chapter 20 electrochemistry 1. electrochemical reactions in electrochemical reactions, electrons are...

Chapter 20Chapter 20ElectrochemistryElectrochemistry

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Electrochemical ReactionsElectrochemical Reactions

In electrochemical reactions, electrons are transferred from one species to another.

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Oxidation NumbersOxidation Numbers

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

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Oxidation and ReductionOxidation and Reduction

A species is oxidized when it loses electrons.◦ Here, zinc loses two electrons to go from

neutral zinc metal to the Zn2+ ion.

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Oxidation and ReductionOxidation and Reduction

A species is reduced when it gains electrons.◦ Here, each of the H+ gains an electron and

they combine to form H2.

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Oxidation and ReductionOxidation and Reduction

What is reduced is the oxidizing agent.◦ H+ oxidizes Zn by taking electrons from it.

What is oxidized is the reducing agent.◦ Zn reduces H+ by giving it electrons.

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Assigning Oxidation Assigning Oxidation NumbersNumbers1. Elements in their elemental

form have an oxidation number of 0.

2. The oxidation number of a monatomic ion is the same as its charge.

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Assigning Oxidation Assigning Oxidation NumbersNumbers3. Nonmetals tend to have

negative oxidation numbers, although some are positive in certain compounds or ions.

◦ Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.

◦ Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

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Assigning Oxidation Assigning Oxidation NumbersNumbers

◦ Fluorine always has an oxidation number of −1.

◦ The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

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Assigning Oxidation Assigning Oxidation NumbersNumbers

4. The sum of the oxidation numbers in a neutral compound is 0.

5. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

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SAMPLE EXERCISE 20.1 What Chemical Reactions Occur in a Battery?

Identify the substances that are oxidized and reduced, and indicate which are oxidizing agents and which are reducing agents.

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Balancing Oxidation-Reduction Balancing Oxidation-Reduction EquationsEquations

Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

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Balancing Oxidation-Reduction Balancing Oxidation-Reduction EquationsEquations

This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

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Half-Reaction MethodHalf-Reaction Method

1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions.

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Half-Reaction MethodHalf-Reaction Method3. Balance each half-reaction.

a. Balance elements other than H and O.

b. Balance O by adding H2O.c. Balance H by adding H+.d. Balance charge by adding electrons.

4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.

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Half-Reaction MethodHalf-Reaction Method5. Add the half-reactions,

subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass.

7. Make sure the equation is balanced according to charge.

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Half-Reaction MethodHalf-Reaction Method

Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)

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Half-Reaction MethodHalf-Reaction Method

First, we assign oxidation numbers.

MnO4− + C2O4

2- Mn2+ + CO2

+7 +3 +4+2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.

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Oxidation Half-ReactionOxidation Half-Reaction

C2O42− CO2

To balance the carbon, we add a coefficient of 2:

C2O42− 2 CO2

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Oxidation Half-ReactionOxidation Half-Reaction

C2O42− 2 CO2

The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.

C2O42− 2 CO2 + 2 e−

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Reduction Half-ReactionReduction Half-Reaction

MnO4− Mn2+

The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.

MnO4− Mn2+ + 4 H2O

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Reduction Half-ReactionReduction Half-Reaction

MnO4− Mn2+ + 4 H2O

To balance the hydrogen, we add 8 H+ to the left side.

8 H+ + MnO4− Mn2+ + 4 H2O

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Reduction Half-ReactionReduction Half-Reaction

8 H+ + MnO4− Mn2+ + 4 H2O

To balance the charge, we add 5 e− to the left side.

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

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Combining the Half-Combining the Half-ReactionsReactions

Now we evaluate the two half-reactions together:

C2O42− 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− Mn2+ + 4

H2O

To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2. 25

Combining the Half-Combining the Half-ReactionsReactions

5 C2O42− 10 CO2 + 10 e−

10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8

H2O

When we add these together, we get:

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2 +10 e−

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Combining the Half-Combining the Half-ReactionsReactions

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2 +10 e−

The only thing that appears on both sides are the electrons. Subtracting them, we are left with:

16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2

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SAMPLE EXERCISE 20.2 Balancing Redox Equations in Acidic Solution

Complete and balance this equation by the method of half-reactions:

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Balancing in Basic Balancing in Basic SolutionSolution

If a reaction occurs in basic solution, one can balance it as if it occurred in acid.

Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.

If this produces water on both sides, you might have to subtract water from each side.

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SAMPLE EXERCISE 20.3 Balancing Redox Equations in Basic Solution

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PRACTICE EXERCISEComplete and balance the following equations for oxidation-reduction reactions that occur in basic solution:

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Voltaic CellsVoltaic Cells

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

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Voltaic CellsVoltaic Cells

We can use that energy to do work if we make the electrons flow through an external device.

We call such a setup a voltaic cell.

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Voltaic CellsVoltaic Cells

A typical cell looks like this.

The oxidation occurs at the anode.

The reduction occurs at the cathode.

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Voltaic CellsVoltaic Cells

Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

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Voltaic CellsVoltaic Cells

Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.◦ Cations move

toward the cathode.◦ Anions move toward

the anode.

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Voltaic CellsVoltaic CellsIn the cell, then,

electrons leave the anode and flow through the wire to the cathode.

As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

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Voltaic CellsVoltaic CellsAs the electrons

reach the cathode, cations in the cathode are attracted to the now negative cathode.

The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

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SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell

is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes.

The oxidation-reduction reaction

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(a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode is consumed in the cell reaction? (c) Which electrode is positive?

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Electromotive Force (emf)Electromotive Force (emf)Water only

spontaneously flows one way in a

waterfall.Likewise, electrons only spontaneously

flow one way in a redox reaction—

from higher to lower potential

energy.

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Electromotive Force (emf)Electromotive Force (emf)The potential difference between

the anode and cathode in a cell is called the electromotive force (emf).

It is also called the cell potential, and is designated Ecell.

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Cell PotentialCell Potential

Cell potential is measured in volts (V).

1 V = 1 JC

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Standard Reduction PotentialsStandard Reduction Potentials

Reduction potentials for

many electrodes have been measured

and tabulated.

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Standard Hydrogen ElectrodeStandard Hydrogen Electrode

Their values are referenced to a standard hydrogen electrode (SHE).

By definition, the reduction potential for hydrogen is 0 V:

2 H+ (aq, 1M) + 2 e− H2 (g, 1 atm)

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Standard Cell PotentialsStandard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell = Ered (cathode) − Ered (anode)

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

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Cell PotentialsCell PotentialsFor the oxidation in this cell,

For the reduction,

Ered = −0.76 V

Ered = +0.34 V

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Cell PotentialsCell Potentials

Ecell = Ered (cathode) − Ered (anode)

= +0.34 V − (−0.76 V)

= +1.10 V

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SAMPLE EXERCISE 20.5 Calculating Ered from Ecellº º

For the Zn-Cu2+ voltaic cell shown in Figure 20.5, we have

Given that the standard reduction potential of Zn2+ to Zn(s) is –0.76 V, calculate the for the reduction of Cu2+ to Cu:

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The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate for the reduction of In3+ to In+.

PRACTICE EXERCISEA voltaic cell is based on the half-reactions

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SAMPLE EXERCISE 20.6 Calculating Ecell from Eredº º

Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction

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PRACTICE EXERCISEUsing data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:

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SAMPLE EXERCISE 20.7 From Half-Reactions to Cell EMF

By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the anode, and (b) the standard cell potential.

A voltaic cell is based on the following two standard half-reactions:

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PRACTICE EXERCISEA voltaic cell is based on a Co2+/Co half-cell and an AgCl/Ag half-cell. (a) What reaction occurs at the anode? (b) What is the standard cell potential?

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Oxidizing and Reducing AgentsOxidizing and Reducing Agents

The greater the difference between the two, the greater the voltage of the cell.

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Oxidizing and Reducing AgentsOxidizing and Reducing Agents

The strongest oxidizers have the most positive reduction potentials.

The strongest reducers have the most negative reduction potentials.

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SAMPLE EXERCISE 20.8 Determining the Relative Strengths of Oxidizing Agents

Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3–

(aq), Ag+(aq), Cr2O72–(aq).

PRACTICE EXERCISEUsing Table 20.1, rank the following species from the strongest to the weakest reducing agent: I–(aq), Fe(s), Al(s).

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Free EnergyFree Energy

G for a redox reaction can be found by using the equation

G = −nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday.1 F = 96,485 C/mol = 96,485 J/V-mol

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Free EnergyFree Energy

Under standard conditions,

G = −nFE

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SAMPLE EXERCISE 20.9 Spontaneous or Not?

Using standard reduction potentials (Table 20.1), determine whether the following reactions are spontaneous under standard conditions.

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PRACTICE EXERCISEUsing the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions:

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Nernst EquationNernst EquationRemember that

G = G + RT ln Q

This means−nFE = −nFE + RT ln Q

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Nernst EquationNernst Equation

Dividing both sides by −nF, we get the Nernst equation:

E = E −RTnF

ln Q

or, using base-10 logarithms,

E = E −2.303 RTnF

log Q

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Nernst EquationNernst Equation

At room temperature (298 K),

Thus the equation becomes

E = E −0.0592n

log Q

2.303 RTF

= 0.0592 V

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Concentration CellsConcentration Cells

Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. The more dilute is the anode so its [Ni2+] can increase. The more concentrated is the cathode so its [Ni2+] can decrease.

• For such a cell, would be 0, but Q would not.Ecell

• Therefore, as long as the concentrations are different, E will not be 0. (Until they become equal)

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SAMPLE EXERCISE 20.10 Determining G° and K

What are the values of E°, G°, and K when the reaction is written in this way?

(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, G°, and the equilibrium constant, K, at room temperature (T = 298 K) for the reaction

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(a) What is the value of n? (b) Use the data in Appendix E to calculate G°. (c) Calculate K at T = 298 K.

PRACTICE EXERCISEFor the reaction

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SAMPLE EXERCISE 20.11 Voltaic Cell EMF Under Nonstandard Conditions

Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 when [Cr2O7

2–] = 2.0 M, [H+] = 1.0 M, [I–] = 1.0 M, and [Cr3+] = 1.0 10–5 M.

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PRACTICE EXERCISECalculate the emf generated by the cell described in the practice exercise accompanying Sample Exercise 20.6 when [Al3+] = 4.0 10–3 M and [I–] = 0.010 M.

2 Al + 3 I2 2 Al3+ + 6 I1- Eo = 2.20 V

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SAMPLE EXERCISE 20.12 Concentrations in a Voltaic Cell

If the voltage of a Zn–H+ cell (like that in Figure 20.11) is 0.45 V at 25°C when [Zn2+] = 1.0 M and atm, what is the concentration of H+?

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[Zn2+] = 0.10 M

0.542 V

P(H2 )= 1.0 atm

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SAMPLE EXERCISE 20.13 pH of a Concentration Cell

A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has atm and an unknown

concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, atm). At

298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1

through the external circuit to electrode 2. Calculate[H+] for the solution at electrode 1. What is its pH?

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PRACTICE EXERCISEA concentration cell is constructed with two Zn(s)-Zn2+(aq) half-cells. The first half-cell has [Zn2+] = 1.35 M, and the second half-cell has [Zn2+] = 3.75 10–4 M. (a) Which half-cell is the anode of the cell? (b) What is the emf of the cell?

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Electric Current and Electric Current and ElectrolysisElectrolysisCoulombs = Joules/VoltsCoulombs = Amperes X Seconds(q = I * t)Faraday’s Constant = 96,500

Coulombs/mole of e-

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SAMPLE EXERCISE 20.14 Aluminum Electrolysis

Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.

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SAMPLE EXERCISE 20.15 Calculating Energy in Kilowatt-hours

Calculate the number of kilowatt-hours of electricity required to produce 1.0 103 kg of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V.

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PRACTICE EXERCISECalculate the number of kilowatt-hours of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl2 if the applied emf is 5.00 V. Assume that the process is 100% efficient.

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Applications of Applications of Oxidation-Reduction Oxidation-Reduction ReactionsReactions

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BatteriesBatteries

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Alkaline BatteriesAlkaline Batteries

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Hydrogen Fuel CellsHydrogen Fuel Cells

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Corrosion and…Corrosion and…

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……Corrosion PreventionCorrosion Prevention

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SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

The Ksp at 298 K for iron(II) fluoride is 2.4 10–6. (a) Write a half-reaction that gives the likely products of the two-electron reduction of FeF2(s) in water. (b) Use the Ksp value and the standard reduction potential of Fe2+(aq) to calculate the standard reduction potential for the half-reaction in part (a). (c) Rationalize the difference in the reduction potential for the half-reaction in part (a) with that for Fe2+(aq).

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