Transcript Powerpoint - mvhs

Electrochemistry
Brown, LeMay Ch 20
AP Chemistry
Monta Vista High School
20.1: Oxidation-reduction reactions
“Redox”: rxns with a change in oxidation number
• At least one element’s oxidation number will increase
and one will decrease
• Consists of two half-reactions:
oxidation: loss of e-; oxidation number increases
reduction: gain of e-; oxidation number decreases
(“reduces”)
Oxidizing agent or oxidant: causes another substance
to be oxidized; gains eReducing agent or reductant: causes another
substance to be reduced; loses e2
Ex: Determine the half-reactions.
Cr2O72- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2 (g)
+6
-2
-1
+3
Reduction half-reaction:
Cr2O72- (aq) → Cr3+ (aq)
Oxidizing agent
Oxidation half-reaction:
Cl1- (aq)
Reducing agent
3
→ Cl2 (g)
0
20.2: Balancing Redox Reactions
“Method of Half-Reactions” steps:
1. Assign oxidation numbers to all species; based on this,
split the half-reactions of reduction and oxidation.
2. For each, balance all elements except H and O.
3. Balance oxygen by adding H2O to the opposite side.
4. Balance hydrogen by adding H+ to the opposite side.
5. Balance charges by adding e- to side with overall
positive charge.
6. Multiply each half-reaction by an integer so there are
an equal number of e- in each.
7. Add the half reactions; cancel any species; check final
balance.
4
Ex: Balance the following reaction, which takes
place in acidic solution:
Cr2O72- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2 (g)
Reduction:
2
6 e- + 14 H+ (aq) + Cr2O72- (aq) →
Cr3+ (aq) + 7 H2O (l)
Oxidation:
3 [ 2 Cl- (aq) → Cl2 (g)
5
+ 2 ]e-
6 Cl- (aq) → 3 Cl2 (g) + 6 e-
6 e- + 14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7 H2O (l)
6 Cl- (aq) → 3 Cl2 (g) + 6 e14 H+ (aq) + Cr2O72- (aq) + 6 Cl- (aq) →
2 Cr3+ (aq) + 7 H2O (l) + 3 Cl2 (g)
• Since occurs in acidic solution, H+ appears as a reactant
• If reaction occurs in basic solution, balance using same
method, but neutralize H+ with OH- as a last step.
6
Ex: In basic solution,
MnO4- (aq) + Br- (aq) → MnO2 (s) + BrO3- (aq)
+7
-2
-1
+4
-2
+5
-2
MnO4- (aq) → MnO2 (s)
MnO4- (aq) → MnO2 (s) + 2 H2O (l)
4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)
3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)
Reduction:
7
Oxidation: Br- (aq) → BrO3- (aq)
3 H2O (l) + Br- (aq) → BrO3- (aq)
3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq)
3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e2[3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O (l)]
6 e- + 8 H+(aq) + 2 MnO4-(aq) → 2 MnO2(s) + 4 H2O(l)
3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e2 H+(aq) + 2 MnO4- (aq) + Br- (aq) →
2 MnO2 (s) + BrO3- (aq) + H2O (l)
8
Neutralize with OH-:
2 H+(aq) + 2 MnO4-(aq) + Br- (aq) + 2 OH- (aq) →
2 MnO2 (s) + BrO3- (aq) + H2O (l) + 2 OH- (aq)
2 H2O (l) + 2 MnO4-(aq) + Br- (aq) →
2 MnO2 (s) + BrO3- (aq) + H2O (l) + 2 OH- (aq)
Simplify:
H2O (l) + 2 MnO4-(aq) + Br- (aq) →
2 MnO2 (s) + BrO3- (aq) + 2 OH- (aq)
9
Introduction to Electrochemistry
• An electric cell converts chemical energy into electrical energy
– Alessandro Volta invented the first electric cell but got his inspiration from Luigi
Galvani. Galvani’s crucial observation was that two different metals could make the
muscles of a frog’s legs twitch. Unfortunately, Galvani thought this was due to some
mysterious “animal electricity”. It was Volta who recognized this experiment’s potential.
– An electric cell produces very little electricity, so Volta came up with a better design:
• A battery is defined as two or more electric cells connected in
series to produce a steady flow of current
– Volta’s first battery consisted of several bowls of brine (NaCl(aq))
connected by metals that dipped from one bowl to another
– His revised design, consisted of a sandwich of two metals
by paper soaked in salt water.
separated
20.3: Voltaic (or Galvanic) Cells
Alessandro Volta
(1745–1827)
Luigi Galvani
(1737–1798)
• Device that utilizes energy released in a spontaneous
electrochemical reaction by directing e- transfer along an
external pathway, rather than directly between reactants
Electrodes: metals connected by an external circuit AnOx RedCat
Anode: half-cell where oxidation occurs
Cathode: half-cell where reduction occurs
11
Voltaic Cells (aka Galvanic Cell)
• A device that spontaneously produces electricity by spontaneous
redox reactions.
– The reduction half-reaction (SOA) will be above the
oxidation half-reaction (SRA) in the activity series to ensure a
spontaneous reaction.
• Composed of two half-cells; which each consist of a metal rod or
strip immersed in a solution of its own ions or an inert
electrolyte.
– Electrodes: solid conductors connecting the cell to an
external circuit
– Anode: electrode where oxidation occurs (-)
– Cathode: electrode where reduction occurs (+)
Voltaic Cells (aka Galvanic Cell)
-The electrons flow from the anode to the cathode (“a before
c”) through an electrical circuit rather than passing directly
from one substance to another
-A porous boundary separates the two electrolytes while still
allowing ions to flow to maintain cell neutrality. Often the
porous boundary is a salt bridge, containing an inert aqueous
electrolyte (such as Na2SO4(aq) or KNO3(aq)), Or you can use a
porous cup containing one electrolyte which sits in a container
of a second electrolyte.
Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Oxidation half-cell:
Reduction half-cell:
(-)
Zn (s) → Zn2+ (aq) + 2 eCu2+ (aq) + 2 e- → Cu (s)
(+)
e- flow
anode
cathode
NO31-
Na1+
Zn
Cu
salt bridge, saturated
with NaNO3 (aq)
Zn2+ (aq)
Cu2+ (aq)
Notation:
Anode| Anode product|Salt |Cathode|Cathode product
Zn(s)|Zn2+(aq, 1 M)|NaNO3 (saturated)|Cu2+(aq, 1 M)|Cu(s)
14
Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cell Notation for Voltaic Cells
Voltaic cells can be represented using cell notation:
The single line represents a phase boundary (electrode to
electrolyte) and the double line represents a physical boundary
(porous boundary)
For a redox reaction to be spontaneous, the metal with higher
reduction potential should be made cathode.
Write cell notation for Zn-Cu cell
Voltaic Cell Summary
• A voltaic cell consists of two-half cells
separated by a porous boundary with
solid electrodes connected by an external
circuit
• Oxidizing agent undergoes reduction at
the cathode (+ electrode) – cathode
increases in mass
• Reducing agent undergoes oxidation at
the anode (- electrode) – anode
decreases in mass
• Electrons always travel in the external
circuit from anode to cathode
• Internally, cations move toward the
cathode, anions move toward the anode,
keeping the solution neutral
20.4: Electromotive force (emf or E)
• Potential difference between cathode and anode that
causes a flow of e-.
– The result of the difference in electric fields produced at
each electrode
• Also called cell potential or cell voltage, Ecell
• Units = Volt
(1 V = 1 J / 1 C)
• Standard conditions (25°C, 1 atm, 1 M): Eºcell
• Eºcell is unique to each set of half-cells involved
Charles-Augustin de Coulomb
(1736–1806)
18
Standard Reduction Potentials
• Assigned to each reduction reaction
Eºcell = Eºreduction(cathode) - Eº reduction(anode)
• All Eºred are referenced to standard
hydrogen electrode (S.H.E.):
2 H+ (aq, 1 M) + 2 e- → H2 (g, 1 atm)
Eºreduction = 0 V
19
Standard Cells and Cell Potentials
• Standard Cell Potential, E0 cell = the electric potential difference of
the cell (voltage)
E0 cell = E0r cathode – E0r anode
Where E0r is the standard reduction potential, and is a measure of a
standard ½ cell’s ability to attract electrons.
• The higher the E0r , the stronger the OA
• All standard reduction potentials are based on the standard hydrogen ½
cell being 0.00V. This means that all standard reduction potentials that
are positive are stronger OA’s than hydrogen ions and all standard
reduction potentials that are negative are weaker.
• If the E0 cell is positive, the reaction occurring is spontaneous.
• If the E0 cell is negative, the reaction occurring is non-spontaneous
Rules for Calculating Cell Potential
1. Determine which electrode is the cathode. The cathode is the electrode
where reduction occurs- marked by the higher reduction potential of the
metal
i.e. The OA that is closet to the top on the left side of the redox table =
SOA
If required, copy the reduction half-reaction for the strongest oxidizing
agent and its reduction potential
2. Determine which electrode is the anode. The anode is the electrode where
the strongest reducing agent present in the cell reacts.
i.e. The RA that is closet to the bottom on the right side of the redox table
= SRA
If required, copy the oxidation half-reaction (reverse the half-reaction)
3. Determine the overall cell reaction. Balance the electrons for the two half
reactions (but DO NOT change the E0r) and add the half-reaction equations.
4. Determine the standard cell potential, E0cell using the equation:
E0 cell = E0r cathode – E0r anode
Practice Problem # 1
Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
• Anode: Zn (s) → Zn2+ (aq) + 2 e– This is oxidation, but Eºreduction is recorded as the reduction
reaction:
Zn2+ (aq) + 2 e- → Zn (s), where Eºreduction = - 0.76 V
• Cathode:
Cu2+ (aq) + 2 e- → Cu (s)
Eºreduction = + 0.34 V
Eºcell = Eºreduction (cathode) - Eºreduction (anode)
= 0.34 V – (- 0.76 V)
Eºcell = +1.10 V
22
Practice Problem # 2
• Example: What is the standard potential of the cell represented below:
1. Determine the cathode and anode
2. Determine the overall cell reaction
3. Determine the standard cell potential
Some Important Facts to Remember
• Positive voltages = reaction is spontaneous
(cell will produce voltage, DG is -)
• Changing the stoichiometric coefficients of a
half-cell does not affect the value of Eº, since
potential is a measure of the energy per
electrical charge.
• However, having more moles means the
reaction will continue for a greater time.
24
Using the Standard Reduction Potential table
• When Eºred is very negative, reduction is difficult (but
oxidation is easy):
Li1+ (aq) + e- → Li (s) Eºred = - 3.05 V
– Li (s) oxidizes easily, so it is the strongest reducing agent;
lower on the Standard Reduction Potential chart signifies the
anode.
• When Eºred is very positive, reduction is easy:
F2 (g) + 2 e- → 2 F1Eºred = + 2.87 V
– F2 reduces easily, so it is the strongest oxidizing agent;
higher on the chart signifies the cathode.
• Predicting metal “activity series”: most reactive metals
are near bottom
25
20.5: Spontaneity of Redox Reactions
• Gibbs free energy and emf:
DGo = - n F Eo
n = # of moles of e- transferred
Faraday’s constant, F
= 96,500 C/mol of e-
Michael Faraday
(1791-1867)
• 1F = 1 mole of e-, which has a charge of 96,500 C
• Units = (mol e-)(C/mol e-)(V) = (mol e-)(C/mol e-)(J/C) = J
• As Eºcell increases, DGº decreases (becomes more
spontaneous)
26
20.6: Effect of Concentration on Ecell
The Nernst Equation: Used to figure out cell
potential under experimental conditions (E)
DG = DGo + RT ln Q
and
DGo = - n F Eo
(R = 8.314 J/(mol•K))
- n F E = - n F Eº + RT ln Q
RT
EE 
ln Q
nF
0
Walther Nernst
(1864-1941)
27
0.0591
EE 
log Q
n
0
• At equilibrium, DG = 0
RT
E 
ln K
nF
0
0.0591
E 
log K
n
0
since T º = 298 K
DGo
K
Eocell
Reaction
Negative
>1
Positive
Spontaneous
0
1
0
Equilibrium
Positive
<1
Negative
Non-spontaneous
28
Relationship Between Thermodynamic Quantities
Equilibrium
composition
measurements
Calorimetry and
entropy: S0 and
DS0
Electrical
measurements:
E0 and E0cell
E0cell of cells
• Model of a portion of a cell membrane shows a K+ channel
(blue) and a Na+ channel (red). The area outside the cell
(top) is rich in Na+ and low in K+. Inside the cell, the fluids
are relatively rich in K+ and low in Na+.
30
Electrolytic Cells
• The term “electrochemical cell” is often used to refer to a:
– Voltaic Cell – one with a spontaneous reaction
The metal that is higher on activity series, should be made
cathode for a spontaneous reaction
Eocell greater than zero = spontaneous
– Electrolytic cell – one with a nonspontaneous reaction
Eocell less than zero= nonspontaneous
• Why would anyone be interested in a cell that is not spontaneous?
– This would certainly not a good battery choice, but by supplying
electrical energy to a non spontaneous cell, we can force this
reaction to occur.
– This is especially useful for producing substances, particularly
elements. I.e. the zinc sulfate cell discussed above is similar to the
cell used in the industrial production of zinc metal.
Electrolytic Cells
• Electrolytic Cell – a cell in which a nonspontaneous redox
reaction is forced to occur; a combination of two electrodes, an
electrolyte and an external power source.
– Electrolysis – the process of supplying electrical energy to
force a nonspontaneous redox reaction to occur
– The external power source acts as an “electron pump”; the
electric energy is used to do work on the electrons to cause
an electron transfer
Electrons are pulled from the
anode and pushed to the
cathode by the battery or power
supply
Comparing Electrochemical Cells:
Voltaic and Electrolytic
It is best to think of “positive” and “negative” for electrodes as labels, not charges.
20.8-9: Electrolytic cells
• Electrical energy is used to cause a nonspontaneous reaction to occur.
• Current (I): a measure of the charge (Q) per unit
time (t)
Q
I
t
• Units = amps or amperes (A)
1A=1C/1s
André-Marie Ampère
(1775 – 1836)
34
Ex: Na+ (l) + Cl- (l) → Na (l) + Cl2 (g)
Oxidation half-cell:
2 Cl- (l) → Cl2 (g) + 2 eReduction half-cell: Na+ (l) + e- → Na (l)
e- flow
(+)
(-)
Pt
Pt
Cl2 (g)
Na (l)
anode
NaCl (l)
cathode
• In a voltaic cell: anode is (-), cathode is (+)
•35 In an electrolytic cell: anode is (+), cathode is (-)
20.9: Quantitative Electrolysis
Ex: A current of 0.452 A is passed through an
electrolytic cell containing molten NaCl for 1.5
hours.
• Write the electrode reactions.
Oxidation:
2 Cl- (l) → Cl2 (g) + 2 e-
Reduction:
Na+ (l) + e- → Na (l)
• Calculate the mass products formed at each
electrodes
Strategy: calculate the total charge of e- transferred,
and use F to convert to moles.
36
Q
I
or Q  I  t
t
Q  (0.452 A)(5400 sec)  2441 C
 1F
 1 mol Na  22.99 g 
Mass Na  (2441 C)



 1 mol Na 
 96,500 C  1F
 0.58 g
 1F
 1 mol Cl2  70.90 g 

Mass Cl2  (2441 C)


 96,500 C  2F  1 mol Cl2 
37
 0.90 g
Electrolysis of Aqueous Mixtures
•
If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in an
electrolytic cell is slowly increased, what products will form at
each electrode?
o What exists in solution that could be reduced?
Cu2+, Zn2+, and H2O (not Cl-)
Cu2+ + 2 e- → Cu
Eºred = 0.34 V
Zn2+ + 2 e- → Zn
Eºred = - 0.76 V
2 H2O + 2 e- → H2 + 2 OHEºred = - 0.83 V
o
38
Since Eºred (Cu) is the most positive, it is the most probable
reaction, and therefore will occur most easily (at the
cathode):
Cu (s) > Zn (s) > H2 (g)
•
If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in an
electrolytic cell is slowly increased, what products will form at
each electrode?
o What exists in solution that could be oxidized?
Cl- and H2O (not Cu2+ and Zn2+)
2 H2O → O2 + 4 H+ + 4 eEºox = - 1.23 V
2 Cl- → Cl2 + 2 eEºox = - 1.36 V
o
39
Since Eºox (H2O) is the most positive, it is the most probable
reaction, and therefore will occur most easily (at the
anode):
O2 (g) > Cl2 (g)
Electroplating
• The anode is a silver
bar, and the cathode
is an iron spoon.
Anode:
Ag (s) → Ag+ (aq)
Cathode:
Ag+ (aq) → Ag (s)
40