Transcript A.P. Chemistry

A.P. Chemistry
Electrochemistry
Electrochemistry - study of the interchange of chemical &
electrical energy (p. 837)
(Review Ch. 4. Aqueous Chemistry: Redox; ½ Reaction Writing;
Determining Oxidatn #)
Oxidation- Reduction Reactions (p. 838)- a transfer of
electrons from a reducing agent to an oxidizing agent.
Oxidation is Loss of electrons
(OIL RIG)
Reduction is Gain of Electrons
Writing ½ Reactions
Example:
2Na(s) + Cl2(g)  2NaCl (S)
Oxidation: 2Na  2Na+ + 2eReduction: Cl2 + 2e-  2ClExample: 8H+(aq) + MnO4-(aq) + 5Fe+2(aq)  Mn+2(aq) +
5Fe+3(aq) + 4H2O(l)
Oxidation:
Reduction:
Galvanic Cell- a device in which chemical energy is changed to
electrical energy (p. 839) Draw it!
Anode- electrode compartment where oxidation occurs (p.
839)(vowels)
Cathode- electrode compartment where reduction occurs (p.
839)(consonants)
Cell Potential (Ecell) or electromotive force (emf)- the “pull” or
driving force on the electrons in a galvanic cell.
Volt- unit of electrical potential
1 V = 1 Joule of work per 1 coulomb of
charge transfer
Ampere- the unit of electric current equal to
one coulomb of charge per second
1 A = 1C/s
Charge on an electron- 1.60218 x 10-19 C
(coulombs)
Faraday- 1 F = 96485 Coulombs/mol
(on equation sheet, rounded to 96,500)
Electrode Potentials and Half-Cells (p. 840-845)
When a metal comes into contact with a solution containing
its own ions, an equilibrium is set up.
Mx+(aq) + xe-  M(s)
Some reactive metals such as Mg will lose electrons readily
and the equilibrium lies to the left. A large number of
electrons are released which collect on the surface of the
metal giving a negative charge.
Mg+2(aq) + 2e-  Mg(s)
A less reactive metal such as silver will show less tendency to
ionize and the equilibrium lies to the right. Fewer electrons
will collect on the metal and the charge will be much less
negative. In fact, if the aqueous ions remove electrons from
the metal, it will develop a positive charge.
Ag+(aq) + e-  Ag(s)
Non-metals can also be considered, for ex.,
H+(aq) + e-  ½ H2(g)
Whenever an element is placed in contact with a solution
containing its own ions, an electric charge will develop on
the metal or, in the case of a nonmetal, on the inert
conductor placed in solution. The charge is called the
electrode potential and the system is called a half-cell. The
sign and size of the charge will depend on the ability of the
element to lose or gain electrons. See the Table of
Standard Reduction Electrode Potentials (SREP) (on the
back of the Periodic Table) or p. 843 in your textbook.
Standard potentials are given as reduction processes. Two
manipulations are necessary:
One must be reversed in the calculations (reverse the sign,
too)
#electrons lost must equal the # electrons gained. Therefore,
multiply equation by the appropriate integer.
(The Eocell is not changed and is not multiplied
by the integer required to balance the cell
equation (it is an intensive property). (p. 844)
Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
Anode: Zn Zn2+ + 2eCathode: Cu2+ + 2e- Cu
Eocell = E oZn  Zn2+ + Eo Cu2+ Cu
(Eo = Eoox + Eored)
1.10 V = 0.76 + ?
? = 0.34 V
Example: Consider a galvanic cell based on the
following reaction:
Al+3(aq) + Mg(s)  Al(s) + Mg+2(aq)
Write the ½ reactions:
Find the Eo (standard electrode potential values)
values: (look on p. 843)
Give the balanced cell reaction and determine Eo
for the cell.
Example: A galvanic cell is based on the following
reaction:
MnO4-(aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) +
Mn+2(aq) + H2O(aq)
The half-reactions are:
MnO4-(aq) + 5e- + 8H+(aq)  Mn+2(aq) +
4H2O
Eo = 1.51V
ClO4-(aq) + 2H+(aq) + 2e-  ClO3-(aq) + H2O
Eo = 1.19V
Give the balanced cell reaction and determine the
value of Eo for the cell.
Electrochemical Species and Electrode Potentials
(Look at the review sheet: SRP table)
Elements that appear at the top of the
electrochemical series lose electrons most readily
and therefore have the most negative Eo values
and are the best reducing agents. (You can derive
chemical activity from this.)
Elements that appear at the bottom of the series
gain electrons most readily and therefore have
the most positive Eo values and are the best
oxidizing agents.
Electrochemical Cells
An electrochemical cell is the apparatus for generating
electrical energy from a spontaneous REDOX reaction.
Connecting two half-cells that have different electrode
potentials will form an electrochemical cell.
A high resistance voltmeter is used to measure the voltage
since it draws no current and will not affect the reading.
A salt bridge connects the two half-cells. It can be made from
a piece of filter paper soaked in a suitable ionic solution,
often potassium chloride or potassium nitrate. The ionic
solution used in the salt bridge must not interfere with the
equilibrium of the two half-cells; for ex., KCl could not be
used if one of the half-cells contained Ag+ ions as they
would react with the Cl- ions, precipitating the silver out
and reducing the concentration of Ag+ ions. The salt bridge
allows the transfer of ions between the two solutions, thus
completing the circuit. Ions will flow in order to balance the
charge.
Measurement of Electromotive Force (EMF) or Ecell
The EMF or Ecell is the voltage measured when no current
is being drawn from the cell and is determined using a
high resistance voltmeter. It can be calculated by
applying the following relationship to the cell diagram.
Ecell = ERHS - ELHS (same thing as Eox + Ered, once you
reverse the equation from the standard reduction
potential chart and change the sign)
Since the half reactions are reversible they are subject to
change according to Le Chatelier’s Principle. In order to
make more meaningful comparisons, it is necessary to
stipulate a set of standard conditions under which the
electrode potential of a given half-cell is measured.
Standard Electrode Potential (SHE)
The Standard Electrode Potential of a half-cell,
Eocell , is defined as the electrode potential of a
half-cell, measured relative to a standard
hydrogen electrode which has a value of
0.00 V, measured under standard conditions
(temperature @ 25oC (298 K), gases at 1 atm,
all solutions at 1 M concentration). The
diagram below shows the standard hydrogen
electrode being used to determine the Eo for
an electrode being tested.
The practical use of the hydrogen half-cell
for determining Eo values suffers from three
main problems.
• It is difficult to set up the H2(g) at precisely
1 atm pressure.
• It is fragile and non-portable.
• The equilibrium H+(aq) + e-  ½ H2(g)
is only established slowly.
Example:
Fe3+(aq) + Cu(s) Cu2+(aq) + Fe2+(aq)
Fe3+ + e-  Fe2+
Eo = 0.77 V
Cu2+ + 2e- Cu
Eo = 0.34 V
*reverse, therefore,
Cu  Cu2+ + 2eMultiply equation #1
2Fe3+ + 2e-  2Fe2+
Eo = 0.77 V
(NO CHANGE!)
(Now, add them together)
Eo = -0.34 V + 0.77 V =
Cell diagrams have a number of standard features. The reduced species is
always placed on the RHS and this electrode is the cathode. The oxidized
species is always placed on the LHS and this electrode is the anode. (Watch
out: some AP questions have reversed the cell diagrams- you must be able
to identify which is the anode and which is the cathode based on the
processes occurring.)
The vertical line │ represents the different phases present in each electrode.
The double vertical line ║ represents the salt bridge connecting the two
electrodes.
A(electrode) │ A+(solution) ║ C+(solution) │ C(electrode)
Different species in the same phase are separated by a comma, for ex., Fe+2
(aq), Fe+3(aq). The order in which they are written is not important.
The presence of an inert conductor (required when no solid metal is present)
may also be shown inside parentheses ( ). For ex., (Pt) ½ H2(g) , H+(aq)
The electrons will flow toward the more positive half-cell, in this case, from
zinc to copper. As a result, the copper ions gain electrons and are therefore
reduced. A “cell diagram” can be written thus
Zn(s) │Zn+2(aq) ║ Cu+2 (aq) │Cu(s)
Example: (p. 846)
Fe(s) │ Fe2+ (aq) ║ MnO4-(aq), Mn+2 (aq) │ Pt(s)
Problem: Describe completely the galvanic cell based on the following
half-reactions under standard conditions:
Ag+ + e-  Ag
Eo = 0.80 V
Fe+3 + e-  Fe+2
Eo = 0.77 V
1. must have Eocell positive; therefore, 2nd reaction must run in
reverse.
2. Ag+ receives electrons, Fe+2 ion loses electrons, electrons will flow
from the compartment containing Fe+2 to the compartment
containing Ag+
3. Oxidation occurs in the Fe+2 compartment (the anode)
4. Must use an inert electrode in the Fe+2/Fe+3 compartment.
Therefore, line notation is:
The Relationship between Gibbs Free Energy and Ecell
The relationship between Gibbs Free Energy and
the Ecell is summarized by the expression
/\Go = -n F Eocell
Where
F = Faraday constant = 96500 J/V mol
n = number of moles of electrons transferred
Example: Calculate the standard free energy for the
zinc-silver galvanic cell.
Coupling this with the expression below (see Unit
13)
/\Go = -RT ln K
R is the ideal gas constant; there are several
different values of R, depending on the equation
you are using and the units of the other
quantities in the equation; for electrochemistry,
R = 8.314 J/K mol
Example: Calculate the equilibrium constant for the
zinc-silver cell in the previous example.
It is fairly simple to derive the expression
Eocell = RT ln K
nF
At T = 298 K, and substituting for R and F, the
expression can be simplified to
Eocell = 0.0257 ln K
n
The Nernst Equation and Non-standard conditions of temperature and
concentration
The Nernst equation can be used to calculate the voltage generated by
the combination of two half-cells when the conditions are not
standard. One form of the equation is
E = Eocell - RT lnQ
nF
Where R is the gas constant (8.314 J/K mol), T is the Kelvin
temperature, n is the # of electrons transferred between the
species, F is the Faraday constant, Eocell is the voltage generated IF
the conditions WERE standard, ln is the natural logarithm, and Q is
the reaction quotient.
(To review logarithm and natural logarithm, read p. A4-A6 in the
textbook)
(The reaction quotient is defined as the concentrations of the product
ions, raised to their stoichiometric powers and multiplied together,
divided by the concentrations of the reactant ions, raised to their
stoichiometric powers and multiplied together. Any pure solids or
liquids are omitted from the reaction quotient. See Unit 13)
This equation allows us to calculate the cell
voltage at any concentration of reactants and
products and at any temperature. Commonly
it is only the concentrations of ions that have
changed, and at 298 K, the expression can be
simplified to
(Nernst eqx.)
E = Eocell - 0.0257 lnQ =
n
Eocell – 0.0592 V log Q
n
Example: Determine the measured emf
(electromotive force) for the zinc-silver cell in
the previous example if the concentration of
the zinc in the zinc half-cell is 1.10 M and the
concentration of the silver in the silver halfcell is 0.09 M.
Example: The Danielle cell composed of a zinc
half-cell and a copper half-cell has a measured
voltage of 1.25 V. The standard cell potential is
1.10 V. If the concentration of the copper halfcell is 0.60 M, what is the concentration of the
zinc half-cell?
As cell voltage (E) declines with reactants
converting to products, E eventually reaches
zero. Zero potential means the reaction is at
equilibrium.
If the cell is at equilibrium, Eocell = 0,
Q must equal K,
and then,
logK = n Eo
0.0592
Example:
If the reaction below is carried out using solutions that 5 M [Zn+2] and 0.3 M [Cu+2] at
298 K, what is the actual cell voltage?
Zn(s) + Cu2+(aq)  Cu(s) + Zn +2(aq)
First, work out the Eocell assuming standard conditions
Zn(s) │ Zn+2(aq) ║ Cu+2(aq) │ Cu(s)
Zn+2(aq) + 2e-  Zn(s)
Cu+2(aq) + 2e-  Cu(s)
Eo = -0.76 V
Eo = +0.34 V
Eocell = ER - EL = +0.34 - (-0.76) = 1.1 V
Then, calculate Q. Since zinc and copper metals are solids, they are omitted from Q.
Q = [Zn+2] = 5 = 16.7
[Cu+2] 0.3
Two electrons are transferred between the zinc and the copper, so n = 2. Plug
everything in
E = 1.1 V 0.0257 ln(16.7) = 1.06 V
2
Electrolysis
Electrolysis is the process in which electrical
energy is used to cause a non-spontaneous
REDOX reaction to occur. It is the opposite of
an electrochemical cell. An electrolytic cell is
one that is used to carry out electrolysis.
Anions are attracted towards the anode where
they undergo a process of oxidation. Electrons
flow from the anode to the cathode where
cations undergo a process of reduction.
Quantitative Aspects of Electrolysis
The amount of a substance produced in an electrolytic cell can
be calculated using Faraday’s Law.
Calculate the number of Faradays passed in the electrolysis by
using the expression
Number of Faradays = Current (in amps) x Time (in seconds)
96500
Then use the stoichiometry of the electrode process to
determine the mass of product deposited at the electrode.
Remembering that a process that produces one mole of a
product by the transfer of one electron will require one
Faraday to produce that one mole, a process that produces
one mole of product by the transfer of two electrons will
require two Faradays to produce that one mole, etc.
Current = charge/time = coulombs/seconds = amps
Problem: a car battery is listed as 675 “cold
cranking amps,” which means that it can
produce 675 amps for 30 seconds. If the halfreaction is
Pb(s) + SO42-  PbSO4 + 2eThen how much lead is consumed in the 30
seconds?
Problem: An aqueous copper (II) chloride
solution is electrolyzed for a period of 156
minutes using a current of 9.00 A. If inert
electrodes are used in the process, how many
grams of copper were removed from the
solution?
Problem: How long, in hours, does it take to
remove all of the polonium from an aqueous
PoCl4 solution that contains 1958 g of PoCl4
using a current of 6.80 A?
Example: Calculate the mass of bromine liquid
produced in the electrolysis of an aqueous
solution of sodium bromide if a current of
2.50 A is passed through the solution for 3.0
hr.
Example: A 1.00 M aqueous solution of
chromium (III) bromide will be electrolyzed to
“chrome plate” a faucet. How many hours will
it take to deposit 75.0 g of chromium metal
from the solution when a current of 3.00 A is
running through the solution?