Transcript Electrochemistry - Manchester High School

Electrochemistry
The first of the
BIG FOUR
Introduction of Terms
 Electrochemistry- using chemical changes
to produce an electric current or using
electric current to produce a chemical
change
 Current- movement of electrons or a
transfer of electrons
 Conductor- contains charge carriers
1. Metallic conductor- wire where the metal ions
float around the electrons
2. Electro - ions present from ionization or
dissociation of a dissolved electrolyte.
Voltaic Cells
A. Definition: An electrochemical cell in
which a spontaneous reaction generates
an electric current.
B. This type of cell is spontaneous.
C. Consists of two half cells; a portion of
the cell contains one half of the reaction
D. Oxidation - ANODE
“an ox”
Reduction - CATHODE “red cat”
Illustration
Galvanic Cell Animation
Notation for Voltaic Cells
A. Anode || Cathode (the double line represents the
salt bridge
B. Reactants, products || reactants, products
C. | - represents a phase boundary
D. Examples:
1. Zn(s) | Zn2+(aq) || Cu2+ (aq)| Cu(s)
2. Zn(s) | Zn2+(aq)|| H+ (aq) | H2 (g) | Pt
3. Tl + Sn2+ → Tl 3++ Sn
4. Zn + Fe3+ → Zn2+ + Fe2+
E. A gas is involved, Pt is usually the electrode
1. || Cl2 (g) | Cl- | Pt
Notation for Voltaic Cells
2 H+(aq) + 2e- g H2(g)
Zn(s) g Zn2+ + 2eTl(s) │Tl+(aq) ║Sn2+ │Sn(s)
Zn(s) │Zn2+(aq) ║Fe3+(aq), Fe2+(aq) │Pt
Electromotive Force (EMF)
A. Definition- The maximum potential difference
between the electrodes of a voltaic cell.
B. Wmax = ∆G = -nFEcell
where F is a faraday = 9.6485 x 104 C
∆G is Gibb’s Free Energy (thermodynamics)
C. The maximum cell potential is directly related to the
free energy difference between the reactants and the
products in the cell.
D. Example 1: Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+
Answer: -1.5 x 105 J
Example 2
 The emf of a voltaic cell with the reaction:
Hg22+ (aq) + H2 ↔ 2 Hg(l) + 2 H+ (aq)
is 0.650 V. Calculate the maximum
electrical work of this cell when 0.500 g of
H2 is consumed.
Answer: -2 x 96, 480 x 0.650 = -1.25 x 105 J
.500g (1/2.02)(-1.25 x 105J/mol)= -3.09 x 104J
Standard Cell EMF’s
A. A cell emf is a measure of the driving
force of the cell reaction.
B. Anode + Cathode
1. Ox potential- a measure in V of the tendency
for a species to lose electrons in the oxidation
half reaction
2. Red. Potential- a measure in V of the
tendency for a species to gain electrons in the
reduction half reaction.
Calculating Ecell
1.
2.
Equation: Ecell = Eox + Ered
Turning Ered into a Eox
a) Reverse the half cell reaction (to a ox.
Rxn)
i. Sn2+ + 2e- → Sn
Sn → Sn2+ + 2eb) Reverse the sign on the potential value
i. +0.14 V
Strengths of Oxidizing & Reducing
Agents
1.
2.
Oxidizing Agents – these are the substances
reduced, so the strongest oxidizing agents
have the highest positive Ered. (F2 is the
strongest oxidizing agent while Li+ is the
weakest oxidizing agent).
Reducing Agents- these are the substances being
oxidized so they appear on the right side of
the yield sign. The strongest reducing agents
correspond with the highest negative Ered. (Li
is the strongest reducing agent and F- is the
weakest.
Examples
3.
3+
Fe ,
Cl2, H2O2
4. Cu, H2, Al
Nernst Equation
1. Calculate cell potentials when
conditions are not standard
2. Calculate cell potentials of half
reactions
3. Predict whether the cell voltage
will decrease or increase
depending on Q.
The Equation
EE
0.0592
log Q
cell 
n
Q and E are inversely related
Q>1 : then the log value is positive and E decreases
Q<1 : then the log value is negative and E increases
Controls the voltage of the cell
Electrolytic Cells
A. Definition: An electrochemical
cell in which an electric current
drives an otherwise
nonspontaneous reaction.
B. Electrolysis- the process by which
a direct electrical current causes a
chemical reaction to occur.
1. Electrolysis of MOLTEN Salts
(Down’s Cell)
The battery acts as an
electron pump, pulling
electrons from the
anode and pushing them
to the cathode.
Overall cell reaction:
2Na+ + 2Cl- → 2Na + Cl2
2 Cl- → Cl2 + 2 eWithdrawing
electrons so
it’s positive
2 Na+ + 2 e- → 2 Na
Anode
Cathode
Oxidation
Reduction
Adding
electrons so
it’s negative
Examples:
a) KCl
Anode: 2 Cl- → Cl2 + 2 eCathode: 2 e- + 2 K+ → 2 K
 KOH
Anode: 4 OH- → O2 + 2H2O + 4 eCathode: 4 K+ + 4 e- → 4 K
2. Electrolysis of Aqueous
Solutions
a. There are two reactions possible at each
electrode for an aqueous solution.
 Water can be oxidized to O2 at the
anode:
Anode: 2 H2O → O2 + 4 H+ + 4 e♦ Water can be reduced to H2 at the
cathode:
Cathode: 2 H2O + 2 e- → H2 + 2 OH-
b. Chloride Solutions
At the anode there are 2 possible half reactions:
- →O Cl
- positive potential, -1.36
has
the
more
we
2 ClSince
+
2
e
V
2
2
would expect to see O2 produced
at the
+
2 Hanode.
O
→
O
+
4
H
+
4
e
This does
2
2 not happen. There is a
1.23much
V higher voltage needed to produce the
O2, in excess of the expected value
At the (overvoltage).
cathode there
2 possible
half reactions:
Theare
overvoltage
is greater
than the production of Cl2 so the chlorine is
2 Na+
+ 2ate2 Na
-2.71 V
produced
the →
anode.
2 H2O + 2 e- → H2 + 2 OH- -0.83 V
Overall Balanced Reaction:
2 H2O + 2 Cl- → Cl2 + H2 + 2 OH-
c. Sulfuric acid solutions/ SO42- ions
Possible Anode Reactions:
- as -2.71 V
In this
2 SO42→case,
S2Othe82-sulfate
+ 2isenot
easily oxidized as chloride in the
- -1.23 V
so+this
time
the
2 H2Oprevious
→ Oexample,
+
4
H
+
4
e
2
water is oxidized as expected.
Possible Cathode Reactions:
2 H+ + 2 e- → H2
2 H2O + 2 e- → H2 + 2 OHOverall Balanced Reaction:
2 H2O → O2 + 2 H2
Examples
1. Half reactions of CuSO4 (aq)
Anode:
2 SO42- → S2O82- + 2 e2 H2O → O2 + 4 H+ + 4 eCathode:
-2.71 V
-1.23 V
Cu2+ + 2e- → Cu
+0.34 V
2 H2O + 2 e- → H2 + 2 OH-0.83V
Overall Balanced Reaction:
2 Cu2+ + 2 H2O → 2 Cu + O2 + 4H+
(acidic)
Examples
2. Aqueous AgNO3 (NO3- is not oxidized)
Anode:
2 H2O → O2 + 4 H+ + 4 e-
-1.23 V
Cathode:
Ag+ + e- → Ag
+0.80 V
2 H2O + 2 e- → H2 + 2 OH- -0.83V
Overall Balanced Reaction:
4 Ag+ + 2 H2O → Ag + O2 + 4 H+ (acidic)
*NOTE:
It is difficult to reduce elements
in group IA (alkali metals) in
the presence of water (II A
too!). You have the formation
of H2 instead.
Answers to Homework
 Anode: 2 Cl- g
Cl2 + 2e Cathode: 2e- + Ca2+ g Ca
 Anode: 4 OH- → O2 + 2H2O + 4 e Cathode: 4e- + 4 Cs+ g
4 Cs
 Anode: 2 H2O → O2 + 4 H+ + 4 e-
 Cathode: 2 H2O + 2 e- → H2 + 2 OH Overall Balanced Reaction:
2 H2O → O2 + 2 H2
Answers to Homework
 Anode: 2 Br- g
Br2 + 2e Cathode: 2 H2O + 2 e- → H2 + 2 OH Overall Reaction:
2 Br- + 2 H2O g Br2 + H2 + 2 OH-
Stoichiometry of Electrolysis
A. Electrical Units
1. ampere- base unit of current (rate of flow)
2. Coulomb – SI unit of electrical charge
(amp•sec): a current of 1 amp flowing for 1
second.
3. Faraday- 1 mole of electrons = 6.02x1023 e1F = 96,500 C (96,480 book #) = 96,500 amp•sec
Examples
 1.
 2.
 3.
Example problems:
1.
How many grams of copper is plated out when a
current of 10.0 amps is passed for 30.0 minutes?
10.0am ps 30.0 min 60.0 sec
1m olE
1m olCu 63.546g
X
X
X
X
X
1
1
1 min
96,500C 2.0m olE
1m ol
2. How long must a current of 5.00 A be applied to a
solution of Ag+ to produce 10.5 g silver metal?
10.5 g
1m olAg
1m olE 96,500C 1am psec
1
X
X
X
X
X
1
107.868g 1m olAg 1m olE
1C
5.00A