Transcript Electrochemistry

Electrochemistry
Oxidation-Reduction (Redox) Reactions
TRANSFER OF ELECTRONS
Oxidation: Electrons are lost (product side
of equation), charge increases. The
substance that is oxidized is the reducing
agent.
Reduction: Electrons are gained (reactant
side of equation), charge decreases. The
substance that is reduced is the oxidizing
agent.
Oxidation-Reduction (Redox) Reactions
4 Fe(s) + 3 O2(g)
2 Fe2O3(s) + 3 C(s)
2 Fe2O3(s)
4 Fe(s) + 3 CO2(g)
Rusting of iron:
an oxidation of Fe
Manufacture of iron: a
reduction of Fe3+
Oxidation-Reduction (Redox) Reactions
Oxidation: The loss of one or more electrons by a
substance, whether element, compound, or ion.
Reduction: The gain of one or more electrons by a
substance, whether element, compound, or ion.
Oxidation-Reduction (Redox) Reactions
Oxidation Number (State): A value that indicates whether
an atom is neutral, electron-rich, or electron-poor.
Rules for Assigning Oxidation Numbers
1. An atom in its elemental state has an oxidation
number of 0.
Na
H2
Br2
Oxidation number 0
S
Ne
Oxidation-Reduction (Redox) Reactions
2. An atom in a monatomic ion has an oxidation number
identical to its charge.
Na+
Ca2+
Al3+
Cl–
O2–
+1
+2
+3
–1
–2
Oxidation-Reduction (Redox) Reactions
3. An atom in a polyatomic ion or in a molecular compound
usually has the same oxidation number it would have if it
were a monatomic ion.
a) Hydrogen can be either +1 or –1.
1–
H
+1
O
H
Ca
–1
–2
H
–1
+2
b) Oxygen usually has an oxidation number of –2.
H
+1
O
–2
H
+1
H
+1
O
–1
O
–1
H
+1
Oxidation-Reduction (Redox) Reactions
3.
c) Halogens usually have an oxidation number of -1.
H
Cl
+1
–1
Cl
+1
O
–2
Cl
+1
Oxidation-Reduction (Redox) Reactions
4. The sum of the oxidation numbers is 0 for a neutral
compound and is equal to the net charge for a
polyatomic ion.
2(+1) + x + 3(–2) = 0 (net charge)
H2SO3
+1
–2
x
Cr2O72–
x = +4
2(x) + 7(-2) = –2 (net charge)
x = +6
x
–2
Identifying Redox Reactions
Reducing Agent
•
•
•
•
Causes reduction
Loses one or more electrons
Undergoes oxidation
Oxidation number of atom increases
Oxidizing Agent
•
•
•
•
Causes oxidation
Gains one or more electrons
Undergoes reduction
Oxidation number of atom decreases
Identifying Redox Reactions
Reducing Agent
oxidation
0
4 Fe(s)
+3
+
3 O2(g)
2 Fe2 O3 (s)
–2
0
Oxidizing Agent
reduction
Identifying Redox Reactions
Reducing Agent
oxidation
0
2 Fe2O3
(s) + 3 C (s)
4 Fe(s)
0
+3
Oxidizing Agent
+4
reduction
+ 3 C O2 (g)
In the redox reaction indicated above,
which is the oxidizing agent?
a) MnO4–
b) Fe2+
c) H+
d) H2O
e) None of the above
In the redox reaction indicated above,
which is the oxidizing agent?
a) MnO4–
b) Fe2+
c) H+
d) H2O
e) None of the above
Worked Example 7.9 Identifying Oxidizing and Reducing Agents
Assign oxidation numbers to all atoms, tell in each case which substance is undergoing oxidation and which
reduction, and identify the oxidizing and reducing agents.
a.
b.
Strategy and Solution
a. The elements Ca and H2 have oxidation numbers of 0; Ca2+ is +2 and H+ is +1. Ca is oxidized, because its
oxidation number increases from 0 to +2, and H+ is reduced, because its oxidation number decreases from
+1 to 0. The reducing agent is the substance that gives away electrons, thereby going to a higher oxidation
number, and the oxidizing agent is the substance that accepts electrons, thereby going to a lower oxidation
number. In the present case, calcium is the reducing agent and H + is the oxidizing agent.
Worked Example 7.9 Identifying Oxidizing and Reducing Agents
Continued
b. Atoms of the neutral element Cl2 have an oxidation number of 0; the monatomic ions have oxidation
numbers equal to their charge:
Fe2+ is oxidized because its oxidation number increases from +2 to +3, and Cl 2 is reduced because its
oxidation number decreases from 0 to –1. Fe2+ is the reducing agent, and Cl2 is the oxidizing agent.
The Activity Series of the Elements
The elements that
are higher up in
the table are more
likely to be
oxidized.
Thus, any element
higher in the
activity series will
reduce the ion of
any element lower
in the activity
series.
The Activity Series of the Elements
Cu(s) + 2 Ag+(g)
Cu2+(aq) + 2 Ag(s)
Which one of these reactions will occur?
2 Ag(s) + Cu2+(g)
2 Ag+(aq) + Cu(s)
Silver cation will oxidize copper, becoming solid silver, and liberating copper (II)
ions. Looking at the diagram to the left, which one of the four reactions below
will occur?
a)
b)
c)
d)
e)
Silver cation will oxidize copper, becoming solid silver, and liberating copper (II)
ions. Looking at the diagram to the left, which one of the four reactions below
will occur?
a)
b)
c)
d)
e)
Worked Example 7.10 Predicting the Products of a Redox Reaction
Predict whether the following redox reactions will occur:
a.
b.
Strategy
Look at Table 7.5 to find the relative reactivities of the
elements.
Solution
a. Zinc is above mercury in the activity series, so this
reaction will occur.
b. Copper is below hydrogen in the activity series, so
this reaction will not occur.
Balancing Redox Reactions by the
Half-Reaction Method
Balancing Redox Reactions by the
Half-Reaction Method
Balance the following net ionic equation in acidic
solution:
I–(aq) + Cr2O72–(aq)
Cr3+(aq) + IO3–(aq)
Balancing Redox Reactions by the
Half-Reaction Method
• Write the two unbalanced half-reactions.
Cr2O72–(aq)
Cr3+(aq)
I–(aq)
IO3–(aq)
Balancing Redox Reactions by the
Half-Reaction Method
• Balance both half-reactions for all atoms except O
and H.
Cr2O72–(aq)
I–(aq)
2 Cr3+(aq)
IO3–(aq)
Balancing Redox Reactions by the
Half-Reaction Method
• Balance each half-reaction for O by adding H2O,
and then balance for H by adding H+.
14 H+(aq) + Cr2O72–(aq)
3 H2O(l) + I–(aq)
2 Cr3+(aq) + 7 H2O(l)
IO3–(aq) + 6 H+(aq)
Balancing Redox Reactions by the
Half-Reaction Method
• Balance each half-reaction for charge by adding
electrons to the side with greater positive charge.
6 e- + 14 H+(aq) + Cr2O72–(aq)
3 H2O(l) + I–(aq)
2 Cr3+(aq) + 7 H2O(l)
IO3–(aq) + 6 H+(aq) + 6 e–
Balancing Redox Reactions by the
Half-Reaction Method
• Multiply each half-reaction by a factor to make the
electron count the same in both half-reactions.
reduction:
6 e– +14 H+(aq) + Cr2O72–(aq)
oxidation:
3 H2O(l) + I–(aq)
2 Cr3+(aq) + 7 H2O(l)
IO3–(aq) + 6 H+(aq) + 6 e–
Balancing Redox Reactions by the
Half-Reaction Method
• Add the two balanced half-reactions together and
cancel species that appear on both sides of the
equation.
reduction:
6 e– +14 H+(aq) + Cr2O72–(aq)
oxidation:
3 H2O(l) + I–(aq)
2 Cr3+(aq) + 7 H2O(l)
IO3–(aq) + 6 H+(aq) + 6 e–
8 H+(aq) + I–(aq) + Cr2O72–(aq)
IO3–(aq) + 2 Cr3+(aq) + 4 H2O(l)
Example #1
1. MnO4-(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)
2. H+(aq) +Cr2O72-(aq) + C2H5OH(l) 
Cr3+(aq) + CO2(g) + H2O(l)
Balancing Redox Reactions by the
Half-Reaction Method
Balance the following net ionic equation in basic
solution:
MnO4–(aq) + Br–(aq)
MnO2(s) + BrO3–(aq)
Balancing Redox Reactions by the
Half-Reaction Method
• Write the two unbalanced half-reactions.
Br–(aq)
MnO4–(aq)
BrO3–(aq)
MnO2(s)
Balancing Redox Reactions by the
Half-Reaction Method
• Balance both half-reactions for all atoms except
O and H.
Br–(aq)
MnO4–(aq)
BrO3–(aq)
MnO2(s)
Balancing Redox Reactions by the
Half-Reaction Method
• Balance each half-reaction for O by adding H2O, and then
balance for H by adding H+.
3 H2O(l) + Br–(aq)
4 H+(aq) + MnO4–(aq)
BrO3–(aq) + 6 H+(aq)
MnO2(s) + 2 H2O(l)
Balancing Redox Reactions by the
Half-Reaction Method
• Balance each half-reaction for charge by adding electrons
to the side with greater positive charge.
3 H2O(l) + Br–(aq)
3 e– + 4 H+(aq) + MnO4–(aq)
BrO3–(aq) + 6 H+(aq) + 6 e–
MnO2(s) + 2 H2O(l)
Balancing Redox Reactions by the
Half-Reaction Method
• Multiply each half-reaction by a factor to make the electron
count the same in both half-reactions.
3 H2O(l) + Br–(aq)
2
3 e– + 4 H+(aq) + MnO4–(aq)
BrO3–(aq) + 6 H+(aq) + 6 e–
MnO2(s) + 2 H2O(l)
Balancing Redox Reactions by the
Half-Reaction Method
• Add the two balanced half-reactions together and cancel
species that appear on both sides of the equation.
3 H2O(l) + Br–(aq)
6 e– + 8 H+(aq) + 2 MnO4–(aq)
BrO3–(aq) + 6 H+(aq) + 6 e–
2 MnO2(s) + 4 H2O(l)
2 H+(aq) + 2 MnO4–(aq) + Br–(aq)
2 MnO2(s) + H2O(l) + BrO3–(aq)
Balancing Redox Reactions by the
Half-Reaction Method
• Since the reaction occurs in a basic solution,
“neutralize” the excess H+ by adding OH– and
cancel any water (if possible).
2 H2O
2 OH–(aq) + 2 H+(aq) + 2 MnO4–(aq) + Br–(aq)
2 MnO2(s) + H2O(l) + BrO3–(aq) + 2 OH–(aq)
H2O(l) + 2 MnO4–(aq) + Br–(aq)
2 MnO2(s) + BrO3–(aq) + 2 OH–(aq)
Worked Example Balancing an Equation for a Reaction in Base
Aqueous sodium hypochlorite (NaOCl; household bleach) is a strong oxidizing agent that reacts with chromite ion
[Cr(OH)4–] in basic solution to yield chromate ion (CrO42–) and chloride ion. The net ionic equation is
Balance the equation using the half-reaction method.
Strategy
Follow the steps outlined in Figure 7.4.
Solution
Steps 1 and 2. The unbalanced net ionic equation shows
that chromium is oxidized (from +3 to +6) and chlorine is
reduced (from +1 to –1). Thus, we can write the following
half-reactions:
Step 3. The half-reactions are already balanced for atoms
other than O and H.
Figure 7.4
Using the half-reaction method to balance redox
equations for reactions in acidic solution.
Worked Example Balancing an Equation for a Reaction in Base
Continued
Step 4. Balance both half-reactions for O by adding H2O to the sides with less O, and then balance both for H by
adding H+ to the sides with less H:
Step 5. Balance both half-reactions for charge by adding electrons to the sides with the greater positive charge:
Next, multiply the half-reactions by factors that make the electron count in each the same. The oxidation halfreaction must be multiplied by 2, and the reduction half-reaction must be multiplied by 3 to give 6 e– in both:
Step 6. Add the balanced half-reactions:
Worked Example Balancing an Equation for a Reaction in Base
Continued
Now, cancel the species that appear on both sides of the equation:
Finally, since we know that the reaction takes place in basic solution, we must add 2 OH – ions to both sides of the
equation to neutralize the 2 H+ ions on the right giving 2 additional H2O. The final net ionic equation, balanced for
both atoms and charge, is
Example #2
1. CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2 (s)
2. S2-(aq) + MnO4-(aq)  S(s) + MnO2 (s)
Galvanic Cells
Electrochemistry: The area of chemistry concerned with
the interconversion of chemical and electrical energy.
In a Galvanic (Voltaic) Cell: A spontaneous chemical
reaction generates an electric current.
In an Electrolytic Cell: An electric current drives a
nonspontaneous reaction.
How to draw a Galvanic Cell
The oxidation reaction occurs at the anode. The reduction
reaction occurs at the cathode.
You will be give the unbalanced net ionic reaction or a list
of the substances present (line notation). From the
information given you need to decide what half reactions
occur in each beaker. Draw each beaker with its
substances present. The electrons leave the anode and
travel to the cathode. Remember to draw a salt bridge.
If the substance reduced produces a gas you will have a
platinum electrode and an inverted test tube to catch the
gas. Use platinum as electrode for redox involving ions;
otherwise, use the metal given.
Example #3
Draw the galvanic cell for the reaction:
1. Cu(s) + Ag+(aq)  Ag(s) + Cu2+(aq)
2. Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
3. Pt(s) |Fe2+(aq), Fe3+(aq) |Ag+(aq), Ag(s)|
Cell Potentials and Free-Energy
Changes for Cell Reactions
Electromotive Force (emf): The force (or electrical
potential) that pushes the negatively charged electrons
away from the anode (– electrode) and pulls them toward
the cathode (+ electrode).
It is also called the cell potential (eo) or the cell voltage.
Cell Potentials and Free-Energy
Changes for Cell Reactions
1J=1C×1V
joule
SI unit of energy
volt
SI unit of electric potential
coulomb
Electric charge
1 coulomb is the amount of charge transferred
when a current of 1 ampere flows for 1 second.
Cell Potentials and Free-Energy
Changes for Cell Reactions
faraday or Faraday’s constant
the electric charge on 1 mol of electrons
96,500 C/mol e–
∆G = –nFe
free-energy change
or ∆G° = –nFeo
cell potential
number of moles of electrons
transferred in the reaction
Calculating electromotive force,
emf or ecell
From table of standard reduction potentials, write
the half reactions according to the way the
reaction is written.
Flip the sign of eo for the species being oxidized.
Add the two values.
The reaction with 2H+(aq) and H2(g) is known at the standard
hydrogen electrode (SHE). The voltages for all half
reactions are based on this reaction.
e
o for
SHE is 0.
Example #4
Determine the electromotive force, for the
following reactions.
1. Ag(s) + Cu2+(aq) Cu(s) + Ag+(aq)
2. Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
ecell and spontaneity
If ecell is positive the reaction is spontaneous
as it is written (runs forward).
If ecell is negative it is not spontaneous as it
is written (runs backward).
If ecell is zero, it is at equilibrium.
Look at the previous example, are these
reactions spontaneous?
To Decide which way a galvanic
cell will run from eo
If you are given the substances in each cell
without any indication which is the
oxidation and which is the reduction you
will look up the reactions and the numbers
in the eo table. Decide which reaction you
can reverse and still have a positive
number after you add the two values.
Example #5
Draw the galvanic cell based on the
following half-reactions under standard
conditions.
Ag+ +e-  Ag
Fe3+ + e-  Fe2+
Example #6
Determine the spontaneous reaction that will
occur from the following half reactions.
1. Sn2+ + 2e-  Sn
I2 + 2e-  2I2. Au3+ + 3e-  Au
Cu2+ + 2e-  Cu
Example #7
Using eo values, predict whether 1M HNO3
will dissolve gold to form a 1M Au3+
solution.
ecell and DG
DGo = -nF eo
n: number of moles of electrons
F: Faradays constant, 96500C/mole
Joule = V C (Volt-Coulomb)
Named for Michael Faraday.
Example #8
Using eo values calculate DGo for the
reaction:
Fe(s) + Cu2+(aq) Cu(s) + Fe2+(aq)
Example #9
Write the balanced oxidation reduction
reaction for the following:
Zn + 2H+  H2 + Zn2+
A. Calculate DGo for the reaction using the
DGof (kJ/mol) values below:
DGof
Zn H+ H2 Zn2+
0
0
0
-146.68
B. Calculate eo using DGo in part a.
C. Compare to eo from Table.
Effects of concentration on emf.
DGo = -nF
e
o
and
DGo = -RTlnK
R = 8.314 J/K mol
F = 96500 C/mol
Example #10
Calculate the emf for the reaction:
2Al(s) + 3Mn2+(aq)  2Al3+(aq) + 3Mn(s)
Calculate the equilibrium constant for this
reaction.
Predict whether the reaction will be
spontaneous at 30oC when:
A. [Al3+] = 2M, [Mn2+] = 1M
B. [Al3+] = 2M, [Mn2+] = 3M
Example #11: Putting
Everything Together
A galvanic cell is composed of aqueous Cr2O72-, Cr3+, Fe3+ and Fe2+ ions.
At 25°C it runs at a pH of 2.75.
a. Write the balanced spontaneous reaction for the galvanic cell.
b. Draw a diagram of the galvanic cell. Identify the anode and the cathode.
Indicate the direction of electron flow.
c. Determine the εo for this reaction.
d. Calculate ΔGo and K values for this reaction.
e. If the concentrations of all the species present are:
[Cr2O72-]= 0.025M [Cr3+]= 0.45M [Fe3+]= 2.3M [Fe2+]= 1.75M
Calculate the ε value. Is the reaction spontaneous under these conditions?
Example #11: Putting
Everything Together
A galvanic cell is composed of aqueous Cr2O72-, Cr3+, Fe3+ and Fe2+ ions.
At 25°C it runs at a pH of 2.75. Acidic balance
a. Write the balanced spontaneous reaction for the galvanic cell.
14H+ + Cr2O72- + 6Fe2+  2Cr3+ + 6Fe3+ + 7H2O
b. Draw a diagram of the galvanic cell. Identify the anode and the cathode.
Indicate the direction of electron flow.*
c. Determine the εo for this reaction. 0.56V
d. Calculate ΔGo and K values for this reaction.
-3,242,400J, 6.86 x 1056 or e130.87
e. If the concentrations of all the species present are:
[Cr2O72-]= 0.025M [Cr3+]= 0.45M [Fe3+]= 2.3M [Fe2+]= 1.75M
Calculate the ε value. Is the reaction spontaneous under these conditions?
0.16V, yes (don’t forget about H+ concentration)
What is electrolysis
When a current of electricity is forced
through the cell to produce a chemical
reaction for which the emf is negative.
Electrical energy  Chemical Energy
*Opposite of Galvanic Cell which is
Chemical Energy  Electrical Energy
Electrolysis half reactions
Write the half reactions for the electrolysis of
molten NaCl.
Molten NaCl is NaCl liquid (melted). The first
reaction is the Na+ becoming Na0 and the
second is Cl- becoming Cl20. These types will
always take the ions from the solid and form
their elements. If they are part of
H O N Cl Br I F they will get a subscript of 2.
Example #11
Write the half reactions for the electrolysis of
molten KBr.
Electrolysis in Aqueous Solutions
Sometimes the water reaction will give a
larger ecell value than the metal (especially
group 1 and some group 2 metals). The
larger ecell value is the more favorable
reaction thus the water reaction will occur
instead of the reduction of the metal and
oxidation of nonmetal. This means a lower
voltage needs to be applied for water
reaction.
Let’s Inspect Water Reaction
2H2O  O2 + 4H+ + 4e2H2O + 2e-  H2 + 2OHWater Reaction:
2H2O  2H2 + O2
-1.23V
-0.83V
ox
red
ε° = -2.06V
Compare to Aqueous: Ex: MgCl2
Mg2+ + 2e-  Mg
-2.37V
red
2Cl-  Cl2 + 2e-1.37V
ox
Overall: Mg2+ + 2Cl-  Mg + Cl2
ε° = -3.74V
Water Reaction costs less and is therefore more favorable.
Example #12
An aqueous solution of sodium chloride has
an electrical current running through it.
Write the half reactions that will occur at
each electrode.
Formulas and relationships
Amps x seconds = Coulombs
1 mole of electrons = 96500C
Electrolysis calculations work kind of like
stoichiometry problems from Chem 1.
Example #13
How long must a current of 5A be applied to
a solution of Ag+ to produce 10.5g of silver
metal?
Example #13
How long must a current of 5A be applied to
a solution of Ag+ to produce 10.5g of silver
metal?
Ag+ + e-  Ag
10.5g Ag/107.87g/mol = 0.097 mol Ag
0.097 mol Ag*(1 mol e-/1 mol Ag) = 0.097 mol e0.097 mol e-*(96500C/1 mol e-) = 9393C
1A*1s = 1C
5A*s = 9393C
s = 1879 s
Example #14
A 3A current runs through an Au+3 solution
for 1 hour. What mass of gold metal will
plate out on the cathode during that time?
Example #14
A 3A current runs through an Au+3 solution
for 1 hour. What mass of gold metal will
plate out on the cathode during that time?
Au+3 + 3e-  Au
1A*1s = 1C
3A*3600s = 10800C
10800C*(1 mol e-/96500C) = 0.112 mol e0.112 mol e- * (1 mol Au/3 mol e-) = 0.0373 mol Au
0.0373 mol Au *(196.97g/mol) = 7.35g Au
Example #15
Electrolysis of a molten metal bromide,
MBr3, using a 7A current for 30 minutes
deposits 2.263g of the metal. What is the
metal? (to find the identity of the metal
you need to find its atomic weight)
Example #15
Electrolysis of a molten metal bromide,
MBr3, using a 7A current for 30 minutes
deposits 2.263g of the metal. What is the
metal? (to find the identity of the metal
you need to find its atomic weight)
M+3 + 3e-  M
1A*1s = 1C
7A*1800s = 12600C
12600C*(1 mol e-/96500C) = 0.131 mol e0.131 mol e- * (1 mol M/3 mol e-) = 0.0435 mol M
2.263g M/0.0435 mol M = 51.995g/mol  Cr