20 lecture.ppt [호환 모드]2011-11-19 6 assigning oxidation numbers 4. the sum of the oxidation...
TRANSCRIPT
2011-11-19
1
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
© 2012 Pearson Education, Inc.
2011-11-19
2
Oxidation Numbers(산화수)
In order to keep ptrack of what loses electrons and what gains them, we assign oxidation numbers.
Electrochemistry
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
3
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.
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
4
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
© 2012 Pearson Education, Inc.
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).
© 2012 Pearson Education, Inc.
2011-11-19
5
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).
© 2012 Pearson Education, Inc.
2011-11-19
6
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
© 2012 Pearson Education, Inc.
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
© 2012 Pearson Education, Inc.
2011-11-19
7
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.
© 2012 Pearson Education, Inc.
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
© 2012 Pearson Education, Inc.
2011-11-19
8
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.
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
9
The Half-Reaction Method
Electrochemistry
Consider the reaction between MnO4− and C2O4
2−:
MnO4−(aq) + C2O4
2−(aq) ⎯⎯→ Mn2+(aq) + CO2(aq)
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
10
Electrochemistry
Oxidation Half-Reaction
C O 2− → COC2O42 ⎯⎯→ CO2
To balance the carbon, we add a coefficient of 2:
Electrochemistry
C2O42− ⎯⎯→ 2CO2
© 2012 Pearson Education, Inc.
2011-11-19
11
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−
© 2012 Pearson Education, Inc.
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
© 2012 Pearson Education, Inc.
2011-11-19
12
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
© 2012 Pearson Education, Inc.
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
© 2012 Pearson Education, Inc.
2011-11-19
13
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:
© 2012 Pearson Education, Inc.
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−
© 2012 Pearson Education, Inc.
2011-11-19
14
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
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
15
Voltaic Cells(볼타전지)
In spontaneous poxidation-reduction (redox) reactions, electrons are transferred and energy is released.
Electrochemistry
© 2012 Pearson Education, Inc.
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).
© 2012 Pearson Education, Inc.
2011-11-19
16
Voltaic Cells• A typical cell looks like this.• The oxidation occurs at the anode(산화전극).• The reduction occurs at the cathode(환원전극).
Electrochemistry
© 2012 Pearson Education, Inc.
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
© 2012 Pearson Education, Inc.
2011-11-19
17
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
© 2012 Pearson Education, Inc.
2011-11-19
18
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).
© 2012 Pearson Education, Inc.
Electrochemistry
2011-11-19
19
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).
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
20
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.
© 2012 Pearson Education, Inc.
Cell Potential
Cell potential is measured in volts (V)(볼트)Cell potential is measured in volts (V)(볼트).
1 V = 1 JC
Electrochemistry
© 2012 Pearson Education, Inc.
2011-11-19
21
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:
2 H+(aq, 1M) + 2e− ⎯→ H2(g, 1 atm)© 2012 Pearson Education, Inc.
2011-11-19
22
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).
© 2012 Pearson Education, Inc.
Cell Potentials(전지전위)
• For the oxidation in this cell,
Ered = −0.76 V°
Electrochemistry
© 2012 Pearson Education, Inc.
2011-11-19
23
• 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
© 2012 Pearson Education, Inc.
2011-11-19
24
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
© 2012 Pearson Education, Inc.
Oxidizing and Reducing Agents
The greater theThe greater the difference between the two, the greater the voltage of the cell.
Electrochemistry
© 2012 Pearson Education, Inc.
2011-11-19
25
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
© 2012 Pearson Education, Inc.
Free Energy
Under standard conditions,
ΔG° = −nFE°
Electrochemistry
© 2012 Pearson Education, Inc.
2011-11-19
26
Nernst Equation
• Remember that• Remember thatΔG = ΔG° + RT ln Q
• This meansnFE = nFE° + RT ln Q
Electrochemistry
−nFE = −nFE° + RT ln Q
© 2012 Pearson Education, Inc.
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
© 2012 Pearson Education, Inc.
2011-11-19
27
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
© 2012 Pearson Education, Inc.
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.
© 2012 Pearson Education, Inc.
2011-11-19
28
Applications of ppOxidation-Reduction
Reactions
Electrochemistry
© 2012 Pearson Education, Inc.
Batteries
Electrochemistry
© 2012 Pearson Education, Inc.
2011-11-19
29
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
© 2012 Pearson Education, Inc.
2011-11-19
30
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
© 2012 Pearson Education, Inc.
2011-11-19
31
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
© 2012 Pearson Education, Inc.
2011-11-19
32
…Corrosion Prevention부식방지
Electrochemistry
© 2012 Pearson Education, Inc.
Electrochemistry
2011-11-19
33
NaCl 수용액의전기분해
Electrochemistry
활성전극을이용한전해전지: 니켈도금
Electrochemistry
2011-11-19
34
Electrochemistry
ElectrochemistryCharles M. Hall (1863-1914)
2011-11-19
35
용융된알루미늄은빙정석(Na3AlF6)이나Al2O3보다밀도가높아전지의아래부분에서채집된다.
산화전극: C(s) + 2O2- CO2(g) + 4e-
환원전극: 3e- + Al3+(l) Al(l)
Electrochemistry