electrochemistry chapter 19 electron transfer reactions electron transfer reactions are oxidation-...
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Electrochemistry
Chapter 19
Electron Transfer ReactionsElectron Transfer ReactionsElectron Transfer ReactionsElectron Transfer Reactions• Electron transfer reactions are oxidation-reduction or redox
reactions.
• Results in the generation of an electric current (electricity)
or be caused by imposing an electric current.
• Therefore, this field of chemistry is often called
ELECTROCHEMISTRY.
2Mg (s) + O2 (g) 2MgO (s)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
19.1
Electrochemical processes are oxidation-reduction reactions in which:
• the energy released by a spontaneous reaction is converted to electricity or
• electrical energy is used to cause a nonspontaneous reaction to occur
0 0 2+ 2-
Terminology for Redox ReactionsTerminology for Redox ReactionsTerminology for Redox ReactionsTerminology for Redox Reactions
• OXIDATIONOXIDATION—loss of electron(s) by a species; —loss of electron(s) by a species; increase in oxidation number; increase in oxygen.increase in oxidation number; increase in oxygen.
• REDUCTIONREDUCTION—gain of electron(s); decrease in —gain of electron(s); decrease in oxidation number; decrease in oxygen; increase oxidation number; decrease in oxygen; increase in hydrogen.in hydrogen.
• OXIDIZING AGENTOXIDIZING AGENT—electron acceptor; species is —electron acceptor; species is reduced. (an agent facilitates something; ex. reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)Travel agents don’t travel, they facilitate travel)
• REDUCING AGENTREDUCING AGENT—electron donor; species is —electron donor; species is oxidized.oxidized.
• OXIDATIONOXIDATION—loss of electron(s) by a species; —loss of electron(s) by a species; increase in oxidation number; increase in oxygen.increase in oxidation number; increase in oxygen.
• REDUCTIONREDUCTION—gain of electron(s); decrease in —gain of electron(s); decrease in oxidation number; decrease in oxygen; increase oxidation number; decrease in oxygen; increase in hydrogen.in hydrogen.
• OXIDIZING AGENTOXIDIZING AGENT—electron acceptor; species is —electron acceptor; species is reduced. (an agent facilitates something; ex. reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)Travel agents don’t travel, they facilitate travel)
• REDUCING AGENTREDUCING AGENT—electron donor; species is —electron donor; species is oxidized.oxidized.
You can’t have one… without the other!
• Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons.
• You can’t have 2 oxidations only or 2 reductions only in the same equation. Reduction has to occur at the cost of oxidation
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Review of Oxidation numbers
The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
4.4
4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.
5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
HCO3-
O = -2 H = +1
3x(-2) + 1 + ? = -1
C = +4
Oxidation numbers of all the atoms in HCO3
- ?
4.4
Balancing Redox Equations
19.1
1. Write the unbalanced equation for the reaction in ionic form.
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Fe2+ + Cr2O72- Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
Oxidation:
Cr2O72- Cr3+
+6 +3
Reduction:
Fe2+ Fe3++2 +3
3. Balance the atoms other than O and H in each half-reaction.
Cr2O72- 2Cr3+
Balancing Redox Equations
4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O72- 2Cr3+ + 7H2O
14H+ + Cr2O72- 2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe2+ Fe3+ + 1e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe2+ 6Fe3+ + 6e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
19.1
Balancing Redox Equations
7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. You should also cancel like species.
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6Fe2+ 6Fe3+ + 6e-Oxidation:
Reduction:
14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O
8. Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
19.1
9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. You should combine H+ and OH- to make H2O.
•To obtain a useful current, To obtain a useful current, we separate the oxidizing we separate the oxidizing and reducing agents so that and reducing agents so that electron transfer occurs thru electron transfer occurs thru an external wire. an external wire.
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
Zn
Zn2+ ions
Cu
Cu2+ ions
wire
saltbridge
electrons
CHEMICAL CHANGE --->CHEMICAL CHANGE --->ELECTRIC CURRENTELECTRIC CURRENT
CHEMICAL CHANGE --->CHEMICAL CHANGE --->ELECTRIC CURRENTELECTRIC CURRENT
This is accomplished in a This is accomplished in a GALVANIC, VOLTAIC,GALVANIC, VOLTAIC, or an or an ELECTROCHEMICAL ELECTROCHEMICAL cell.cell.
A group of such cells is called a A group of such cells is called a batterybattery..http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf
Galvanic Cells
19.2
spontaneousredox reaction
anodeoxidation
cathodereduction
- +
The potential difference between the two electrodes of a voltaic cell provides the driving force that pushes electrons through the external circuit.This potential difference is callled,
• cell voltage
• electromotive (causing electron motion) force (emf)
• cell potential
It’s denoted as Ecell and is measured in Volts (V).
For any cell rxn happening spontaneously, the cell potential will be positive.
Galvanic Cells
19.2
Line notation:
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode cathode
A common dry cell battery (for watches, calculators etc.)
NiO2 , H2O, OH- are reacting materials.
Anode: Cd, Cathode:Ni
Mercury battery, used in calculators
Standard Electrode Potentials
Standard reduction potential (E0red) is the voltage
associated with a reduction reaction at an electrode when all solutes are 1 M, 25°C and all gases are at 1 atm.
E0 = 0 V
Standard hydrogen electrode (SHE) is a reference cell to determine E0
red of the half cells.
2e- + 2H+ (1 M) H2 (1 atm)
Reduction Reaction
The superscript ° indicates standard state conditions.
Whenever we assign a potential to a half-reaction, we write the reaction as a reduction.
Standard Electrode Potentials
19.3
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
2e- + 2H+ (1 M) H2 (1 atm)
Zn (s) Zn2+ (1 M) + 2e-Anode (oxidation):
Cathode (reduction):
Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)
19.3
• E0 is for the reaction as written
• The more positive E0red, the
greater the driving force for reduction.
• The half-cell reactions are reversible
• The sign of E0 changes when the reaction is reversed
• Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0
19.3E0 = 0 + 0.76 V = 0.76 Vcell
Standard emf (E0 )cell
Zn2+ (1 M) + 2e- Zn E0 = -0.76 V
E0 = EH /H + EZn /Zn cell0 0
+ 2
Standard Electrode Potentials
E0 = Ecathode + Eanodecell0 0
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
If the reaction is backwards, be sure to flip the sign!
+2
So Eo Zn/Zn = + 0.76 V+2
Standard Electrode Potentials
19.3
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) 2H+ (1 M) + 2e-Anode (oxidation):
Cathode (reduction):
H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)
E0 = Ecathode + Eanodecell0 0
E0 = 0.34 Vcell
Ecell = ECu /Cu + E H /H+ 2+ 2
0 0 0
0.34 = ECu /Cu + - 00 2+
ECu /Cu = 0.34 V2+0
In a voltaic cell,the cathode reaction is always the one that has the more positive value for E0
red since the greater driving force of the cathode half rxn will force the anode rxn to occur “in reverse,” as an oxidation.
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V
Cd is the stronger oxidizer
Cd will oxidize Cr
2e- + Cd2+ (1 M) Cd (s)
Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):
Cathode (reduction):
2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)
x 2
x 3
E0 = Ecathode + Eanodecell0 0
E0 = -0.40 + (+0.74) cell
E0 = 0.34 V cell
19.3
19.4
Spontaneity of Redox Reactions
G0 = -nFEcell0
n = number of moles of electrons in reaction
F = 96,500J
V • mol = 96,500 C/mol
Factors affecting the cell potential:
• Specific rxns that occur at the cathode and anode
• Concentrations of reactants and products
• temperature
Effect of concentration on cell EMF
• As a voltaic cell discharges, concentrations of reactants and products change.
• The emf drops until E=0, at which the cell is “dead”, the reactants and products are at equilibrium.
How will you calculate the emf of a voltaic cell generated at non-
standard conditions?
• Use the Nernst equation (Walther Nernst, 1864-1941, a German chemist)
The Effect of Concentration on Cell Emf
G0 = -nFE 0
Nernst equation
At 298
19.5
-0.0257 V
nln QE0E = -
0.0592 Vn
log QE0E =
• n: the number of electrons transferred in the rxn
• F: Faraday’s constant (96,500 J/V-mol)
• Q: Rxn quotient, determined in the same w/ Kc except that the concentrations used in Q are those that exist in the rxn mixture at a given moment.
• E: non-standard voltaic cell potential
Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
2e- + Fe2+ 2Fe
Cd Cd2+ + 2e-Oxidation:
Reduction:n = 2
E0 = -0.44 + -(-0.40)
E0 = -0.04 V
E0 = EFe /Fe + ECd /Cd0 0
2+ 2+
-0.0257 V
nln QE0E =
-0.0257 V
2ln -0.04 VE =
0.0100.60
E = 0.013
E > 0 Spontaneous
19.5
• Calculate the emf at 298 K generated by the cell containing K2Cr2O7 and H2SO4 solutions in one compartment and a solution of KI in another compartment(a metallic conductor that will not react w/ either solution is suspended in each solution) when [Cr2O7
2-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0M, and [Cr3+] = 1.0x10-5M.
Cr2O72− + 14H+ + 6I− 2Cr3+ + 7H2O + 3I2
Cr2O72− + 14H+ + 6e− 2Cr3+ + 7H2O E° = 1.33V
I2 + 2e− 2I− E° = 0.54V
Answer: .89V
CONCENTRATION CELLS
• The cell operates until the
concentrations of reacting ions in two compartments become equal, at which point the cell has reached equilibrium and is “dead.”
• Anode: Cu Cu2+ (.01M) + 2e-
• Cathode: Cu2+ (1M) + 2e- Cu
• Overall: Cu2+ (1M) Cu2+ (.01M)
• At 298K;
• E= 0- .0592/2 log .01/1 = .0592V
How can you increase the cell potential of a concentration cell?
• By making the difference between concentrations of the ions higher.
2) Temperature
• Temp ↑,
• voltaic cell potential ↓
Electrolysis• Voltaic cells are based on spontaneous
rxns.
• It’s possible to cause a nonspontaneous rxn to occur by using a source of electrical energy.This process is called electrolysis.
• Electrolysis occur in an electrolytic cell.
19.8
Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.
electrolytic cell
An electrolytic cell consists of two electrodes in a molten salt or solution.
Source of direct electrical current: pushes e- into cathode & pulls e- from anode
(Where reduction occurs)
(Where oxidation occurs)
-+
-+
The Downs Cell for the Electrolysis of Molten Sodium Chloride
Copyright©2000 by Houghton Mifflin Company. All rights
reserved.
(a) A Standard Galvanic Cell (b) A Standard Electrolytic Cell
based on a spontaneous rxn:
Zn + CuZn + Cu2+2+ ZnZn2+2+ + Cu + CuZn + CuZn + Cu2+2+ Zn Zn2+2+ + Cu + Cu
Chemical energy turns into electrical energy
electrical energy turns into chemical energy
• By means of the movement of both anions and cations, the electrical current is carried in the electrolyte.
Electrolysis of molten salts
Overall electrolytic cell rxn: AlCl3(l) Al(s) + 3/2Cl2(g)
C(graphite)electrode
Al deposits
C(graphite)electrode
Al+3
Molten AlCl3
Cathode:Al+3(l)+3e-Al(s)
In the electrolysis of molten salts:
at the anode: at the cathode:
The elemental form of the anion is obtained
The elemental form of the cation is obtained
Because of high melting points of metals, electrolysis of molten salts requires very high temperatures.
Exercise
Which substances will be obtained at anode and cathode?
C(graphite)electrode
C(graphite)electrode
Molten MgCl2 & KF
answer
• Tendency to lose e: K>Mg
• At cathode: Mg solid will be obtained.
• Tendency to gain e: F>Cl
• At the anode: Cl2 gas will be obtained.
Electrolysis of Water
19.8
Electrolysis of aqueous solutions• We have to also consider whether the water is
oxidized (to form O2) or reduced (to form H2) rather than the ions of the salt.
Possible reactants: Na+,
Cl-, H2O
• At anode:
2Cl---->Cl2 + 2e- Eo=+1.36V
2H2O---> O2 + 4H+ + 4e- Eo =1.23V
- Cl- will be oxidized at the anode since the activation energy of Cl- is lower.So, it is kinetically favored.
•At cathode:2 H2O + 2 e− H2(g) + 2 OH− Eo= -0.83VNa+ + e− Na(s) Eo= -2.71V-Hydrogen gas will be produced at the cathode since the reduction of water is more favorable.
Tendency to be oxidized at anode:
Anions of halogens (except F-)> H2O > SO4-2, CO3
-2, NO3-
•Tendency to be reduced at cathode:•Cations of semi/Noble metals > H2O > cations of IA &IIA
metals
• The electrolysis of aqueous NaCl(brine) is an important for the production of chlorine and NaOH.
Exercise
• If we electrolyze the aqueous solutions of CuCl2, K2SO4 , KBr, and NaOH is separate containers,which substances will be obtained at anode and cathode for each solution?
answer
Anode Cathode
CuCl2 - Cl2(g) Cu(s)
K2SO4 - O2(g) H2(g)
KBr - Br2(l) H2(g)
NaOH O2(g) H2(g)
Quantitative aspects of electrolysis
Q (charge in Coulombs) = I (current in Amperes) x time (sec)
1 mole e- = 96,500 C = 1 Faraday (F)
19.8
1 amp = 1 Coulomb / sec
(Faradays)
Faraday’s Law: the amount of a substance reduced/oxidized at each electrode during electrolysis is directly proportional to the quantity of electricity (the number of electrons) passed into the cell.(Mass α Q)
Q = I x t
How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?
Anode:
Cathode: Ca2+ (l) + 2e- Ca (s)
2Cl- (l) Cl2 (g) + 2e-
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
mol Ca = 0.452Cs
x 1.5 hr x 3600shr 96,500 C
1 mol e-
x2 mol e-
1 mol Cax
= 0.0126 mol Ca
= 0.50 g Ca
19.8
Quantitative aspects of electrolysis
19.8
(Faradays)
The steps relating the quantity of electrical charge used in electrolysis to the amounts of substances oxidized/reduced:
exercise• How many seconds does it take to produce 11.2 L of
hydrogen gas measured at 760 mm-Hg and 27 °C by electrolysis of water using a current of 1.5 A?
answer
• 59,000s
exercise• Two electrolytic cells including molten MgCl2 and
salt of X+n ions separately are connected in series.When some electrical current is passed through these cells for a while, 7.2 g solid Mg is collected in the first electrolytic cell and 64.8 g of solid X is collected in the second one.What is the formula of the compound that is formed between X+n ions and sulfate ions? (Mg: 24g/mol, X: 108g/mol)
answer
• n=1
In the electrolytic cells connected in series:
• The amount of electrical current that passes through the cells is the same. Therefore, the number of moles of electrons exchanged in the electrolytic cells is the same.
Corrosion
19.7
Corrosion rxn are spontaneous redox rxns in which a metal is attacked by some substance in its environment and converted to an unwanted compound.
Fe2O3.nH2O
w/o the moisture, corrosion can’t happen since it acts as a kind of salt bridge.
Cathodic Protection of an Iron Storage Tank
19.7
Electroplating
cathode
e-
In electroplating, electrodes participate in electrolysis.
Electroplating uses electrolysis to deposit a thin layer of one metal on another metal in order to improve beauty or resistance to corrosion.
Charging a BatteryCharging a BatteryWhen you charge a battery, you are When you charge a battery, you are forcing the electrons backwards (from forcing the electrons backwards (from the + to the -). To do this, you will the + to the -). To do this, you will need a higher voltage backwards than need a higher voltage backwards than forwards. This is why the ammeter in forwards. This is why the ammeter in your car often goes slightly higher your car often goes slightly higher while your battery is charging, and then while your battery is charging, and then returns to normal.returns to normal.
In your car, the battery charger is In your car, the battery charger is called an alternator. If you have a called an alternator. If you have a dead battery, it could be the dead battery, it could be the battery needs to be replaced OR battery needs to be replaced OR the alternator is not charging the the alternator is not charging the battery properly.battery properly.
Batteries
19.6
Leclanché cell
Dry cell
Zn (s) Zn2+ (aq) + 2e-Anode:
Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)+
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
Batteries
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-Anode:
Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
Mercury Battery
19.6
Batteries
19.6
Anode:
Cathode:
Lead storagebattery
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l)4
Pb (s) + SO2- (aq) PbSO4 (s) + 2e-4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l)4
Batteries
19.6Solid State Lithium Battery
Batteries
19.6
A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning
Anode:
Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)
2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-
2H2 (g) + O2 (g) 2H2O (l)
Chemistry In Action: Dental Filling Discomfort
Hg2 /Ag2Hg3 0.85 V2+
Sn /Ag3Sn -0.05 V2+
Sn /Ag3Sn -0.05 V2+