New chm 152_unit_8_power_points-sp13

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ELECTROCHEMISTRY I

• VOLTAIC OR GALVANIC CELLS• STANDARD ELECTRODE POTENTIALS• NONSTANDARD CELLS• Chapter 18.1-18.10 (McM)• Chapter 21.2-21.5 (Silberberg)

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Chapter 21

Electrochemistry: Chemical Change and Electrical Work

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Electrochemistry: Chemical Change and Electrical Work

21.3 Cell Potential: Output of a Voltaic Cell

21.4 Free Energy and Electrical Work

21.5 Electrochemical Processes in Batteries

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Goals & Objectives

• See the following Learning Objectives on pages 894.

• Understand these Concepts:• 21.4-12.

• Master these Skills:• 21.2-7.

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Electrochemical Cells

A voltaic cell uses a spontaneous redox reaction (DG < 0) to generate electrical energy.- The system does work on the surroundings.

A electrolytic cell 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 electrodes placed in an electrolyte solution.

The anode is the electrode at which oxidation occurs.

The cathode is the electrode at which reduction occurs.

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

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Spontaneous Redox Reactions

A strip of zinc metal 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.

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The Zinc-Copper Cell

• The cell consists of a strip of copper in a 1.0M CuSO4 solution and a strip of zinc in a 1.0M ZnSO4 solution. A wire and salt bridge complete the circuit.

• The copper electrode gains mass and the [Cu2+] decreases while the zinc electrode loses mass and the [Zn2+] increases as the cell operates.

• The initial voltage is 1.10 volts(v).

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Figure 21.3 The spontaneous reaction between zinc and copper(II) ion.

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• Anode reaction:– Zn --> Zn2+ + 2e-

• Cathode reaction:– Cu2+ + 2e- --> Cu

• Overall reaction:– Zn + Cu2+ --> Zn2+ + Cu

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• Shorthand Notation– Zn / Zn2+(1.0M) // Cu2+(1.0M) / Cu

• or– Zn / Zn2+ // Cu2+ / Cu

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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 salt bridge completes the electrical circuit.

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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 mass 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.

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Charges of the Electrodes

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

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

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

The cathode is the positive electrode in a voltaic cell.

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

Oxidation half-reactionZn(s) → Zn2+(aq) + 2e-

Reduction half-reactionCu2+(aq) + 2e- → Cu(s)

Overall (cell) reactionZn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

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

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Oxidation half-reactionZn(s) → Zn2+(aq) + 2e-

After several hours, the Zn anode weighs less as Zn is oxidized to Zn2+.

Reduction half-reactionCu2+(aq) + 2e- → Cu(s)

The Cu cathode gains mass over time as Cu2+ ions are reduced to Cu.

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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.

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Flow of Charge in a Voltaic Cell

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

Electrons flow through the wire from anode to cathode.

Cations move through the salt bridge from the anode solution to the cathode solution.

Zn2+

Anions move through the salt bridge from the cathode solution to the anode solution.

SO42-

By convention, a voltaic cell is shown with the anode on the left and the cathode on the right.

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Active and Inactive Electrodes

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.

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

Inactive electrodes are usually unreactive substances such as graphite or platinum.

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Figure 21.5 A voltaic cell using inactive electrodes.

Reduction half-reactionMnO4

-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)Oxidation half-reaction2I-(aq) → I2(s) + 2e-

Overall (cell) reaction2MnO4

-(aq) + 16H+(aq) + 10I-(aq) → 2Mn2+(aq) + 5I2(s) + 8H2O(l)

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Notation for a Voltaic Cell

Zn(s)│Zn2+(aq)║Cu2+(aq) │Cu(s)

The anode components are written on the left.

The cathode components are written on the right.

The single line shows a phase boundary between the components of a half-cell.

The double line shows that the half-cells are physically separated.

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

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

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graphite I-(aq)│I2(s)║MnO4-(aq), H+(aq), Mn2+(aq) │graphite

Notation for a Voltaic Cell

The inert electrode is specified.

A comma is used to show components that are in the same phase.

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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.

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SOLUTION:

Sample Problem 21.2

Ag+(aq) + e- → Ag(s) [reduction; cathode]

Cr(s) → Cr3+(aq) + 3e- [oxidation; anode]

3Ag+ + Cr(s) → 3Ag(s) + Cr3+(aq)

The half-reactions are:

The balanced overall equation is:

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.

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The Copper-Silver Cell

• The cell consists of a strip of copper in a 1.0M CuSO4 solution and a strip of silver in a 1.0M AgNO3 solution. A wire and salt bridge complete the circuit.

• The copper electrode loses mass and the [Cu2+] increases while the silver electrode gains mass and the [Ag+] decreases as the cell operates.


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The Copper-Silver Cell• Anode reaction:

– Cu --> Cu2+ + 2e-

• Cathode reaction:– e- + Ag+ --> Ag– 2(e- + Ag+ --> Ag)

• Overall reaction:– Cu + 2Ag+ --> Cu2+ + 2Ag

• Shorthand notation:– Cu / Cu2+ // Ag+ / Ag

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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

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The Zinc-SHE Electrode

• The cell consists of a SHE and a strip of Zn in a 1.0M ZnSO4 solution solution. A wire and salt bridge complete the circuit.

• The zinc electrode loses mass and the [Zn2+] increases while the [H+] decreases in the SHE and hydrogen gas is produced as the cell operates.


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Figure 21.7 Determining an unknown E°half-cell with the standard reference (hydrogen) electrode.

Oxidation half-reactionZn(s) → Zn2+(aq) + 2e−

Reduction half-reaction2H3O+(aq) + 2e- → H2(g) + 2H2O(l)

Overall (cell) reactionZn(s) + 2H3O+(aq) → Zn2+(aq) + H2(g) + 2H2O(l)

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The Zinc-SHE Electrode

• Anode:– Zn --> Zn2+ + 2e-

• Cathode:– 2e- + 2H+ --> H2

• Overall:– Zn + 2H+ --> Zn2+ + H2

• Zn / Zn2+ // H+ / H2

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The Copper-SHE Cell

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The Copper-SHE Cell

• Anode:– H2 --> 2H+ + 2e-

• Cathode:– 2e- + Cu2+ --> Cu

• Overall:– H2 + Cu2+ --> 2H+ + Cu

• Cell voltage = 0.337v

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Table 21.2 Selected Standard Electrode Potentials (298 K)

Half-Reaction E°(V)

+2.87

−3.05

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

−0.23

−0.44

−0.83

−2.71

strength of reducing agent

stre

ngth

of o

xidi

zing

age

nt

F2(g) + 2e− 2F−(aq)

Cl2(g) + 2e− 2Cl−(aq)

MnO2(g) + 4H+(aq) + 2e− Mn2+(aq) + 2H2O(l)

NO3-(aq) + 4H+(aq) + 3e− NO(g) + 2H2O(l)

Ag+(aq) + e− Ag(s)

Fe3+(g) + e− Fe2+(aq)

O2(g) + 2H2O(l) + 4e− 4OH−(aq)

Cu2+(aq) + 2e− Cu(s)

N2(g) + 5H+(aq) + 4e− N2H5+(aq)

Fe2+(aq) + 2e− Fe(s)

2H2O(l) + 2e− H2(g) + 2OH−(aq)

Na+(aq) + e− Na(s)

Li+(aq) + e− Li(s)

2H+(aq) + 2e− H2(g)

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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.

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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 metals to give up their electrons.

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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 voltage of the cell or the electromotive forces (emf).

Ecell > 0 for a spontaneous process.

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Standard Electrode Potentials

The standard electrode potential (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)

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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:

Br2(aq) + Zn(s) → Zn2+(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]Zn(s) → Zn2+(aq) + 2e- [oxidation; anode] E°zinc = -0.76 V

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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

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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.

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Using half-reactions to write a spontaneous redox reaction:

Sn2+(aq) + 2e- → Sn(s) E°tin = -0.14 VAg+(aq) + e- → Ag(s) E°silver = 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:

Sn(s) → Sn2+(aq) + 2e- E°tin = -0.14 V

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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°.

Sn(s) → Sn2+(aq) + 2e- E°tin = -0.14 V2Ag+(aq) + 2e- → 2Ag(s) E°silver = 0.80 V

Step 3: Add the reactions together, cancelling common species. Calculate E°cell = E°cathode - E°anode.

Sn(s) + 2Ag+(aq) → 2Ag(s) + Sn2+(aq) E°cell = 0.94 V

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

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Writing Spontaneous Cell Reactions

• Write the spontaneous cell reaction and Eo

cell expected when a zinc-silver cell is set up in the usual fashion.

• Write the shorthand notation for this cell.• Draw the cell indicating the anode,

cathode, salt bridge, direction of electron flow, and direction of ion flow from the salt bridge.

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• Write the spontaneous cell reaction and Eocell

expected when an Fe3+/Fe2+-MnO4-/Mn2+

cell is set up in the usual fashion.• Fe3+ (0.771v) Fe2+ MnO4

- (1.51v) Mn2+

• Write the shorthand notation for this cell.

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• Write the spontaneous cell reaction and Eo

cell expected when a cell using the following half-reactions is set up in the usual fashion.

• NO3- (0.96v) NO

• H3AsO4 (0.58v) H3AsO3

• Write the shorthand notation for this cell.

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Electrochemical Processes in Batteries

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.

A battery consists of self-contained voltaic cells arranged in series, so their individual voltages are added.

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Alkaline battery.Figure 21.15

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

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Silver button battery.Figure 21.16

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 VThe mercury battery uses HgO as the oxidizing agent instead of Ag2O and has cell potential of 1.3 V.

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Figure 21.17 Lithium battery.

Anode (oxidation):

3.5Li(s) → 3.5Li+ + 3.5e-

Cathode (reduction):

AgV2O5.5 + 3.5Li- + 3.5e- → Li3.5V2O5.5


AgV2O5.5 + 3.5Li(s) → Li3.5V2O5.5

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

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Lead-acid battery.Figure 21.18

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

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Anode (oxidation): Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e-


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

The reactions in a lead-acid battery:

Overall (cell) reaction (recharge):

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

The cell generates electrical energy when it discharges as a voltaic cell.

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Nickel-metal hydride batteryFigure 21.19

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)


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

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Lithium-ion battery.Figure 21.20

Anode (oxidation):

LixC6(s) → xLi+ + xe- + C6(s)


Li1-xMn2O4(s) + xLi+ + xe- → LiMn2O4(s)


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.

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

Fuel Cells

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

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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

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Walther Hermann Nernst

• Walther Hermann Nernst (June 25, 1864 – November 18, 1941) was a German physicist who is known for his theories behind the calculation of chemical

• affinity as embodied in the third law of thermodynamics, for which he won the 1920 Nobel Prize in chemistry. He is also known for developing the Nernst equation.

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Nonstandard Cells

• The Nernst equation– E = Eo - (0.0592/n)logQ– where n = moles of electrons transferred– and Q = reaction quotient

• For Cu2+ + e- --> Cu+ Eo = +0.153• E = Eo - (0.0592/1)log([Cu+]/[Cu2+])

– n = 1– Q = [Cu+]/[Cu2+]

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The Nernst Equation

• For the above cell when [Cu2+]=[Cu+]=1.0M

• E = Eo - (0.0592/1)log(1)• E = Eo

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Nonstandard Cells

• Calculate the potential for the Cu2+/Cu+ cell at 25oC when [Cu2+] = 2.0x10-3M and [Cu+]=6.0x10-3M. Eo = 0.153v

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Nonstandard Cells

• Determine the electrode potential for the hydrogen electrode when

• [H+] = 1.0x10-3M and • the p(hydrogen gas) = 0.50atm.

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Nonstandard Cells

• Determine the initial potential for the following cell reaction when

• the [Fe3+]=(1.0x10-2), • the [Fe2+] = 0.10M, • the [Sn4+]=1.0M and • the [Sn2+]=0.10M• Fe3+ (0.771v) Fe2+ Sn4+ (0.15v) Sn2+

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