Cu}}^{2+}\left(aq\right)\end{array}\)

The equation for the reduction half-reaction had to be doubled so the number electrons “gained” in the reduction half-reaction equaled the number of electrons “lost” in the oxidation half-reaction.

When a clean piece of copper metal is placed into a clear solution of silver nitrate (a), an oxidation-reduction reaction occurs that results in the exchange of Cu2+ for Ag+ ions in solution. As the reaction proceeds (b), the solution turns blue (c) because of the copper ions present, and silver metal is deposited on the copper strip as the silver ions are removed from solution. (credit: modification of work by Mark Ott)

This figure includes three photographs. In a, a test tube containing a clear, colorless liquid is shown with a loosely coiled copper wire outside the test tube to its right. In b, the wire has been submerged into the clear colorless liquid in the test tube and the surface of the wire is darkened. In c, the liquid in the test tube is a bright blue-green color, the wire in the solution appears dark near the top, and a grey “fuzzy” material is present at the bottom of the test tube on the lower portion of the copper coil, giving a murky appearance to the liquid near the bottom of the test tube.

Galvanic or voltaic cells involve spontaneous electrochemical reactions in which the half-reactions are separated ([link]) so that current can flow through an external wire. The beaker on the left side of the figure is called a half-cell, and contains a 1 M solution of copper(II) nitrate [Cu(NO3)2] with a piece of copper metal partially submerged in the solution. The copper metal is an electrode. The copper is undergoing oxidation; therefore, the copper electrode is the anode. The anode is connected to a voltmeter with a wire and the other terminal of the voltmeter is connected to a silver electrode by a wire. The silver is undergoing reduction; therefore, the silver electrode is the cathode. The half-cell on the right side of the figure consists of the silver electrode in a 1 M solution of silver nitrate (AgNO3). At this point, no current flows—that is, no significant movement of electrons through the wire occurs because the circuit is open. The circuit is closed using a salt bridge, which transmits the current with moving ions. The salt bridge consists of a concentrated, nonreactive, electrolyte solution such as the sodium nitrate (NaNO3) solution used in this example. As electrons flow from left to right through the electrode and wire, nitrate ions (anions) pass through the porous plug on the left into the copper(II) nitrate solution. This keeps the beaker on the left electrically neutral by neutralizing the charge on the copper(II) ions that are produced in the solution as the copper metal is oxidized. At the same time, the nitrate ions are moving to the left, sodium ions (cations) move to the right, through the porous plug, and into the silver nitrate solution on the right. These added cations “replace” the silver ions that are removed from the solution as they were reduced to silver metal, keeping the beaker on the right electrically neutral. Without the salt bridge, the compartments would not remain electrically neutral and no significant current would flow. However, if the two compartments are in direct contact, a salt bridge is not necessary. The instant the circuit is completed, the voltmeter reads +0.46 V, this is called the cell potential. The cell potential is created when the two dissimilar metals are connected, and is a measure of the energy per unit charge available from the oxidation-reduction reaction. The volt is the derived SI unit for electrical potential

\(\text{volt}=V=\frac{\text{J}}{\text{C}}\)

In this equation, A is the current in amperes and C the charge in coulombs. Note that volts must be multiplied by the charge in coulombs (C) to obtain the energy in joules (J).

In this standard galvanic cell, the half-cells are separated; electrons can flow through an external wire and become available to do electrical work.

This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a blue solution and is labeled below as “1 M solution of copper (II) nitrate ( C u ( N O subscript 3 ) subscript 2 ).” The beaker on the right contains a colorless solution and is labeled below as “1 M solution of silver nitrate ( A g N O subscript 3 ).” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. The plug in the left beaker is labeled “Porous plug.” At the center of the diagram, the tube is labeled “Salt bridge ( N a N O subscript 3 ). Each beaker shows a metal strip partially submerged in the liquid. The beaker on the left has an orange brown strip that is labeled “C u anode negative” at the top. The beaker on the right has a silver strip that is labeled “A g cathode positive” at the top. A wire extends from the top of each of these strips to a rectangular digital readout indicating a reading of positive 0.46 V that is labeled “Voltmeter.” An arrow points toward the voltmeter from the left which is labeled “Flow of electrons.” Similarly, an arrow points away from the voltmeter to the right which is also labeled “Flow of electrons.” A curved arrow extends from the C u strip into the surrounding solution. The tip of this arrow is labeled “C u superscript 2 plus.” A curved arrow extends from the salt bridge into the beaker on the left into the blue solution. The tip of this arrow is labeled “N O subscript 3 superscript negative.” A curved arrow extends from the solution in the beaker on the right to the A g strip. The base of this arrow is labeled “A g superscript plus.” A curved arrow extends from the colorless solution to salt bridge in the beaker on the right. The base of this arrow is labeled “N O subscript 3 superscript negative.” Just right of the center of the salt bridge on the tube an arrow is placed on the salt bridge that points down and to the right. The base of this arrow is labeled “N a superscript plus.” Just above this region of the tube appears the label “Flow of cations.” Just left of the center of the salt bridge on the tube an arrow is placed on the salt bridge that points down and to the left. The base of this arrow is labeled “N O subscript 3 superscript negative.” Just above this region of the tube appears the label “Flow of anions.”

When the electrochemical cell is constructed in this fashion, a positive cell potential indicates a spontaneous reaction and that the electrons are flowing from the left to the right. There is a lot going on in [link], so it is useful to summarize things for this system:

  • Electrons flow from the anode to the cathode: left to right in the standard galvanic cell in the figure.
  • The electrode in the left half-cell is the anode because oxidation occurs here. The name refers to the flow of anions in the salt bridge toward it.
  • The electrode in the right half-cell is the cathode because reduction occurs here. The name refers to the flow of cations in the salt bridge toward it.
  • Oxidation occurs at the anode (the left half-cell in the figure).
  • Reduction occurs at the cathode (the right half-cell in the figure).
  • The cell potential, +0.46 V, in this case, results from the inherent differences in the nature of the materials used to make the two half-cells.
  • The salt bridge must be present to close (complete) the circuit and both an oxidation and reduction must occur for current to flow.

There are many possible galvanic cells, so a shorthand notation is usually used to describe them. The cell notation (sometimes called a cell diagram) provides information about the various species involved in the reaction. This notation also works for other types of cells. A vertical line, │, denotes a phase boundary and a double line, ‖, the salt bridge. Information about the anode is written to the left, followed by the anode solution, then the salt bridge (when present), then the cathode solution, and, finally, information about the cathode to the right. The cell notation for the galvanic cell in [link] is then

\(\text{Cu}\left(s\right)│{\text{Cu}}^{2+}\left(aq\text{, 1}\phantom{\rule{0.2em}{0ex}}M\right)║{\text{Ag}}^{\text{+}}\left(aq\text{, 1}\phantom{\rule{0.2em}{0ex}}M\right)│\text{Ag}\left(s\right)\)

Note that spectator ions are not included and that the simplest form of each half-reaction was used. When known, the initial concentrations of the various ions are usually included.

One of the simplest cells is the Daniell cell. It is possible to construct this battery by placing a copper electrode at the bottom of a jar and covering the metal with a copper sulfate solution. A zinc sulfate solution is floated on top of the copper sulfate solution; then a zinc electrode is placed in the zinc sulfate solution. Connecting the copper electrode to the zinc electrode allows an electric current to flow. This is an example of a cell without a salt bridge, and ions may flow across the interface between the two solutions.

Some oxidation-reduction reactions involve species that are poor conductors of electricity, and so an electrode is used that does not participate in the reactions. Frequently, the electrode is platinum, gold, or graphite, all of which are inert to many chemical reactions. One such system is shown in [link]. Magnesium undergoes oxidation at the anode on the left in the figure and hydrogen ions undergo reduction at the cathode on the right. The reaction may be summarized as

\(\begin{array}{}\\ \underset{¯}{\begin{array}{l}\text{oxidation:}\phantom{\rule{4.1em}{0ex}}\text{Mg}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)+{\text{2e}}^{\text{−}}\\ \text{reduction:}\phantom{\rule{0.2em}{0ex}}{\text{2H}}^{\text{+}}\left(aq\right)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)\end{array}}\\ \text{overall:}\phantom{\rule{0.45em}{0ex}}\text{Mg}\left(s\right)+{\text{2H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)+{\text{H}}_{2}\left(g\right)\end{array}\)

The cell used an inert platinum wire for the cathode, so the cell notation is

\(\text{Mg}\left(s\right)│\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)║{\text{H}}^{\text{+}}\left(aq\right)│{\text{H}}_{2}\left(g\right)│\text{Pt}\left(s\right)\)

The magnesium electrode is an active electrode because it participates in the oxidation-reduction reaction. Inert electrodes, like the platinum electrode in [link], do not participate in the oxidation-reduction reaction and are present so that current can flow through the cell. Platinum or gold generally make good inert electrodes because they are chemically unreactive.

Using Cell Notation
Consider a galvanic cell consisting of

\(\text{2Cr}\left(s\right)+{\text{3Cu}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Cr}}^{3+}\left(aq\right)+\text{3Cu}\left(s\right)\)

Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?

Solution
By inspection, Cr is oxidized when three electrons are lost to form Cr3+, and Cu2+ is reduced as it gains two electrons to form Cu. Balancing the charge gives

\(\begin{array}{}\\ \underset{¯}{\begin{array}{l}\text{oxidation:}\phantom{\rule{4.75em}{0ex}}\text{2Cr}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Cr}}^{3+}\left(aq\right)+{\text{6e}}^{\text{−}}\\ \text{reduction:}\phantom{\rule{0.2em}{0ex}}{\text{3Cu}}^{2+}\left(aq\right)+{\text{6e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{3Cu}\left(s\right)\end{array}}\\ \text{overall:}\phantom{\rule{0.3em}{0ex}}\text{2Cr}\left(s\right)+{\text{3Cu}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Cr}}^{3+}\left(aq\right)+\text{3Cu}\left(s\right)\end{array}\)

Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so: \(\text{Cr}\left(s\right)│{\text{Cr}}^{3+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}║\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}\left(aq\right)│\text{Cu}\left(s\right).\) Oxidation occurs at the anode and reduction at the cathode.

Using Cell Notation

Consider a galvanic cell consisting of

\(5{\text{Fe}}^{2+}\left(aq\right)+{\text{MnO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{8H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{5Fe}}^{3+}\left(aq\right)+{\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{2}\text{O}\left(l\right)\)

Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?

Solution
By inspection, Fe2+ undergoes oxidation when one electron is lost to form Fe3+, and MnO4− is reduced as it gains five electrons to form Mn2+. Balancing the charge gives

\(\begin{array}{}\\ \underset{¯}{\begin{array}{l}\text{oxidation:}\phantom{\rule{9.65em}{0ex}}{\text{5(Fe}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Fe}}^{3+}\left(aq\right)+{\text{e}}^{\text{−}}\right)\\ \text{reduction:}\phantom{\rule{1.65em}{0ex}}{\text{MnO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{8H}}^{\text{+}}\left(aq\right)+{\text{5e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{2}\text{O}\left(l\right)\end{array}}\\ {\text{overall: 5Fe}}^{2+}\left(aq\right)+{\text{MnO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{8H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{5Fe}}^{3+}\left(aq\right)+{\text{Mn}}^{2+}\left(aq\right)+{\text{4H}}_{2}\text{O}\left(l\right)\end{array}\)

Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode. No concentrations were specified so: \(\text{Pt}\left(s\right)│\phantom{\rule{0.2em}{0ex}}{\text{Fe}}^{2+}\left(aq{\text{), Fe}}^{3+}\left(aq\right)║{\text{MnO}}_{4}{}^{\text{−}}\left(aq{\text{), H}}^{\text{+}}\left(aq{\text{), Mn}}^{2+}\left(aq\right)│\text{Pt}\left(s\right).\) Oxidation occurs at the anode and reduction at the cathode.

Check Your Learning
Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.

Answer:

From the information given in the problem:

\(\begin{array}{}\\ \underset{¯}{\begin{array}{l}\text{anode (oxidation):}\phantom{\rule{5.3em}{0ex}}\text{Zn}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Zn}}^{2+}\left(aq\right)+{\text{2e}}^{\text{−}}\\ \text{cathode (reduction):}\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}\left(aq\right)+{\text{2e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Cu}\left(s\right)\end{array}}\\ \text{overall:}\phantom{\rule{4.7em}{0ex}}\text{Zn}\left(s\right)+{\text{Cu}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Zn}}^{2+}\left(aq\right)+\text{Cu}\left(s\right)\end{array}\)

Using cell notation:

\(\text{Zn}\left(s\right)│{\text{Zn}}^{2+}\left(aq\right)║{\text{Cu}}^{2+}\left(aq\right)│\text{Cu}\left(s\right).\)

The oxidation of magnesium to magnesium ion occurs in the beaker on the left side in this apparatus; the reduction of hydrogen ions to hydrogen occurs in the beaker on the right. A nonreactive, or inert, platinum wire allows electrons from the left beaker to move into the right beaker. The overall reaction is: \(\text{Mg}+{\text{2H}}^{\text{+}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}+{\text{H}}_{2},\) which is represented in cell notation as: \(\text{Mg}\left(s\right)│\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}║\phantom{\rule{0.2em}{0ex}}{\text{H}}^{\text{+}}\left(aq\right)│\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)│\phantom{\rule{0.2em}{0ex}}\text{Pt}\left(s\right).\)

This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a colorless solution and is labeled “Solution of M g C l subscript 2.” The beaker on the right contains a colorless solution and is labeled “Solution of H C l.” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. At the center of the diagram, the tube is labeled “Salt bridge. Each beaker shows a metal coils submerged in the liquid. The beaker on the left has a thin grey coiled strip that is labeled “M g coil.” The beaker on the right has a black wire that is oriented horizontally and coiled up in a spring-like appearance that is labeled “Non reactive platinum wire.” A wire extends across the top of the diagram that connects the ends of the M g strip and platinum wire just above the opening of each beaker. This wire is labeled “Conducting wire.” At the center of this wire above the two beakers near the center of the diagram is a right pointing arrow with the label “e superscript negative” at its base. Bubbles appear to be rising from the coiled platinum wire in the beaker. These bubbles are labeled “Bubbles of H subscript 2.” An arrow points down and to the right from a plus sign at the upper right region of the salt bride. An arrow points down and to the left from a negative sign at the upper left region of the salt bride. A curved arrow extends from the grey plug at the left end of the salt bridge into the surrounding solution in the left beaker. The label “Oxidation M g right pointing arrow M g superscript 2 plus plus 2 e superscript negative” appears beneath the left beaker. The label “Reduction 2 H superscript plus plus 2 e superscript negative right pointing arrow H subscript 2” appears beneath the right beaker.

Key Concepts and Summary

Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire. One half-cell, normally depicted on the left side in a figure, contains the anode. Oxidation occurs at the anode. The anode is connected to the cathode in the other half-cell, often shown on the right side in a figure. Reduction occurs at the cathode. Adding a salt bridge completes the circuit allowing current to flow. Anions in the salt bridge flow toward the anode and cations in the salt bridge flow toward the cathode. The movement of these ions completes the circuit and keeps each half-cell electrically neutral. Electrochemical cells can be described using cell notation. In this notation, information about the reaction at the anode appears on the left and information about the reaction at the cathode on the right. The salt bridge is represented by a double line, ‖. The solid, liquid, or aqueous phases within a half-cell are separated by a single line, │. The phase and concentration of the various species is included after the species name. Electrodes that participate in the oxidation-reduction reaction are called active electrodes. Electrodes that do not participate in the oxidation-reduction reaction but are there to allow current to flow are inert electrodes. Inert electrodes are often made from platinum or gold, which are unchanged by many chemical reactions.

Chemistry End of Chapter Exercises

Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed.

(a) \(\text{Mg}\left(s\right)+{\text{Ni}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)+\text{Ni}\left(s\right)\)

(b) \(2{\text{Ag}}^{\text{+}}\left(aq\right)+\text{Cu}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}\left(aq\right)+\text{2Ag}\left(s\right)\)

(c) \(\text{Mn}\left(s\right)+{\text{Sn(NO}}_{3}{\right)}_{2}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mn(NO}}_{3}{\right)}_{2}\left(aq\right)+\text{Au}\left(s\right)\)

(d) \(3{\text{CuNO}}_{3}\left(aq\right)+{\text{Au(NO}}_{3}{\right)}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3Cu(NO}}_{3}{\right)}_{2}\left(aq\right)+\text{Au}\left(s\right)\)

(a) \(\text{Mg}\left(s\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)║{\text{Ni}}^{\text{2+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}\text{Ni}\left(s\right);\) (b) \(\text{Cu}\left(s\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{2+}\left(aq\right)║{\text{Ag}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}\text{Ag}\left(s\right);\) (c) \(\text{Mn}\left(s\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}{\text{Mn}}^{2+}\left(aq\right)║{\text{Sn}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}\text{Sn}\left(s\right);\) (d) \(\text{Pt}\left(s\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}{\text{Cu}}^{\text{+}}\left(aq{\text{), Cu}}^{2+}\left(aq\right)║{\text{Au}}^{\text{3+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}│\phantom{\rule{0.2em}{0ex}}\text{Au}\left(s\right)\)

Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions.

(a) \(\text{Mg}\left(s\right)│{\text{Mg}}^{2+}\left(aq\right)║{\text{Cu}}^{2+}\left(aq\right)│\text{Cu}\left(s\right)\)

(b) \(\text{Ni}\left(s\right)│\phantom{\rule{0.2em}{0ex}}{\text{Ni}}^{2+}\left(aq\right)║{\text{Ag}}^{\text{+}}\left(aq\right)│\phantom{\rule{0.2em}{0ex}}\text{Ag}\left(s\right)\)

For the cell notations in the previous problem, write the corresponding balanced reactions.

(a) \(\text{Mg}\left(s\right)+{\text{Cu}}^{2+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Mg}}^{2+}\left(aq\right)+\text{Cu}\left(s\right);\) (b) \(2{\text{Ag}}^{\text{+}}\left(aq\right)+\text{Ni}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ni}}^{2+}\left(aq\right)+\text{2Ag}\left(s\right)\)

Balance the following reactions and write the reactions using cell notation. Ignore any inert electrodes, as they are never part of the half-reactions.

(a) \(\text{Al}\left(s\right)+{\text{Zr}}^{4+}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Al}}^{3+}\left(aq\right)+\text{Zr}\left(s\right)\)

(b) \({\text{Ag}}^{\text{+}}\left(aq\right)+\text{NO}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ag}\left(s\right)+{\text{NO}}_{3}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}\left(\text{acidic solution}\right)\)

(c) \({\text{SiO}}_{3}{}^{2-}\left(aq\right)+\text{Mg}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Si}\left(s\right)+\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\phantom{\rule{5em}{0ex}}\text{(basic solution)}\)

(d) \({\text{ClO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{MnO}}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cl}}^{\text{−}}\left(aq\right)+{\text{MnO}}_{4}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}\text{(basic solution)}\)

Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.

Species oxidized = reducing agent: (a) Al(s); (b) NO(g); (c) Mg(s); and (d) MnO2(s); Species reduced = oxidizing agent: (a) Zr4+(aq); (b) Ag+(aq); (c) \({\text{SiO}}_{3}{}^{2-}\left(aq\right)\); and (d) \({\text{ClO}}_{3}{}^{\text{−}}\left(aq\right)\)

From the information provided, use cell notation to describe the following systems:

(a) In one half-cell, a solution of Pt(NO3)2 forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO3)2 solution with all solute concentrations 1 M.

(b) The cathode consists of a gold electrode in a 0.55 M Au(NO3)3 solution and the anode is a magnesium electrode in 0.75 M Mg(NO3)2 solution.

(c) One half-cell consists of a silver electrode in a 1 M AgNO3 solution, and in the other half-cell, a copper electrode in 1 M Cu(NO3)2 is oxidized.

Why is a salt bridge necessary in galvanic cells like the one in [link]?

Without the salt bridge, the circuit would be open (or broken) and no current could flow. With a salt bridge, each half-cell remains electrically neutral and current can flow through the circuit.

An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.

An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.

Active electrodes participate in the oxidation-reduction reaction. Since metals form cations, the electrode would lose mass if metal atoms in the electrode were to oxidize and go into solution. Oxidation occurs at the anode.

The mass of three different metal electrodes, each from a different galvanic cell, were determined before and after the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes. The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, given the label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass. Make an educated guess as to which electrodes were active and which were inert electrodes, and which were anode(s) and which were the cathode(s).

Glossary

active electrode
electrode that participates in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidation-reduction reaction
anode
electrode in an electrochemical cell at which oxidation occurs; information about the anode is recorded on the left side of the salt bridge in cell notation
cathode
electrode in an electrochemical cell at which reduction occurs; information about the cathode is recorded on the right side of the salt bridge in cell notation
cell notation
shorthand way to represent the reactions in an electrochemical cell
cell potential
difference in electrical potential that arises when dissimilar metals are connected; the driving force for the flow of charge (current) in oxidation-reduction reactions
galvanic cell
electrochemical cell that involves a spontaneous oxidation-reduction reaction; electrochemical cells with positive cell potentials; also called a voltaic cell
inert electrode
electrode that allows current to flow, but that does not otherwise participate in the oxidation-reduction reaction in an electrochemical cell; the mass of an inert electrode does not change during the oxidation-reduction reaction; inert electrodes are often made of platinum or gold because these metals are chemically unreactive.
voltaic cell
another name for a galvanic cell

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