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Redox Reactions and Electrochemistry. Redox reactions Galvanic cells Standard reduction potential Cell potential Free energy Effect of concentration Battery Corrosion Electrolysis. 2Fe ( s ) + O 2 ( g ) 2FeO ( s ). 2Fe 2Fe 2+ + 4e -. O 2 + 4e - 2O 2-. - PowerPoint PPT Presentation
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Redox Reactions and Electrochemistry
• Redox reactions
• Galvanic cells
• Standard reduction potential
• Cell potential
– Free energy
– Effect of concentration
• Battery
• Corrosion
• Electrolysis
I. Oxidization–Reduction (Redox) Reaction
2Fe 2Fe2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-):Fe is oxidized and Fe is the reducing agent
Reduction half-reaction (gain e-):
O2 is reduced and O2 is the oxidizing agent
2Fe (s) + O2 (g) 2FeO (s)
Electrochemical processes are redox reactions in which:• energy released by a spontaneous reaction is converted to electricity• electrical energy is used to cause a nonspontaneous reaction
0 0 2+ 2-
A. Oxidation number/state
• Free elements have an oxidation number (ON) =0• Monatomic ions, the ON is equal to the charge on the ion.
• The ON of oxygen is usually –2. In H2O2 and O22-:–1, in O2
- : -1/2
• The ON of hydrogen is +1 except when it is bonded to metals in
binary compounds. In these cases, its oxidation number is –1.• Fluorine is always –1, Group IA metals are +1, IIA metals are +2
in compounds.• The sum of the ON of all the atoms in a molecule or ion is equal to
the charge on the molecule or ion.
Ex. Oxidation numbers of all the atoms in HSO3–
O =-2 H= +1 S=? 1 +3x (-2)+S= -1
S=4
B. Balance redox reaction
1. By inspection
2. Ion-electron method (half-reaction method)
By balancing the number of electrons lost and gainedduring the reaction
Cu+ H+ Cu2+ + H22
Sn2+ (aq) + Fe3+ (aq) Sn4+(aq) + Fe2+(aq)
Oxidation half reaction
Reduction half reaction
Sn2+ (aq) Sn4+(aq) + 2 e–
Fe3+ (aq) + e– Fe2+(aq)
2 2
( ) x 2
2 Fe3+ (aq) + 2e– 2Fe2+(aq)
Ion-electron method
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+
1. Write the unbalanced equation for the reaction ion ionic form.
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Ion-electron method
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
Ion-electron method
7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.
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
9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.
Summary of balancing redox reaction in acidic solutions
1. Divide reaction into two incomplete half-reactions2. Balance each half-reaction by doing the following:
a. Balance all elements, except O and H
b. Balance O by adding H2O
c. Balance H by adding H+d. Balance charge by adding e– as needed
3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to get a common multiple
4. The overall reaction is the sum of the half-reactions5. The oxidation half-reaction is the one that produces electrons as
products, and the reduction half-reaction is the one that uses electrons as reactants
Summary of balancing redox reaction in basic solutions
Steps 1-3 are the same as the reaction in acidic solution4. Since we are under basic conditions, we neutralize any H+ ions
in solution by adding an equal number of OH– to both sides5. The overall reaction is the sum of the half-reactions
Ex. Balance I– + OCl– I2 + Cl– in basic solution?
Oxidation I– I2
Reduction
OCl– Cl–
2 H2O+2 e– + OCl– Cl– + H2O + 2 OH–
2 e– + H2O + OCl– Cl– + 2 OH–
2 I– I2 + 2 e–
2 I– + H2O + OCl– I2 + Cl– + 2 OH–
Add up
+ H2O2 H+ +2 OH– + + 2 OH–2e– +
II. Galvanic Cells
• Consider reaction Zn(s) + Cu(NO3)2 --> Cu(s) + Zn(NO3)2
Spontaneous (activity series, chapter 4)half reaction: Zn(s) --> Zn2+ + 2e– Cu2+ +2 e– --> Cu(s)
• Galvanic cell: a device uses the spontaneous redox reaction to produce an electric current
OxidationZn --> Zn2+ + 2e–
Anode
ReductionCu2+ +2 e– --> Cu
Cathode
e– flowe–
Salt bridge
Flow of cation
Flow of anion
Wire
Electrode: anode (oxidation rxn) and cathode (reduction rxn)Salt bridge: contains conc. soln of strong electrolyte (KCl)
II. Galvanic cell
Cell diagram : a conventional notation to represent galvanic cell
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M and [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)anode cathode
1. Anode is on the left, cathode on the right2. | : represent boundary between phases3. || : represent salt bridge
Ex. Draw cell diagram for the following reaction:
Al (s) + Ag+ (aq) Al3+ (s) + Ag (s)
Al (s) | Al3+ || Ag+ | Ag (s)
III. Standard reduction potential
A. Cell voltage, cell potential, electromotive force (emf)
1. The potential energy of the e– is higher at the anode than at the cathode
2. The difference in electrical potential between the anode and cathode is
called as emf
3. Unit: V (volt) = 1 J/C
C: coulomb, unit of charge
4. Standard cell potential, Eocell
Cell potential under standard condition
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)
B. Standard reduction potentialStandard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.
E0 = 0 V
Standard hydrogen electrode (SHE) : half-cell contain 1M H+ and
1 atm H2, used as a reference to determine Eo for other species
2e- + 2H+ (1 M) H2 (1 atm)
Reduction Reaction
B. Standard reduction potential
E0 = 0.76 Vcell
Standard emf (E0cell )
0.76 V = 0 - EZn /Zn 0
2+
EZn /Zn = -0.76 V02+
Zn2+ (1 M) + 2e- Zn E0 = -0.76 V
E0 = Ecathode - Eanodecell0 0
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
E0 = EH /H - EZn /Zn cell0 0
+ 2+2
B. Standard reduction potential
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 – EH /H 2+ +2
0 0 0
0.34 = ECu /Cu - 00 2+
ECu /Cu = 0.34 V2+0
• E0 is for the reaction as written
• The more positive E0 the greater the tendency for the substance to be reduced
• 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
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
Ex.
Calculate Eocell
for the following reactions:
(a) 2 Al (s) + 6 H+ -> 2 Al3+ + 3 H2 (g)
(b) H2O2 + 2 H+ + Cu (s) -> Cu2+ + 2 H2O
6 H+(aq) + 6e- 3 H2(g)
2 Al(s) 2 Al3+ + 6e-Anode (oxidation):
Cathode (reduction):
Ex.
(a) 2 Al (s) + 6 H+ -> 2 Al3+ + 3 H2 (g)
E0 = EH /H - EAl /Al cell0 0
+ 3+2 = 1.66 V
E0 = Ecathode - Eanodecell0 0
H2O2+2 H+ +2e- 2 H2O
Cu(s) Cu2+ + 2e-Anode (oxidation):
Cathode (reduction):
(b) H2O2 + 2 H+ + Cu (s) -> Cu2+ + 2 H2O
= 1.77 V- 0.34 V = 1.43 VE0 = E - ECu /Cu cell0
2+H2O2/H2 O
IV. Cell Potential
• Free energy : spontaneity of redox reactions
G = -nFEcell
G0 = -nFEcell0
n = number of moles of electrons in reaction
F = 96,500J
V • mol = 96,500 C/mol
G0 = -RT ln K = -nFEcell0
Ecell0 =
RTnF
ln K(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)ln K=
=0.0257 V
nln KEcell
0
=0.0592 V
nlog KEcell
0
A. Free energy and spontaneity of redox reaction
G = -nFEcell
G0 = -nFEcell0
=0.0257 V
nln KEcell
0
=0.0592 V
nlog KEcell
0
Ex.
2e- + Fe2+ Fe
2Ag 2Ag+ + 2e-Oxidation:
Reduction:
=0.0257 V
nln KEcell
0
E0 = -0.44 – (0.80) E0 = -1.24 V
0.0257 Vx nE0 cellexpK =
n = 2
0.0257 Vx 2-1.24 V
= exp
K = 1.23 x 10-42
E0 = EFe /Fe – EAg /Ag0 0
2+ +
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)
B. Concentration effect on cell potential
G = G0 + RT ln Q G = -nFE G0 = -nFE 0
-nFE = -nFE0 + RT ln Q
E = E0 - ln QRTnF
Nernst equation
At 298 K
-0.0257 V
nln QE0E = -
0.0592 Vn
log QE0E =
Ex.
2e- + Fe2+ Fe
Cd Cd2+ + 2e-Oxidation:
Reduction:n = 2
= -0.44 – (-0.40) V = -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
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)
[Cd2+][Fe2+]
Q =
Ex.
2e- +2 Ag+ 2 Ag
Ni Ni2+ + 2e-Oxidation:
Reduction:n = 2
= 0.80 – (-0.25) V = 1.05V E0 = EAg /Ag – E Ni /Ni0 0
+ 2+
A redox reaction for the oxidation of nickel by silver ion is set up with an initial concentration of 5.0 M silver ion and 0.050 M nickel ion. What is the cell EMF at 298K?
-0.0257 V
nln QE0E =
-0.0257 V
2ln 1.05 VE =
0.0505.02 E = 1.13
[Ni2+][Ag+]2
Q =
Ni + 2 Ag+ Ni2+ + 2 Ag
V. Battery
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)
About 1.5 V
V. Battery
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
1.35 V
V. Battery
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
Total 12V
VII. Electrolysis
Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.
B. Quantitative aspect of electrolysis
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
Ex. It will take the greates amount electricity to produce 2 mol of
the ____ metal by electrolysis.
(1) potassium (2) calcium (3) aluminum (4) silver
Ex.
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
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?