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# 24.2: Galvanic cells and Electrodes

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• Contributed by Stephen Lower
• Professor Emeritus (Chemistry) at Simon Fraser University

It is physically impossible to measure the potential difference between a piece of metal and the solution in which it is immersed. We can, however, measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials.

This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through.

If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn2+ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu2+ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:

$Zn_(s) + Cu^2+ \rightarrow Zn^2+ + Cu_(s)$

but this time, the oxidation and reduction steps (half reactions) take place in separate locations:

 left electrode: Zn(s) → Zn2+ + 2e– oxidation right electrode: Cu2+ + 2e–→ Cu(s) reduction

#### Electrochemical cells allow measurement and control of a redox reaction

The reaction can be started and stopped by connecting or disconnecting the two electrodes. If we place a variable resistance in the circuit, we can even control the rate of the net cell reaction by simply turning a knob. By connecting a battery or other source of current to the two electrodes, we can force the reaction to proceed in its non-spontaneous, or reverse direction. By placing an ammeter in the external circuit, we can measure the amount of electric charge that passes through the electrodes, and thus the number of moles of reactants that get transformed into products in the cell reaction.

Electric charge q is measured in coulombs. The amount of charge carried by one mole of electrons is known as the Faraday, which we denote by F. Careful experiments have determined that 1 F = 96467 C. For most purposes, you can simply use 96,500 Coulombs as the value of the faraday. When we measure electric current, we are measuring the rate at which electric charge is transported through the circuit. A current of one ampere corresponds to the flow of one coulomb per second.

### Charge Transport within the Cell

For the cell to operate, not only must there be an external electrical circuit between the two electrodes, but the two electrolytes (the solutions) must be in contact. The need for this can be understood by considering what would happen if the two solutions were physically separated. Positive charge (in the form of Zn2+) is added to the electrolyte in the left compartment, and removed (as Cu2+) from the right side, causing the solution in contact with the zinc to acquire a net positive charge, while a net negative charge would build up in the solution on the copper side of the cell. These violations of electroneutrality would make it more difficult (require more work) to introduce additional Zn2+ ions into the positively-charged electrolyte or for electrons to flow into right compartment where they are needed to reduce the Cu2+ ions, thus effectively stopping the reaction after only a chemically insignificant amount has taken place.

In order to sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells. This means that we must provide a path for ions to move directly from one cell to the other. This ionic transport involves not only the electroactive species Cu2+ and Zn2+, but also the counterions, which in this example are nitrate, NO3. Thus an excess of Cu2+ in the left compartment could be alleviated by the drift of these ions into the right side, or equally well by diffusion of nitrate ions to the left. More detailed studies reveal that both processes occur, and that the relative amounts of charge carried through the solution by positive and negative ions depends on their relative mobilities, which express the velocity with which the ions are able to make their way through the solution. Since negative ions tend to be larger than positive ions, the latter tend to have higher mobilities and carry the larger fraction of charge.

In the simplest cells, the barrier between the two solutions can be a porous membrane, but for precise measurements, a more complicated arrangement, known as a salt bridge, is used. The salt bridge consists of an intermediate compartment filled with a concentrated solution of KCl and fitted with porous barriers at each end. The purpose of the salt bridge is to minimize the natural potential difference, known as the junction potential, that develops (as mentioned in the previous section) when any two phases (such as the two solutions) are in contact. This potential difference would combine with the two half-cell potentials so as introduce a degree of uncertainty into any measurement of the cell potential. With the salt bridge, we have two liquid junction potentials instead of one, but they tend to cancel each other out.

### Cell description conventions

In order to make it easier to describe a given electrochemical cell, a special symbolic notation has been adopted. In this notation the cell we described above would be

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

There are several other conventions relating to cell notation and nomenclature that you are expected to know:

• The anode is where oxidation occurs, and the cathode is the site of reduction. In an actual cell, the identity of the electrodes depends on the direction in which the net cell reaction is occurring.
• If electrons flow from the left electrode to the right electrode (as depicted in the above cell notation) when the cell operates in its spontaneous direction, the potential of the right electrode will be higher than that of the left, and the cell potential will be positive.
• "Conventional current flow" is from positive to negative, which is opposite to the direction of the electron flow. This means that if the electrons are flowing from the left electrode to the right, a galvanometer placed in the external circuit would indicate a current flow from right to left.

### Electrodes and Electrode Reactions

An electrode reaction refers to the net oxidation or reduction process that takes place at an electrode. This reaction may take place in a single electron-transfer step, or as a succession of two or more steps. The substances that receive and lose electrons are called the electroactive species.

Fig. 4: Electron transfer at an anode

This process takes place within the very thin interfacial region at the electrode surface, and involves quantum-mechanical tunneling of electrons between the electrode and the electroactive species. The work required to displace the H2O molecules in the hydration spheres of the ions constitutes part of the activation energy of the process.

In the example of the Zn/Cu cell we have been using, the electrode reaction involves a metal and its hydrated cation; we call such electrodes metal-metal ion electrodes. There are a number of other kinds of electrodes which are widely encountered in electrochemistry and analytical chemistry.

#### Ion-ion Electrodes

Many electrode reactions involve only ionic species, such as $$Fe^2+$$ and $$Fe^3+$$. If neither of the electroactive species is a metal, some other metal must serve as a conduit for the supply or removal of electrons from the system. In order to avoid complications that would arise from electrode reactions involving this metal, a relatively inert substance such as platinum is commonly used. Such a half cell would be represented as

Pt(s) | Fe3+(aq), Fe2+(aq) || …

and the half-cell reaction would be

$Fe^2+(aq) \rightarrow Fe^3+ (aq) + e^-$

The reaction occurs at the surface of the electrode (Fig 4 above). The electroactive ion diffuses to the electrode surface and adsorbs (attaches) to it by van der Waals and Coulombic forces . In doing so, the waters of hydration that are normally attached to any ionic species must be displaced. This process is always endothermic, sometimes to such an extent that only a small fraction of the ions be able to contact the surface closely enough to undergo electron transfer, and the reaction will be slow. The actual electron-transfer occurs by quantum-mechanical tunnelling .

#### Gas Electrodes

Some electrode reactions involve a gaseous species such as $$H_2$$, $$O_2$$, or $$Cl_2$$. Such reactions must also be carried out on the surface of an electrochemically inert conductor such as platinum. A typical reaction of considerable commercial importance is

$Cl^-(aq) \rightarrow ½ Cl_2(g) + e^-$

Similar reactions involving the oxidation of $$Br_2$$ or $$I_2$$ also take place at platinum surfaces.

#### Insoluble–salt Electrodes

A typical electrode of this kind consists of a silver wire covered with a thin coating of silver chloride, which is insoluble in water. The electrode reaction consists in the oxidation and reduction of the silver:

$AgCl(s) + e^– → Ag(s) + Cl^–(aq)$

The half cell would be represented as

$… || Cl^– (aq) | AgCl (s) | Ag (s)$

Although the usefulness of such an electrode may not be immediately apparent, this kind of electrode finds very wide application in electrochemical measurements, as we shall see later.

### Reference Electrodes

In most electrochemical experiments our interest is concentrated on only one of the electrode reactions. Since all measurements must be on a complete cell involving two electrode systems, it is common practice to employ a reference electrode as the other half of the cell. The major requirements of a reference electrode are that it be easy to prepare and maintain, and that its potential be stable. The last requirement essentially means that the concentration of any ionic species involved in the electrode reaction must be held at a fixed value. The most common way of accomplishing this is to use an electrode reaction involving a saturated solution of an insoluble salt of the ion. One such system, the silver-silver chloride electrode has already been mentioned:

$Ag | AgCl(s) | Cl^–(aq) || …$

$Ag(s) + Cl^–(aq) →AgCl(s) + e^–$

This electrode usually takes the form of a piece of silver wire coated with AgCl. The coating is done by making the silver the anode in an electrolytic cell containing HCl; the Ag+ ions combine with Cl ions as fast as they are formed at the silver surface.

The other common reference electrode is the calomel electrode; calomel is the common name for mercury(I) chloride. Such a half cell would be represented as

$Hg | Hg^2+(aq) | KCl || …$

and the half-cell reaction would be

$Hg(l) + Cl^– → ½ HgCl2(s) + e^–$

The potentials of both of these electrodes have been very accurately determined against the hydrogen electrode. The latter is seldom used in routine electrochemical measurements because it is more difficult to prepare; the platinum surface has to be specially treated by preliminary electrolysis. Also, there is need for a supply of hydrogen gas which makes it somewhat cumbersome and hazardous.

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.

• A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge or some other path that allows ions to pass between the two sides in order to maintain electroneutrality.
• The conventional way of representing an electrochemical cell of any kind is to write the oxidation half reaction on the left and the reduction on the right. Thus for the reaction

Zn(s) + Cu2+ → Zn2+ + Cu(s)

we write

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

in which the single vertical bars represent phase boundaries. The double bar denotes a liquid-liquid boundary which in laboratory cells consists of a salt bridge or in ion-permeable barrier. If the net cell reaction were written in reverse, the cell notation would become

Cu(s) | Cu2+(aq) || Zn 2+(aq) | Zn (s)

Remember: the Reduction process is always shown on the Right.

• The transfer of electrons between an electrode and the solution takes place by quantum-mechanical tunneling at the electrode surface. The energy required to displace water molecules from the hydration shell of an ion as it approaches the electrode surface constitutes an activation energy which can slow down the process. Even larger activation energies (and slower reactions) occur when a molecule such as O2 is formed or consumed.

### Contributors

Stephen Lower, Professor Emeritus ( Simon Fraser U. ) Chem1 Virtual Textbook

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Chapter 17. Electrochemistry

# 17.2 Galvanic Cells

### 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 ( Figure 1 ). 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.

$\beginarray{lr @{{}\longrightarrow{}} l} \textoxidation: & \textCu(s) & \textCu^2+(aq)\;+\;2\texte^- \\[0.5em] \textreduction: & 2\;\times\;(\textAg^+(aq)\;+\;\texte^- & \textAg(s))\;\;\;\;\;\;\;\textor\;\;\;\;\;\;\;2\textAg^+(aq)\;+\;2\texte^-\;\longrightarrow\;\textAg(s) \\[0.5em] \hline \\[-0.25em] \textoverall: & 2\textAg^+(aq)\;+\;\textCu(s) & 2\textAg(s)\;+\;\textCu^2+(aq) \endarray$

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.

Galvanic or voltaic cells involve spontaneous electrochemical reactions in which the half-reactions are separated ( Figure 2 ) 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

$\textvolt = V = \frac\textJ\textC$

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).

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 Figure 2 , 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 Figure 2 is then

$\textCu(s)\mid\textCu^2+(aq\text,\;1\;M)\parallel\textAg^+(aq\text,\;1\;M)\mid\textAg(s)$

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 Figure 3 . 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

$\beginarray{lr @{{}\longrightarrow{}} l} \textoxidation: & \textMg(s) & \textMg^2+(aq)\;+\;2\texte^- \\[0.5em] \textreduction: & 2\textH^+(aq)\;+\;2\texte^- & \textH_2(g) \\[0.5em] \hline \\[-0.25em] \textoverall: & \textMg(s)\;+\;2\textH^+(aq) & \textMg^2+(aq)\;+\;\textH_2(g) \endarray$

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

$\textMg(s)\mid\textMg^2+(aq)\parallel\textH^+(aq)\mid\textH_2(g)\mid\textPt(s)$

The magnesium electrode is an active electrode because it participates in the oxidation-reduction reaction. Inert electrodes, like the platinum electrode in Figure 3 , 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.

### Example 1

Using Cell Notation
Consider a galvanic cell consisting of

$2\textCr(s)\;+\;3\textCu^2+(aq)\;\longrightarrow\;2\textCr^3+(aq)\;+\;3\textCu(s)$

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

$\beginarray{lr @{{}\longrightarrow{}} l} \textoxidation: & 2\textCr(s) & 2\textCr^3+(aq)\;+\;6\texte^- \\[0.5em] \textreduction: & 3\textCu^2+(aq)\;+\;6\texte^- & 3\textCu(s) \\[0.5em] \hline \\[-0.25em] \textoverall: & 2\textCr(s)\;+\;3\textCu^2+(aq) & 2\textCr^3+(aq)\;+\;3\textCu(s) \endarray$

Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so: $\textCr(s)\mid\textCr^3+(aq)\parallel\textCu^2+(aq)\mid\textCu(s)$. Oxidation occurs at the anode and reduction at the cathode.

Using Cell Notation

Consider a galvanic cell consisting of

$5\textFe^2+(aq)\;+\;\textMnO_4^\;\;-(aq)\;+\;8\textH^+(aq)\;\longrightarrow\;5\textFe^3+(aq)\;+\;\textMn^2+(aq)\;+\;4\textH_2\textO(l)$

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

$\beginarray{lr @{{}\longrightarrow{}} l} \textoxidation: & 5(\textFe^2+(aq) & \textFe^3+(aq)\;+\;\texte^-) \\[0.5em] \textreduction: & \textMnO_4^\;\;-(aq)\;+\;8\textH^+(aq)\;+\;5\texte^- & \textMn^2+(aq)\;+\;4\textH_2\textO(l) \\[0.5em] \hline \\[-0.25em] \textoverall: & 5\textFe^2+(aq)\;+\;\textMnO_4^\;\;-(aq)\;+\;8\textH^+(aq) & 5\textFe^3+(aq)\;+\;\textMn^2+(aq)\;+\;4\textH_2\textO(l) \endarray$

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: $\textPt(s)\mid\textFe^2+(aq)\text,\;\textFe^3+(aq)\parallel\textMnO_4^\;\;-(aq)\text,\;\textH^+(aq)\text,\;\textMn^2+(aq)\mid\textPt(s)$. Oxidation occurs at the anode and reduction at the cathode.

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.

From the information given in the problem:

$\beginarray{lr @{{}\longrightarrow{}} l} \textanode\;(oxidation): & \textZn(s) & \textZn^2+(aq)\;+\;2\texte^- \\[0.5em] \textcathode\;(reduction): & \textCu^2+(aq)\;+\;2\texte^- & \textCu(s) \\[0.5em] \hline \\[-0.25em] \textoverall: & \textZn(s)\;+\;\textCu^2+(aq) & \textZn^2+(aq)\;+\;\textCu(s) \endarray$

Using cell notation:
$\textZn(s)\mid\textZn^2+(aq)\parallel\textCu^2+(aq)\mid\textCu(s)$.

# 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

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

(a) $\textMg(s)\;+\;\textNi^2+(aq)\;\longrightarrow\;\textMg^2+(aq)\;+\;\textNi(s)$

(b) $2\textAg^+(aq)\;+\;\textCu(s)\;\longrightarrow\;\textCu^2+(aq)\;+\;2\textAg(s)$

(c) $\textMn(s)\;+\;\textSn(NO_3)_2(aq)\;\longrightarrow\;\textMn(NO_3)_2(aq)\;+\;\textAu(s)$

(d) $3\textCuNO_3(aq)\;+\;\textAu(NO_3)_3(aq)\;\longrightarrow\;3\textCu(NO_3)_2(aq)\;+\;\textAu(s)$

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

(a) $\textMg(s)\mid\textMg^2+(aq)\parallel\textCu^2+(aq)\mid\textCu(s)$

(b) $\textNi(s)\mid\textNi^2+(aq)\parallel\textAg^+(aq)\mid\textAg(s)$

3. For the cell notations in the previous problem, write the corresponding balanced reactions.
4. 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) $\textAl(s)\;+\;\textZr^4+(aq)\;\longrightarrow\;\textAl^3+(aq)\;+\;\textZr(s)$

(b) $\textAg^+(aq)\;+\;\textNO(g)\;\longrightarrow\;\textAg(s)\;+\;\textNO_3^\;\;-(aq)\;\;\;\;\;\;\;\text(acidic\;solution)$

(c) $\textSiO_3^\;\;2-(aq)\;+\;\textMg(s)\;\longrightarrow\;\textSi(s)\;+\;\textMg(OH)_2(s)\;\;\;\;\;\;\;\text(basic\;solution)$

(d) $\textClO_3^\;\;-(aq)\;+\;\textMnO_2(s)\;\longrightarrow\;\textCl^\;\;-(aq)\;+\;\textMnO_4^\;\;-(aq)\;\;\;\;\;\;\;\text(basic\;solution)$

5. Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.
6. 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.

7. Why is a salt bridge necessary in galvanic cells like the one in Figure 2 ?
8. 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.
9. 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.
10. 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

### Solutions

Answers to Chemistry End of Chapter Exercises

1. (a) $\textMg(s)\mid\textMg^2+(aq)\parallel\textNi^2+(aq)\mid\textNi(s)$; (b) $\textCu(s)\mid\textCu^2+(aq)\parallel\textAg^+(aq)\mid\textAg(s)$; (c) $\textMn(s)\mid\textMn^2+(aq)\parallel\textSn^2+(aq)\mid\textSn(s)$; (d) $\textPt(s)\mid\textCu^+(aq)\text,\;Cu^2+(aq)\parallel\textAu^3+(aq)\mid\textAu(s)$

3. (a) $\textMg(s)\;+\;\textCu^2+(aq)\;\longrightarrow\;\textMg^2+(aq)\;+\;\textCu(s)$; (b) $2\textAg^+(aq)\;+\;\textNi(s)\;\longrightarrow\;\textNi^2+(aq)\;+\;2\textAg(s)$

5. 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) $\textSiO_3^\;\;2-(aq)$; and (d) $\textClO_3^\;\;-(aq)$

7. 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.

9. 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.