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Chapter 21: CHEM 1B: Electrochemistry: GENERAL and Electrical Work

Instructor: Dr. Orlando E. Raola Santa Rosa Junior College

21-1 Chapter 21

Electrochemistry: Chemical Change and Electrical Work

21-2 Electrochemistry: Chemical Change and Electrical Work 21.1 Reactions and Electrochemical Cells

21.2 Voltaic Cells: Using Spontaneous Reactions to Generate

21.3 Cell Potential: Output of a Voltaic Cell

21.4 Free Energy and Electrical Work

21.5 Electrochemical Processes in Batteries

21.6 : An Environmental Voltaic Cell

21.7 Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions

21-3 Overview of Redox Reactions

Oxidation is the loss of and reduction is the gain of electrons. These processes occur simultaneously. Oxidation results in an increase in O.N. while reduction results in a decrease in O.N.

The takes electrons from the substance being oxidized. The oxidizing agent is therefore reduced.

The takes electrons from the substance being oxidized. The reducing agent is therefore oxidized.

21-4 Figure 21.1 A summary of redox terminology, as applied to the reaction of with .

0 +1 +2 0 + 2+ Zn(s) + 2H (aq) → Zn (aq) + H2(g)

21-5 Half-Reaction Method for Balancing Redox Reactions

The half-reaction method divides a redox reaction into its oxidation and reduction half-reactions. - This reflects their physical separation in electrochemical cells.

This method does not require assigning O.N.s.

The half-reaction method is easier to apply to reactions in acidic or basic .

21-6 Steps in the Half-Reaction Method

• Divide the skeleton reaction into two half-reactions, each of which contains the oxidized and reduced forms of one of the species. • Balance the and charges in each half-reaction. – First balance atoms other than O and H, then O, then H. – Charge is balanced by adding electrons (e-) to the left side in the reduction half-reaction and to the right side in the oxidation half-reaction. • If necessary, multiply one or both half-reactions by an integer so that – number of e- gained in reduction = number of e- lost in oxidation • Add the balanced half-reactions, and include states of matter.

21-7 Balancing Redox Reactions in Acidic

2- - 3+ Cr2O7 (aq) + I (aq) → Cr (aq) + I2(s) Step 1: Divide the reaction into half-reactions. 2- 3+ Cr2O7 → Cr - I → I2 Step 2: Balance the atoms and charges in each half-reaction.

2- 3+ For the Cr2O7 /Cr half-reaction: Balance atoms other than O and H: 2- 3+ Cr2O7 → 2Cr

Balance O atoms by adding H2O :

2- 3+ Cr2O7 → 2Cr + 7H2O

21-8 + Balance H atoms by adding H : + 2- 3+ 14H + Cr2O7 → 2Cr + 7H2O Balance charges by adding electrons:

- + 2- 3+ 6e + 14H + Cr2O7 → 2Cr + 7H2O 2- This is the reduction half-reaction. Cr2O7 is reduced, and is the oxidizing agent. The O.N. of Cr decreases from +6 to +3.

- For the I /I2 half-reaction: Balance atoms other than O and H: - 2I → I2 There are no O or H atoms, so we balance charges by adding electrons:

- - 2I → I2 + 2e This is the oxidation half-reaction. I- is oxidized, and is the reducing agent. The O.N. of I increases from -1 to 0.

21-9 Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e- lost in the oxidation equals the number of e- gained in the reduction. The reduction half-reaction shows that 6e- are gained; the oxidation half-reaction shows only 2e- being lost and must be multiplied by 3:

- - 3(2I → I2 + 2e ) - - 6I → 3I2 + 6e Step 4: Add the half-reactions, canceling substances that appear on both sides, and include states of matter. Electrons must always cancel.

- + 2- 3+ 6e + 14H + Cr2O7 → 2Cr + 7H2O - - 6I → 3I2 + 6e - + 2- 3+ 6I (aq) + 14H (aq) + Cr2O7 (aq) → 3I2(s) + 7H2O(l) + 2Cr (aq)

21-10 Balancing Redox Reactions in Basic Solution

+ + An acidic solution contains H ions and H2O. We use H ions to balance H atoms.

- A basic solution contains OH ions and H2O. To balance H atoms, we proceed as if in acidic solution, and then add one OH- ion to both sides of the equation.

For every OH- ion and H+ ion that appear on the same

side of the equation we form an H2O .

Excess H2O molecules are canceled in the final step, when we cancel electrons and other common species.

21-11 Sample Problem 21.1 Balancing a Redox Reaction in Basic Solution

PROBLEM: Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid dioxide. Balance the skeleton ionic equation for the reaction

between NaMnO4 and Na2C2O4 in basic solution: - 2- 2- MnO4 (aq) + C2O4 (aq) → MnO2(s) + CO3 (aq) [basic solution] PLAN: We follow the numbered steps as described in the text, and proceed through step 4 as if this reaction occurs in acidic solution. Then we add the appropriate number of OH- ions

and cancel excess H2O molecules. SOLUTION: Step 1: Divide the reaction into half-reactions. - 2- 2- MnO4 → MnO2 C2O4 → CO3

21-12 Sample Problem 21.1 Step 2: Balance the atoms and charges in each half-reaction. Balance atoms other than O and H: - 2- 2- MnO4 → MnO2 C2O4 → 2CO3

Balance O atoms by adding H2O molecules: - 2- 2- MnO4 → MnO2 + 2H2O 2H2O + C2O4 → 2CO3

+ Balance H atoms by adding H ions: + - 2- 2- + 4H + MnO4 → MnO2 + 2H2O 2H2O + C2O4 → 2CO3 + 4H

Balance charges by adding electrons: - + - 2- 2- + - 3e + 4H + MnO4 → MnO2 + 2H2O 2H2O + C2O4 → 2CO3 + 4H + 2e [reduction] [oxidation]

21-13 Sample Problem 21.1

Step 3: Multiply each half-reaction, if necessary, by an integer so that the number of e- lost in the oxidation equals the number of e- gained in the reduction. - + - x 2 6e + 8H + 2MnO4 → 2MnO2 + 4H2O 2- 2- + - x 3 6H2O + 3C2O4 → 6CO3 + 12H + 6e

Step 4: Add the half-reactions, canceling substances that appear on both sides. - + - 6e + 8H + 2MnO4 → 2MnO2 + 4H2O 2- 2- + - 2 6H2O + 3C2O4 → 6CO3 +4 12H + 6e - 2- 2- + 2MnO4 + 2H2O + 3C2O4 → 2MnO2 + 6CO3 + 4H

21-14 Sample Problem 21.1

Basic. Add OH- to both sides of the equation to neutralize H+, and

cancel H2O.

- 2- - 2- + - 2MnO4 + 2H2O + 3C2O4 + 4OH → 2MnO2 + 6CO3 + [4H + 4OH ] - 2- - 2- 2MnO4 + 2H2O + 3C2O4 + 4OH → 2MnO2 + 6CO3 + 2 4H2O

Including states of matter gives the final balanced equation:

- 2- - 2- 2MnO4 (aq) + 3C2O4 (aq) + 4OH (aq) → 2MnO2(s) + 6CO3 (aq) + 2H2O(l)

21-15 Electrochemical Cells

A voltaic cell uses a spontaneous redox reaction (DG < 0) to generate electrical energy. - The system does work on the surroundings. A uses electrical energy to drive a nonspontaneous reaction (DG > 0). - The surroundings do work on the system. Both types of cell are constructed using two placed in an solution. The is the at which oxidation occurs. The is the electrode at which reduction occurs.

21-16 Figure 21.2 General characteristics of (A) voltaic and (B) electrolytic cells.

21-17 Spontaneous Redox Reactions

A strip of zinc in a solution of Cu2+ ions will react spontaneously:

Cu2+(aq) + 2e- → Cu(s) [reduction] Zn(s) → Zn2+(aq) + 2e- [oxidation] Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)

Zn is oxidized, and loses electrons to Cu2+.

Although e- are being transferred, electrical energy is not generated because the reacting substances are in the same container.

21-18 Figure 21.3 The spontaneous reaction between zinc and (II) ion.

21-19 Construction of a Voltaic Cell

Each half-reaction takes place in its own half-cell, so that the reactions are physically separate.

Each half-cell consists of an electrode in an electrolyte solution.

The half-cells are connected by the external circuit. A completes the electrical circuit.

21-20 Figure 21.4A A voltaic cell based on the zinc-copper reaction.

Oxidation half-reaction Reduction half-reaction Zn(s) → Zn2+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s)

Overall (cell) reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 21-21 Operation of the Voltaic Cell

Oxidation (loss of e-) occurs at the anode, which is therefore the source of e-. Zn(s) → Zn2+(aq) + 2e- Over time, the Zn(s) anode decreases in and the [Zn2+] in the electrolyte solution increases.

Reduction (gain of e-) occurs at the cathode, where the e- are used up. Cu2+(aq) + 2e- → Cu(s) Over time, the [Cu2+] in this half-cell decreases and the mass of the Cu(s) cathode increases.

21-22 Figure 21.4B A voltaic cell based on the zinc-copper reaction.

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Oxidation half-reaction Reduction half-reaction Zn(s) → Zn2+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s)

After several , the The Cu cathode gains Zn anode weighs less as mass over time as Cu2+ 2+ Zn is oxidized to Zn . ions are reduced to Cu.

21-23 Charges of the Electrodes

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

The anode produces e- by the oxidation of Zn(s). The anode is the negative electrode in a voltaic cell.

Electrons flow through the external from the anode to the cathode, where they are used to reduce Cu2+ ions.

The cathode is the positive electrode in a voltaic cell.

21-24 The Salt Bridge

The salt bridge completes the electrical circuit and allows ions to flow through both half-cells.

As Zn is oxidized at the anode, Zn2+ ions are formed and enter the solution. Cu2+ ions leave solution to be reduced at the cathode. The salt bridge maintains electrical neutrality by allowing excess Zn2+ ions to enter from the anode, and excess negative ions to enter from the cathode. A salt bridge contains nonreacting cations and anions, + - often K and NO3 , dissolved in a gel.

21-25 Flow of Charge in a Voltaic Cell

Electrons flow through the wire from anode to cathode.

Zn(s) → Zn2+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s)

2+ 2- Zn SO4 Cations move through the salt Anions move through the salt bridge from the anode solution bridge from the cathode solution to the cathode solution. to the anode solution. By convention, a voltaic cell is shown with the anode on the left and the cathode on the right.

21-26 Active and Inactive Electrodes

An active electrode is an active component in its half- cell and is a reactant or product in the overall reaction.

An inactive electrode provides a surface for the reaction and completes the circuit. It does not participate actively in the overall reaction. - Inactive electrodes are necessary when none of the reaction components can be used as an electrode.

Inactive electrodes are usually unreactive substances such as graphite or .

21-27 Figure 21.5 A voltaic cell using inactive electrodes.

Oxidation half-reaction Reduction half-reaction - - - + - 2+ 2I (aq) → I2(s) + 2e MnO4 (aq) + 8H (aq) + 5e → Mn (aq) + 4H2O(l)

Overall (cell) reaction - + - 2+ 2MnO4 (aq) + 16H (aq) + 10I (aq) → 2Mn (aq) + 5I2(s) + 8H2O(l) 21-28 Notation for a Voltaic Cell

The components of each half-cell are written in the same order as in their half-reactions.

The anode components The cathode components are written on the left. are written on the right. Zn(s)│Zn2+(aq)║Cu2+(aq) │Cu(s)

The single line shows a phase The double line shows that the half- boundary between the cells are physically separated. components of a half-cell.

If needed, of dissolved components are given in parentheses. (If not stated, it is assumed that they are 1 M.)

21-29 Notation for a Voltaic Cell

- - + 2+ graphite I (aq)│I2(s)║MnO4 (aq), H (aq), Mn (aq) │graphite

The inert electrode is specified. A comma is used to show components that are in the same phase.

21-30 Sample Problem 21.2 Describing a Voltaic Cell with Diagram and Notation

PROBLEM: Draw a diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with

a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.

PLAN: From the given contents of the half-cells, we write the half- reactions. To determine which is the anode compartment (oxidation) and which is the cathode (reduction), we note the relative electrode charges. Electrons are released into the anode during oxidation, so it has a negative charge. Since Cr is negative, it must be the anode, and Ag is the cathode.

21-31 Sample Problem 21.2

SOLUTION: The half-reactions are: Ag+(aq) + e- → Ag(s) [reduction; cathode] Cr(s) → Cr3+(aq) + 3e- [oxidation; anode] The balanced overall equation is: 3Ag+ + Cr(s) → 3Ag(s) + Cr3+(aq)

The cell notation is given by: Cr(s)│Cr3+(aq)║Ag+(aq)│Ag(s)

The cell diagram shows the anode on the left and the cathode on the right.

21-32 Electrical Potential and the Voltaic Cell When the switch is closed and no reaction is occurring, each half-cell is in an equilibrium state: Zn(s) Zn2+(aq) + 2e- (in Zn metal) Cu(s) Cu2+(aq) + 2e- (in Cu metal) Zn is a stronger reducing agent than Cu, so the position of the Zn equilibrium lies farther to the right. Zn has a higher electrical potential than Cu. When the switch is closed, e- flow from Zn to Cu to equalize the difference in electrical potential The spontaneous reaction occurs as a result of the different abilities of these to give up their electrons.

21-33 Cell Potential

A voltaic cell converts the DG of a spontaneous redox reaction into the kinetic energy of electrons.

The cell potential (Ecell) of a voltaic cell depends on the difference in electrical potential between the two electrodes. Cell potential is also called the of the cell or the electromotive forces (emf).

Ecell > 0 for a spontaneous process.

21-34 Table 21.1 of Some Voltaic Cells

Voltaic Cell Voltage (V) Common alkaline battery 1.5 -acid car battery (6 cells ≈ 12 V) 2.1 Calculator battery () 1.3 Lithium-ion laptop battery 3.7 (~5000 cells in 6-ft eel 0.15 = 750 V) of giant squid 0.070 (across cell membrane)

21-35 Figure 21.6 Measuring the standard cell potential of a zinc- copper cell.

The standard cell potential is designated E°cell and is measured at a specified with no current flowing and all components in their standard states.

21-36 Standard Electrode Potentials

The standard (E°half-cell) is the potential of a given half-reaction when all components are in their standard states.

By convention, all standard electrode potentials refer to the half-reaction written as a reduction.

The standard cell potential depends on the difference between the abilities of the two electrodes to act as reducing agents.

E°cell = E°cathode (reduction) - E°anode (oxidation)

21-37 The Standard Hydrogen Electrode

Half-cell potentials are measured relative to a standard reference half-cell.

The standard hydrogen electrode has a standard

electrode potential defined as zero (E°reference = 0.00 V).

This standard electrode consists of a Pt electrode with H2 gas at 1 atm bubbling through it. The Pt electrode is immersed in 1 M strong acid.

+ - 2H (aq; 1 M) + 2e H2(g; 1 atm) E°ref = 0.00V

21-38 Figure 21.7 Determining an unknown E°half-cell with the standard reference (hydrogen) electrode. Oxidation half-reaction Zn(s) → Zn2+(aq) + 2e−

Reduction half-reaction + - 2H3O (aq) + 2e → H2(g) + 2H2O(l)

Overall (cell) reaction + 2+ Zn(s) + 2H3O (aq) → Zn (aq) + H2(g) + 2H2O(l) 21-39 Sample Problem 21.3 Calculating an Unknown E°half-cell from E°cell

PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: 2+ - Br2(aq) + Zn(s) → Zn (aq) + 2Br (aq) E°cell = 1.83 V.

Calculate E°bromine, given that E°zInc = -0.76 V

PLAN: E°cell is positive, so the reaction is spontaneous as written. By dividing the reaction into half-reactions, we

see that Br2 is reduced and Zn is oxidized; thus, the zinc half-cell contains the anode. We can use the equation for

E°cell to calculate E°bromine.

SOLUTION: - - Br2(aq) + 2e → 2Br (aq) [reduction; cathode] 2+ - Zn(s) → Zn (aq) + 2e [oxidation; anode] E°zinc = -0.76 V

21-40 Sample Problem 21.3

E°cell = E°cathode − E°anode

1.83 = E°bromine – (-0.76)

1.83 + 0.76 = E°bromine

E°bromine = 1.07 V

21-41 Comparing E°half-cell values

Standard electrode potentials refer to the half-reaction as a reduction.

E° values therefore reflect the ability of the reactant to act as an oxidizing agent.

The more positive the E° value, the more readily the

reactant will act as an oxidizing agent. The more negative the E° value, the more readily the

product will act as a reducing agent.

21-42 Table 21.2 Selected Standard Electrode Potentials (298 K)

Half-Reaction E°(V) − − F2(g) + 2e 2F (aq) +2.87 − − Cl2(g) + 2e 2Cl (aq) +1.36 + − 2+ MnO2(g) + 4H (aq) + 2e Mn (aq) + 2H2O(l) +1.23 - + − agent reducing of strength NO3 (aq) + 4H (aq) + 3e NO(g) + 2H2O(l) +0.96 Ag+(aq) + e− Ag(s) +0.80 Fe3+(g) + e− Fe2+(aq) +0.77 − − O2(g) + 2H2O(l) + 4e 4OH (aq) +0.40 Cu2+(aq) + 2e− Cu(s) +0.34 + − 2H (aq) + 2e H2(g) 0.00 + − + N2(g) + 5H (aq) + 4e N2H5 (aq) −0.23

Fe2+(aq) + 2e− Fe(s) −0.44

strength agent of oxidizing − − 2H2O(l) + 2e H2(g) + 2OH (aq) −0.83 Na+(aq) + e− Na(s) −2.71 Li+(aq) + e− Li(s) −3.05

21-43 Writing Spontaneous Redox Reactions

Each half-reaction contains both a reducing agent and an oxidizing agent. The stronger oxidizing and reducing agents react spontaneously to form the weaker ones.

A spontaneous redox reaction (E°cell > 0) will occur between an oxidizing agent and any reducing agent that lies below it in the emf series (i.e., one that has a less positive value for E°). The oxidizing agent is the reactant from the half-reaction

with the more positive E°half-cell.

21-44 Using half-reactions to write a spontaneous redox reaction:

2+ - Sn (aq) + 2e → Sn(s) E°tin = -0.14 V + - Ag (aq) + e → Ag(s) E° = 0.80 V

Step 1: Reverse one of the half-reactions into an oxidation step so that the difference between the E° values will be positive.

Here the Ag+/Ag half-reaction has the more positive E° value, so it must be the reduction. This half-reaction remains as written. We reverse the Sn2+/Sn half-reaction, but we do not reverse the sign:

2+ - Sn(s) → Sn (aq) + 2e E°tin = -0.14 V

21-45 Step 2: Multiply the half-reactions if necessary so that the number of e- lost is equal to the number or e- gained.

+ - 2Ag (aq) + 2e → 2Ag(s) E°silver = 0.80 V

Note that we multiply the equation but not the value for E°.

Step 3: Add the reactions together, cancelling common species.

Calculate E°cell = E°cathode - E°anode.

2+ - Sn(s) → Sn (aq) + 2e E°tin = -0.14 V + - 2Ag (aq) + 2e → 2Ag(s) E°silver = 0.80 V

+ 2+ Sn(s) + 2Ag (aq) → 2Ag(s) + Sn (aq) E°cell = 0.94 V

E°cell = E°silver – E°tin = 0.80 – (-0.14) = 0.94 V

21-46 Sample Problem 21.4 Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength PROBLEM: (a) Combine the following three half-reactions into three balanced equations for spontaneous reactions (A, B,

and C), and calculate E°cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents. - + - (1) NO3 (aq) + 4H (aq) + 3e → NO(g) + 2H2O(l) E° = 0.96 V + - + (2) N2(g) + 5H (aq) + 4e → N2H5 (aq) E° = -0.23 V + - 2+ (3) MnO2(s) +4H (aq) + 2e → Mn (aq) + 2H2O(l) E° = 1.23 V

PLAN: To write the redox equations, we combine the possible pairs of half-reactions. In each case the half-reaction with the less positive value for E° will be reversed. We make e- lost equal to e- gained, add the half-reactions and calculate

E°cell. We can then rank the relative strengths of the oxidizing and reducing agents by comparing E° values. 21-47 Sample Problem 21.4 SOLUTION: (a) For (1) and (2), equation (2) has the smaller, less positive E° value:

- + - (1) NO3 (aq) + 4H (aq) + 3e → NO(g) + 2H2O(l) E° = 0.96 V + + - (2) N2H5 (aq) → N2(g) + 5H (aq) + 4e E° = -0.23 V We multiply equation (1) by 4 and equation (2) by 3: - + - (1) 4NO3 (aq) + 16H (aq) + 12e → 4NO(g) + 8H2O(l) E° = 0.96 V + + - (2) 3N2H5 (aq) → 3N2(g) + 15H (aq) + 12e E° = -0.23 V

- + + (A) 4NO3 (aq) + 3N2H5 (aq) + H (aq) → 3N2(g) + 4NO(g) + 8H2O(l)

E°cell = 0.96 V – (-0.23 V) = 1.19 V

21-48 Sample Problem 21.4

For (1) and (3), equation (1) has the smaller, less positive E° value:

- + - (1) NO(g) + 2H2O(l) → NO3 (aq) + 4H (aq) + 3e E° = 0.96 V + - 2+ (3) MnO2(s) +4H (aq) + 2e → Mn (aq) + 2H2O(l) E° = 1.23 V

We multiply equation (1) by 2 and equation (3) by 3: - + - (1) 2NO(g) + 4H2O(l) → 2NO3 (aq) + 8H (aq) + 6e E° = 0.96 V + - 2+ (3) 3MnO2(s) +12H (aq) + 6e → 3Mn (aq) + 6H2O(l) E° = 1.23 V

+ 2+ - (B) 3MnO2(s) + 4H (aq) + 2NO(g) → 2Mn (aq) + 2NO3 (aq) + 2H2O(l)

E°cell = 1.23 V – (0.96 V) = 0.27 V

21-49 Sample Problem 21.4

For (2) and (3), equation (2) has the smaller, less positive E° value:

+ + - (2) N2H5 (aq) → N2(g) + 5H (aq) + 4e E° = -0.23 V + - 2+ (3) MnO2(s) +4H (aq) + 2e → Mn (aq) + 2H2O(l) E° = 1.23 V

We multiply equation (3) by 2: + + - (2) N2H5 (aq) → N2(g) + 5H (aq) + 4e E° = -0.23 V + - 2+ (3) 2MnO2(s) +8H (aq) + 4e → 2Mn (aq) + 4H2O(l) E° = 1.23 V

+ + 2+ (C) N2H5 (aq) + 2MnO2(s) + 3H (aq) → N2(g) + 2Mn (aq) + 4H2O(l)

E°cell = 1.23 V – (-0.23 V) = 1.46 V

21-50 Sample Problem 21.4:

(b) We first rank the oxidizing and reducing agents within each

equation, then we can compare E°cell values. - Equation (A) Oxidizing agents: NO3 > N2 + Reducing Agents: N2H5 > NO - Equation (B) Oxidizing agents: MnO2 > NO3 + Reducing Agents: N2H5 > NO

Equation (C) Oxidizing agents: MnO2 > N2 + 2+ Reducing Agents: N2H5 > Mn

Comparing the relative strengths from the E°cell values: - Oxidizing agents: MnO2 > NO3 > N2 + 2+ Reducing agents: N2H5 > NO > Mn

21-51 The Activity Series of the Metals

Metals that can displace H2 from acid are metals that are stronger reducing agents than H2.

+ - 2H (aq) + 2e → H2(g) E° = 0.00V Fe(s) → Fe2+ + 2e- E° = -0.44 V

+ 2+ Fe(s) + 2H (aq) → Fe (aq) + H2(g) E°cell = 0.44 V

The lower the metal is in the list of half-cell potentials,

the more positive its E°half-cell and the stronger it is as a reducing agent.

The larger (more positive) the E°half-cell of a metal, the more active a metal it is.

21-52 The Activity Series of the Metals

Metals that cannot displace H2 from acid are metals that are weaker reducing agents than H2.

+ - 2H (aq) + 2e → H2(g) E° = 0.00V 2Ag(s) → 2Ag+ + 2e- E° = 0.80 V

+ + 2Ag(s) + 2H (aq) → 2Ag (aq) + H2(g) E°cell = -0.80 V

The higher the metal is in the list of half-cell potentials,

the smaller its E°half-cell and the weaker it is as a reducing agent.

The smaller (more negative) the E°half-cell of a metal, the less active a metal it is.

21-53 The Activity Series of the Metals

Metals that can displace H2 from are metals whose half-reactions lie below that of H2O:

- - 2H2O(l) + 2e → H2(g) + 2OH (aq) E° = -0.42 V 2Na(s) → 2Na+(aq) + 2e- E° = -2.17 V

+ - 2Na(s) + 2H2O(l) → 2Na (aq) + H2(g) + 2OH (aq) E°cell = 2.29 V

21-54 The Activity Series of the Metals

We can also predict whether one metal can displace another from solution. Any metal that is lower in the list of electrode potentials (i.e., has a larger E° value) will reduce the ion of a metal higher up the list.

Zn(s) → Zn2+(aq) + 2e- E° = -0.76V Fe2+(aq) + 2e- → Fe(s) E° = -0.44V

2+ 2+ Zn (s) + Fe (aq) → Zn (aq) + Fe(s) E°cell = 0.32 V

21-55 Figure 21.8 The reaction of calcium in water.

Oxidation half-reaction Reduction half-reaction 2+ - - - Ca(s) → Ca (aq) + 2e 2H2O(l) + 2e → H2(g) + 2OH (aq)

Overall (cell) reaction 2+ - Ca(s) + 2H2O(l) → Ca (aq) + H2(g) + 2OH (aq) 21-56 Figure 21.9 A dental “voltaic cell.”

Biting down with a filled tooth on a scrap of aluminum foil will cause

pain. The foil acts as an active anode (E°aluminum = -1.66 V), saliva as the electrolyte, and the filling as an inactive cathode as O2 is reduced to H2O.

21-57 Free Energy and Electrical Work

For a spontaneous redox reaction, DG < 0 and Ecell > 0.

n = mol of e- transferred DG = -nFEcell F is the = 9.65x104 J/V·mol e-

Under standard conditions, DG° = -nFE°cell and

RT 0.0592 V E° = ln K or E° = log K cell nF cell n for T = 298.15 K

21-58 Figure 21.10 The interrelationship of DG°, E°cell, and K.

DG° Reaction Parameters at the Reaction at standard-

DG° K E°cell state conditions < 0 > 1 > 0 spontaneous 0 1 0 at equilibrium > 0 < 1 < 0 nonspontaneous

E°cell K

RT ln K E°cell = nF

21-59 Sample Problem 21.5 Calculating K and DG° from E°cell

PROBLEM: Lead can displace silver from solution, and silver occurs in trace amounts in some ores of lead. Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s) As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ore. Calculate K and DG° at 298.15 K for this reaction. PLAN: We divide the spontaneous redox reaction into the half-reactions

and use values from Appendix D to calculate E°cell. From this we can find K and DG°.

SOLUTION: Writing the half-reactions with their E° values:

(1) Ag+(aq) + e- → Ag(s) E° = 0.80 V (2) Pb2+(aq) + 2e- → Pb(s) E° = -0.13 V

21-60 Sample Problem 21.5

We need to reverse equation (2) and multiply equation (1) by 2: (1) 2Ag+(aq) + 2e- → 2Ag(s) E° = 0.80 V (2) Pb(s) → Pb2+(aq) + 2e- E° = -0.13 V

+ 2+ 2Ag (aq) + Pb(s) → 2Ag(s) + Pb (aq) Ecell = 0.80 – (-0.13) = 0.93 V

RT ln K 0.0592 V E°cell = = log K = 0.93 V nF 2 0.93 V x 2 log K = = 31.42 K = 2.6x1031 0.0592 V

2 mol e- 96.5 kJ DG° = -nFE°cell = - x x 0.93 V mol rxn V·mol e-

= -1.8x102 kJ/mol rxn

21-61 Cell Potential and

Nernst Equation RT Ecell = E°cell - ln Q nF

• When Q < 1, [reactant] > [product], ln Q < 0, so Ecell > E°cell

• When Q = 1, [reactant] = [product], ln Q = 0, so Ecell = E°cell

• When Q > 1, [reactant] < [product], ln Q > 0, so Ecell < E°cell

We can simplify the equation as before for T = 298.15 K:

0.0592 V Ecell = E°cell - log Q n

21-62 Sample Problem 21.6 Using the to Calculate Ecell

PROBLEM: In a test of a new reference electrode, a constructs 2+ + a voltaic cell consisting of a Zn/Zn half-cell and an H2/H half-cell under the following conditions: [Zn2+] = 0.010 M [H+] = 2.5 M P = 0.30 atm H2 Calculate Ecell at 298 K.

PLAN: To apply the Nernst equation and determine Ecell, we must know E°cell and Q. We write the equation for the spontaneous reaction and calculate E°cell from standard electrode potentials. We must convert the given pressure to molarity in order to have consistent units.

SOLUTION: + - (1) 2H (aq) + 2e → H2(g) E° = 0.00 V (2) Zn(s) → Zn2+(aq) + 2e- E° = -0.76 V

+ 2+ 2H (aq) + Zn(s) → H2(g) + Zn (aq) E°cell = 0.00 – (-0.76) = 0.76 V

21-63 Sample Problem 21.6

Converting pressure to molarity: n P 0.30 atm = = = 1.2x10-2 M V RT 0.0821 atm·L x 298.15 K mol·K [H ][Zn2+] 0.012 x 0.010 Q = 2 = = 1.9x10-5 [H+]2 (2.5)2

Solving for Ecell at 25°C (298.15 K), with n = 2: 0.0592 V Ecell = E°cell - log Q n 0.0592 V = 1.10 V - log(1.9x10-5) = 0.76 – (-0.14 V) = 0.90 V 2

21-64 Figure 21.11A The relation between Ecell and log Q for the zinc- copper cell.

2+ 2+ If the reaction starts with [Zn ] < [Cu ] (Q < 1), Ecell is higher than the standard cell potential. As the reaction proceeds, [Zn2+] decreases and [Cu2+] increases, so

Ecell drops. Eventually the system reaches equilibrium and the cell can no longer do work. 21-65 Figure 21.11B The relation between Ecell and log Q for the zinc- copper cell.

A summary of the changes in Ecell as any voltaic cell operates.

21-66 Concentration Cells

A exploits the effect of concentration changes on cell potential.

The cell has the same half-reaction in both cell compartments, but with different concentrations of electrolyte: Cu(s) → Cu2+(aq; 0.10 M) + 2e- [anode; oxidation] Cu2+(aq; 1.0 M) → Cu(s) [cathode; reduction] Cu2+(aq; 1.0 M) → Cu2+(aq; 0.10 M)

As long as the concentrations of the solutions are different, the cell potential is > 0 and the cell can do work.

21-67 Figure 21.12 A concentration cell based on the Cu/Cu2+ half-reaction.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Oxidation half-reaction Cu(s) → Cu2+(aq, 0.1 M) + 2e- Ecell > 0 as long as the half-cell concentrations are different. Reduction half-reaction The cell is no longer able to do work Cu2+(aq, 1.0 M) + 2e- → Cu(s) once the concentrations are equal. Overall (cell) reaction Cu2+(aq,1.0 M) → Cu2+(aq, 0.1 M)

21-68 Sample Problem 21.7 Calculating the Potential of a Concentration Cell PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells. In

half-cell A, the electrolyte is 0.0100 M AgNO3; in half-cell -4 B, it is 4.0x10 M AgNO3. What is the cell potential at 298.15 K?

PLAN: The standard half-cell potentials are identical, so E°cell is zero, and we find Ecell from the Nernst equation. Half-cell A has a higher [Ag+], so Ag+ ions are reduced and plate out on electrode A, which is therefore the cathode. In half-cell B, Ag atoms of the electrode are oxidized and Ag+ ions enter the solution. Electrode B is thus the anode. As for all voltaic cells, the cathode is positive and the anode is negative.

21-69 Sample Problem 21.7

SOLUTION: The [Ag+] decreases in half-cell A and increases in half- cell B, so the spontaneous reaction is: Ag+(aq; 0.010 M) [half-cell A] → Ag+(aq; 4.0x10-4 M) [half-cell B]

[Ag+] E = E° - 0.0592 V log dil cell cell + 1 [Ag ]conc 4.0x10-4 = 0.0 V - 0.0592 log = 0.0828 V 0.010

21-70 Figure 21.13 Laboratory measurement of pH. The operation of a pH meter illustrates an important application of concentration cells. The glass electrode monitors the [H+] of the solution relative to its own fixed internal [H+].

An older style of pH meter Modern pH meters use a includes two electrodes. combination electrode.

21-71 Table 21.3 Some Ions Measured with Ion-Specific Electrodes

Species Detected Typical Sample

+ NH3/NH4 Industrial wastewater, seawater - CO2/HCO3 Blood, groundwater F- Drinking water, urine, soil, industrial stack gases Br- Grain, plant tissue I- Milk, pharmaceuticals - NO3 Soil, fertilizer, drinking water K+ Blood serum, soil, wine H+ Laboratory solutions, soil, natural

21-72 Figure 21.14 Minimocroanalysis.

A microelectrode records electrical impulses of a single in a monkey’s visual cortex. The electrical potential of a nerve cell is due to the difference in concentration of [Na+] and [K+] ions inside and outside the cell. 21-73 Electrochemical Processes in Batteries

A battery consists of self-contained voltaic cells arranged in series, so their individual voltages are added. A primary battery cannot be recharged. The battery is “dead” when the cell reaction has reached equilibrium.

A secondary battery is rechargeable. Once it has run down, electrical energy is supplied to reverse the cell reaction and form more reactant.

21-74 Figure 21.15 Alkaline battery.

- - Anode (oxidation): Zn(s) + 2OH (aq) → ZnO(s) + H2O(l) + 2e - - Cathode (reduction): MnO2(s) + 2H2O(l) + 2e → Mn(OH)2(s) + 2OH (aq) Overall (cell) reaction:

Zn(s) + MnO2(s) + H2O(l) → ZnO(s) + Mn(OH)2(s) Ecell = 1.5 V

21-75 Figure 21.16 Silver button battery.

- - Anode (oxidation): Zn(s) + 2OH (aq) → ZnO(s) + H2O(l) + 2e - - Cathode (reduction): Ag2O(s) + H2O(l) + 2e → 2Ag(s) + 2OH (aq)

Overall (cell) reaction: Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

Ecell = 1.6 V The uses HgO as the oxidizing agent instead of

Ag2O and has cell potential of 1.3 V.

21-76 Figure 21.17 .

The primary lithium battery is widely used in watches, implanted medical devices, and remote-control devices.

Anode (oxidation): 3.5Li(s) → 3.5Li+ + 3.5e- Cathode (reduction): - - AgV2O5.5 + 3.5Li + 3.5e → Li3.5V2O5.5 Overall (cell) reaction:

AgV2O5.5 + 3.5Li(s) → Li3.5V2O5.5

21-77 Figure 21.18 Lead-acid battery.

The lead-acid car battery is a secondary battery and is rechargeable.

21-78 The reactions in a lead-acid battery: The cell generates electrical energy when it discharges as a voltaic cell.

- + - Anode (oxidation): Pb(s) + HSO4 (aq) → PbSO4(s) + H (aq) + 2e Cathode (reduction): + - - PbO2(s) + 3H (aq) + HSO4 (aq) + 2e → PbSO4(s) + 2H2O(l) Overall (cell) reaction (discharge):

PbO2(s) + Pb(s) + H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Ecell = 2.1 V

Overall (cell) reaction (recharge):

2PbSO4(s) + 2H2O(l) → PbO2(s) + Pb(s) + H2SO4(aq)

21-79 Figure 21.19 Nickel-metal hydride battery

- - Anode (oxidation): MH(s) + OH (aq) → M(s) + H2O(l) + e - - Cathode (reduction): NiO(OH)(s) + H2O(l) + e → Ni(OH)2(s) + OH (aq) Overall (cell) reaction:

MH(s) + NiO(OH)(s) → M(s) + Ni(OH)2(s) Ecell = 1.4 V

21-80 Figure 21.20 Lithium-ion battery.

Anode (oxidation): + - LixC6(s) → xLi + xe + C6(s) Cathode (reduction): + - Li1-xMn2O4(s) + xLi + xe → LiMn2O4(s) Overall (cell) reaction:

LixC6(s) + Li1-xMn2O4(s) → LiMn2O4(s)

Ecell = 3.7 V

The secondary (rechargeable) lithium-ion battery is used to power laptop computers, cell phones, and camcorders.

21-81 Fuel Cells

In a , also called a flow cell, reactants enter the cell and products leave, generating through controlled .

Reaction rates are lower in fuel cells than in other batteries, so an is used to decrease the .

21-82 Figure 21.21 Hydrogen fuel cell.

+ - Anode (oxidation): 2H2(g) → 4H (aq) + 4e + - Cathode (reduction): O2(g) + 4H (aq) + 4e → 2H2O(g)

Overall (cell) reaction: 2H2(g) + O2(g) → 2H2O(g) Ecell = 1.2 V

21-83 Corrosion: an Environmental Voltaic Cell

Corrosion is the process whereby metals are oxidized to their oxides and sulfides.

The rusting of is a common form of corrosion.

- is not a direct product of the reaction between Fe and

O2, but arises through a complex electrochemical process. - Rusting requires moisture, and occurs more quickly at low pH, in ionic solutions, and when the iron is in contact with a less active metal.

21-84 The Rusting of Iron The loss of iron: Fe(s) → Fe2+(aq) + 2e- [anodic region; oxidation] + - O2(g) + 4H (aq) + 4e → 2H2O(l) [cathodic region; reduction] + 2+ 2Fe(s) + O2(g) + 4H (aq) → 2Fe (aq) + 2H2O(l) [overall]

The rusting process: Overall reaction: 2+ + 2Fe (aq) + ½O2(g) + (2 + n)H2O(l) → Fe2O3·nH2O(s) + 4H (aq)

H+ ions are consumed in the first step, so lowering the pH increases the overall rate of the process. H+ ions act as a catalyst, since they are regenerated in the second part of the process.

21-85 Figure 21.22 The corrosion of iron.

21-86 Figure 21.23 Enhanced corrosion at sea.

The high ion concentration of seawater enhances the corrosion of iron in hulls and anchors.

21-87 Figure 21.24 The effect of metal-metal contact on the corrosion of iron.

Fe in contact with Cu corrodes Fe in contact with Zn does not faster. corrode. The process is known as cathodic protection. 21-88 Figure 21.25 The use of sacrificial to prevent iron corrosion.

In cathodic protection, an active metal, such as zinc, , or aluminum, acts as the anode and is sacrificed instead of the iron. 21-89 Electrolytic Cells

An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction. Cu(s) → Cu2+(aq) + 2e- [anode; oxidation] Sn2+(aq) + 2e- → Sn(s) [cathode; reduction]

2+ 2+ Cu(s) + Sn (aq) → Cu (aq) + Sn(s) E°cell = -0.48 V and ΔG° = 93 kJ

As with a voltaic cell, oxidation occurs at the anode and reduction takes place at the cathode. An external source supplies the cathode with electrons, which is negative, and removes then from the anode, which is positive. Electrons flow from cathode to anode.

21-90 Figure 21.26 The tin-copper reaction as the basis of a voltaic and an electrolytic cell. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Sn(s) → Sn2+(aq) + 2e- Cu(s) → Cu2+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s) Sn2+(aq) + 2e- → Sn(s) Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq) Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq) voltaic cell electrolytic cell

21-91 Figure 21.27 The processes occurring during the discharge and recharge of a lead-acid battery.

VOLTAIC (discharge)

Switch

ELECTROLYTIC (recharge)

21-92 Table 21.4 Comparison of Voltaic and Electrolytic Cells

Electrode

Cell Type DG Ecell Name Process Sign Voltaic < 0 > 0 Anode Oxidation - Voltaic < 0 > 0 Cathode Reduction +

Electrolytic > 0 < 0 Anode Oxidation +

Electrolytic > 0 < 0 Cathode Reduction -

21-93 Products of

Electrolysis is the splitting (lysing) of a substance by the input of electrical energy.

During electrolysis of a pure, molten salt, the cation will be reduced and the anion will be oxidized. During electrolysis of a mixture of molten salts - the more easily oxidized species (stronger reducing agent) reacts at the anode, and - the more easily reduced species (stronger oxidizing agent) reacts at the cathode.

21-94 Sample Problem 21.8 Predicting the Electrolysis Products of a Molten Salt Mixture PROBLEM: A chemical engineer melts a naturally occurring mixture of

NaBr and MgCl2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction.

PLAN: We need to determine which metal and nonmetal will form more easily at the electrodes. We list the ions as oxidizing or reducing agents. If a metal holds its electrons more tightly than another, it has a higher ionization energy (IE). Its cation will gain electrons more easily, and it will be the stronger oxidizing agent. If a nonmetal holds its electrons less tightly than another, it has a lower (EN). Its anion will lose electrons more easily, and it will be the reducing agent.

21-95 Sample Problem 21.8 SOLUTION: Possible oxidizing agents: Na+, Mg2+ Possible reducing agents: Br-, Cl- Mg is to the right of Na in Period 3. IE increases from left to right across the period, so Mg has the higher IE and gives up its electrons less easily. The Mg2+ ion has a greater attraction for e- than the Na+ ion.

Mg2+(l) + 2e- → Mg(l) [cathode; reduction]

Br is below Cl in Group 7A. EN decreases down the group, so Br accepts e- less readily than Cl. The Br- ion will lose its e- more easily, so it is more easily oxidized.

- - 2Br (l) → Br2(g) + 2e [anode; oxidation]

2+ - The overall cell reaction is: Mg (l) + 2Br (l) → Mg(l) + Br2(g)

21-96 Figure 21.28 The .

Overall (cell) reaction

2H2O(l) → H2(g) + O2(g)

Oxidation half-reaction Reduction half-reaction + - - - 2H2O(l) → 4H (aq) + O2(g) + 4e 2H2O(l) + 4e → 2H2(g) + 2OH (aq)

21-97 Electrolysis of Aqueous Salt Solutions

When an aqueous salt solution is electrolyzed - The strongest oxidizing agent (most positive electrode potential) is reduced, and - The strongest reducing agent (most negative electrode potential) is oxidized. Overvoltage is the additional voltage needed (above that predicted by E° values) to produce gases at metal electrodes.

Overvoltage needs to be taken into account when predicting the products of electrolysis for aqueous solutions.

Overvoltage is 0.4 – 0.6 V for H2(g) or O2(g).

21-98 Summary of the Electrolysis of Aqueous Salt Solutions

• Cations of less active metals (Au, Ag, Cu, Cr, Pt, Cd) are reduced to the metal. • Cations of more active metals are not reduced.

H2O is reduced instead. • Anions that are oxidized, because of

overvoltage from O2 formation, include the halides, except for F-. • Anions that are not oxidized include F- and

common oxoanions. H2O is oxidized instead.

21-99 Sample Problem 21.9 Predicting the Electrolysis Products of Aqueous Salt Solutions PROBLEM: What products form at which electrode during electrolysis of aqueous solution of the following salts?

(a) KBr (b) AgNO3 (c) MgSO4 PLAN: We identify the reacting ions and compare their electrode potentials with those of water, taking the 0.4 – 0.6 V overvoltage into account. The reduction half-reaction with the less negative E° occurs at the cathode, while the oxidation half-reaction with the less positive E° occurs at the anode. SOLUTION: (a) KBr K+(aq) + e- → K(s) E° = -2.93 - - 2H2O(l) + 2e → H2(g) + 2OH (aq) E° = -0.42V Despite the overvoltage, which makes E for the reduction of water + between -0.8 and -1.0 V, H2O is still easier to reduce than K , so H2(g) forms at the cathode.

21- 100 Sample Problem 21.9

- - 2Br (aq) → Br2(l) + 2e E° = 1.07 V + - 2H2O(l) → O2(g) + 4H (aq) + 4e E° = 0.82 V The overvoltage makes E for the oxidation of water between 1.2 - and 1.4 V. Br is therefore easier to oxidize than water, so Br2(g) forms at the anode.

+ - (b) AgNO3 Ag (aq) + e → Ag(s) E° = 0.80 V - - 2H2O(l) + 2e → H2(g) + 2OH (aq) E° = -0.42V

As the cation of an inactive metal, Ag+ is a better oxidizing agent

than H2O, so Ag(s) forms at the cathode.

- NO3 cannot be oxidized, because N is already in its highest (+5) . Thus O2(g) forms at the anode:

+ - 2H2O(l) → O2(g) + 4H (aq) + 4e

21- 101 Sample Problem 21.9

2+ - (c) MgSO4 Mg (aq) + 2e → Mg(s) E° = -2.37 V - - 2H2O(l) + 2e → H2(g) + 2OH (aq) E° = -0.42V

2+ Mg is a much weaker oxidizing agent than H2O, so H2(g) forms at the cathode.

2- SO4 cannot be oxidized, because S is already in its highest (+6) oxidation state. Thus O2(g) forms at the anode:

+ - 2H2O(l) → O2(g) + 4H (aq) + 4e

21- 102 of Electrolysis

Faraday’s law of electrolysis states that the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell.

The current flowing through the cell is the amount of charge per unit time. Current is measured in amperes.

Current x time = charge

21- 103 Figure 21.29 A summary diagram for the stoichiometry of electrolysis.

MASS (g) of substance CURRENT oxidized or (A) reduced

M (g/mol) time (s)

AMOUNT (mol) AMOUNT (mol) of substance CHARGE of electrons oxidized or (C) transferred reduced balanced Faraday half-reaction constant (C/mol e-)

21- 104 Sample Problem 21.10 Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM: A technician plates a faucet with 0.86 g of Cr metal by

electrolysis of aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed? PLAN: To find the current, we divide charge by time, so we need to find the charge. We write the half-reaction for Cr3+ reduction to get the amount (mol) of e- transferred per of Cr. We convert mass of Cr needed to amount (mol) of Cr. We can then use the Faraday constant to find charge and current. mass (g) of Cr needed divide by M mol of Cr 3 mol e- = 1 mol Cr mol e- transferred Charge (C) current (A) 1 mol e- = 9.65x104 C divide by time in s 21- 105 Sample Problem 21.10

SOLUTION: Cr3+(aq) + 3e- → Cr(s)

- 0.86 g Cr x 1 mol Cr x 3 mol e = 0.050 mol e- 52.00 g Cr 1 mol Cr

4 Charge (C) = 0.050 mol e- x 9.65x10 C = 4.8x103 C 1 mol e-

charge (C) 4.8x103 C 1 min Current (A) = = x = 6.4 C/s = 6.4 A time (s) 12.5 min 60 s

21- 106 Chemical Connections Figure B21.1 The

Mitochondria are subcellular particles outside the cell nucleus that control the -transport chain, an essential part of energy production in living organisms.

21- 107 Chemical Connections Figure B21.2 The main energy-yielding steps in the electron- transport chain (ETC).

21- 108 Chemical Connections Figure B21.3 Coupling electron transport to proton transport to ATP synthesis.

21- 109