Transcript Electrochemistry

Electrochemistry
(Applications of Redox)
Unit Essential Questions
What does electrochemistry study?
How are cell potentials calculated?
Review- Redox
Redox = oxidation/reduction reaction.
 What occurs during a redox reaction?
– Changes in oxidation states; oxidation =
losing e-, reduction = gaining e-.
 OIL- RIG
– Oxidation Involves Loss
– Reduction Involves Gain
 LEO-GER
– Lose Electrons Oxidation
– Gain Electrons Reduction

Practice
Which of the reactions below exhibits
oxidation and which exhibits reduction?
Cu+2(aq) + 2e-  Cu(s) reduction
Zn(s)  Zn+2(aq) + 2e- oxidation
What is the reducing agent? Zn(s)
+2
Cu
What is the oxidizing agent?
Electrochemical Cells
The two half reactions can be combined
from the previous example:
Cu+2(aq) + Zn(s)  Cu(s) + Zn+2(aq)
– Notice the e- are not shown because
they cancelled out on both sides.
– The e- are directly transferred from the
copper to the zinc.
– These e- can be used to do work if they
are indirectly transferred between
substances!
Electrochemical Cells
 Two
half reactions are separated in two
different beakers with a wire connecting
them.
– Wire allows the current (e-) to travel
between beakers.
– Flow of e- through the wire can be used to
do work.
 Electrochemical cells can be used to
produce electricity from a redox reaction or
can use electricity to produce a redox
reaction.
Section 17.1
Galvanic (Voltaic) Cells
What is a Galvanic Cell?
A
type of electrochemical cell that allows
chemical energy to be changed into
electrical energy.
 Chemical energy comes from change in
oxidation states (redox reaction).
 Wire used in the galvanic cell allows the
electrical energy to be used for work.
Making a Galvanic Cell
 Continue
with previous example:
Cu+2(aq) + Zn(s)  Cu(s) + Zn+2(aq)
 Zinc metal is placed in one beaker and
copper metal is placed in the other.
– These are the electrodes- the part that
conducts e- in the redox reaction.
 Zinc sulfate solution is added to the zinc
metal and copper (II) sulfate is added to the
copper metal.
– Called electrode compartments.
 Electrodes
are connected with a wire so the
reaction can start. But nothing happens!
Why?
 Charge would build up, and solutions
(electrode compartments) must remain
neutral!
Negative
b/c lose
Cu+2 ions
Zn
Cu
Positive
b/c form
Zn+2 ions
Zn+2
SO4-2
Cu+2
SO4-2
Galvanic Cell
e-
e-
K+
Cu
Zn
e-
SO4
-2
NO3-
Zn+2
Cu+2
SO4-2
Salt bridge
is added
to
maintain
neutrality.
Salt Bridge
 Any
electrolyte can be used to keep charges
neutral, as long as the ions don’t interfere
with the redox reaction.
 KNO3 used in previous example.
– K+ ions flow into the copper beaker (to
make up for the negative charge).
– NO3- ions flow into the zinc beaker (to
make up for the positive charge).
 Salt bridge allows the circuit to be complete
and the redox reaction to occur.
Galvanic Cell Components
A voltmeter can be attached to the wire
between the electrodes to measure the current
that can do work.
 Oxidation half reaction is always shown in the
left beaker and the reduction half reaction is
always shown in the right beaker.
– Anode = oxidation half reaction,
– Cathode = reduction half reaction
– Also have anode and cathode
compartments.

Galvanic Cell
e-
e-
K+
Cu
Zn
e-
SO4
-2
NO3-
Zn+2
Cu+2
SO4-2
anode
cathode
Notice the etravel from the
anode to the
cathode!
Inert Electrodes
 Solid
electrodes that do not participate in
the redox reaction can be used.
 These electrodes are only there to
complete the circuit/allow the electricity to
flow.
– They supply/accept e- as needed.
– Graphite and platinum are often used.
Alternative Galvanic Cell
* A porous disk is
used instead of a
salt bridge to
maintain the
circuit/neutrality.
* Porous disk
allows ions to
flow between
solutions.
Cell Potential
 Oxidizing
agent ‘pulls’ the e- from
reducing agent.
 The pull/driving force = cell potential
Ecell, or electromotive force (emf).
 Unit
= volts (V)
– 1 volt = 1 joule of work/coulomb of
charge
 Measured with a voltmeter
Section 1 Homework
Pg. 830 #15, 25
Section 17.2
Standard Reduction Potential
Standard Reduction Potentials
 Show
the number of volts produced from
a half reaction.
 All values are based on the reduction half
reaction (thus the name standard
reduction potential).
 This value is for substances in their
standard states: 25°C(298K), 1M, 1atm,
and pure solid electrodes.
 Standard hydrogen electrode- all other
values based off of H!
AP Practice Question
2BrO3-(aq) + 12H+(aq) + 10e-  Br2(aq) + H2O (l)
Which of the following statements is correct for
the above reaction?
a)
b)
c)
d)
The BrO3- is oxidized at the anode.
Br goes from a -1 to a 0 oxidation state.
Br2 is oxidized at the anode.
The BrO3- is reduced at the cathode.
Standard Hydrogen Electrode
 This
is the reference
all other
oxidations/reductions
are compared to.
Eº = 0
+
H
 º indicates standard
Cl
states of 25ºC,
1 atm, 1 M
1 M HCl
solutions.
H2 in
0.76V
Since H = 0,
this value is
assigned to the
oxidation of Zn.
H2 in
Anode
Zn+2
SO4-2
1 M ZnSO4
Cathode
H+
Cl1 M HCl
Cell Potential
+ Cu+2 (aq)  Zn+2(aq) + Cu(s)
 The total cell potential is the sum of the
potential at each electrode.
 Zn(s)

E ºcell = E ºZn Zn+2 + E ºCu+2 Cu
 Look
up reduction potentials in a table.
– Since potentials are given in terms of
reduction, the sign is switched for the
oxidation reaction since the reaction is
the reverse of reduction.
Cell Potential
+ Cu+2 (aq)  Zn+2(aq) + Cu(s)
 Reduction potentials are as follows:
Zn+2 + 2e-  Zn E = -0.76V
Cu+2 + 2e-  Cu E = 0.34V
 But Zn is not being reduced- it’s being
oxidized! So the sign must be changed:
E ºcell = 0.76V + 0.34V = 1.10V
 Alternate formula: E ºcell = E ºcathode - E ºanode
 Cell potentials are always > 0 because they
run spontaneously in the direction that
produces a positive potential.
 Zn(s)
Cell Potential
 Note
that even if a half reaction must be
multiplied by an integer to keep e- transfer
equal, the standard reduction potential
associated with that half reaction IS NOT
multiplied by that same integer!
 This is because the standard reduction
potential depends upon the reduction that is
occurring, not how many times it occurs.
 Like how the density of a sample of a
substance is always the same, regardless of
the size. It only depends on identity!
Practice: Cell Potential
 Determine
the cell potential for a galvanic
cell based on the redox reaction.
Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq)
Fe+3(aq) + e- Fe+2(aq)
Cu+2(aq)+2e- Cu(s)
Eºcell = 0.43V
Eº = 0.77 V
Eº = 0.34 V
Cell/Line Notation
 Shorthand
way of representing galvanic
cells. Format:
solidAqueousAqueoussolid
 Anode on the left,cathode on the right.
 Single line separates different phases at
each electrode.
 Double line indicates porous disk or salt
bridge.
 Concentrations in ( ) may be added after
the aqueous species, if known.
Cell/Line Notation
 Examples:
(1) Mg(s)Mg+2(aq)Al+3(aq)Al(s)
(2) Zn(s)Zn+2(1M)Cu+2(1M)Cu(s)
 If an inert electrode is used:
Ag+(aq) + Fe+2(aq)  Fe+3(aq) + Ag(s)
*Pt electrode used for Fe+2 or Fe+3
Pt(s)Fe+2(aq),Fe+3(aq)Ag+(aq)Ag(s)
Practice: Cell/Line Notation
 Given
the cell reaction below, write the
cell/line notation.
Ni(s) + 2Ag+(aq)  Ni+2(aq) + 2Ag(s)
Ni(s)Ni+2(aq)Ag+ (aq)Ag
Summing up Galvanic Cells


1)
2)
3)
4)
Reaction always runs spontaneously in
the direction that produces a positive
cell potential (E ºcell > 0).
Four things for a complete description:
Cell Potential
Direction of flow
Designation of anode and cathode
Species present in all componentselectrodes and ions.
Section 2 Homework
I will do #31, part 25, with you.
Pg. 830 #27, 31
Review: Balancing Redox
Rxns. Using Half Rxn. Method
 Write separate half reactions.
 For each half reaction balance all species
except H and O.
 Balance O by adding H2O to one side.
 Balance H by adding H+ to one side.
 Balance charge by adding e- to the more
positive side.
Review: Balancing Redox
Rxns. Using Half Rxn. Method
 Multiply equations by a number to make
electrons equal.
 Add equations together and cancel
identical species. Reduce coefficients to
smallest whole numbers.
 Check that charges and elements are
balanced.
In Basic Solution
enough OH- to both sides to
neutralize the H+.
 Any H+ and OH- on the same side form
water. Cancel out any H2O’s on both
sides.
 Simplify coefficients, if necessary.
 Add
AP Practice Question
What is the coefficient of H+ when the
following reaction is balanced?
MnO4-(aq) + H+(aq) + C2O4-2(aq) 
Mn+2(aq) + H2O(l) + CO2(g)
a) 16
b) 2
c) 8
d) 5
Homework
Balance the following equation in acidic and
basic solution:
NO3- + Mn  NO + Mn+2
Balanced equations:
3Mn + 8H+ + 2NO3-  2NO + 4H2O + 3Mn+2
3Mn + 4H2O + 2NO3-  2NO + 8OH- + 3Mn+2
Section 17.3
Cell Potential, Electrical Work,
and Free Energy
Linking Cell Potential to ΔG
 Cell
potential is directly related to the
difference in free energy between
reactants and products.
 This is shown in the following equation:
ΔG° = -nFE
n
= moles of e- in redox/half reaction
 F = Faraday’s constant = 96,485C/mole E = cell potential (V = J/C)
 Units
of G = J
Coulomb = unit of
electric charge
Linking Cell Potential to ΔG
 ΔG°
 Why
= -nFE
negative?
– Galvanic cells are spontaneous!
 If ΔG° was positive, the redox reaction
would be nonspontaneous.
 ΔG° can be calculated for half reactions
and for entire redox reactions.
Sample Calculation
Calculate ΔG° for the reaction:
Cu+2(aq) + Fe(s)  Cu(s) + Fe+2(aq)
Look
up half reactions and determine
individual E values. Then calculate E for the
cell.
– E = 0.78V = 0.78J/C
Make sure e- are correct!
ΔG° = -(2mole-)(96,485C)(0.78J) = 1.5x105J
moleC
Process
is spontaneous.
Practice Problem
Consider the following reaction:
Mn+2(aq) + IO4-(aq)  IO3-(aq) + MnO4-(aq)
the value of E cell and ΔG°. Be
sure to balance the e- (you can use the half
reactions and balance them from there).
E cell = 0.09V
ΔG° = -90kJ
Calculate
Section 3 Homework
Pg. 831 #: 37
Section 17.4
Dependence of Cell Potential on
Concentration
Qualitative Understanding
 The
following reaction is under standard
conditions:
Cu(s) + 2Ce+4(aq)  Cu+2(aq) + 2Ce+3(aq)
 What if [Ce+4] was greater than 1.0M?
– Use LeChatelier’s principle!
– This shifts the rxn. forward, which
means more products are formed and
therefore more e- are
flowing/transferring to allow them to
form.
Another Way of Looking at It
Cu(s) + 2Ce+4(aq)  Cu+2(aq) + 2Ce+3(aq)
What if [Ce+4] was greater than 1.0M?
– Use LeChatelier’s principle!
– The rxn. shifting more towards the
products means the rxn. is more
spontaneous, so –ΔG increases.
– Recall that –ΔG = -nFE°cell, so if -ΔG
is increasing, so must E°cell because n
and F are constant for this rxn.
Practice
For the cell reaction:
2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
E°cell = 0.48V
Predict whether E°cell is larger or smaller
than E°cell for the following cases.
(a)[Al+3]
= 2.0M, [Mn+2] = 1.0M
(b)[Al+3] = 1.0M, [Mn+2] = 3.0M
smaller
larger
Quantitative Understanding
•
We will not cover the quantitative side of
how cell potential changes with
concentration, but I want to mention the
Nernst Equation:
Ecell = E°cell –(RT)lnQ = E°cell –
nF(0.0592)logQ
n
•
•
Lets you calculate cell potential under nonstandard conditions.
Not tested on the AP exam.
Homework
 Pg.
832 # 51
Section 17.7
Electrolysis
Electrolytic Cells
Electrolytic cells use electrolysis and are the
opposite of galvanic cells.
– Reactions are not spontaneous.
 Electrolysis: a voltage bigger than the cell
potential is applied to the cell to force the non
spontaneous redox reaction to occur.
– Can be used to decompose compounds.
• Water can be broken into hydrogen &
oxygen.
 Can also be used for (electro)plating.
– Forces metal ions in solution to plate out on
electrodes in their solid form.

Galvanic
Cell
Cell
potential
= 1.10V
e
1.10
e-
Zn
Cu
1.0 M
Zn+2
Anode
1.0 M
Cu+2
Cathode
Voltage
applied > 1.10V
Electrolytic
Cell

e-
A battery
>1.10V
Zn
1.0 M Zn+2
Cathode
e

Cu
1.0 M Cu+2
Anode
Electrolytic Cells
 Relationship
exists between current, charge,
and time: I = q/t
– I = current (A, amp)
– q = charge (C)
– t = time (s)
 This formula can be used in conjunction
with other conversion factors to solve for
various items with respect to electrolytic
cells.
Electrolytic Cells & Stoichiometry
 The
previous formula and stoichiometry
can be used to determine the amount of
chemical change that occurs when
current is applied for a certain amount of
time.
 Answers the following questions:
– How much of a substance will be
produced?
– How long will it take?
– How much current is needed?
Electrolytic Cells & Stoichiometry
 No
set way to solve these problems each time!
In addition to the formula, the following
conversion factors may be used:
 96,485C/mol e- (this is 1 Faraday)
 Molar mass
 Moles of electrons to moles of other species
involved in the redox rxn.
– Make sure moles in the half rxn. are correct!
Example
If liquid titanium (IV) chloride (acidified with HCl)
is electrolyzed by a current of 1.000amp for
2.000h, how many grams of titanium will be
produced?
Have: amps & time  can use both to find q
from previously discussed formula:
I = q/t  1.000A = q/2.000h  convert h to s
1.000A = q/7,200.s
q = 7,200.As  q = 7,200.C  s  q = 7,200.C
s
*Now use stoichiometry to solve for grams!
Example Cont.
If liquid titanium (IV) chloride (acidified with
HCl) is electrolyzed by a current of
1.000amp for 2.000h, how many grams of
titanium will be produced?
Use
Faraday’s constant to get moles of e-,
write balanced half rxn. to get moles of eand moles of Ti: Ti+4  Ti + 4e-
7,200.C x 1mole- x 1molTi x 47.88gTi
96,485C 4mole- 1molTi
= 0.8932g
Practice Problem
What mass of copper is plated out when a
current of 10.0amps is passed for 30.0min
through a solution containing Cu+2?
Solve for q: 10.0A = q/1,800s  q = 18,000C
 18,000C (1mole-)(1molCu)(63.55gCu)
96,485C 2mole- 1molCu

=5.93gCu
Homework
 In
class- 77(a)
– Given mass and I; from mass we can
find charge, q. Then use formula to
solve for time.
 Pg. 834 #77(b&c),79(a&b)