Post on 11-Jan-2016
Prof. T.L. Heise, CHE 116
1 Chapter TwentyChapter Twenty
Copyright © Tyna L. Heise 2001 - 2002Copyright © Tyna L. Heise 2001 - 2002
All Rights ReservedAll Rights Reserved
Prof. T.L. Heise, CHE 116
2 ElectrochemistryElectrochemistryRedox reactionsRedox reactions
Oxidation is the loss of electronsOxidation is the loss of electrons
Reduction is the gain of electronsReduction is the gain of electrons
The transfer of electrons that occurs is The transfer of electrons that occurs is exothermic and can be used to exothermic and can be used to
produce electricityproduce electricity
Prof. T.L. Heise, CHE 116
3 ElectrochemistryElectrochemistryAt other times the reactions you want At other times the reactions you want to occur may be nonspontaneous and to occur may be nonspontaneous and require the addition of electricity to require the addition of electricity to
drive the reaction.drive the reaction.
Electrochemistry is the branch of Electrochemistry is the branch of chemistry that deals with the chemistry that deals with the
relationships between electricity and relationships between electricity and chemical reactionschemical reactions
Prof. T.L. Heise, CHE 116
4Oxidation - Oxidation - Reductions Reductions ReactionReactionDetermination of oxidation reduction Determination of oxidation reduction
reactions occurs by examining the reactions occurs by examining the following:following:
assign oxidation numbersassign oxidation numberslook for the change in oxidation look for the change in oxidation statesstatesdon’t forget, something that don’t forget, something that oxidizes is also the reducing agent, oxidizes is also the reducing agent, something that reduces is also the something that reduces is also the oxidizing agentoxidizing agent
Prof. T.L. Heise, CHE 116
5Oxidation - Oxidation - Reductions Reductions ReactionReactionSample Exercise: Identify the oxidizing Sample Exercise: Identify the oxidizing
and reducing agents in the following and reducing agents in the following oxidation-reduction equation:oxidation-reduction equation:
2H2H22O(l) + Al(s) + MnOO(l) + Al(s) + MnO44--(aq) (aq)
Al(OH)Al(OH)44--(aq) + MnO(aq) + MnO22(s)(s)
Prof. T.L. Heise, CHE 116
6Oxidation - Oxidation - Reductions Reductions ReactionReactionSample Exercise: Identify the oxidizing Sample Exercise: Identify the oxidizing
and reducing agents in the following and reducing agents in the following oxidation-reduction equation:oxidation-reduction equation:
+1 -2 0 +7 -2+1 -2 0 +7 -2
2H2H22O(l) + Al(s) + MnOO(l) + Al(s) + MnO44--(aq) (aq)
+3 -2 +1 +4 -2 +3 -2 +1 +4 -2
Al(OH)Al(OH)44--(aq) + MnO(aq) + MnO22(s)(s)
Prof. T.L. Heise, CHE 116
7Oxidation - Oxidation - Reductions Reductions ReactionReactionSample Exercise: Identify the oxidizing Sample Exercise: Identify the oxidizing
and reducing agents in the following and reducing agents in the following oxidation-reduction equation:oxidation-reduction equation:
+1 -2 0 +7 -2+1 -2 0 +7 -2
2H2H22O(l) + Al(s) + MnOO(l) + Al(s) + MnO44--(aq) (aq)
+3 -2 +1 +4 -2 +3 -2 +1 +4 -2
Al(OH)Al(OH)44--(aq) + MnO(aq) + MnO22(s)(s)
Prof. T.L. Heise, CHE 116
8Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionWhenever we balance a chemical Whenever we balance a chemical
equation, we must obey the conservation equation, we must obey the conservation of mass.of mass.
As we balance redox reactions we will see As we balance redox reactions we will see an additional rule: the balancing of charge an additional rule: the balancing of charge as well.as well.
Prof. T.L. Heise, CHE 116
9Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReaction1. Assign oxidation numbers to all 1. Assign oxidation numbers to all
elementselements
2. Bridge the elements that are changing 2. Bridge the elements that are changing in oxidation statein oxidation state
3. Identify the number of electrons being 3. Identify the number of electrons being lost and gained and balance them.lost and gained and balance them.
4. Balance all remaining elements besides 4. Balance all remaining elements besides H and OH and O
5. Balance O by adding H5. Balance O by adding H22O to the O to the deficient sidedeficient side
Prof. T.L. Heise, CHE 116
10Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReaction6. Balance H by adding H6. Balance H by adding H++ to the deficient to the deficient
sideside
7. Check to make sure your charge for 7. Check to make sure your charge for each side is the sameeach side is the same
Prof. T.L. Heise, CHE 116
11Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
A) Cu(s) + NOA) Cu(s) + NO33--(aq) (aq) Cu Cu2+2+(aq) + NO(aq) + NO22(g)(g)
Prof. T.L. Heise, CHE 116
12Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
A) A) 0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2
Cu(s) + NOCu(s) + NO33--(aq) (aq) Cu Cu2+2+(aq) + NO(aq) + NO22(g)(g)
Prof. T.L. Heise, CHE 116
13Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
A) A) Leo = RA 2e -Leo = RA 2e -
0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2
Cu(s) + NOCu(s) + NO33--(aq) (aq) Cu Cu2+2+(aq) + NO(aq) + NO22(g)(g)
Ger = OA 1e -Ger = OA 1e -
Prof. T.L. Heise, CHE 116
14Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
A) A) Leo = RA 2e -Leo = RA 2e -
0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2
Cu(s) + Cu(s) + 22NONO33--(aq)(aq)CuCu2+2+(aq) + (aq) + 22NONO22(g)(g)
Ger = OA 2(1e -)Ger = OA 2(1e -)
Prof. T.L. Heise, CHE 116
15Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance the Sample Exercise: Complete and balance the
following oxidation-reduction equations in following oxidation-reduction equations in acidic solutions.acidic solutions.
A) A) Leo = RA 2e -Leo = RA 2e -
0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2
Cu(s)+Cu(s)+22NONO33--(aq)(aq)CuCu2+2+(aq)+(aq)+22NONO22(g)(g)+2H+2H22OO
Ger = OA 2(1e -)Ger = OA 2(1e -)
Prof. T.L. Heise, CHE 116
16Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance the Sample Exercise: Complete and balance the
following oxidation-reduction equations in following oxidation-reduction equations in acidic solutions.acidic solutions.
A) A) Leo = RA 2e -Leo = RA 2e -
0 +5 -2 +2 +4 -20 +5 -2 +2 +4 -2
Cu +Cu +22NONO33--(aq)(aq)+4H+4H++CuCu2+2+(aq)+(aq)+22NONO22(g)(g)+2H+2H22OO
Ger = OA 2(1e -)Ger = OA 2(1e -)
Prof. T.L. Heise, CHE 116
17Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
B) MnB) Mn2+2+ + NaBiO + NaBiO33 Bi Bi3+3+ + MnO + MnO44--
Prof. T.L. Heise, CHE 116
18Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
B) B) +2 +5 -2 +3 +7 -2 +2 +5 -2 +3 +7 -2
MnMn2+2+ + BiO + BiO33-- Bi Bi3+3+ + MnO + MnO44
--
Prof. T.L. Heise, CHE 116
19Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
B) B) Leo = RA 5e -Leo = RA 5e -
+2 +5 -2 +3 +7 -2+2 +5 -2 +3 +7 -2
MnMn2+2+ + BiO + BiO33 - - Bi Bi3+3+ + MnO + MnO44--
Ger = OA 2e -Ger = OA 2e -
Prof. T.L. Heise, CHE 116
20Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
B) B) Leo = RA 2(5e -)Leo = RA 2(5e -)
+2 +5 -2 +3 +7 -2+2 +5 -2 +3 +7 -2
22MnMn2+2+ + BiO + BiO33 -- Bi Bi3+3+ + + 22MnOMnO44--
Ger = OA 2e -Ger = OA 2e -
Prof. T.L. Heise, CHE 116
21Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
B) B) Leo = RA 2(5e -)Leo = RA 2(5e -)
+2 +5 -2 +3 +7 -2+2 +5 -2 +3 +7 -2
22MnMn2+2+ + + 55BiOBiO33 -- 55BiBi3+3+ + + 22MnOMnO44--
Ger = OA 5(2e -)Ger = OA 5(2e -)
Prof. T.L. Heise, CHE 116
22Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample Exercise: Complete and balance Sample Exercise: Complete and balance
the following oxidation-reduction the following oxidation-reduction equations in acidic solutions.equations in acidic solutions.
B)B)
22MnMn2+ 2+ + + 55BiOBiO33- - + + 14H14H++
55BiBi3+ 3+ + + 22MnOMnO44- - + + 7H7H22OO
Prof. T.L. Heise, CHE 116
23Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionAt times the solution in which a redox At times the solution in which a redox
reaction occurs is a basic one. The reaction occurs is a basic one. The balancing of an equation in a basic balancing of an equation in a basic solution is slightly different. Instead of solution is slightly different. Instead of balancing with Hbalancing with H22O and HO and H++, you must , you must balance with Hbalance with H22O and OHO and OH--..
Prof. T.L. Heise, CHE 116
24Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
A) NOA) NO22-- + Al + Al NH NH33 + Al(OH) + Al(OH)44
--
Prof. T.L. Heise, CHE 116
25Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
A) A) +3 -2 0 -3 +1 +3 -2 +1 +3 -2 0 -3 +1 +3 -2 +1
NONO22-- + Al + Al NH NH33 + Al(OH) + Al(OH)44
--
Prof. T.L. Heise, CHE 116
26Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
A) A) Ger = OA 6e - Ger = OA 6e -
+3 -2 0 -3 +1 +3 -2 +1+3 -2 0 -3 +1 +3 -2 +1
NONO22-- + Al + Al NH NH33 + Al(OH) + Al(OH)44
--
Leo = RA 3e -Leo = RA 3e -
Prof. T.L. Heise, CHE 116
27Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
A) A) Ger = OA 6e - Ger = OA 6e -
+3 -2 0 -3 +1 +3 -2 +1+3 -2 0 -3 +1 +3 -2 +1
NONO22-- + + 22Al Al NH NH33 + + 22Al(OH)Al(OH)44
--
Leo = RA 2(3e -)Leo = RA 2(3e -)
Prof. T.L. Heise, CHE 116
28Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
A) A)
NONO22-- + + 22Al + Al + 5H5H22OO + + OHOH--
NH NH33 + + 22Al(OH)Al(OH)44--
Prof. T.L. Heise, CHE 116
29Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
B) Cr(OH)B) Cr(OH)33 + ClO + ClO-- CrO CrO442-2- + Cl + Cl22
Prof. T.L. Heise, CHE 116
30Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
B) B) +3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0
Cr(OH)Cr(OH)33 + ClO + ClO-- CrO CrO442-2- + Cl + Cl22
Prof. T.L. Heise, CHE 116
31Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
B) B) Leo = RA 3e -Leo = RA 3e -
+3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0
Cr(OH)Cr(OH)33 + ClO + ClO-- CrO CrO442-2- + Cl + Cl22
Ger = OA 1e -Ger = OA 1e -
Prof. T.L. Heise, CHE 116
32Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
B) B) Leo = RA 3e -Leo = RA 3e -
+3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0
Cr(OH)Cr(OH)33 + + 66ClOClO-- CrO CrO442-2- + + 33ClCl22
Ger = OA 6(1e -)Ger = OA 6(1e -)
Prof. T.L. Heise, CHE 116
33Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
B) B) Leo = RA 2(3e -)Leo = RA 2(3e -)
+3 -2 +1 +1 -2 +6 -2 0+3 -2 +1 +1 -2 +6 -2 0
22Cr(OH)Cr(OH)33 + + 66ClOClO-- 22CrOCrO442-2- + + 33ClCl22
Ger = OA 6(1e -)Ger = OA 6(1e -)
Prof. T.L. Heise, CHE 116
34Balancing Oxidation Balancing Oxidation
- Reductions - Reductions ReactionReactionSample exercise: Complete and balance Sample exercise: Complete and balance
the following equations for oxidation and the following equations for oxidation and reduction reactions that occur in basic reduction reactions that occur in basic solutions.solutions.
B) B) 22Cr(OH)Cr(OH)33 + + 66ClOClO-- 22CrOCrO44
2-2- + + 33ClCl2 2 + 2H+ 2H22O + 2OHO + 2OH--
Prof. T.L. Heise, CHE 116
35Voltaic CellsVoltaic Cells
The energy released in a spontaneous The energy released in a spontaneous redox reaction can be used to perform redox reaction can be used to perform electrical work.electrical work.
This task is accomplished in a voltaic cell, This task is accomplished in a voltaic cell, a device in which the transfer of electrons a device in which the transfer of electrons takes place through an external pathway takes place through an external pathway rather than directly between reactants.rather than directly between reactants.
Prof. T.L. Heise, CHE 116
36 Voltaic CellsVoltaic Cells
The The spontaneous spontaneous reaction reaction occurs in a occurs in a single cell, single cell, but splitting but splitting the cell into the cell into two halves two halves allows for allows for the external the external pathway of pathway of electrons to electrons to be created.be created.
Prof. T.L. Heise, CHE 116
37Voltaic CellsVoltaic Cells
Prof. T.L. Heise, CHE 116
38Voltaic CellsVoltaic Cells
Prof. T.L. Heise, CHE 116
39Voltaic CellsVoltaic Cells
Parts of Voltaic Cell:Parts of Voltaic Cell:the two solid metals connected by the the two solid metals connected by the external circuit are called external circuit are called electrodeselectrodes..the electrode at which oxidation occurs the electrode at which oxidation occurs is called the anodeis called the anodethe electrode at which reduction occurs the electrode at which reduction occurs is called the cathode.is called the cathode.
Prof. T.L. Heise, CHE 116
40Voltaic CellsVoltaic Cells
Parts of Voltaic Cell:Parts of Voltaic Cell:the anode metal is the substance that the anode metal is the substance that undergoes oxidation to form more undergoes oxidation to form more positive ions.positive ions.the positive metallic ion in the beaker the positive metallic ion in the beaker that contains the cathode is the substance that contains the cathode is the substance that undergoes reduction to form more of that undergoes reduction to form more of the metal electrode.the metal electrode.
Prof. T.L. Heise, CHE 116
41Voltaic CellsVoltaic Cells
Parts of Voltaic Cell:Parts of Voltaic Cell:As the oxidation half cell gets a larger As the oxidation half cell gets a larger concentration of positive ions, negative concentration of positive ions, negative ions flow through the semipermeable ions flow through the semipermeable membrane that forms the salt bridge to membrane that forms the salt bridge to balance the charges in the beaker.balance the charges in the beaker.As the reduction half cell gets a smaller As the reduction half cell gets a smaller concentration of positive ions, positive concentration of positive ions, positive ions flow through the membrane to ions flow through the membrane to replace the loss.replace the loss.
Prof. T.L. Heise, CHE 116
42Voltaic CellsVoltaic Cells
Parts of Voltaic Cell:Parts of Voltaic Cell:in any voltaic cell the electrons flow in any voltaic cell the electrons flow from the anode (-) to the cathode (+) from the anode (-) to the cathode (+) through the external circuitthrough the external circuitThe collection of lost electrons at the The collection of lost electrons at the anode causes it to be negative, and the anode causes it to be negative, and the cathode is positive so it can attract the cathode is positive so it can attract the electrons and pull them through the wire.electrons and pull them through the wire.
Prof. T.L. Heise, CHE 116
43Voltaic CellsVoltaic Cells
Prof. T.L. Heise, CHE 116
44Voltaic CellsVoltaic Cells
Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are
Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--
ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)
a) Indicate which reaction occurs at the a) Indicate which reaction occurs at the anodeanode and the and the cathodecathode
Prof. T.L. Heise, CHE 116
45Voltaic CellsVoltaic Cells
Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are
Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--
ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)
Prof. T.L. Heise, CHE 116
46Voltaic CellsVoltaic Cells
Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are
Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--
ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)
b) Which electrode is consumed first in the b) Which electrode is consumed first in the cell reaction?cell reaction?
Prof. T.L. Heise, CHE 116
47Voltaic CellsVoltaic Cells
Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are
Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--
ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)
b) Which electrode is consumed first in the b) Which electrode is consumed first in the cell reaction?cell reaction?
The anode is consumedThe anode is consumed
Prof. T.L. Heise, CHE 116
48Voltaic CellsVoltaic Cells
Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are
Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--
ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)
c) Which electrode is positive?c) Which electrode is positive?
Prof. T.L. Heise, CHE 116
49Voltaic CellsVoltaic Cells
Sample exercise: The two half-reactions in Sample exercise: The two half-reactions in a voltaic cell area voltaic cell are
Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e(aq) + 2e--
ClOClO33--(aq) + 6H(aq) + 6H++ + 6e + 6e-- Cl Cl--(aq) + 3H(aq) + 3H22O(l)O(l)
c) Which electrode is positive?c) Which electrode is positive?
The cathode is positiveThe cathode is positive
Prof. T.L. Heise, CHE 116
50Voltaic CellsVoltaic Cells
An atomic view of what occurs in a single An atomic view of what occurs in a single cell in which both oxidation and reduction cell in which both oxidation and reduction are occurring may give better insight.are occurring may give better insight.
Prof. T.L. Heise, CHE 116
51Voltaic CellsVoltaic Cells
A molecular view allows a better A molecular view allows a better understanding of what is occurring in each understanding of what is occurring in each half cell.half cell.
Prof. T.L. Heise, CHE 116
52Cell EMFCell EMF
The reason that electrons flow The reason that electrons flow spontaneously is do to the “driving force” spontaneously is do to the “driving force” behind certain redox reactions.behind certain redox reactions.
The ‘driving force’ is a difference in The ‘driving force’ is a difference in potential energy.potential energy.The anode has a higher potential The anode has a higher potential energy, and electrons flow to the cathode energy, and electrons flow to the cathode to achieve a lower potential energy.to achieve a lower potential energy.As in many natural wonders, the flow As in many natural wonders, the flow from higher concentration to lower from higher concentration to lower continuescontinues
Prof. T.L. Heise, CHE 116
53Cell EMFCell EMF
The difference in potential energy per The difference in potential energy per charge is measured in volts.charge is measured in volts.
1 V (volt) is the potential difference 1 V (volt) is the potential difference required to impart 1 J (joule) of energy to required to impart 1 J (joule) of energy to a charge of 1 C (coulomb)a charge of 1 C (coulomb)the potential difference is called an the potential difference is called an electromotive force or emfelectromotive force or emfemf, or Eemf, or Ecellcell, is also called the cell , is also called the cell potentialpotential
Prof. T.L. Heise, CHE 116
54Cell EMFCell EMF
The difference in potential energy per The difference in potential energy per charge is measured in volts.charge is measured in volts.
The emf of a particular cell is The emf of a particular cell is dependent on the reaction that is dependent on the reaction that is occurring there.occurring there.We will always assume standard We will always assume standard conditions, and the cell potential should conditions, and the cell potential should be denoted as such, Ebe denoted as such, E°°
cellcell
Prof. T.L. Heise, CHE 116
55Cell EMFCell EMF
Standard reduction potential:Standard reduction potential:the cell potential will depend on the the cell potential will depend on the particular cathode and anode usedparticular cathode and anode usedthe cell potential is the difference the cell potential is the difference between two electrode potentialsbetween two electrode potentials
EEcellcell = E = Ecathodecathode + E + Eanodeanode
the amount of moles of each DOESthe amount of moles of each DOES NOT change the potentials of cellsNOT change the potentials of cells
Prof. T.L. Heise, CHE 116
56Cell EMFCell EMF
Standard reduction potential:Standard reduction potential:when finding the potentials, a reduction when finding the potentials, a reduction table will be used.table will be used.Finding the potential for cathodes is Finding the potential for cathodes is straightforward, simply read the tablestraightforward, simply read the tableFinding the potential for anodes means Finding the potential for anodes means reading the table BACKWARDS and the reading the table BACKWARDS and the sign of the potential should be switchedsign of the potential should be switchedTable 20.1 is a standard reduction table Table 20.1 is a standard reduction table
Prof. T.L. Heise, CHE 116
57Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:
0 +3 +2 +20 +3 +2 +2
Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq) (aq)
Prof. T.L. Heise, CHE 116
58Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:
0 +3 +2 +20 +3 +2 +2
Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq) (aq)
Prof. T.L. Heise, CHE 116
59Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction: overall reaction: Ger = cathodeGer = cathode
0 +3 +2 +20 +3 +2 +2
Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq)(aq)
Leo = anodeLeo = anode
Prof. T.L. Heise, CHE 116
60Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction: overall reaction: Ger = cathodeGer = cathode
0 +3 +2 +20 +3 +2 +2
Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq)(aq)
Leo = anodeLeo = anode
EEcellcell = E = Ecathodecathode + E + Eanodeanode
Prof. T.L. Heise, CHE 116
61Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction: overall reaction: Ger = cathode Ger = cathode
0 +3 +2 +20 +3 +2 +2
Ni(s) + 2FeNi(s) + 2Fe3+3+(aq) (aq) 2Fe 2Fe2+2+(aq) + Ni(aq) + Ni+2+2(aq)(aq)
Leo = anodeLeo = anode
EEcellcell = E = Ecathodecathode + E + Eanodeanode
= +0.77 + +0.28= +0.77 + +0.28
= +1.05 V= +1.05 V
Prof. T.L. Heise, CHE 116
62Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq) (aq)
Prof. T.L. Heise, CHE 116
63Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:
0 0 +3 -10 0 +3 -1
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq) (aq)
Prof. T.L. Heise, CHE 116
64Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:
leo = anodeleo = anode
0 0 +3 -10 0 +3 -1
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq) (aq)
ger = cathodeger = cathode
Prof. T.L. Heise, CHE 116
65Cell EMFCell EMF
Sample exercise: Calculate the standard Sample exercise: Calculate the standard emf for a cell that employs the following emf for a cell that employs the following overall reaction:overall reaction:
leo = anodeleo = anode
0 0 +3 -10 0 +3 -1
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq) (aq)
ger = cathodeger = cathode
Ecell = +0.54 + +1.66 = 2.20 VEcell = +0.54 + +1.66 = 2.20 V
Prof. T.L. Heise, CHE 116
66Cell EMFCell EMF
The more positive the potential difference, The more positive the potential difference, the greater the driving force.the greater the driving force.
Prof. T.L. Heise, CHE 116
67 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Any reaction that can occur in a voltaic cell Any reaction that can occur in a voltaic cell to produce a positive emf must be to produce a positive emf must be spontaneous.spontaneous.
EEcellcell = E = Eredred + E + Eoxox
a + emf is spontaneousa + emf is spontaneousa - emf is not spontaneousa - emf is not spontaneous
Prof. T.L. Heise, CHE 116
68 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
A) IA) I22(s) + 5Cu(s) + 5Cu2+2+(aq) + 6H(aq) + 6H22O(l) O(l)
2IO2IO33--(aq) + 5Cu(s) + 12H(aq) + 5Cu(s) + 12H++(aq)(aq)
Prof. T.L. Heise, CHE 116
69 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
A) IA) I22(s) + 5Cu(s) + 5Cu2+2+(aq) + 6H(aq) + 6H22O(l) O(l)
2IO2IO33--(aq) + 5Cu(s) + 12H(aq) + 5Cu(s) + 12H++(aq)(aq)
Red: CuRed: Cu2+2+(aq) (aq) Cu(s) Cu(s)
Ox: IOx: I22(s) + 6H(s) + 6H22O O 2IO 2IO33--(aq) + 12H(aq) + 12H++(aq)(aq)
Prof. T.L. Heise, CHE 116
70 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
A) IA) I22(s) + 5Cu(s) + 5Cu2+2+(aq) + 6H(aq) + 6H22O(l) O(l)
2IO2IO33--(aq) + 5Cu(s) + 12H(aq) + 5Cu(s) + 12H++(aq)(aq)
Red: CuRed: Cu2+2+(aq) (aq) Cu(s) Cu(s)
Ox: IOx: I22(s) + 6H(s) + 6H22O O 2IO 2IO33--(aq) + 12H(aq) + 12H++(aq)(aq)
Ecell = +0.34 + -1.195Ecell = +0.34 + -1.195
Prof. T.L. Heise, CHE 116
71 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
A) IA) I22(s) + 5Cu(s) + 5Cu2+2+(aq) + 6H(aq) + 6H22O(l) O(l)
2IO2IO33--(aq) + 5Cu(s) + 12H(aq) + 5Cu(s) + 12H++(aq)(aq)
Red: CuRed: Cu2+2+(aq) (aq) Cu(s) Cu(s)
Ox: IOx: I22(s) + 6H(s) + 6H22O O 2IO 2IO33--(aq) + 12H(aq) + 12H++(aq)(aq)
Ecell = +0.34 + -1.195 = -0.855 NOTEcell = +0.34 + -1.195 = -0.855 NOT
Prof. T.L. Heise, CHE 116
72 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
B) HgB) Hg2+2+(aq) + 2I(aq) + 2I--(aq) (aq) Hg(l) + I Hg(l) + I22(s)(s)
Prof. T.L. Heise, CHE 116
73 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
B) HgB) Hg2+2+(aq) + 2I(aq) + 2I--(aq) (aq) Hg(l) + I Hg(l) + I22(s)(s)
Red: HgRed: Hg2+2+(aq) (aq) Hg(l) Hg(l)
Ox: 2IOx: 2I--(aq) (aq) I I22(s) (s)
Prof. T.L. Heise, CHE 116
74 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
B) HgB) Hg2+2+(aq) + 2I(aq) + 2I--(aq) (aq) Hg(l) + I Hg(l) + I22(s)(s)
Red: HgRed: Hg2+2+(aq) (aq) Hg(l) Hg(l)
Ox: 2IOx: 2I--(aq) (aq) I I22(s) (s)
Ecell = +0.854 + -0.536 Ecell = +0.854 + -0.536
Prof. T.L. Heise, CHE 116
75 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Using the standard Sample exercise: Using the standard reduction potentials, determine which of the reduction potentials, determine which of the following reactions are spontaneous.following reactions are spontaneous.
B) HgB) Hg2+2+(aq) + 2I(aq) + 2I--(aq) (aq) Hg(l) + I Hg(l) + I22(s)(s)
Red: HgRed: Hg2+2+(aq) (aq) Hg(l) Hg(l)
Ox: 2IOx: 2I--(aq) (aq) I I22(s) (s)
Ecell = +0.854 + -0.536 = +0.318 SPONEcell = +0.854 + -0.536 = +0.318 SPON
Prof. T.L. Heise, CHE 116
76 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Looking at the standard reduction table, a Looking at the standard reduction table, a general rule can be established:general rule can be established:
any metal in the series is able to any metal in the series is able to spontaneously react with any ion beneath spontaneously react with any ion beneath ititthere is also a relationship between emf there is also a relationship between emf and Gibbs Free Energy and Gibbs Free Energy (remember a negative (remember a negative G is spontaneous)G is spontaneous)
G = -nFEG = -nFE
Prof. T.L. Heise, CHE 116
77 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
G = -nFEG = -nFE
n = number of electrons n = number of electrons transferred in reactiontransferred in reaction
F = Faraday’s constantF = Faraday’s constant
1 F = 96,500 J/V-mol1 F = 96,500 J/V-mol
E = emf of cellE = emf of cell
units of units of G are J/molG are J/mol
Prof. T.L. Heise, CHE 116
78 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Consider the following Sample exercise: Consider the following reaction:reaction:
3Ni3Ni2+2+(aq) + 2Cr(OH)(aq) + 2Cr(OH)33(s) + 10OH(s) + 10OH-- 3Ni(s) + 2CrO 3Ni(s) + 2CrO44
2-2-(aq) + 8H(aq) + 8H22O(l)O(l)
a) what is the value of n for this reaction?a) what is the value of n for this reaction?
Prof. T.L. Heise, CHE 116
79 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Consider the following Sample exercise: Consider the following reaction:reaction:
3Ni3Ni2+2+(aq) + 2Cr(OH)(aq) + 2Cr(OH)33(s) + 10OH(s) + 10OH-- 3Ni(s) + 2CrO 3Ni(s) + 2CrO44
2-2-(aq) + 8H(aq) + 8H22O(l)O(l)
a) what is the value of n for this reaction?a) what is the value of n for this reaction?
Each NiEach Ni2+2+ gains 2e gains 2e-- and there are 3 and there are 3 ions, so 6e- are involvedions, so 6e- are involved
n = 6n = 6
Prof. T.L. Heise, CHE 116
80 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Consider the following Sample exercise: Consider the following reaction:reaction:
3Ni3Ni2+2+(aq) + 2Cr(OH)(aq) + 2Cr(OH)33(s) + 10OH(s) + 10OH-- 3Ni(s) + 2CrO 3Ni(s) + 2CrO44
2-2-(aq) + 8H(aq) + 8H22O(l)O(l)
b) Calculate b) Calculate GG
G = -nFEG = -nFE
Prof. T.L. Heise, CHE 116
81 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Consider the following Sample exercise: Consider the following reaction:reaction:
3Ni3Ni2+2+(aq) + 2Cr(OH)(aq) + 2Cr(OH)33(s) + 10OH(s) + 10OH-- 3Ni(s) + 2CrO 3Ni(s) + 2CrO44
2+2+(aq) + 8H(aq) + 8H22O(l)O(l)
b) Calculate b) Calculate GG
G = -nFE n = 6G = -nFE n = 6
F = 96,500 J/V-molF = 96,500 J/V-mol
E = -0.28 + +0.13= -0.15 VE = -0.28 + +0.13= -0.15 V
Prof. T.L. Heise, CHE 116
82 Spontaneity of Spontaneity of Redox ReactionsRedox Reactions
Sample exercise: Consider the following Sample exercise: Consider the following reaction:reaction:
3Ni3Ni2+2+(aq) + 2Cr(OH)(aq) + 2Cr(OH)33(s) + 10OH(s) + 10OH-- 3Ni(s) + 2CrO 3Ni(s) + 2CrO44
2+2+(aq) + 8H(aq) + 8H22O(l)O(l)
b) Calculate b) Calculate GG
G = -nFE G = -nFE -(6)(96,500 J/V-mol)(-0.15 V) -(6)(96,500 J/V-mol)(-0.15 V)+86850 J/mol+86850 J/mol
NOT SponNOT Spon
Prof. T.L. Heise, CHE 116
83Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFAs a voltaic cell is discharged, the reactants As a voltaic cell is discharged, the reactants
of the reaction are consumed and the of the reaction are consumed and the products are generated, so the products are generated, so the concentrations of these substances change.concentrations of these substances change.
The emf progressively drops until E = 0, at The emf progressively drops until E = 0, at which point we say the cell is dead. At that which point we say the cell is dead. At that point the concentrations in the cell are at point the concentrations in the cell are at equilibrium.equilibrium.
Prof. T.L. Heise, CHE 116
84Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFThe dependence of cell emf on concentration The dependence of cell emf on concentration
can be obtained from the dependence of the can be obtained from the dependence of the free energy change on concentrationfree energy change on concentration
Nernst Equation: E = ENernst Equation: E = E°° - RT (lnQ) - RT (lnQ) nF nF
= E= E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
Prof. T.L. Heise, CHE 116
85Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFThe dependence of cell emf on concentration The dependence of cell emf on concentration
can be obtained from the dependence of the can be obtained from the dependence of the free energy change on concentrationfree energy change on concentration
Nernst Equation: E = ENernst Equation: E = E°° - RT (lnQ) - RT (lnQ) nF nF
* Q is the reaction* Q is the reaction = E= E°° - 0.0592 V (log Q) - 0.0592 V (log Q)quotient quotient n n
Prof. T.L. Heise, CHE 116
86Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
Prof. T.L. Heise, CHE 116
87Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
Prof. T.L. Heise, CHE 116
88Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
EE°° = +0.536 + +1.66 = 2.22 V = +0.536 + +1.66 = 2.22 V
Prof. T.L. Heise, CHE 116
89Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
n = 6 en = 6 e--
Prof. T.L. Heise, CHE 116
90Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
Q = [AlQ = [Al33+]+]22[I[I--]]66 = [4.0 x 10 = [4.0 x 10-3-3]]22[0.010][0.010]66 = 1.6 x 10= 1.6 x 10-17-17
Prof. T.L. Heise, CHE 116
91Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
= 2.22 - 0.0592 (log 1.6 x 10= 2.22 - 0.0592 (log 1.6 x 10-17-17)) 6 6
Prof. T.L. Heise, CHE 116
92Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Calculate the emf Sample exercise: Calculate the emf
generated by the cellgenerated by the cell
2Al(s) + 3I2Al(s) + 3I22(s) (s) 2Al 2Al3+3+(aq) + 6I(aq) + 6I--(aq)(aq)
when [Alwhen [Al3+3+] = 4.0 x 10] = 4.0 x 10-3-3M and [IM and [I--] = 0.010 M.] = 0.010 M.
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
= 2.22 - 0.0592 (log 1.6 x 10= 2.22 - 0.0592 (log 1.6 x 10-17-17)) 6 6
= 2.36 V = 2.36 V
Prof. T.L. Heise, CHE 116
93Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFConcentration cells: a voltaic cell with an Concentration cells: a voltaic cell with an
emf of ZERO can be constructed if the same emf of ZERO can be constructed if the same species is used as both the anode and the species is used as both the anode and the cathode if concentrations are the same; cathode if concentrations are the same; using two different concentrations of the using two different concentrations of the same species will create a non-zero cell.same species will create a non-zero cell.
A cell based solely on the emf generated A cell based solely on the emf generated because of a difference in a concentration is because of a difference in a concentration is called a concentration cellcalled a concentration cell
Prof. T.L. Heise, CHE 116
94Effect of Effect of
Concentration on Concentration on Cell EMFCell EMF
Prof. T.L. Heise, CHE 116
95Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFThe standard emf for this cell would have to The standard emf for this cell would have to
zero, but the cell operates due to the zero, but the cell operates due to the differences in concentration.differences in concentration.
The driving force for the cell is provided by The driving force for the cell is provided by the difference in concentration.the difference in concentration.
Operation of the cell proceeds to equalize Operation of the cell proceeds to equalize the concentrations: the more dilute the concentrations: the more dilute concentration acts as the anode, the solid concentration acts as the anode, the solid metal electrode will oxidize to turn into metal electrode will oxidize to turn into more ions.more ions.
Prof. T.L. Heise, CHE 116
96Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFThe more concentrated solution acts as the The more concentrated solution acts as the
cathode, where the larger number of ions cathode, where the larger number of ions reduce to plate the electrode and thereby reduce to plate the electrode and thereby reduce there number. reduce there number.
The nernst equation can also be used here to The nernst equation can also be used here to calculate the Ecalculate the Ecellcell under non-standard under non-standard conditions.conditions.
Anode: X(s) Anode: X(s) X X++(dilute) + e(dilute) + e--
Cathode: XCathode: X++(conc.) + e(conc.) + e-- X(s) X(s)
Prof. T.L. Heise, CHE 116
97Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFAnode: X(s) Anode: X(s) X X++(dilute) + e(dilute) + e--
Cathode: XCathode: X++(conc.) + e(conc.) + e-- X(s) X(s)
XX++(conc.) + X(s) (conc.) + X(s) X X++(dilute) + X(s)(dilute) + X(s)
Q = Q = X X++(dilute) / X(dilute) / X++(conc.) (conc.)
Prof. T.L. Heise, CHE 116
98Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is
constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.
a) Which half cell is the anode of the cell?a) Which half cell is the anode of the cell?
Prof. T.L. Heise, CHE 116
99Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is
constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.
a) Which half cell is the anode of the cell?a) Which half cell is the anode of the cell?
The cell with [ZnThe cell with [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M is M is the anode.the anode.
Prof. T.L. Heise, CHE 116
100Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is
constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.
b) What is the emf of the cell?b) What is the emf of the cell?
Prof. T.L. Heise, CHE 116
101Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is
constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.
b) What is the emf of the cell?b) What is the emf of the cell?
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
Prof. T.L. Heise, CHE 116
102Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is
constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.
b) What is the emf of the cell?b) What is the emf of the cell?
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
= 0 - 0.0592 (log 3.75x10= 0 - 0.0592 (log 3.75x10-4-4/1.35)/1.35) 2 2
Prof. T.L. Heise, CHE 116
103Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: A concentration cell is Sample exercise: A concentration cell is
constructed with two Zn(s) constructed with two Zn(s) Zn Zn2+2+(aq) half-(aq) half-cells. The first cell has [Zncells. The first cell has [Zn2+2+] = 1.35 M and ] = 1.35 M and the second cell has [Znthe second cell has [Zn2+2+] = 3.75 x 10] = 3.75 x 10-4-4 M. M.
b) What is the emf of the cell?b) What is the emf of the cell?
E = EE = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
= 0 - 0.0592 (log 3.75x10= 0 - 0.0592 (log 3.75x10-4-4/1.35)/1.35) 2 2
= 0.105 V= 0.105 V
Prof. T.L. Heise, CHE 116
104Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFCell emf and Chemical Equilibrium: the Cell emf and Chemical Equilibrium: the
nernst equation helps us understand why nernst equation helps us understand why the emf drops as the cell is discharged: as the emf drops as the cell is discharged: as the reactants are converted into products, the reactants are converted into products, the value of Q increases, so the value of E the value of Q increases, so the value of E decreases.decreases.
The cell emf eventually reaches 0.The cell emf eventually reaches 0.
When When G = 0, E = 0, and an EG = 0, E = 0, and an Ecellcell = 0 shows a = 0 shows a cell at equilibrium.cell at equilibrium.
Prof. T.L. Heise, CHE 116
105Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFIf 0 = EIf 0 = E°° - 0.0592 V (log Q) - 0.0592 V (log Q) n n
Then log K = nEThen log K = nE°° 0.0592 0.0592
Prof. T.L. Heise, CHE 116
106Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Using standard reduction Sample exercise: Using standard reduction
potentials, calculate the equilibrium potentials, calculate the equilibrium constant at 25°C for the reactionconstant at 25°C for the reaction
BrBr22(l) + 2Cl(l) + 2Cl--(aq) (aq) Cl Cl22(g) + 2Br(g) + 2Br--(aq)(aq)
Prof. T.L. Heise, CHE 116
107Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Using standard reduction Sample exercise: Using standard reduction
potentials, calculate the equilibrium potentials, calculate the equilibrium constant at 25°C for the reactionconstant at 25°C for the reaction
BrBr22(l) + 2Cl(l) + 2Cl--(aq) (aq) Cl Cl22(g) + 2Br(g) + 2Br--(aq)(aq)
log K = nElog K = nE°° 0.0592 0.0592
Prof. T.L. Heise, CHE 116
108Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Using standard reduction Sample exercise: Using standard reduction
potentials, calculate the equilibrium potentials, calculate the equilibrium constant at 25°C for the reactionconstant at 25°C for the reaction
BrBr22(l) + 2Cl(l) + 2Cl--(aq) (aq) Cl Cl22(g) + 2Br(g) + 2Br--(aq)(aq)
log K = nElog K = nE°° 0.0592 0.0592
= 2(+1.065 + -1.359)= 2(+1.065 + -1.359) 0.0592 0.0592
Prof. T.L. Heise, CHE 116
109Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Using standard reduction Sample exercise: Using standard reduction
potentials, calculate the equilibrium potentials, calculate the equilibrium constant at 25°C for the reactionconstant at 25°C for the reaction
BrBr22(l) + 2Cl(l) + 2Cl--(aq) (aq) Cl Cl22(g) + 2Br(g) + 2Br--(aq)(aq)
log K = nElog K = nE°° 0.0592 0.0592
= 2(+1.065 + -1.359)= 2(+1.065 + -1.359) 0.0592 0.0592
= -9.93= -9.93
Prof. T.L. Heise, CHE 116
110Effect of Effect of
Concentration on Concentration on Cell EMFCell EMFSample exercise: Using standard reduction Sample exercise: Using standard reduction
potentials, calculate the equilibrium potentials, calculate the equilibrium constant at 25°C for the reactionconstant at 25°C for the reaction
BrBr22(l) + 2Cl(l) + 2Cl--(aq) (aq) Cl Cl22(g) + 2Br(g) + 2Br--(aq)(aq)
K = 10K = 10(-9.93)(-9.93)
= 1.2 x 10= 1.2 x 10-10-10
Prof. T.L. Heise, CHE 116
111BatteriesBatteries
A battery is a portable, self-contained A battery is a portable, self-contained electrochemical power source that consists electrochemical power source that consists of one or more voltaic cells.of one or more voltaic cells.
The common 1.5 V battery is a single voltaic The common 1.5 V battery is a single voltaic cell, where a 12 V battery uses multiple cell, where a 12 V battery uses multiple voltaic cells in one case.voltaic cells in one case.
When cells are connected in series, with the When cells are connected in series, with the cathode of one attached to the anode to cathode of one attached to the anode to another, the battery produces a voltage that another, the battery produces a voltage that is the sum of the emfs of the single cells.is the sum of the emfs of the single cells.
Prof. T.L. Heise, CHE 116
112BatteriesBatteries
Batteries connected in series...
Prof. T.L. Heise, CHE 116
113BatteriesBatteries
Commerical batteries that has specific Commerical batteries that has specific performance characteristics can require performance characteristics can require considerable ingenuity.considerable ingenuity.
The emf of a battery is determined by the The emf of a battery is determined by the substances used as the cathode and anode, substances used as the cathode and anode, and the usable life of the battery depends on and the usable life of the battery depends on the quantities of the substances packaged.the quantities of the substances packaged.
Keep in mind the anode and cathode need to Keep in mind the anode and cathode need to be separated by a porous barrier similar to be separated by a porous barrier similar to a salt bridge.a salt bridge.
Prof. T.L. Heise, CHE 116
114BatteriesBatteries
The materials used to construct the battery The materials used to construct the battery must be stable under the conditions in which must be stable under the conditions in which it is to be used and must be chosen to it is to be used and must be chosen to minimize health and environmental minimize health and environmental concerns upon use and disposal.concerns upon use and disposal.
Different applications require batteries with Different applications require batteries with different properties.different properties.
Prof. T.L. Heise, CHE 116
115BatteriesBatteries
Lead-Acid BatteryLead-Acid Battery
Battery used in cars is one of the most Battery used in cars is one of the most common batteries, with over 100 million common batteries, with over 100 million produced annually.produced annually.
A 12-V battery consists of six voltaic cells in A 12-V battery consists of six voltaic cells in series, each producing 2-V.series, each producing 2-V.
Cathode: lead dioxide packed on a Cathode: lead dioxide packed on a metal gridmetal grid
Anode: leadAnode: lead
Prof. T.L. Heise, CHE 116
116BatteriesBatteries
Lead-Acid BatteryLead-Acid Battery
Both electrodes are immersed in sulfuric Both electrodes are immersed in sulfuric acid.acid.
Both reactants are solids so there is no need Both reactants are solids so there is no need to separate the cell into anode and cathode to separate the cell into anode and cathode compartments.compartments.
To keep the electrodes from touching, wood To keep the electrodes from touching, wood or fiber glass spacers are placed between or fiber glass spacers are placed between themthem
Prof. T.L. Heise, CHE 116
117BatteriesBatteries
Lead-Acid BatteryLead-Acid Battery
Using solids offers another advantage, no Using solids offers another advantage, no concentration changes occur and the emf concentration changes occur and the emf stays mostly constant, fluctuations occuring stays mostly constant, fluctuations occuring with variations in [Hwith variations in [H22SOSO44].].
One major advantage with this battery is One major advantage with this battery is the ability to be rechargedthe ability to be recharged
Prof. T.L. Heise, CHE 116
118BatteriesBatteries
Prof. T.L. Heise, CHE 116
119BatteriesBatteries
Alkaline Battery:Alkaline Battery:
The most common non-rechargeable battery The most common non-rechargeable battery is the alkaline battery. is the alkaline battery. More than 10More than 101010 are produced annually. are produced annually.
The anode of this battery consists of The anode of this battery consists of powdered zinc metal immobilized in a gel in powdered zinc metal immobilized in a gel in contact with a concentrated solution of contact with a concentrated solution of KOH.KOH.
The cathode is a mixture of MnOThe cathode is a mixture of MnO22 and and graphitegraphite
Prof. T.L. Heise, CHE 116
120BatteriesBatteries
Prof. T.L. Heise, CHE 116
121BatteriesBatteries
Nickel-Cadmium, Nickel-Metal-Hydride, Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteriesand Lithium-Ion Batteries
All of these batteries are lightweight, easily All of these batteries are lightweight, easily rechargeable batteries for cell phones, rechargeable batteries for cell phones, notebook computers and video recorders.notebook computers and video recorders.
Prof. T.L. Heise, CHE 116
122BatteriesBatteries
Nickel-CadmiumNickel-Cadmium
Cadmium is oxidized while nickel Cadmium is oxidized while nickel oxyhydroxide is reducedoxyhydroxide is reduced
The solid reaction products adhere to the The solid reaction products adhere to the electrodes, which permits the reactions to be electrodes, which permits the reactions to be reversed during charging.reversed during charging.
Drawbacks: cadmium is a toxic heavy metalDrawbacks: cadmium is a toxic heavy metal
Prof. T.L. Heise, CHE 116
123BatteriesBatteries
Nickel-Metal-HydrideNickel-Metal-Hydride
Nickel oxyhydroxide is still reduced, but Nickel oxyhydroxide is still reduced, but now the anode is a metal alloy that has the now the anode is a metal alloy that has the ability to absorb hydrogen atoms.ability to absorb hydrogen atoms.
The hydrogen atoms are released as waterThe hydrogen atoms are released as water
Prof. T.L. Heise, CHE 116
124BatteriesBatteries
Lithium-Ion BatteriesLithium-Ion Batteries
Lithium is a very light-weight element , and Lithium is a very light-weight element , and the technology is unique.the technology is unique.
Li+ ions insert themselves into certain Li+ ions insert themselves into certain layered solids; during discharge lithium ions layered solids; during discharge lithium ions migrate between two different layered migrate between two different layered materials that serve as the anode and materials that serve as the anode and cathodecathode
Prof. T.L. Heise, CHE 116
125BatteriesBatteries
Fuel CellsFuel Cells
The direct production of electricity from The direct production of electricity from fuels by a voltaic cell could, in principle, fuels by a voltaic cell could, in principle, yield a higher rate of conversion of the yield a higher rate of conversion of the chemical energy of the reaction.chemical energy of the reaction.
Voltaic cells that perform this conversion Voltaic cells that perform this conversion using conventional fuels are called fuel cells.using conventional fuels are called fuel cells.
Fuel cells are not true batteries because they Fuel cells are not true batteries because they are not self-contained.are not self-contained.
Prof. T.L. Heise, CHE 116
126BatteriesBatteries
Fuel CellsFuel Cells
The most promising The most promising fuel cell system fuel cell system involves the reaction of involves the reaction of HH22 and O and O22 to form to form HH22O(l) as the only O(l) as the only product.product.
Cathode: 4e- + OCathode: 4e- + O22 + 2H + 2H22O O 4OH- 4OH-
Anode: 2HAnode: 2H22 + 4OH- + 4OH- 4H 4H22O + 4e- O + 4e-
Prof. T.L. Heise, CHE 116
127CorrosionCorrosion
Undesirable redox reactions that lead to the Undesirable redox reactions that lead to the corrosion of metals are the other side to corrosion of metals are the other side to spontaneous reactions.spontaneous reactions.
For nearly all metals, oxidation is a For nearly all metals, oxidation is a thermodynamically favorable process in air thermodynamically favorable process in air at room temperature.at room temperature.
With certain metals, when oxidation occurs With certain metals, when oxidation occurs a protective oxide layer is formed - example: a protective oxide layer is formed - example: aluminumaluminum
Prof. T.L. Heise, CHE 116
128CorrosionCorrosion
Corrosion of ironCorrosion of iron
From an economic standpoint, this is a From an economic standpoint, this is a major problem: 20% of the iron produced is major problem: 20% of the iron produced is used to replace iron objects that have been used to replace iron objects that have been discarded because of rust damage.discarded because of rust damage.
The rusting of iron requires both oxygen The rusting of iron requires both oxygen and water.and water.
Other factors can accelerate rusting - pH, Other factors can accelerate rusting - pH, salts, contact with metals and stresssalts, contact with metals and stress
Prof. T.L. Heise, CHE 116
129CorrosionCorrosion
Corrosion of ironCorrosion of iron
Corrosion can occur anywhere, but the best Corrosion can occur anywhere, but the best place is the spot where the iron has the most place is the spot where the iron has the most access to oxygen.access to oxygen.
Prof. T.L. Heise, CHE 116
130CorrosionCorrosion
Preventing the Corrosion of IronPreventing the Corrosion of Iron
Covering with paint or another metal can Covering with paint or another metal can protect the iron from corrosion.protect the iron from corrosion.
Paint offers a layer against the iron coming Paint offers a layer against the iron coming into contactinto contact
A second metal that is easier to oxidize then A second metal that is easier to oxidize then iron will corrode first offering protection.iron will corrode first offering protection.
Prof. T.L. Heise, CHE 116
131CorrosionCorrosion
Prof. T.L. Heise, CHE 116
132CorrosionCorrosion
Prof. T.L. Heise, CHE 116
133ElectrolysisElectrolysis
It is possible to use electrical energy to It is possible to use electrical energy to cause nonspontaneous redox reactions to cause nonspontaneous redox reactions to occur.occur.
Processes which are driven by an outside Processes which are driven by an outside source of electrical energy are called source of electrical energy are called electrolysis reactions and take place in electrolysis reactions and take place in electrolytic cells.electrolytic cells.
Prof. T.L. Heise, CHE 116
134ElectrolysisElectrolysis
Electrolytic cellsElectrolytic cellsconsists of two electrodes in a molten consists of two electrodes in a molten salt or a solution.salt or a solution.driven by a battery or some other driven by a battery or some other source of direct electrical current.source of direct electrical current.the battery acts as an electron pump, the battery acts as an electron pump, pushing electrons into one electrode and pushing electrons into one electrode and pulling them from the otherpulling them from the otherthe electrodes are inert, acting only as the electrodes are inert, acting only as the site for oxidation and reductionthe site for oxidation and reduction
Prof. T.L. Heise, CHE 116
135ElectrolysisElectrolysis
Electrolytic cellsElectrolytic cells
reduction reduction still occurs still occurs
at the at the
cathodecathode
oxidation oxidation still occurs still occurs
at the at the
anodeanode
Prof. T.L. Heise, CHE 116
136ElectrolysisElectrolysis
Electrolytic cellsElectrolytic cells
Na+ Na+ reduces at reduces at
cathode to cathode to form Naform Na
Cl- oxidizes Cl- oxidizes at anode to at anode to form Clform Cl22
Prof. T.L. Heise, CHE 116
137ElectrolysisElectrolysis
Electrolytic cellsElectrolytic cellsthe electrode connected to the negative the electrode connected to the negative terminal is now the cathode due to the terminal is now the cathode due to the battery supplying the electrons for battery supplying the electrons for reductionreductionthe electrode connected to the positive the electrode connected to the positive terminal is now the anode to draw terminal is now the anode to draw electrons off the metal and cause the electrons off the metal and cause the metal to oxidize and form ionsmetal to oxidize and form ions
Prof. T.L. Heise, CHE 116
138ElectrolysisElectrolysis
Electrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutionsbecause of the high melting points of because of the high melting points of ionic substances, the electrolysis of ionic substances, the electrolysis of molten salts requires very high molten salts requires very high temperaturestemperatureswe can produce ions from soluble salts we can produce ions from soluble salts at room temperatures by dissolving the at room temperatures by dissolving the salt in watersalt in waterthe electrolysis of an aqueous solution the electrolysis of an aqueous solution is complicated by the presence of wateris complicated by the presence of water
Prof. T.L. Heise, CHE 116
139ElectrolysisElectrolysis
Electrolysis of Aqueous SolutionsElectrolysis of Aqueous Solutionswe must consider if the water is we must consider if the water is oxidized or reducedoxidized or reducedif the water is oxidized, Oif the water is oxidized, O22 and H and H++ ions ions are formedare formedif the water is reduced, Hif the water is reduced, H22 and OH and OH-- ions are formedions are formedthe more positive the value of Ethe more positive the value of E°°
redred, the , the more favorable the reduction ismore favorable the reduction is
Prof. T.L. Heise, CHE 116
140ElectrolysisElectrolysis
Electrolysis with Active ElectrodesElectrolysis with Active Electrodesseveral practical applications of several practical applications of electrochemistry are based on active electrochemistry are based on active electrodeselectrodeselectroplating involves using electroplating involves using electrolysis to deposit a thin layer of one electrolysis to deposit a thin layer of one metal on top of another in order to metal on top of another in order to improve beauty or resistance to improve beauty or resistance to corrosioncorrosion
Prof. T.L. Heise, CHE 116
141ElectrolysisElectrolysis
Electrolysis with Active ElectrodesElectrolysis with Active Electrodes
Prof. T.L. Heise, CHE 116
142ElectrolysisElectrolysis
Quantitative Aspects of ElectrolysisQuantitative Aspects of Electrolysisfor any half reaction the amount of a for any half reaction the amount of a substance that is reduced or oxidized in substance that is reduced or oxidized in an electrolytic cell is directly an electrolytic cell is directly proportional to the number of electrons proportional to the number of electrons passed into the cellpassed into the cellFig 20.31Fig 20.31
Prof. T.L. Heise, CHE 116
143ElectrolysisElectrolysis
Sample exercise: The half reaction for Sample exercise: The half reaction for formation of magnesium metal upon formation of magnesium metal upon electrolysis of molten MgClelectrolysis of molten MgCl22 is is MgMg2+2+ + 2e + 2e-- Mg. Calculate the mass of Mg. Calculate the mass of magnesium formed upon passage of magnesium formed upon passage of current of 60.0 A for a period of current of 60.0 A for a period of 4.00 x 104.00 x 1033 s. s.
Prof. T.L. Heise, CHE 116
144ElectrolysisElectrolysis
Sample exercise: MgSample exercise: Mg2+2+ + 2e + 2e-- Mg. Calculate Mg. Calculate the mass of magnesium formed upon the mass of magnesium formed upon passage of current of 60.0 A for a period of passage of current of 60.0 A for a period of 4.00 x 104.00 x 1033 s. s.
Coulombs= A x sCoulombs= A x s
= 60.0 A x 4.00 x 10= 60.0 A x 4.00 x 1033 s s
= 2.4 x 10= 2.4 x 1055 C C
Prof. T.L. Heise, CHE 116
145ElectrolysisElectrolysis
Sample exercise: MgSample exercise: Mg2+2+ + 2e + 2e-- Mg. Calculate the Mg. Calculate the mass of magnesium formed upon passage of mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10current of 60.0 A for a period of 4.00 x 1033 s. s.
2.4 x 102.4 x 1055 C 1 mole e C 1 mole e-- 1 mole Mg 24.305 g Mg 1 mole Mg 24.305 g Mg 96,500 C 2 mole e 96,500 C 2 mole e-- 1 mole Mg 1 mole Mg
Prof. T.L. Heise, CHE 116
146ElectrolysisElectrolysis
Sample exercise: MgSample exercise: Mg2+2+ + 2e + 2e-- Mg. Calculate the Mg. Calculate the mass of magnesium formed upon passage of mass of magnesium formed upon passage of current of 60.0 A for a period of 4.00 x 10current of 60.0 A for a period of 4.00 x 1033 s. s.
2.4 x 102.4 x 1055 C 1 mole e C 1 mole e-- 1 mole Mg 24.305 g Mg 1 mole Mg 24.305 g Mg 96,500 C 2 mole e 96,500 C 2 mole e-- 1 mole Mg 1 mole Mg
30.2 g Mg30.2 g Mg
Prof. T.L. Heise, CHE 116
147ElectrolysisElectrolysis
Work done by electrical current can be Work done by electrical current can be calculated now based on the same formula as calculated now based on the same formula as Gibbs Free Energy.Gibbs Free Energy.
wwmaxmax = -nFE * voltaic cells = -nFE * voltaic cells
w = nFEw = nFEextext *electrolytic cells*electrolytic cells