5 Steps to a 5: AP Chemistry 2017 (2016)

STEP 4

Review the Knowledge You Need to Score High

CHAPTER 16

Electrochemistry

IN THIS CHAPTER

Summary: Electrochemistry is the study of chemical reactions that produce electricity, and chemical reactions that take place because electricity is supplied. Electrochemical reactions may be of many types. Electroplating is an electrochemical process. So are the electrolysis of water, the production of aluminum metal, and the production and storage of electricity in batteries. All these processes involve the transfer of electrons and redox reactions.

Keywords and Equations

A table of half-reactions is given in the exam booklet and in the back of this book.

I = current (amperes)

E ° = standard reduction potential

G ° = standard free energy

q = charge (coulombs)

K = equilibrium constant

Faraday’s constant, F = 96,500 coulombs per mole of electrons

Gas constant, R = 8.31 volt coulomb mol−1 K–1

Redox Reactions

Electrochemical reactions involve redox reactions. In the chapter on Reactions and Periodicity, we discussed redox reactions, but here is a brief review: Redox is a term that stands for reduction and oxidation. Reduction is the gain of electrons, and oxidation is the loss of electrons. For example, suppose a piece of zinc metal is placed in a solution containing Cu2+ . Very quickly, a reddish solid forms on the surface of the zinc metal. That substance is copper metal. At the molecular level, the zinc metal is losing electrons to form Zn2+ and Cu2+ is gaining electrons to form copper metal. These two processes can be shown as:

The electrons that are being lost by the zinc metal are the same electrons that are being gained by the cupric ion. The zinc metal is being oxidized, and the cupric cation is being reduced.

Something must cause the oxidation (taking of the electrons), and that substance is called the oxidizing agent (the reactant being reduced). In the example above, the oxidizing agent is Cu2+ . The reactant undergoing oxidation is called the reducing agent, because it is furnishing the electrons used in the reduction half-reaction. Zinc metal is the reducing agent above. The two half-reactions, oxidation and reduction, can be added together to give you the overall redox reaction. The electrons must cancel—that is, there must be the same number of electrons lost as electrons gained:

In these redox reactions, like the electrochemical reactions we will show you, there is a simultaneous loss and gain of electrons. In the oxidation reaction (commonly called a half-reaction) electrons are being lost, but in the reduction half-reaction those very same electrons are being gained. So, in redox reactions electrons are being exchanged as reactants are being converted into products. This electron exchange may be direct, as when copper metal plates out on a piece of zinc, or it may be indirect, as in an electrochemical cell (battery). In this chapter, we will show you both processes and the calculations associated with each.

The balancing of redox reactions is beginning to appear on the AP exam, so we have included the half-reaction method of balancing redox reactions in the Appendixes, just in case you are having trouble with the technique in your chemistry class.

The definitions for oxidation and reduction given above are the most common and the most useful ones. A couple of others might also be useful: Oxidation is the gain of oxygen or loss of hydrogen and involves an increase in oxidation number. Reduction is the gain of hydrogen or loss of oxygen and involves a decrease in oxidation number.

Electrochemical Cells

In the example above, the electron transfer was direct, that is, the electrons were exchanged directly from the zinc metal to the cupric ions. But such a direct electron transfer doesn’t allow for any useful work to be done by the electrons. Therefore, in order to use these electrons, indirect electron transfer must be done. The two half-reactions are physically separated and connected by a wire. The electrons that are lost in the oxidation half-reaction are allowed to flow through the wire to get to the reduction half-reaction. While those electrons are flowing through the wire they can do useful work, like powering a calculator or a pacemaker. Electrochemical cells use indirect electron transfer to produce electricity by a redox reaction, or they use electricity to produce a desired redox reaction.

Figure 16.1   A galvanic cell.

Galvanic (Voltaic) Cells

Galvanic (voltaic) cells produce electricity by using a redox reaction. Let’s take that zinc/copper redox reaction that we studied before (the direct electron transfer reaction) and make it a galvanic cell by separating the oxidation and reduction half-reactions. (See Figure 16.1 .)

Instead of one container, as before, two will be used. A piece of zinc metal will be placed in one, a piece of copper metal in another. A solution of aqueous zinc sulfate will be added to the beaker containing the zinc electrode and an aqueous solution of copper(II) sulfate will be added to the beaker containing the copper metal. The zinc and copper metals will form the electrodes of the cell, the solid portion of the cell that conducts the electrons involved in the redox reaction. The solutions in which the electrodes are immersed are called the electrode compartments . The electrodes are connected by a wire and . . . nothing happens. If the redox reactions were to proceed, the beaker containing the zinc metal would build up a positive charge due to the zinc cations being produced in the oxidation half-reaction. The beaker containing the copper would build up a negative charge due to the loss of the copper(II) ions. The solutions (compartments) must maintain electrical neutrality. To accomplish this, a salt bridge will be used. A salt bridge is often an inverted U-tube that holds a gel containing a concentrated electrolyte solution, such as KNO3 in this example. Any electrolyte could be used as long as it does not interfere with the redox reaction. The anions in the salt bridge will migrate through the gel into the beaker containing the zinc metal, and the salt-bridge cations will migrate in the opposite direction. In this way, electrical neutrality is maintained. In electrical terms, the circuit has been completed and the redox reaction can occur. The zinc electrode is being oxidized in one beaker, and the copper(II) ions in the other beaker are being reduced to copper metal. The same redox reaction is happening in this indirect electron transfer as happened in the direct one:

Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)

The difference is that the electrons are now flowing through a wire from the oxidation half-reaction to the reduction half-reaction. And electrons flowing through a wire is electricity, which can do work. If a voltmeter was connected to the wire connecting the two electrodes, a current of 1.10 V would be measured. This galvanic cell shown in Figure 16.1 is commonly called a Daniell cell.

In the Daniell cell, shown in Figure 16.1 , note that the compartment with the oxidation half-reaction is on the left and the compartment undergoing reduction is on the right. This is a convention that you will have to follow. The AP graders look for this. The electrode at which oxidation is taking place is called the anode , and the electrolyte solution in which it is immersed is called the anode compartment . The electrode at which reduction takes place is called the cathode , and its solution is the cathode compartment . The anode is labeled with a negative sign (−), while the cathode has a positive sign (+). The electrons flow from the anode to the cathode.

Remember: Oxidation is an anode process.

Sometimes the half-reaction(s) involved in the cell lack a solid conductive part to act as the electrode, so an inert (inactive) electrode , a solid conducting electrode that does not take part in the redox reaction, is used. Graphite and platinum are commonly used as inert electrodes.

Note: The electrode must be a conductor onto which a wire may be attached. It can never be an ion in solution.

Cell Notation

Cell notation is a shorthand notation representing a galvanic cell. To write the cell notation in Figure 16.1 :

  1. Write the chemical formula of the anode:Zn(s)
  2. Draw a single vertical line to represent the phase boundary between the solid anode and the solution:Zn(s) |
  3. Write the reactive part of the anode compartment with its initial concentration (if known) in parentheses (assume 1M in this case): Zn(s)|Zn2+ (1 M )
  4. Draw a double vertical line to represent the salt bridge connecting the two electrode compartments:Zn(s)|Zn2+ (1 M )||
  5. Write the reactive part of the cathode compartment with its initial concentration (if known) shown in parentheses:Zn(s)|Zn2+ (1 M )||cu2+ (1 M )
  6. Draw a single vertical line representing the phase boundary between the solution and the solid cathode:Zn(s)|Zn2+ (1 M )||cu2+ (1 M )
  7. Finally, write the chemical formula of the cathode:Zn(s)|Zn2+ (1 M )||cu2+ (1 M )|cu(s)

If an inert electrode is used because one or both redox half-reactions do not have a suitable conducting electrode material associated with the reaction, the inert electrode is shown with its phase boundary. If the electrode components are in the same phase, they are separated by commas; if not, a vertical phase boundary line. For example, consider the following redox reaction:

Ag+ (aq) + Fe2+ (aq) → Fe3+ (aq) + Ag(s)

The oxidation of the ferrous ion to ferric doesn’t involve a suitable electrode material, so an inert electrode, such as platinum, would be used. The cell notation would then be:

Pt(s)|Fe2+ (aq), Fe3+ (aq)||Ag+ (aq)|Ag(s)

Cell Potential

In the discussion of the Daniell cell, we indicated that this cell produces 1.10 volts. This voltage is really the difference in potential between the two half-cells. There are half-cell potentials associated with all half-cells. A list of all possible combinations of half-cells would be tremendously long. Therefore, a way of combining desired half-cells has been developed. The cell potential (really the half-cell potentials) depends on concentration and temperature, but initially we’ll simply look at the half-cell potentials at the standard temperature of 298 K (25°C) and all components in their standard states (1 M concentration of all solutions, 1 atmosphere pressure for any gases, and pure solid electrodes). All the half-cell potentials are tabulated as the reduction potentials, that is, the potentials associated with the reduction reaction. The hydrogen half-reaction has been defined as the standard and has been given a value of exactly 0.00 V. All the other half-reactions have been measured relative to it, some positive and some negative. The table of standard reduction potentials provided on the AP exam is shown in Table 16.1 , on the next page.

Here are some things to be aware of in looking at this table:

  • All reactions are shown in terms of the reduction reaction relative to the standard hydrogen electrode.
  • The more positive the value of the voltage associated with the half-reaction (E°), the more readily the reaction occurs.
  • The strength of the oxidizing agent increases as the value becomes more positive, and the strength of the reducing agent increases as the value becomes more negative.

This table of standard reduction potentials can be used to write the overall cell reaction and to calculate the standard cell potential (E °) , the potential (voltage) associated with the cell at standard conditions. There are a few things to remember when using these standard reduction potentials to generate the cell reaction and cell potential:

  1. The standard cell potential for a galvanic cell is a positive value,E ° > 0.
  2. Because one half-reaction must involve oxidation, one of the half-reactions shown in the table of reduction potentials must be reversed to indicate the oxidation. If the half-reaction is reversed, the sign of the standard reduction potential must be reversed. However, this is not necessary to calculate the standard cell potential.
  3. Because oxidation occurs at the anode and reduction at the cathode, the standard cell potential can be calculated from the standard reduction potentials of the two half-reactions involved in the overall reaction by using the equation:

But remember, both E °cathode and E °anode are shown as reduction potentials, used directly from the table without reversing.

Once the standard cell potential has been calculated, the reaction can be written by reversing the half-reaction associated with the anode and adding the half-reactions together, using appropriate multipliers if needed to ensure that the numbers of electrons lost and gained are equal.

Suppose a galvanic cell was to be constructed utilizing the following two half-reactions taken from Table 16.1 :

Table 16.1 Standard Reduction Potentials in Aqueous Solution at 25 °C

First, the cell voltage can be calculated using:

Since the cell potential must be positive (a galvanic cell), there is only one arrangement of −0.25 and 0.80 volts than can result in a positive value:

This means that the Ni electrode is the anode and must be involved in oxidation, so its reduction half-reaction must be reversed, changing the sign of the standard half-cell potential, and added to the silver half-reaction. Note that the silver half-reaction must be multiplied by two to equalize electron loss and gain, but the half-cell potential remains the same:

Note that the same cell potential is obtained as using: 

If, for example, you are given the cell notation, you could use this method to determine the cell potential. In this case, the cell notation would be: Ni | Ni2+ || Ag+ | Ag |.

Electrolytic Cells

Electrolytic cells use electricity from an external source to produce a desired redox reaction. Electroplating and the recharging of an automobile battery are examples of electrolytic cells.

Figure 16.2 , on the next page, shows a comparison of a galvanic cell and an electrolytic cell for the Sn/Cu system. On the left-hand side of Figure 16.2 , the galvanic cell is shown for this system. Note that this reaction produces 0.48 V. But what if we wanted the reverse reaction to occur, the nonspontaneous reaction? This can be accomplished by applying a voltage in excess of 0.48 V from an external electrical source. This is shown on the right-hand side of Figure 16.2 . In this electrolytic cell, electricity is being used to produce the nonspontaneous redox reaction.

Quantitative Aspects of Electrochemistry

One of the most widely used applications of electrolytic cells is in electrolysis , the decomposition of a compound. Water may be decomposed into hydrogen and oxygen. Aluminum oxide may be electrolyzed to produce aluminum metal. In these situations, several questions may be asked: How long will it take; how much can be produced; what current must be used? Given any two of these quantities, the third may be calculated. To answer these questions, the balanced half-reaction must be known. Then the following relationships can be applied:

1 Faraday = 96,500 coulombs per mole of electrons
(F = 96,500 C/mol e or 96,500 J/V mol)
1 ampere = 1 coulomb/second (A = C/s)

Knowing the amperage and how long it is being applied (seconds), the coulombs can be calculated. Then the coulombs can be converted into moles of electrons, and the moles of electrons can be related to the moles (and then grams) of material being electrolyzed through the balanced half-reaction.

Figure 16.2   Comparison of a galvanic cell (left) and an electrolytic cell (right).

For example, if liquid titanium(IV) chloride (acidified with HCl) is electrolyzed by a current of 1.000 amp for 2.000 h, how many grams of titanium will be produced?

Answer:

Calculation of E °cell also allows for the calculation of two other useful quantities—the Gibbs free energy (ΔG °) and the equilibrium constant (K ).

The Gibbs free energy is the best single thermodynamic indicator of whether a reaction will be spontaneous (review the Thermodynamics chapter). The Gibbs free energy for a reaction can be calculated from the E ° of the reaction using the following equation:

where F is Faraday’s constant of 96,500 C/mol e = 96,500 J/V mol.

If the redox reaction is at equilibrium, E ° = 0, the equilibrium constant may be calculated by:

Let’s apply these relationships. Determine ΔG ° and K for the following reaction:

Answer:

For this reaction, two electrons are transferred from the Ni to the Ag. Thus, n is 2 for this reaction. The value of F (96,500 J/V mol) is given on the exam, so you will not need to memorize it.

The first answer is:

The second answer is:

This gives a K of about 1035 (actually K = 3 × 1035 ). In many cases, the approximate value will be all you need for the AP exam.

Nernst Equation

Thus far, all of our calculations have been based on the standard cell potential or standard half-cell potentials—that is, the standard state conditions that were defined previously. However, many times the cell is not at standard conditions—commonly the concentrations are not 1 M . The actual cell potential, E cell , can be calculated by the use of the Nernst equation :

where R is the ideal gas constant, T is the Kelvin temperature, n is the number of electrons transferred, F is Faraday’s constant, and Q is the reaction quotient discussed in the Equilibrium chapter. The second form, involving log Q , is the more useful form. If one knows the cell reaction, the concentrations of ions, and E °cell , then the actual cell potential can be calculated. Another useful application of the Nernst equation is in calculating the concentration of one of the reactants from cell-potential measurements. Knowing the actual cell potential and E °cell allows calculation of Q , the reaction quotient. Knowing Q and all but one of the concentrations allows the calculation of the unknown concentration. Another application of the Nernst equation is in concentration cells. A concentration cell is an electrochemical cell in which the same chemical species is used in both cell compartments, but differing in concentration. Because the half-reactions are the same, E °cell = 0.00 V. Simply substituting the appropriate concentrations into the reaction quotient allows calculation of the actual cell potential.

When using the Nernst equation on a cell reaction in which the overall reaction is not supplied, only the half-reactions and concentrations, there are two equivalent methods to work the problem. The first way is to write the overall redox reaction based upon E ° values, and then apply the Nernst equation. If E cell turns out to be negative, it indicates that the reaction is not a spontaneous one (an electrolytic cell), or that the reaction is written backwards if it supposed to be a galvanic cell. If it is supposed to be a galvanic cell, all you need to do is reverse the overall reaction and change the sign on E cell to positive. The other method involves using the Nernst equation with the individual half-reactions, then combining them depending on whether or not it is a galvanic cell. The only disadvantage to the second method is that you must use the Nernst equation twice. Either method should lead you to the correct answer.

Let’s practice. Calculate the potential of a half-cell containing 0.10 M K2 Cr2 O7 (aq), 0.20 M Cr3+ (aq), and 1.0 × 10−4 M H+ (aq).

Answer:

The following half-reaction is given on the AP exam:

Experiments

Electrochemical experiments are normally concerned with standard cell voltages. Measurements of the cell potential are essential and require a voltmeter (potentiometer). These measurements may be taken from different combinations of half-cells, or from measurements before and after changes of some aspect of the cell were made.

Using measurements of different half-cell combinations, a set of “standard” reduction potentials may be constructed. This set will be similar to a table of standard reduction potentials. The solutions used in the half-cells must be of known concentration. These solutions are produced by weighing reagents and diluting to volume. The measurements will require a balance and a volumetric flask. It is also possible to produce known concentrations by diluting solutions. This method requires a pipette and a volumetric flask. Review Chapter 13 on Solutions and Colligative Properties for solution techniques.

Common Mistakes to Avoid

  1. Be sure your units cancel to give the unit wanted in your final answer.
  2. Be sure to round your answer off to the correct number of significant figures.
  3. Remember that oxidation is the loss of electrons and reduction the gain, and that in redox reactions the same number of electrons is lost and gained.
  4. When diagramming an electrochemical cell, be sure the electrons go from anode to cathode.
  5. Be sure that for a galvanic cell, the cell potential is greater than 0.
  6. In cell notation, be sure to write anode, anode compartment, salt bridge, cathode compartment, cathode in this specific order.
  7. When using a multiplier to equalize electron loss and gain in reduction half-cell potentials,do not use the multiplier on the voltage of the half-cell.

 Review Questions

Use these questions to review the content of this chapter and practice for the AP Chemistry exam. First are 14 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. Following those is a multipart free-response question like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

Multiple-Choice Questions

Answer the following questions in 20 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.

Choose one of the following for questions 1 and 2.

(A) There is no change in the voltage.

(B) The voltage becomes zero.

(C) The voltage increases.

(D) The voltage decreases, but stays positive.

The following reaction takes place in a voltaic cell:

Zn(s) + Cu2+ (1 M ) → Cu(s) + Zn2+ (1 M )

The cell has a voltage that is measured and found to be +1.10 V.

1 . What happens to the cell voltage when the copper electrode is made smaller?

2 . What happens to the cell voltage when the salt bridge is filled with deionized water instead of 1 M KNO3 ?

3 . MnO4  (aq) + H+ (aq) +  → Mn2+ (aq) + H2 O(l) + CO2 (g)

What is the coefficient of H+ when the above reaction is balanced?

(A) 16

(B) 2

(C) 8

(D) 5

4 . S2 O3 2− (aq) + OH (aq) → SO4 2− (aq) + H2 O(l) + e

After the above half-reaction is balanced, which of the following are the respective coefficients of OH and SO4 2− in the balanced half-reaction?

(A) 8 and 3

(B) 6 and 2

(C) 10 and 2

(D) 5 and 2

5 . How many moles of Pt may be deposited on the cathode when 0.80 F of electricity is passed through a 1.0 M solution of Pt4+ ?

(A) 1.0 mol

(B) 0.60 mol

(C) 0.20 mol

(D) 0.80 mol

6 . Cr2 O7 2− (aq) + 14 H+ (aq) + 3 S2− (aq) → 2 Cr3+ (aq) + 3 S(s) + 7 H2 O(l)

For the above reaction, pick the true statement from the following.

(A) The S2– is reduced by Cr2 O7 2− .

(B) The oxidation number of chromium changes from +7 to +3.

(C) The oxidation number of sulfur remains –2.

(D) The S2– is oxidized by 

7 . H+ (aq) + NO3  (aq) + e → NO(g) + H2 O(g)

What is the coefficient for water when the above half-reaction is balanced?

(A) 3

(B) 4

(C) 2

(D) 1

8 . Co2+ + 2 e → Co E ° = –0.28 V
Cd2+ + 2 e → Cd E ° = –0.40 V

Given the above standard reduction potentials, estimate the approximate value of the equilibrium constant for the following reaction:

Cd + Co2+ → Cd2+ + Co

(A) 10–4

(B) 10–2

(C) 104

(D) 1016

9 . A sample of silver is to be purified by electrorefining. This will separate the silver from an impurity of gold. The impure silver is made into an electrode. Which of the following is the best way to set up the electrolytic cell?

(A) an impure silver cathode and an inert anode

(B) an impure silver cathode and a pure gold anode

(C) a pure silver cathode with an impure silver anode

(D) a pure gold cathode with an impure silver anode

10 . 2 Fe3+ + Zn → Zn2+ + 2 Fe2+

The reaction shown above was used in an electrolytic cell. The voltage measured for the cell was not equal to the calculated E ° for the cell. Which of the following could cause this discrepancy?

(A) The anion in the anode compartment was chloride instead of nitrate as in the cathode compartment.

(B) One or more of the ion concentrations was not 1 M .

(C) Both of the solutions were at 25°C instead of 0°C.

(D) The solution in the salt bridge was Na2 SO4 instead of KNO3 .

11 . How many grams of mercury could be produced by electrolyzing a 1.0 M Hg(NO3 )2 solution with a current of 2.00 A for 3.00 h?

(A) 22.4 g

(B) 201 g

(C) 11.2 g

(D) 44.8 g

12 . An electrolysis cell was constructed with two platinum electrodes in a 1.00 M aqueous solution of KCl. An odorless gas evolves from one electrode, and a gas with a distinctive odor evolves from the other electrode. Choose the correct statement from the following list.

(A) The gas with the distinctive odor was evolved at the anode.

(B) The odorless gas was oxygen.

(C) The gas with the distinctive odor was evolved at the cathode.

(D) The odorless gas was evolved at the anode.

13 . 2 BrO3  (aq) + 12 H+ (aq) + 10 e → Br2 (aq) + 6 H2 O(l)

Which of the following statements is correct for the above reaction?

(A) The BrO3  undergoes oxidation at the anode.

(B) Br goes from a –1 oxidation to a 0 oxidation state.

(C) Br2 is oxidized at the anode.

(D) The BrO3  undergoes reduction at the cathode.

14 . 2 M(s) + 3 Zn2+ (aq) →
2 M3+ (aq) + 3 Zn2+ (aq)         E ° = 0.90 V
Zn2+ (aq) + 2e → Zn(s)        E ° = –0.76 V

Using the above information, determine the standard reduction potential for the following reaction:

M3+ (aq) + 3 e → M(s)

(A) 0.90 V

(B) –1.66 V

(C) 0.00 V

(D) –0.62 V

Answers and Explanations for the Multiple-Choice Questions

1 . A —The size of the electrode is not important.

2 . B —The salt bridge serves as an ion source to maintain charge neutrality. Deionized water would not be an ion source, so the cell could not operate.

3 . A —The balanced equation is:

4 . C —The balanced equation is:

5 . C —It takes 4 mol of electrons (4 F ) to change the platinum ions to platinum metal. The calculation would be: (0.80 F ) (1 mol Pt/4 F ) = 0.20 mol Pt.

6 . D —The dichromate ion oxidizes the sulfide ion to elemental sulfur, as the sulfide ion reduces the dichromate ion to the chromium(III) ion. Chromium goes from +6 to +3, while sulfur goes from –2 to 0. The hydrogen remains at +1, so it is neither oxidized nor reduced.

7 . C —The balanced chemical equation is:

8 . C —Using the equation:

You should realize that the log K = 4 gives K = 104 . The actual value is K = 1.1 × 104 .

9 . C —The impure silver must be oxidized so it will go into solution. Oxidation occurs at the anode. Reduction is required to convert the silver ions to pure silver. Reduction occurs at the cathode. The cathode must be pure silver; otherwise, it could be contaminated with the cathode material.

10 . B —If the voltage was not equal to E °, then the cell was not standard. Standard cells have 1 M concentrations and operate at 25°C with a partial pressure of each gas equal to 1 atm. No gases are involved in this reaction, so the cell must be operating at a different temperature or a different concentration (or both).

11 . A —

You can estimate the answer by replacing 96,500 with 100,000 and 200.6 with 200.

12 . A —The gases produced are hydrogen (at the cathode) and chlorine (at the anode). Hydrogen is odorless, while chlorine has a distinctive odor.

13 . D —The bromate ion, BrO3  , is gaining electrons, so it is being reduced. Reduction always occurs at the cathode.

14 . B —The half-reactions giving the overall reaction must be:

Thus, –0.76 + ? = 0.90, giving ? = 1.66 V. The half-reaction under consideration is the reverse of the one used in this combination, so the sign of the calculated voltage must be reversed. Do not make the mistake of multiplying the voltages when the half-reactions were multiplied to equalize the electrons.

Free-Response Question

You have 15 minutes to answer the following multipart question. You may use a calculator and the tables in the back of the book.

Question 1

V = voltmeter

The above galvanic cell is constructed with a cobalt electrode in a 1.0 M Co(NO3 )2 solution in the left compartment and a silver electrode in a 1.0 M AgNO3 in the right compartment. The salt bridge contains a KNO3 solution. The cell voltage is positive.

(a) What is the balanced net ionic equation for the reaction, and what is the cell potential?

(b) Which electrode is the anode? Justify your answer.

(c) If some solid Co(NO3 )2 is added to the cobalt compartment, what will happen to the voltage? Justify your answer.

(d) If the cell operates until equilibrium is established, what will the potential be? Justify your answer.

Answer and Explanation for the Free-Response Question

(a) The cell reaction is:

Co(s) + 2 Ag+ (aq) → Co2+ (aq) + 2 Ag(s)

Give yourself 1 point if you got this correct. The physical states are not necessary.

The calculation of the cell potential may be done in different ways. Here is one method:

Give yourself 1 point for the correct answer regardless of the method used. The most common mistake is to multiply the silver voltage by 2. You do not get the point for an answer of 1 V.

(b) The cobalt is the anode.

You get 1 point for this statement.

The reason Co is the anode is because the Co is oxidized.

You get 1 point for this statement or if you state that the Co loses electrons.

(c) The voltage would decrease. The excess Co2+ , from the Co(NO3 )2 , would impede the reactions from proceeding as written.

Give yourself 1 point for saying decrease. Give yourself 1 point for the explanation.

(d) At equilibrium the cell voltage would be 0 V.

This is worth 1 point.

At equilibrium no work is done, so the potential must be zero.

Or give yourself 1 point for this answer.

There are a total of 7 points possible on this question.

 Rapid Review

  • In redox reactions, electrons are lost and gained. Oxidation is the loss of electrons, and reduction is the gain of electrons.
  • The same number of electrons is lost and gained in redox reactions.
  • Galvanic (voltaic) cells produce electricity through the use of a redox reaction.
  • The anode is the electrode at which the oxidation half-reaction takes place. The anode compartment is the solution in which the anode is immersed.
  • The cathode is the electrode at which reduction takes place, and the cathode compartment is the solution in which the cathode is immersed.
  • A salt bridge is used in an electrochemical cell to maintain electrical neutrality in the cell compartments.
  • Be able to diagram an electrochemical cell.
  • The cell notation is a shorthand way of representing a cell. It has the form:
    anode|anode compartment||cathode compartment|cathode
  • Standard reduction potentials are used to calculate the cell potential under standard conditions. All half-reactions are shown in the reduction form.
  • For a galvanic cell .
  • . Know how to use this equation to calculate  .
  • Electrolytic cells use an external source of electricity to produce a desired redox reaction.
  • Review how to diagram an electrolytic cell.
  • The following relationships can be used to calculate quantitative changes that occur in an electrochemical cell, especially an electrolytic one: 1F = 96,500 C per mole of electron (F = 96,500 C/mol e = 96,500 J/V mol) and 1 amp = 1 C/s (A = C/s).
  • The standard cell potential can be used to calculate the Gibbs free energy for the reaction: ΔG° = −nFE °cell . Know how to use this equation.
  • The standard cell potential can also be used to calculate the equilibrium constant for a reaction: log . Know how to use this equation.