Transcript Predicting Spontaneous Reactions

Predicting Spontaneous
Reactions
Standard Electrode Potentials
Standard 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.
Reduction Reaction
2e- + 2H+ (1 M)
H2 (1 atm)
E0 = 0 V
Standard hydrogen electrode (SHE)
19.3
•
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
19.3
Reduction Potentials
• The half-reaction with the more positive
reduction potential ( higher on the chart) is
reduced and the other half-reaction is
forced to oxidize.
Example: What spontaneous reaction
occurs if Cl2 and Br2 are added to a
solution of Cl- and Br-?
Standard Reduction Potentials (in Volts), 25oC
Reaction
Eo
F2 + 2e- ---> 2F-
+2.87
Co3+ + e- ---> Co2+
+1.80
PbO2 + 4H+ + SO42- + 2e- --->
PbSO4(s) + 2H2O
+1.69
MnO4- + 8H+ + 5e- ---> Mn2+ +
4H2O
+1.49
PbO2 + 4H+ + 2e- ---> Pb2+ + 2H2O +1.46
Cl2 + 2e- ---> 2Cl-
+1.36
Cr2O72- + 14H+ + 6e- ---> 2Cr3+ +
7H2O
+1.33
O2 + 4H+ + 4e- ---> 2H2O
+1.23
Br2 + 2e- ---> 2Br-
+1.07
Example Continued:
Cl2 + 2 e-  2 Cl+ 1.36 V
Br2 + 2 e-  2 Br+1.09 V
Cl2 is more positive so it is reduced and Br2
is oxidized.
Write Br2 to show oxidation reaction:
2 Br-  Br2 + 2e-1.09 V
Cl2 + 2 e-  2 Cl+ 1.36 V
Cl2 + 2 Br-  Br2 + 2 Cl.27 = EoCell
Spontaneous Reactions
• E0Cell = reduction + oxidation potential
• A positive Eocell means the reaction is
spontaneous in that direction. A negative
Eocell means the reverse reaction is
spontaneous.
Standard Electrode Potentials
0 = 0.34 V
Ecell
0
0 = E0
Ecell
cathode + Eanode
0
0 = E 0 2+
Ecell
Cu /Cu + E H
2
/H+
0 2+
0.34 = ECu
/Cu + - 0
0 2+
ECu
/Cu = 0.34 V
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm)
Cathode (reduction): 2e- + Cu2+ (1 M)
H2 (1 atm) + Cu2+ (1 M)
2H+ (1 M) + 2eCu (s)
Cu (s) + 2H+ (1 M)
19.3
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 Cd is the stronger oxidizer
Cr3+ (aq) + 3e-
Cr (s)
Anode (oxidation):
E0 = -0.74 V
Cr3+ (1 M) + 3e- x 2
Cr (s)
Cathode (reduction): 2e- + Cd2+ (1 M)
2Cr (s) + 3Cd2+ (1 M)
Cd will oxidize Cr
Cd (s)
x3
3Cd (s) + 2Cr3+ (1 M)
0
0 = E0
Ecell
cathode + Eanode
0 = -0.40 + (+0.74)
Ecell
0 = 0.34 V
Ecell
19.3
Free Energy Change & Eocell
• When a reaction takes place in a voltaic
cell it performs work
W = nFEcell
n = moles of e- transferred
F = Faradays constant (96,485 C/mole e-)
Ecell = cell voltage
Work and Ecell
• The maximum amount of work that a
system cam do is equal to the negative of
the change in Gibbs Free energy.
-DG =welectric = nFEcell
DG = -nFEcell
DGo = -nFEocell (standard conditions)
If Ecell = 0, the system is in equilibrium
Concentration and Ecell
• The cell potential gradually drops as the
reactants are consumed.
DG = DGo + RT ln Q
R = gas constant
T = temp (kelvin)
Q = reaction quotient (original molarities)
Concentration and Ecell
• Substituting for DG and DGo
-nFEcell = -nFEocell + RT ln Q
nF
The Nernst Equation
Ecell = Eocell – RT ln Q
nF
Ecell = Eocell - .0592 V log Q (base 10 log)
n
Spontaneity of Redox Reactions
19.4
What is the equilibrium constant for the following reaction
at 250C? Fe2+ (aq) + 2Ag (s)
Fe (s) + 2Ag+ (aq)
0
Ecell
=
0.0257 V
ln K
n
Oxidation:
Reduction:
2e-
+
0
0
E0 = EFe
2+/Fe + EAg
2Ag
2Ag+ + 2e-
Fe2+
Fe
n=2
/Ag +
E0 = -0.44 + -0.80
E0 = -1.24 V
0
Ecell
xn
-1.24 V x 2
= exp
K = exp
0.0257 V
0.0257 V
K = 1.23 x 10-42
19.4
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)
Oxidation:
Reduction:
Cd
2e-
+
Cd2+ + 2e-
Fe2+
2Fe
n=2
0
0
E0 = EFe
2+/Fe + ECd /Cd 2+
E0 = -0.44 + -(-0.40)
E0 = -0.04 V
0.0257 V
ln Q
n
0.010
0.0257 V
ln
E = -0.04 V 2
0.60
E = 0.013
E = E0 -
E>0
Spontaneous
19.5
Charging a Battery
When you charge a battery, you are
forcing the electrons backwards (from
the + to the -). To do this, you will
need a higher voltage backwards than
forwards. This is why the ammeter in
your car often goes slightly higher
while your battery is charging, and
then returns to normal.
In your car, the battery charger is
called an alternator. If you have a
dead battery, it could be the
battery needs to be replaced OR
the alternator is not charging the
battery properly.
Batteries
Dry cell
Leclanché cell
Zn (s)
Anode:
Cathode:
2NH+4 (aq) + 2MnO2 (s) + 2e-
Zn (s) + 2NH4 (aq) + 2MnO2 (s)
Zn2+ (aq) + 2eMn2O3 (s) + 2NH3 (aq) + H2O (l)
Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)
19.6
Batteries
Mercury Battery
Anode:
Cathode:
Zn(Hg) + 2OH- (aq)
HgO (s) + H2O (l) + 2eZn(Hg) + HgO (s)
ZnO (s) + H2O (l) + 2eHg (l) + 2OH- (aq)
ZnO (s) + Hg (l)
19.6
Batteries
Lead storage
battery
Anode:
Cathode:
Pb (s) + SO2-4 (aq)
PbSO4 (s) + 2e-
PbO2 (s) + 4H+ (aq) + SO24 (aq) + 2e
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2(aq)
4
PbSO4 (s) + 2H2O (l)
2PbSO4 (s) + 2H2O (l)
19.6
Batteries
Solid State Lithium Battery
19.6
Batteries
A fuel cell is an
electrochemical cell
that requires a
continuous supply of
reactants to keep
functioning
Anode:
Cathode:
2H2 (g) + 4OH- (aq)
O2 (g) + 2H2O (l) + 4e2H2 (g) + O2 (g)
4H2O (l) + 4e4OH- (aq)
2H2O (l)
19.6
Corrosion
19.7
Cathodic Protection of an Iron Storage Tank
19.7
Electrolysis is the process in which electrical energy is used
to cause a nonspontaneous chemical reaction to occur.
19.8
Electrolysis of Water
19.8
Chemistry In Action: Dental Filling Discomfort
2+
Hg2 /Ag2Hg3 0.85 V
2+
Sn /Ag3Sn -0.05 V
2+
Sn /Ag3Sn -0.05 V
Electrolysis and Mass Changes
charge (Coulombs) = current (Amperes) x time (sec)
1 mole e- = 96,500 C = 1 Faraday
1 amp = 1 Coulomb / sec
19.8
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:
2Cl- (l)
Ca2+ (l) + 2eCa2+ (l) + 2Cl- (l)
Cl2 (g) + 2eCa (s)
Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
C
s 1 mol e- 1 mol Ca
mol Ca = 0.452
x 1.5 hr x 3600 x
x
s
hr 96,500 C 2 mol e= 0.0126 mol Ca
= 0.50 g Ca
19.8