#### Transcript Chapter 20: Electrochemistry

```A redox reaction is one in which the reactants’ oxidation
numbers change.
What are the oxidation numbers of
the metals in the reaction below?
+8 oxidized
+8
+6
+6
+4
+4
+2
+2
0
0
The iron’s charge has become
more negative, so it is reduced.
-2
-2
-4
-4
The tin’s charge has become
more positive, so it is oxidized.
-6
-6
+3
+2
+2
+4
2 FeCl3 + SnCl2  2 FeCl2 + SnCl4
Fe
reduced
Sn
Which reactant is oxidized and which is reduced in the following?
0
+2
-1
+4
-1
1) Cl2 + SnCl2  SnCl4
Synthesis
oxidized
reduced
0
+1
+2
0
2) Cu + 2AgNO3  Cu(NO3)2 + 2Ag
Single-replacement
oxidized
Note that a redox
reduced
reaction can also be
+5
another type of
-2
-1
0
3) 2KClO3  2KCl + 3O2
Decomposition reaction.
oxidized
reduced
KClO3
+1 + x + 3(–2) = 0
x = +5
Balancing Redox Reactions
+7
Rxn:
+3
MnO4 + C2O4 
–
2–
Mn2+
+4
+ CO2
(acidic conditions)
1. Determine the oxidation number of the redox active species.
2. Split the redox reaction into two half reactions (remember
OIL RIG) and add in the #e– being transferred.
Red:
2 8 H+ + 5e– + MnO4–  Mn2+ + 4 H2O
Ox:
+
5 C2O42–  2 CO2 + 2e–
16 H+ + 2 MnO4– + 5 C2O42–  2 Mn2+ + 10 CO2 + 8 H2O
3. Balance the oxygens with H2O and the hydrogens with H+.
4. Multiply both half-reactions by a factor that will make the
#e– transferred equal.
5. Add the two half-reactions to get the final balanced redox rxn.
Balancing Redox Reactions Under Basic Conditions
1. Balance reaction using acidic conditions.
16 H+ + 2 MnO4– + 5 C2O42–  2 Mn2+ + 10 CO2 + 8 H2O
+16 OH–
+16 OH–
8 16 H2O
2. Add OH– to both sides to neutralize the H+.
3. Cancel out excess water and rewrite reaction equation.
8 H2O + 2 MnO4– + 5 C2O42–  2 Mn2+ + 10 CO2 + 16 OH–
Voltaic (alt. Galvanic) Electrochemical Cells
ZnZn2+Cu2+Cu
AnodeCathode
0.92
Ecell = EA + EC
Ecell = 0.763V
+ 0.153 V
0.916 V
Salt bridge
EA = +0.763 V
Oxidation takes place
at the anode.
EC = +0.153 V
Reduction takes place
at the cathode.
Example: What are the standard cell potentials for the folllowing?
Al3+ + 3e–  Al
Eo= –1.66 V
Br2 + 2e–  2Br–
Ag+ + e–  Ag
Eo= +0.80 V
Eo= +1.06 V
1. AlAl3+Ag+Ag
Ox: Al  Al3+ + 3e–
Red:
Ag+
+ e–  Ag
+1.66 V
+0.80 V
+2.46 V
2. AgAg+Br2Br–
Ox: Ag  Ag+ + e–
–0.80 V
Red: Br2 + e–  2 Br– +1.06 V
+0.26 V
Note: Voltaic cells ALWAYS have positive cell emf’s (voltages).
Some terminology
emf– Electromotive force; force causing e– to move
SHE– Standard hydrogen electrode; Eo = 0 V by definition.
Standard conditions– Solution concentrations = 1 M and gas
pressures = 1 atm
Oxidizing agent– is reduced during redox rxn
Reducing agent– is oxidized during redox rxn
Faraday’s constant = 96485 C/mol e– transferred
R = 8.314 J/mol·K
1 A = 1 C/s
1 V = 1 J/C
(C  Coulomb = a quantity of charge)
Table of Standard Reduction Potentials
Good oxidizing agent = large,
positive reduction potential
 SHE
Good reducing agent = large,
negative reduction potential
A voltaic cell generates current as a result of a spontaneous
redox reaction. The equations relating E, DG and K are:
DG = -nFE
K e
 DG
RT
e
 nFE
RT
lnK = -DG/RT = nFE/RT
R = 8.314 J/mol·K, T = temperature in Kelvin, F = 96485 C/mol
n = # electrons transferred in reaction
20.58 What are DG and K for: 2VO2+ + 4H+ + 2Ag  VO2+ + 2H2O + 2Ag+
Ecell = EC + EA = 1.00 V + (–0.799 V) = 0.201 V
n = 2 e–
(Remember 1 V = 1 J/C)
DG = -nFE = -(2 e–)(96485 C/mol e–)(0.201 V) = -38787 J/mol
K = e–DG/RT = e(38787 J/mol)/(8.314 Jmol*298K) = e15.7 = 6294875
Operating under non-standard conditions:
0.0592  [products]
o

log
Nernst Eq. = Ecell  Ecell 
 [reactants] 
n
20.50 What is the cell potential for the following if [Ce4+]=2.0M,
[Ce3+] = 0.010 M and [Cr3+] = 0.010 M?
3Ce4+ + Cr(s)  3Ce3+ + Cr3+
RED
n = 3e–
OX
Eo = EC + EA = +1.61 V + +0.74 V = 2.35 V
E = 2.35 V – (0.0592/3)log
[0.010 M]3[0.010 M]
[2.0 M]3
= 2.53 V
Because [reactants]>>[products] AND the reaction has a large
positive Eo, the emf has increased to produce more product.
Electrolysis: Decomposition of a compound by passing
electricity through it.
20.80 Metallic magnesium can be made by the electrolysis of
molten MgCl2. What mass of Mg is formed by passing a
current of 5.25 A through molten MgCl2 for 2.50 days?
24 hrs 60 min 60 s
• Calculate total time:2.50days x
x
x
 216000s
day
hr
min
• Calculate total charge passed = 5.25 C/s * 216000 s = 1134000 C
• Divide by 2 because it takes 2e–/equiv Mg2+= 567000 C
• Calculate mol Mg = 567000 C /(96485 C/mol) = 5.877 mol Mg(s)
• Calculate grams Mg = 5.877 mol Mg(s) * 24.305 g/mol = 142.8 g
Ans: 143 g of Mg will be formed
```