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Ch t 20

Lecture Presentation

Chapter 20

Electrochemistry(전기화학)

John D. BookstaverSt. Charles Community College

Cottleville, MO© 2012 Pearson Education, Inc.

Electrochemical Reactions

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

Electrochemistry

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Oxidation Numbers(산화수)

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

Electrochemistry

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Oxidation and Reduction(산화와환원)

Electrochemistry

• 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 Reduction

• A species is reduced when it gains electrons.

Electrochemistry

– Here, each of the H+ gains an electron, and they combine to form H2.

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Oxidation and Reduction

• What is reduced is the oxidizing agent

Electrochemistry

• 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 Numbers(산화수할당하기)

1 Elements in their elemental form have1. 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(전하).

Electrochemistry

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Assigning Oxidation Numbers

3. Nonmetals(비금속) tend to have negative oxidation numbers althoughnegative oxidation numbers, although some are positive in certain compounds or ions.

– Oxygen has an oxidation number of −2, except in the peroxide ion(과산화물이온), which has an oxidation number of −1.

Electrochemistry

– Hydrogen is −1 when bonded to a metal, and +1 when bonded to a nonmetal (비금속).

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Electrochemistry

Assigning Oxidation Numbers

3. Nonmetals tend to have negative oxidation numbers although some areoxidation numbers, although some are positive in certain compounds or ions.

– Fluorine always has an oxidation number of −1.

– The other halogens have an oxidation number of −1 when they are negative.

Electrochemistry

number of 1 when they are negative. They can have positive oxidation numbers, however; most notably in oxyanions(산소산음이온).

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Assigning Oxidation Numbers

4. The sum of the oxidation numbers in a t l d i 0neutral compound is 0.

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

Electrochemistry

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Balancing Oxidation-Reduction Equations (산화-환원반응식맞추기)

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

Electrochemistry

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Balancing Oxidation-Reduction Equations

This method involves treating (on paperThis method 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

ll i

Electrochemistry

overall reaction.

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The Half-Reaction Method(반쪽반응법)

1 Assign oxidation numbers to1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions.

Electrochemistry

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The Half-Reaction Method

3. 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

l t

Electrochemistry

electrons.

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The Half-Reaction Method

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

5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced

Electrochemistry

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

according to charge.

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The Half-Reaction Method

Electrochemistry

Consider the reaction between MnO4− and C2O4

2−:

MnO4−(aq) + C2O4

2−(aq) ⎯⎯→ Mn2+(aq) + CO2(aq)

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The Half-Reaction Method

First we assign oxidation numbers:First, we assign oxidation numbers:

MnO4− + C2O4

2− ⎯⎯→ Mn2+ + CO2

+7 +3 +4+2

Electrochemistry

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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Electrochemistry

Oxidation Half-Reaction

C O 2− → COC2O42 ⎯⎯→ CO2

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

Electrochemistry

C2O42− ⎯⎯→ 2CO2

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Oxidation Half-Reaction

C O 2− → 2COC2O42 ⎯⎯→ 2CO2

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

Electrochemistry

e ec o s o e g s de

C2O42− ⎯⎯→ 2CO2 + 2e−

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Reduction Half-Reaction

MnO − → Mn2+MnO4 ⎯⎯→ Mn2

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

Electrochemistry

e g s de

MnO4− ⎯⎯→ Mn2+ + 4H2O

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Reduction Half-Reaction

MnO − → Mn2+ + 4H OMnO4 ⎯⎯→ Mn2 + 4H2O

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

Electrochemistry

8H+ + MnO4− ⎯⎯→ Mn2+ + 4H2O

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Reduction Half-Reaction

8H+ + MnO − → Mn2+ + 4H O8H + MnO4 ⎯⎯→ Mn2 + 4H2O

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

Electrochemistry

5e− + 8H+ + MnO4− ⎯⎯→ Mn2+ + 4H2O

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Combining the Half-Reactions

Now we evaluate the two half-reactions together:together:

C2O42−⎯⎯→ 2CO2 + 2e−

5e− + 8H+ + MnO4− ⎯⎯→ Mn2+ + 4H2O

Electrochemistry

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

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Combining the Half-Reactions

5C O 2− → 10CO + 10e−5C2O42 ⎯⎯→ 10CO2 + 10e

10e− + 16H+ + 2MnO4− ⎯⎯→ 2Mn2+ + 8H2O

When we add these together, we get:

2

Electrochemistry

10e− + 16H+ + 2MnO4− + 5C2O4

2− ⎯⎯→2Mn2+ + 8H2O + 10CO2 +10e−

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Combining the Half-Reactions

10e− + 16H+ + 2MnO4− + 5C2O4

2− ⎯⎯→2M 2+ 8H O 10CO 102Mn2+ + 8H2O + 10CO2 +10e−

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

Electrochemistry

16H+ + 2MnO4− + 5C2O4

2− ⎯⎯→2Mn2+ + 8H2O + 10CO2

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Balancing in Basic Solution

• If a reaction occurs in a basic solutionIf a reaction occurs in a 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

Electrochemistry

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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Voltaic Cells(볼타전지)

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

Electrochemistry

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Voltaic Cells

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

• We call such a setup a

Electrochemistry

We call such a setup a voltaic cell(볼타전지).

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Voltaic Cells• A typical cell looks like this.• The oxidation occurs at the anode(산화전극).• The reduction occurs at the cathode(환원전극).

Electrochemistry

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Voltaic CellsOnce even one electron flows from the anode (산화전극) to the cathode(환원전극), the charges in each beaker would not be balanced and the flow of electrons would stop.

Electrochemistry

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Electrochemistry

Voltaic Cells• Therefore, we use a salt bridge(염다리), usually a U-shaped

tube that contains a salt solution, to keep the charges balanced.

C (환원전극)– Cations move toward the cathode(환원전극).– Anions move toward the anode(산화전극).

Electrochemistry

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Voltaic Cells• In the cell, then,

electrons leave the anode(산화전극) and flow through the wire to the cathode(환원전극).

• As the electrons leave the anode(산화전극), the

Electrochemistry

cations formed dissolve into the solution in the anode compartment (산화전극칸).

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Electrochemistry

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Voltaic Cells• As the electrons reach

the cathode(환원전극), ( 극),cations in the cathode are attracted to the now negative cathode.

• The electrons are taken by the cation, and the neutral

Electrochemistry

metal is deposited on the cathode(환원전극).

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Electromotive Force (emf)기전력

• Water only spontaneously flows one way in a waterfall.

• Likewise, electrons only spontaneously flow one way in a

d ti f

Electrochemistry

redox reaction—from higher to lower potential energy.

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Electromotive Force (emf)

• The potential difference between the• The potential difference between the anode and cathode in a cell(전지) is called the electromotive force (emf)(기전력).

• It is also called the cell potential

Electrochemistry

(전지전위) and is designated Ecell.

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Cell Potential

Cell potential is measured in volts (V)(볼트)Cell potential is measured in volts (V)(볼트).

1 V = 1 JC

Electrochemistry

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Standard Reduction Potentials표준환원전위

Reduction potentials(환원전위)potentials(환원전위) for many electrodes have been measured and tabulated.

Electrochemistry

Standard Hydrogen Electrode표준수소전극

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

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

Electrochemistry

for hydrogen is 0 V:

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Standard Cell Potentials표준전지전위

The cell potential(전지전위) at standardThe cell potential(전지전위) at standard conditions can be found through this equation:

Ecell° = Ered (cathode) − Ered (anode)° °

Electrochemistry

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

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Cell Potentials(전지전위)

• For the oxidation in this cell,

Ered = −0.76 V°

Electrochemistry

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• For the oxidation in this cell,Ered = −0.76 V°

• For the reduction, Ered = +0.34 V°

Electrochemistry

Cell Potentials

Ecell° = Ered° (cathode) −Ered° (anode)= +0.34 V − (−0.76 V)= +1.10 V

Electrochemistry

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Oxidizing and Reducing Agents산화제와환원제

• The strongest oxidizers have the most positivehave the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.

Electrochemistry

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Oxidizing and Reducing Agents

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

Electrochemistry

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Free Energy(자유에너지)

ΔG for a redox reaction(산화환원반응) can be found by using the equation

ΔG = −nFE

where n is the number of moles of electrons transferred and F is the

Electrochemistry

electrons transferred, and F is the Faraday constant (Faraday상수)1 F = 96,485 C/mol = 96,485 J/V-mol

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Free Energy

Under standard conditions,

ΔG° = −nFE°

Electrochemistry

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Nernst Equation

• Remember that• Remember thatΔG = ΔG° + RT ln Q

• This meansnFE = nFE° + RT ln Q

Electrochemistry

−nFE = −nFE° + RT ln Q

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Nernst Equation

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

E = E° − RTnF ln Q

or, using base-10 logarithms,

Electrochemistry

or, using base 10 logarithms,

E = E° − 2.303RTnF log Q

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Nernst Equation

At room temperature (298 K)At room temperature (298 K),

Thus, the equation becomes

2.303RTF = 0.0592 V

Electrochemistry

Thus, the equation becomes

E = E° − 0.0592n log Q

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Concentration Cells(농도전지)

• Notice that the Nernst equation implies that a cell could be created that has the same substance at

Electrochemistry

both electrodes.• 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.

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Applications of ppOxidation-Reduction

Reactions

Electrochemistry

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Batteries

Electrochemistry

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환원전극: PbO2(s) + HSO4

-(aq) + 3H+(aq) + 2e PbSO4(s) + 2H2O(l)산화전극 Pb(s) + HSO4

-(aq) PbSO4(s) + H+(aq) + 2e PbO2(s) + Pb(s) +2HSO4

-(aq) + 2H+(aq) 2PbSO4(s) + 2H2O(l)

2PbSO4(s) + 2H2O(l) PbO2(s) + Pb(s) +2HSO4-(aq) + 2H+(aq)

충전

Electrochemistry

Alkaline Batteries

Electrochemistry

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환원전극: MnO2(s) + 2H2O(l) + 2e 2MnO(OH)(s) + 2OH-

산화전극 Zn(s) + 2OH-(aq) Zn(OH)2(s) + 2e

Electrochemistry

Hydrogen Fuel Cells수소연료전지

Electrochemistry

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환원전극: O2(s) + 4H++ 4e- 2H2O(l)산화전극 2H2(g) 4H+ + 4e-

2H2(g) + O2(s) 2H2O(l)2 2 2

Electrochemistry

Corrosion(부식) and…

Electrochemistry

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…Corrosion Prevention부식방지

Electrochemistry

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Electrochemistry

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NaCl 수용액의전기분해

Electrochemistry

활성전극을이용한전해전지: 니켈도금

Electrochemistry

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Electrochemistry

ElectrochemistryCharles M. Hall (1863-1914)

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용융된알루미늄은빙정석(Na3AlF6)이나Al2O3보다밀도가높아전지의아래부분에서채집된다.

산화전극: C(s) + 2O2- CO2(g) + 4e-

환원전극: 3e- + Al3+(l) Al(l)

Electrochemistry