Transcript Electrochemistry

Electrochemistry
The Study of the Interchange of Chemical and
Electrical Energy
Galvanic or Voltaic Cells
The reactants and products of some
oxidation-reduction reactions can be physically
separated so that the electron transfer can only
occur via a wire.
The device or apparatus that is used to
convert chemical energy to electrical energy is
called galvanic (or voltaic) cell.
Galvanic or Voltaic Cells
As the electrical current passes through the
wire, it can be used to run a device, such as a
motor, light bulb, voltmeter, etc. As a result, the
electrochemical reaction can be used to provide
useful work.
Reaction of Zn with
2+
Cu
If a strip of zinc metal is immersed in a solution of
copper(II)sulfate, a reaction will spontaneously occur.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
As the reaction proceeds, some of the metallic zinc
dissolves into solution, and the blue copper(II) ion
plates out as elemental copper.
Reaction of Zn with
2+
Cu
As each zinc
atom dissolves,
it provides two
electrons to
the copper
ions.
Reaction of Zn with
2+
Cu
As the reaction proceeds, a thin black layer
of Cu is formed on the zinc surface. The blue
color of the Cu2+ ion fades as it is reduced.
Reaction of Zn with
2+
Cu
The transfer of electrons occurs directly on
the zinc surface. As a result, the movement of
electrons cannot be utilized.
Construction of a galvanic cell will allow the
electron transfer to occur via a wire. In this way,
the electrical current can be used to do work.
Galvanic Cells
In the galvanic cell for the reaction, the
oxidizing reagent (Cu2+) and the reducing agent
(Zn) are physically separated into two half-cells.
The overall net-ionic equation for the
reaction is:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Galvanic Cells
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
The half-reactions are:
Zn(s)  Zn2+(aq) + 2 e- (oxidation)
Cu2+(aq) + 2 e-  Cu(s) (reduction)
Galvanic Cells
In galvanic cells, the components of two
half-reactions are physically separated into two
beakers. The two beakers can then be
connected by a wire and a salt bridge so that the
electron transfer can occur.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
The half-reactions are:
Zn(s)  Zn2+(aq) + 2 e- (oxidation)
Cu2+(aq) + 2 e-  Cu(s) (reduction)
One half-cell will contain metallic Zn in a
solution of zinc ion, and the other half-cell will
contain metallic copper in a solution of copper
(II) ion.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn(s)  Zn2+(aq) + 2 e- (oxidation)
Oxidation takes place at the anode, so the
strip of zinc metal will serve as the anode. It will
be immersed in an aqueous solution of a zinc
salt, such as zinc sulfate. The sulfate ions are
inert, and are just spectator ions.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Cu2+(aq) + 2 e-  Cu(s) (reduction)
Reduction takes place at the cathode, so the
copper strip will serve as the cathode. It will be
immersed in an aqueous solution of a copper
(II) salt, such as copper(II)sulfate. The sulfate
ion is inert, and will serve as a spectator.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
The Salt Bridge
The salt bridge, porous cup, or glass frit
allows the flow of ions. This is necessary in
order to maintain a neutral charge in each halfcell.
The Salt Bridge
The salt bridge, porous cup, or glass frit
allows the flow of ions.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Each
combination of
half-cells
produces a
characteristic
voltage.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Oxidation at the Anode
Reduction at the Cathode
Voltage or EMF
The voltage of the galvanic cell is also called
the cell potential, or the electromotive force ( emf). It
is related to the driving force of the reaction.
The units of cell potential are volts (V). A
volt is exactly 1 joule of work per coulomb of
charge transferred.
Cell Potential
There can be no reduction without
oxidation, and thus, each galvanic cell needs two
half-cells in order to produce a voltage.
Scientists have devised a system to measure
the potential (voltage) of any half-cell relative to
a standard half cell. The potential of the
standard half-cell is set at 0.00 volts.
The Hydrogen Half-Cell
The half-cell
consists of an
inert platinum
electrode
immersed in 1M
strong acid.
Hydrogen gas is
bubbled over the
electrode.
Pt electrode
Standard Reduction Potentials
Each half-cell is connected to a standard
hydrogen electrode. All cells contain solutions
which are 1.00M, and all gases are at a pressure
of 1.00 atmospheres.
Since the voltage of the hydrogen electrode
is set at zero, the voltage of the galvanic cell
represents the assigned voltage of the other halfcell.
Standard Reduction Potentials
The emf of the
standard Zn halfcell is 0.76 volts
relative to the
hydrogen halfcell. In this cell,
Zn is being
oxidized, and H+
reduced.
Standard Reduction Potentials
The half-reactions are:
Zn(s)  Zn2+(aq) + 2e2H+(aq) + 2e-  H2(g)
Since the potential of the hydrogen reaction
is set at zero, the potential for the oxidation of
zinc is the measured value of 0.76 volts.
Standard Reduction Potentials
All half-reactions are tabulated as reductions.
Since the potential for the oxidation of zinc is
+0.76 volts, the reduction potential is -.76 volts.
Zn2+(aq) + 2e- Zn(s) Eo = -0.76V
Standard Reduction Potentials
In this way, the reduction potentials for all
half-cells are obtained relative to the hydrogen
electrode. The values of reduction potentials are
listed from highest potential to lowest.
Cell Potentials
When a galvanic cell is constructed, one halfreaction is a reduction reaction, and the other is
an oxidation reaction.
When a reduction half-reaction is reversed to
make it an oxidation reaction, the sign on its cell
potential is reversed.
Cell Potentials
The reaction which will occur spontaneously is
the oxidation and reduction that produces the
most positive cell potential.
Cell Potentials - Problem
 Determine the balanced reaction and the
standard cell potential for a galvanic cell with the
following half-cells:
 Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
H2O2 + 2H+ + 2 e-  2 H2O
Cell Potentials - Problem
 Determine the balanced reaction and the
standard cell potential for a galvanic cell with the
following half-cells:
 Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
H2O2 + 2H+ + 2 e-  2 H2O
1. Look up the standard reduction potentials for
both half-reactions
Cell Potentials - Problem

Determine the balanced reaction and the
standard cell potential for a galvanic cell with
the following half-cells:

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
H2O2 + 2H+ + 2 e-  2 H2O
1.33V
1.78V
2. Reverse one half-reaction so that the net cell
potential is the largest positive number.
Cell Potentials - Problem

Determine the balanced reaction and the
standard cell potential for a galvanic cell with
the following half-cells:
reverse

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O
H2O2 + 2H+ + 2 e-  2 H2O
1.33V
1.78V
2. Reverse one half-reaction so that the net cell
potential is the largest positive number.
Cell Potentials - Problem

Determine the balanced reaction and the
standard cell potential for a galvanic cell with
the following half-cells:

2 Cr3+ + 7 H2O  Cr2O72- + 14 H+ + 6 e- -1.33V
H2O2 + 2H+ + 2 e-  2 H2O
1.78V
3. Multiply half-reactions so that the electrons
lost = electrons gained. Reduction
potentials are not multiplied.
Cell Potentials - Problem


Determine the balanced reaction and the standard cell
potential for a galvanic cell with the following halfcells:
2 Cr3+ + 7 H2O  Cr2O72- + 14 H+ + 6 e- -1.33V
(H2O2 + 2H+ + 2 e-  2 H2O )3
1.78V
3. Multiply half-reactions so that the electrons
lost = electrons gained. Reduction potentials are
not multiplied.
Cell Potentials - Problem

Determine the balanced reaction and the
standard cell potential for a galvanic cell with
the following half-cells:

2 Cr3+ + 7 H2O  Cr2O72- + 14 H+ + 6 e- -1.33V
3H2O2 + 6H+ + 6 e-  6 H2O
1.78V
4. Add the two half-reactions and their cell
potentials.
Cell Potentials - Problem


Determine the balanced reaction and the standard cell
potential for a galvanic cell with the following halfcells:
2
1
8 +
3+
22 Cr + 7 H2O  Cr2O7 + 14 H + 6 e- -1.33V
3H2O2 + 6H+ + 6 e-  6 H2O
1.78V
3H2O2 + 2 Cr3+ + 1 H2O  Cr2O72- + 8 H+
Eo=0.45V
Line Notation
There is a system of notation, called line
notation, used to describe a galvanic cell. For the
cell pictured below:
The anode is written
on the left, the
cathode on the right.
Line Notation
The anode is
written on the left, the
cathode on the right.
A single vertical
line represents a phase
boundary, and a pair
of vertical lines
indicate a salt bridge or
porous disk.
Line Notation
The notation for this cell is:
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)
Line Notation
Note that
Zn is
written on
the left
because it is
the anode.
The notation for this cell is:
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)
Line Notation
Note that the
standard
hydrogen
electrode is
written on the
right because it
is the cathode.
The notation for this cell is:
Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s)
Cell Potential and Free Energy
An electrochemical cell produces a voltage as
a result of the driving force for electron transfer.
As a result, cell potentials are directly related to
∆G for the reaction.
Cell Potential and Free Energy
For standard conditions,
∆Go = -nFEo
where n is the moles of electrons transferred,
and F = 96,485 coulumbs/mol e- (Faraday’s
constant), and a volt equals 1 joule/coulomb.
Cell Potential and Free Energy
For standard conditions,
∆Go = -nFEo
A positive cell potential (spontaneous
reaction) yields a negative value for ∆G.
Cell Potential and Concentration
Standard conditions dictate that all solutions
be 1M, and all gases have a pressure of 1 atm.
Cell potential will vary with concentration.
As a galvanic cell produces voltage, the
concentrations in each half-cell change, and the
voltage gradually decreases to zero.
Cell Potential and Concentration
The cell potential of a non-standard cell can
be calculated using the Nernst equation:
E = Eo-RT ln(Q)
n
where T is temperature in Kelvins and
Q is the reaction quotient.
Cell Potential and Concentration
For standard temperature (25oC), the
equation becomes:
E = Eo-0.0591 log(Q)
n
where n is the number of moles of electrons
transferred.
Cell Potential and K
At equilibrium, the cell potential is 0.0 volts,
and Q=K. Using the Nernst Equation at 25oC:
E = Eo-0.0591 log(Q)
n
0 = Eo- 0.0591 log(K)
n
log K = nEo/0.0591
Cell Potential and K
log K = nEo/0.0591
Since redox reactions often have very large
equilibrium constants, measuring cell potential is
often the only way to obtain the value of K.
Problem

Write the chemical reaction and calculate the
equilibrium constant for the following galvanic
cell under standard conditions.
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
1. Write the half-reactions.
Since the anode is on the left, copper(I) is
oxidized to form copper(II). Gold(III) ion is
reduced to elemental gold at the cathode.
Problem

Write the chemical reaction and calculate the
equilibrium constant for the following galvanic
cell under standard conditions.
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
1. Write the half-reactions.
Cu1+(aq)  Cu2+(aq) + 1 eAu3+(aq) + 3 e-  Au(s)
Problem
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
2. Look up the reduction potentials.
Cu1+(aq)  Cu2+(aq) + 1 e- Eo = -(0.16V)
Au3+(aq) + 3 e-  Au(s)
Eo = 1.50V
Problem
Pt(s) | Cu1+(aq), Cu2+(aq) || Au3+(aq) | Au(s)
3. Combine the reactions.
3[Cu1+(aq)  Cu2+(aq) + 1 e-] Eo = -(0.16V)
Au3+(aq) + 3 e-  Au(s)
Eo = 1.50V
3 Cu1+(aq) + Au3+(aq)  3 Cu2+(aq) + Au(s)
Eo = 1.34V
Problem
3 Cu1+(aq) + Au3+(aq)  3 Cu2+(aq) + Au(s)
Eo = 1.34V
4. Calculate the value of K using Eo.
log K = nEo/0.0591
log K = (3) (1.34)/.0591 = 68.0
K = 1 x 1068
Electrolysis
An electrolytic cell uses
electricity to produce a
non-spontaneous
chemical reaction.
Examples include the
electrolysis of water to
produce hydrogen and
oxygen.
Electrolysis
Electrolysis is also used in electroplating, in
which a metal such as silver is formed on the
surface of a less expensive metal.
Typically, an electrical current is passed
through a solution of the ion to be deposited.
The current is expressed in amperes (amp),
indicated with the symbol A.
Electrolysis
1 amp = 1 coulomb/second
If the current (in A) is multiplied by the time
the current flows (in s), the total number of
coulombs of charge is obtained.
Coulombs of charge = amps x seconds
Electrolysis
Coulombs of charge = amps x seconds
Coulombs can be converted to moles of an
element deposited as follows:
coulombs  moles e-  moles of element
96,485 C/mol e-
divide by n
Problem: Electrolysis

How many grams of copper will be plated out if
a current of 5.00 amps is passed through a Cu2+
solution for an hour. Assume an excess of Cu2+.