Electrochemistry. 1. Determine the oxidation states of each element in the following compounds. (Reference: Ex. 4:16) a. N 2 N: b.

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1 Name: Electrochemistry Two of the most common types of chemical reactions are acid-base reactions in which protons are transferred between two reactants and oxidation-reduction reactions in which electrons are transferred between two reactants. This unit focuses on the latter of these two types of reactions oxidationreduction reactions or electron transfer processes. To understand oxidation-reduction reactions, it is first important to understand the concept of an oxidation state or oxidation number. The oxidation state or oxidation number is the imaginary charge which an atom would have if the shared electrons among covalently bonded atoms were considered to be shared equally among bonded atoms (Masterton, pp ). Chemists have developed rules for the assignment of oxidation numbers to the elements within an atom, an ion and a compound. The rules provide a hierarchical system by which the oxidation state of an element can be determined. They are stated on pages and restated here for easy reference. Rules for Assigning Oxidation Numbers 1. The oxidation number of an element in the elemental state is zero. 2. The oxidation number of a one-atom ion is the same as the charge. 3. The oxidation number of certain elements is the same in nearly all compounds. For instance, F atoms in compounds: -1; alkali metals: +1; alkali earth metals: +2; O atoms in compounds: -2; H atoms in compounds: +1 Exceptions: O atoms in peroxides have oxidation numbers of -1; H atoms in hydrides have oxidation numbers of The sum of all positive and negative oxidation numbers in an ion or molecule is equal to the charge of that species. 1. Determine the oxidation states of each element in the following compounds. (Reference: Ex. 4:16) a. N 2 N: b. Cl - Cl: c. OH - O: H: d. MnO 2 Mn: O: e. H 2 SO 4 H: S: O: f. KMnO 4 K: Mn: O: g. Mg(ClO 3 ) 2 Mg: Cl: O: h. Mg(ClO) 2 Mg: Cl: O: i. Al 2 (SO 4 ) 3 Al: S: O: j. NaH 2 PO 4 Na: H: P: O: k. Na 2 CO 3 Na: C: O: l. NH 4 F N: H: F: m. Sr(OH) 2 Sr: O: H: Page 1

2 2. In an oxidation-reduction (redox for short) reaction, there is a transfer of electrons between reactants. This transfer of electrons results in a change in oxidation number. Oxidation is an increase in oxidation number associated with a loss of electrons. Reduction is a decrease in oxidation number associated with a gain of electrons. (This is in part remembered by the mnemonics OIL RIG and LEO GER.) Oxidation Reduction The oxidation process cannot take place without the reduction process. If one substance is losing electrons (being oxidized), then another substance must gain the electrons (be reduced). And for one substance to gain electrons (be reduced) there must be another substance willing to lose electrons (be oxidized). In this sense, oxidation-reduction reactions involve assistants or agents - oxidizing agents and reducing agents. The oxidizing agent is the substance which assists another substance in losing electrons; the oxidizing agent provides this assist by gaining the electrons that are lost. In this sense, the oxidizing agent is the substance which is reduced. By the same reasoning, the reducing agent is the substance which is oxidized; by losing electrons, the reducing agent assists another substance in gaining electrons. (Reference: Masterton, p. 87) Determine oxidation states for the reactants and products in the following reactions and then identify which substance is being oxidized and which substance is being reduced. (Note that the equations are not balanced.) a. FeCl3 + H2S ----> FeCl2 + HCl + S Substance being oxidized: Oxidizing agent: Substance being reduced: Reducing agent: b. Zn + H > Zn 2+ + H2 Substance being oxidized: Oxidizing agent: Substance being reduced: Reducing agent: c. Al + H2SO4 ----> Al2(SO4)3 + H2 Substance being oxidized: Oxidizing agent: Substance being reduced: Reducing agent: d. Cu + Br2 + OH > Cu2O + Br - + H2O Substance being oxidized: Oxidizing agent: Substance being reduced: Reducing agent: Page 2

3 Name: 3. For each of the redox reactions shown in #2 above, use oxidation states to determine the # of electrons being gained or lost by the oxidized and reduced element. Then use this information to write and balance the chemical equations. a. FeCl3 + H2S ----> FeCl2 + HCl + S Element being oxidized: Element being reduced: # of e - lost: # of e - gained: Balanced Eq'n: FeCl3 + H2S ----> FeCl2 + HCl + S b. Zn + H > Zn 2+ + H2 Element being oxidized: Element being reduced: # of e - lost: # of e - gained: Balanced Eq'n: KClO3 ----> KCl + O2 c. Al + H2SO4 ----> Al2(SO4)3 + H2 Element being oxidized: Element being reduced: # of e - lost: # of e - gained: Balanced Eq'n: Al + H2SO4 ----> Al2(SO4)3 + H2 d. Cu + Br2 + OH > Cu2O + Br - + H2O Element being oxidized: Element being reduced: # of e - lost: # of e - gained: Balanced Eq'n: Cu + Br2 + OH > Cu2O + Br - + H2O 4. The method involved above is a useful method for balancing the easiest chemical equations for redox reactions. The more difficult redox reactions need to be balanced using the half-reaction method discussed in Chapter 4.4. The method is summarized here as follows: Step 1: Step 2: Step 3: Use oxidation numbers to identify which substance is oxidized and which is reduced. Then, write separate equations for the two half-reactions. Balance one of the half-equations. a. Balance the elements other than H and O. b. Balance the oxidation numbers by adding electrons (for reduction, add e - to the reactant side; for oxidation, add e - to the product side) c. Balance charge by adding H + ions (for acidic solutions) or OH - ions (for basic solutions). d. Balance H atoms by adding H 2 O molecules to one of the sides. e. Check to make sure that O atoms are balanced. Balance the other half-equation. Step 4: Multiply each half-reaction by an integer so that the electrons (e - ) in each equation balance. Then combine the two half-equations and cancel identical species wherever possible. Page 3

4 Use the half-reaction method to balance the chemical equations for the following redox reactions. a. Cr 2 O Hg Cr 3+ + Hg 2+ b. Zn + VO 3 - VO 2+ + Zn 2+ c. Zn + NO 3 - NH Zn 2+ d. S 2 O BiO 3 - Bi 3+ + SO 4 2- e. P 4 + IO 3 - H 2 PO I 2 f. MnO H 2 O 2 Mn 2+ + O 2 g. H 2 SO 3 + I 2 + H 2 O H 2 SO 4 + HI Page 4

5 Name: h. H 2 O 2 + SO 2 H + + SO 4 2- i. FeCl 3 + SnCl 2 FeCl 2 + SnCl 4 j. SbCl 5 + KI SbCl 3 + KCl + I 2 k. CdS + I 2 + HCl CdCl 2 + HI + S l. KClO 3 + FeSO 4 + H 2 SO 4 KCl + Fe 2 (SO 4 ) 3 + H 2 O m. K 2 Cr 2 O 7 + HCl + H 2 S KCl + CrCl 3 + S + H 2 O n. Ag + HNO 3 + Ca(ClO) 2 + CaCl 2 AgCl + Ca(NO 3 ) 2 + H 2 O Page 5

6 o. HgS + HCl + HNO 3 HgCl 2 + S + NO + H 2 O p. Na 2 Cr 2 O 7 + HNO 3 + H 2 O 2 H 3 CrO 8 + NaNO 3 + H 2 O q. K 2 MnO 4 + HNO 3 KMnO 4 + MnO 2 + KNO 3 + H 2 O 5. A voltaic cell is a device which utilizes a redox reaction in order to convert chemical energy to electrical energy. In such a device the two half reactions oxidation and reduction are separated in such a manner that the oxidation occurs in one compartment of the voltaic cell and the reduction occurs in a separate compartment. These two compartments are known as half cells. The two half cells are connected by a wire in order to allow electrons lost by the oxidation to travel through the wire to the compartment where reduction occurs. The wire is typically attached to a piece of metal known as an electrode. The electrode where oxidation occurs is known as the anode. Because electrons are lossed or released at this electrode, it is commonly referred to as the negative terminal of the voltaic cell. The electrode where reduction occurs is known as the cathode. Because electrons are drawn through the wire to this half cell, it is commonly referred to as the positive terminal. As the two half reactions take place in the half cells, the movement of electrons through the wire from the terminal to the + terminal, a net charge begins to accumulate in each terminal. This net charge tends to counteract the flow of electrons and eventually brings the reaction to a halt. To prevent this, the two half cells are also connected by a salt bridge. The salt bridge provides a permeable membrane which allows a flow of ions between the half cells so as to prevent this buildup of net charge. The diagram at the right is an example of a voltaic cell. The copper electrode is immersed in a solution of copper ions. The zinc electrode is immersed in a solution of zinc ions. The Cu 2+ ions are reduced at the copper electrode. Zinc atoms are oxidized to form Zn 2+ ions at the zinc electrode. The copper electrode is the cathode and the zinc electrode is the anode. Page 6

7 Name: There is a handy line notation which can be used to describe an electrochemical cell. The line notation to describe the above voltaic cell is Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) The anode compartment is listed on the left and the cathode departments are listed on the right; the compartments are separated by a double vertical line ( ). The above line notation corresponds to the following half reactions: Oxidation: Zn (s) Zn 2+ (aq) + 2 e - Reduction: 2 e - + Cu 2+ (aq) Cu (s) Match the following line notations to the balanced oxidation-reduction half reactions and vice versa. (Reference: Masterton, p. 483) For each voltaic cell, diagram the cell. In your diagram, show each half-cell, label the anode and the cathode, draw the salt bridge and show the direction of electron flow through the wire. a. Line Notation: Zn (s) Zn 2+ (aq) I 2(s) I - (aq) Oxidation: Reduction: Diagram: b. Line Notation: Al (s) Al 3+ (aq) Zn 2+ (aq) Zn (s) Oxidation: Reduction: Diagram: Page 7

8 c. Line Notation: Al (s) Al 3+ (aq) Zn 2+ (aq) Zn (s) Oxidation: Reduction: Diagram: d. Line Notation: Oxidation: Al (s) Al 3+ (aq) + 3 e - Reduction: 2e - + Cu 2+ (aq) Cu (s) Diagram: e. Line Notation: Oxidation: Cu (s) Cu 2+ (aq) + 2 e - Reduction: e - + Ag + (aq) Ag (s) Diagram: Page 8

9 Name: 6. Some substances are more easily reduced or oxidized than others. The relative tendency of a substance to be reduced can be and has been measured. The two-column table of data below lists several substances undergoing a reduction half-reaction. The reactions are written in order according to the relative tendency of the reduction reaction to occur. Substances listed nearest to the top of the leftmost column have the greatest tendency to undergo reduction. Substances listed nearest the bottom of the rightmost column have the lowest tendency to undergo reduction. Thus, fluorine (F 2 ) is easily reduced and lithium ions (Li + ) are NOT easily reduced. The relative tendency of substances to be oxidized can also be determined from this table. Substances on the product side of each reaction are typically oxidized; simply imagine each reaction occurring in reverse. Substances on the product side nearest to the bottom of the rightmost column have the greatest tendency to be oxidized. Similarly, substances on the product side nearest to the top of the leftmost column are least easily oxidized. Use Table 18.1 (from p. 487) (or a similar table below) to answer the following questions. a. Rank these substances according to their ability to be reduced, from least able to most able. F 2, I 2, Li + : Cu 2+, Al 3+, Zn 2+ : b. Rank these substances according to their ability to be oxidized, from least able to most able. Na, Fe, Al : Ag, Fe, Zn : c. Which of these metals would most likely oxidize in air and turn into an oxide: gold (Au), silver (Ag), iron (Fe), tin (Sn)? least likely? Page 9

10 7. Table 18.1 (in our textbook) (or Table 17.1 on the previous page) not only orders substances according to their relative ability to be reduced, it also includes numerical information describing that reduction tendency. The numerical values are known as standard reduction potentials and given the symbol E red. The standard of standard reduction potential means that the measurements were taken under standard conditions concentrations of solutions are 1.0 M and partial pressures of gases are 1.0 atm. All measurements are expressed relative to the standard hydrogen electrode. Those substances found above hydrogen on the table (and on the reactant side) are more easily reduced than hydrogen ions (H + ); and those substances found below hydrogen on the table are less easily reduced than hydrogen ions (H + ). The unit of measurement is the Volt. The volt indicates the Joules of energy released by the half reaction per Coulomb of charge which is gained; a Coulomb of charge is equivalent to 6.25 x electrons. As can be seen, it takes a WHOLE lot of electrons being gained by a substance in order for a mere Joule of energy to be released. Using the table allows one to determine the reduction potential for any oxidation-reduction reaction. The table only lists E for reduction half-reaction. Yet every oxidation-reduction reaction also includes both the reduction half-reaction (whose E is listed) and a oxidation half-reaction. The trick is to recognize that any oxidation reaction is simply the reverse of the reduction reaction listed in the table. The oxidation potential has a E ox which is the opposite sign as the reduction potential value which is listed in the table. So for the oxidation of copper to copper(ii) ions, the E ox is V. Cu e - Cu E red = 0.34 V Cu Cu e - E ox = V Use this information to calculate the cell potentials (E cell ) for the following voltaic cells. Write the two half reactions (or the line notation), list the E values, and calculate the overall cell potentials. (Reference: Masterton, pp , Example 18.2, Example 18.3) a. Line Notation: Al (s) Al 3+ (aq) Zn 2+ (aq) Zn (s) Oxidation: E ox = V Reduction: E red = V E cell = V b. Line Notation: Mg (s) Mg 2+ (aq) Al 3+ (aq) Al (s) Oxidation: E ox = V Reduction: E red = V E cell = V c. Line Notation: Ag (s) Ag + (aq) Cu 2+ (aq) Cu (s) Oxidation: E ox = V Reduction: E red = V E cell = V d. Line Notation: Sn (s) Sn 2+ (aq) Cu 2+ (aq) Cu (s) Oxidation: E ox = V Reduction: E red = V E cell = V Page 10

11 Name: The reactions which occur within a voltaic cell can be determined by knowing the contents of the cell and the standard reduction potentials of those contents. The principle is: Reactions which are found highest in the table of reduction potentials are the ones most likely to undergo reduction. Reactions found lowest in the table of reduction potentials are more likely to occur as an oxidation reaction. Use this principle and the E values from Table 18.1 (or Table 17.1) to answer the following questions. 8. a. (Ag, Ag +, Cd, Cd 2+ ) is oxidized. Please circle. (Ag, Ag +, Cd, Cd 2+ ) is reduced. Please circle. b. The (left, right) half-cell is the cathode. The (left, right) half-cell is the anode. c. Draw arrow on diagram to indicate the dir'n of electron flow through the wire. d. Oxidation half-equation: Reduction half-equation: e. The E cell value is: 9. a. (Pb, Pb 2+, Mg, Mg 2+ ) is oxidized. Please circle. (Pb, Pb 2+, Mg, Mg 2+ ) is reduced. Please circle. b. The (left, right) half-cell is the cathode. The (left, right) half-cell is the anode. c. Draw arrow on diagram to indicate the dir'n of electron flow through the wire. d. Oxidation half-equation: Reduction half-equation: e. The E cell value is: 10. a. (Zn, Zn 2+, Ni, Ni 2+ ) is oxidized. Please circle. (Zn, Zn 2+, Ni, Ni 2+ ) is reduced. Please circle. b. The (left, right) half-cell is the cathode. The (left, right) half-cell is the anode. c. Draw arrow on diagram to indicate the dir'n of electron flow through the wire. d. Oxidation half-equation: Reduction half-equation: e. The E cell value is: Page 11

12 11. Suppose that the following pairs of metals are placed in separate solutions containing their respective ions to form a voltaic cell. For each pair, identify the anode and corresponding halfequation, the cathode and the corresponding half-equation, and the E cell value. Use the E values in Table 18.1 (Table 17.1). Finally, use standard line notation to describe the voltaic cell. a. Zinc (Zn) and aluminum (Al) Cathode half-equation: Anode half-equation: E cell = V PSYW: Standard line notation: b. Magnesium (Mg) and iron (Fe) Cathode half-equation: Anode half-equation: E cell = V PSYW: Standard line notation: c. Silver (Ag) and gold (Au) Cathode half-equation: Anode half-equation: E cell = V PSYW: Standard line notation: d. Magnesium (Mg) and gold (Au) Cathode half-equation: Anode half-equation: E cell = V PSYW: Standard line notation: e. Lithium (Li) and cadmium (Cd) Cathode half-reaction: Anode half-reaction: E cell = V PSYW: Standard line notation: Page 12

13 Name: For the following questions, write the half-equations from the given balanced chemical equations (or vice versa). Then use the reduction potential values in Table 18.1 (or Table 17.1) to determine the overall cell potential (E cell ). Finally, indicate if the reaction is spontaneous or nonspontaneous as written. 12. Equation: 2 Al 3+ (aq) + 3 Mn (s) 2 Al (s) + 3 Mn 2+ (aq) Oxidation: E ox = Reduction: E red = E cell = V Spontaneous or Nonspontaneous? (circle) 13. Equation: 2 Al 3+ (aq) + 3 Mg (s) 2 Al (s) + 3 Mg 2+ (aq) Oxidation: E ox = Reduction: E red = E cell = V Spontaneous or Nonspontaneous? (circle) 14. Equation: Oxidation: ½ I 2(s) + 3 H 2 O (l) IO 3(g) H + (aq) + 5 e- E ox = Reduction: Zn 2+ (aq) + 2 e - Zn (s) E red = E cell = V Spontaneous or Nonspontaneous? (circle) 15. Equation: Oxidation: 4 OH - (aq) O 2(g) + 2 H 2 O (l) + 4 e- E ox = Reduction: Ba 2+ (aq) + 2 e - Ba (s) E red = E cell = V Spontaneous or Nonspontaneous? (circle) 16. Equation: NO 3- (aq) + 4 H + (aq) + Cr (s) NO (g) + 2 H 2 O (l) + Cr 3+ (aq) Oxidation: E ox = Reduction: E red = E cell = V Spontaneous or Nonspontaneous? (circle) Page 13

14 The process of using an electric current to drive an otherwise nonspontaneous redox reaction into occurring is known as electrolysis. Electrolysis involves the passage of an electric current through a solution or molten liquid of an ionic compound. By inserting two electrodes into the solution (or liquid) and connecting them to the positive and negative terminals of the electrical power source, an electrolysis reaction occurs. The electrode connected to the negative terminal draws positive ions towards it. The positive ions (cations) gain electrons at this location and so become reduced. The electrode connected to the positive terminal attracts negative ions (anions) towards it. The negative ions lose electrons and become oxidized. Such a cell is known as an electrolytic cell and operates quite differently than a voltaic cell. In an electrolytic cell, the anode is still the location where oxidation occurs; only this oxidation halfreaction occurs at the so-called positive terminal. Similarly, the reduction occurring at the anode location is occurring at the electrode connected to the so-called negative terminal. Electrolysis has very useful industrial applications, providing a means of electroplating a metal on the surface of a solid and the separation of metals from its ores. (Reference: Chapter 18.5) 17. Suppose a solution contains the Cr 3+, Cu 2+ and Al 3+ ions. As an electric current is applied to the solution, which ion would be the first to be reduced to solid form? Which ion would be the last to be reduced to its solid form? 18. Suppose that a current is passed through a strongly concentrated solution of sodium chloride. Use reduction potential values in Table 18.1 (or Table 17.1) to determine the half-equations occurring at the anode and the cathode. HINT: consider all the major species present in the solution. Anode reaction: Cathode reaction: 19. Repeat question #18 for a strongly concentrated solution of copper(ii) chloride. HINT: consider all the major species present in the solution. Anode reaction: Cathode reaction: 20. There is a good deal of stoichiometry associated with an electrolytic process. A current flow for some amount of time provides electrons which serve to turn a metal from the reduced state (ion form) to the oxidized state (elemental form). The unit for current is the Ampere or amp; an Ampere is equivalent to a Coulomb per second; and Coulombs is equivalent to a mole of electrons. And finally, an electron is a stoichiometric quantity in the half reaction equation. Use these relationships to solve the following problems. (Reference: p. 496; Example 18.8) a. If a current of 2.5 Amperes is passed through an aqueous solution of CuCl 2 for 22.0 minutes, then what mass of copper will be produced? Page 14

15 Name: b. An electrolysis process is being used to plate out aluminum from an aqueous solution of aluminum iodide. What current would be required to plate out 0.50 grams in a time period of 10.0 minutes? 21. Oxygen is an element that is readily reduced. As such, it is an effective oxidizing agent and commonly causes metals to be oxidized. The oxidation of a metal by oxygen is known as corrosion. With the exception of gold, all the metals in Table 18.1 (or Table 17.1) are less easily reduced than oxygen. Thus it is common to find metallic elements in an oxidized state in nature. The oxidation or corrosion of a metal reduces its attractive appeal and its structural integrity. Steel structures are composed of an alloy of iron. Iron is very subject to corrosion in air (a.k.a., rusting). Unlike other metals, the iron oxide rust on a steel surface tends to flake off the surface, causing the corrosion to penetrate the surface to the interior of the steel. To protect steel from corrosion, a process known as cathodic protection is used. A more active (or more easily oxidized) metal is connected by a wire to the steel surface. Being more easily oxidized, this volunteer metal supplies electrons to the iron in steel to keep in a reduced state as long as possible. Use your table of reduction potentials to decide which of the following metals would provide cathodic protection for the iron in steel. Circle all that apply. Silver (Ag) Magnesium (Mg) Tin (Sn) Gold (Au) Aluminum (Al) 22. The standard cell potential calculations presume standard conditions concentrations of ions in aqueous solutions of 1.0 M, temperatures of 298 K, and partial pressures of gases of 1.0 atm. Deviations from standard conditions results in changes in the cell potentials. That is, the cell potential for standard conditions (E ) is not necessarily equal to the cell potential value at nonstandard conditions (E ). LeChatelier s principle can be used to predict whether E is greater than, less than or equal to E. Consider the following voltaic cell: Al (s) Al 3+ (aq) Mn 2+ (aq) Mn (s) E = 0.48 V Determine if E is greater than or less than E for the stated conditions. (Reference: Section 18.4) a. [Al 3+ ] = 1.0 M, [Mn 2+ ] = 2.0 M E E ( >, <, = ) b. [Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M E E ( >, <, = ) c. [Al 3+ ] = 1.0 M, [Mn 2+ ] = 0.50 M E E ( >, <, = ) d. [Al 3+ ] = 0.20 M, [Mn 2+ ] = 1.0 M E E ( >, <, = ) Page 15

16 23. The dependence of cell potential upon concentrations of aqueous solutions (and partial pressures of gases) at 298 K is predicted by the Nernst equation: E = E - (0.0591/n) * log(q) where E is the cell potential at non-standard conditions, E is the cell potential at standard conditions (see table of reduction potentials), n is the number of electrons transferred in the balanced redox equation, and Q is the reaction quotient (a subject covered in the equilibrium unit). Use the Nernst equation to determine the non-standard cell potentials under the following conditions. PSYW (Reference: Masterton, Table 18.4 and Example 18.6 and Example 18.7) a. Al (s) Al 3+ (aq) Mn 2+ (aq) Mn (s) [Al 3+ ] = 1.0 M, [Mn 2+ ] = 2.0 M b. Al (s) Al 3+ (aq) Mn 2+ (aq) Mn (s) [Al 3+ ] = 2.0 M, [Mn 2+ ] = 2.0 M c. Al (s) Al 3+ (aq) Mn 2+ (aq) Mn (s) [Al 3+ ] = 2.0 M, [Mn 2+ ] = 0.5 M d. Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) [Cu 2+ ] = 1.8 M, [Zn 2+ ] = 1.2 M e. Sn (s) Sn 2+ (aq) Cu 2+ (aq) Cu (s) [Cu 2+ ] = 0.48 M, [Sn 2+ ] = 1.6 M Page 16

17 Name: 24. A zinc-copper battery is constructed as follows at 25 C: Zn (s) Zn 2+ (aq) (0.10 M) Cu 2+ (aq) (2.50 M) Cu (s) The mass of each electrode is g. a. Write the balanced chemical equation for this redox reaction. b. Use the Nernst equation to calculate the cell potential when this battery is first connected. c. Suppose that 10.0 Amps of current has flowed for 10.0 hrs. Calculate the cell potential at this time. (Assume that the two half cells contain 1.00 L of solution.) PSYW d. Calculate the mass of each electrode after 10.0 hrs. PSYW e. How long can this battery deliver a 10.0 Amp current before it goes completely dead? Page 17

18 25. Concentration cells are voltaic cells in which both compartments contain the same components but at different concentrations. The diagram at the right depicts a concentration cell consisting of iron(ii) ions and an iron electrode. Since the reduction potential depends upon concentration, each cell will have a different reduction potential. For the compartment on the left, the lower Fe(II) concentration would cause the reduction process to occur less readily compared to the compartment on the right. Thus reduction occurs in the right compartment and oxidation occurs in the left compartment. Left Fe (s) Fe 2+ (aq) + 2 e - Right Fe 2+ (aq) + 2 e - Fe (s) Electrons will move from the left compartment (anode) to the right compartment (cathode). Overall, there is no net chemical equation, yet one would observe over time that the mass of solid iron in the right compartment would decrease and the mass of solid iron in the right compartment would increase. Diagram the following cell (similar to above diagram) and indicate the half reaction which occurs in each compartment and indicate the direction of electron flow through the wire. Cu (s) Cu 2+ (aq) (1.50 M) Cu 2+ (aq) (2.50 M) Cu (s) 26. As learned earlier, there must be a positive cell potential for a reaction to be spontaneous. The cell potential for a concentration cell can be calculated using the Nernst equation. The E value is 0 V and the reaction quotient (Q) is the ratio of the smaller concentration to the larger concentration. Calculate the cell potential for the following concentration cells. a. Cu (s) Cu 2+ (aq) (1.50 M) Cu 2+ (aq) (2.50 M) Cu (s) b. Cu (s) Cu 2+ (aq) (0.900 M) Cu 2+ (aq) (0.020 M) Cu (s) c. Ag (s) Ag + (aq) (1.850 M) Ag + (aq) (0.120 M) Ag (s) Page 18

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20 13. Obviously voltaic cells and redox reactions are the basis of modern day batteries. One of the first batteries developed was constructed by Italian scientist Alessandro Giuseppe Antonio Anastasio Volta. Plagued as a child by indecisive parents, Volta grew up to become interested in a study of the relationship between chemistry and electricity. In 1800, he developed the so-called voltaic pile in which he stacked alternating layers of zinc metal and cardboard-like wafers soaked in silver ions. A wire connecting the top and the bottom of the layer will produce a significant current due to the large voltage output of the collection of cells. The silver ions are reduced to silver. The zinc solid is oxidized to its ions. Assuming standard conditions, the net voltage output of a single cell (one pair of anode and cathode) is 1.56 volts. e - + Ag + (aq) Ag (s) Zn (s) Zn 2+ (aq) + 2e - E = 0.80 V E = 0.76 V Page 20