Transcript Redox II - Electrochemistry

Galvanic (= voltaic) Cells
• Redox reactions which occur spontaneously
are called galvanic reactions.
• Zn will dissolve in a solution of copper(II)
sulfate to form zinc sulfate:
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
as the reaction starts
a little later
Cu(s)
+ ZnSO4
Galvanic Cells
To take the
energy from a
galvanic
reaction and get
useful work from
it requires
setting up a
galvanic (voltaic)
cell.
A salt bridge
connects the two
solutions of
electrolytes and
allows ions to move.
Electrodes are conductors used to establish
electrical contact with a nonmetallic part of a
circuit.
Galvanic Cells
Oxidation
occurs at
the anode.
In a galvanic
cell, electrons
flow
spontaneously
through the
external circuit
to the cathode.
Reduction
occurs at
the cathode.
Ion movement allows the
circuit to be completed.
Galvanic Cells
What happens if
the external
circuit is
removed?
What happens if
the porous
barrier is
removed?
The useful work from a galvanic cell comes from the electric current
flowing in the external circuit. This current can be used a source of
electric power (for example, a battery is a source of electric power).
Galvanic Cells
Can the external
circuit be
removed?
What happens if
the salt bridge
(electrolyte) is
removed?
If the water is
removed, there
are no means for
the ions to move
and corrosion
will stop.
Corrosion is a galvanic process. Removing one of
the essential components of a corrosion cell can
stop the corrosion process (usually a desirable
thing).
Galvanic Cells – Cell EMF
• Given the equation for a redox reaction, how do we
determine if it is galvanic (how do we know it will
proceed spontaneously)?
• There is a potential difference E (measured in volts)
between the two electrodes of a redox cell. What is a
volt? Moving a charge of +1 coulomb (C) through a
potential difference of 1 volt requires 1 joule of energy:
1V = 1J/C
• This potential difference is called the electromotive
force (emf) of a cell and is symbolized Ecell. It is also
called the cell potential.
• When the cell potential Ecell is positive, the
reaction will be spontaneous.
Ecell is an Intensive Property
1V = 1J/C
• This potential difference is called the electromotive
force (emf) of a cell and is symbolized Ecell. It is also
called the cell potential.
• Because the cell potential Ecell is a ratio of
energy to total charge, it is an intensive
property. (This is different from ΔG.)
Cell EMF
• When the cell emf (cell potential) Ecell is positive,
the reaction will be spontaneous.
• Rather than tabulate all of the possible redox
reactions and their emfs, we tabulate the emfs of
the half-cell reactions in terms of their reduction
potentials under standard conditions relative to the
standard hydrogen electrode (SHE). This means “is
defined to be.”
• SHE: 2H+(aq,1M) + 2e-  H2(g,1 bar) E°≡ 0V
• Standard conditions are for a given temperature
(often 25°C): all gases at 1 bar, aqueous species
have an activity of 1 (which is approximately 1 M).
Cell EMF
• The emfs of the half-cell reactions are
tabulated in terms of their reduction potentials
under standard conditions. These standard
conditions are denoted by the superscript: E°.
• If the table is ordered in terms of reduction
potential (increasing or decreasing), it is then
called an electrochemical series.
• The cell emf is calculated from the reduction
potentials of the two half-reactions of the cell.
Ecell = Ered(reduction half-reaction) – Ered(oxidation half-reaction)
Cell EMF
What is the emf when solid iron and the nitrate ion react
under standard conditions at 25°C to form the ferrous ion
and nitrogen monoxide gas?
E°(cell) = 0.96 – (-0.44) = 1.40V
unbalanced reaction:
Fe(s) + NO3-(aq)  Fe2+(aq) +
NO(g)
reduction half-rxn:
NO3-(aq) + 4H+(aq) + 3e- 
NO(g) + 2H2O(l)
E°(red half-rxn) = 0.96 V
oxidation half-rxn:
Fe(s)  Fe2+(aq) + 2efrom the table, we have
E°(Fe2+(aq) + 2e-  Fe(s))
= -0.44 V
Information from Reduction Potentials
Reduction potentials can be used to determine
• whether a reaction will proceed as written
 which leads you to the galvanic
reaction
• which of two species is more likely to be
reduced (or oxidized)
 which leads you to the anode halfreaction
 which leads you to the cathode halfreaction
• which of a group of species is the most
reactive (size of the reduction potential)
Reduction Potentials - Example
You are told to set up a galvanic cell using a strip of
iron, a strip of zinc, and 1M solutions of iron(II)
nitrate and zinc nitrate at 25°C.
1. Which one of the metals will oxidize?
Fe2+(aq) + 2e-  Fe(s)
Zn2+(aq) + 2e-  Zn(s)
E° = -0.44V
E° = -0.76V
A negative value for the standard reduction potential
means the reduction is NOT favored. Zn has the more
negative reduction potential, so it will oxidize.
Reduction Potentials - Example
You are told to set up a galvanic cell using a strip of
iron, a strip of zinc, and 1M solutions of iron(II)
nitrate and zinc nitrate at 25°C.
2. Determine what the two half-reactions will be.
Red: Fe2+(aq) + 2e-  Fe(s)
Ox: Zn(s)  Zn2+(aq) + 2e-
Reduction Potentials - Example
You are told to set up a galvanic cell using a strip of
iron, a strip of zinc, and 1M solutions of iron(II)
nitrate and zinc nitrate at 25°C.
3. Which half-reaction occurs at the anode?
Oxidation occurs at the anode:
Zn(s)  Zn2+(aq) + 2e-
Reduction Potentials - Example
You are told to set up a galvanic cell using a strip of
iron, a strip of zinc, and 1M solutions of iron(II)
nitrate and zinc nitrate at 25°C.
4. What is the oxidizing agent? What is the reducing
agent?
The oxidizing agent gets reduced: Fe2+(aq)
The reducing agent gets oxidized: Zn(s)
Reduction Potentials - Example
You are told to set up a galvanic cell using a strip of
iron, a strip of zinc, and 1M solutions of iron(II)
nitrate and zinc nitrate at 25°C.
5. What is the emf for the galvanic reaction under
standard conditions?
E°cell= -0.44 - (-0.76)=0.32V
Cell reaction: Fe2+(aq) + Zn(s) Zn2+(aq)+ Fe(s)
Reduction Potentials - Example
You are told to set up a galvanic cell using a strip of
iron, a strip of zinc, and 1M solutions of iron(II)
nitrate and zinc nitrate at 25°C.
6. Why does plating zinc onto steel protect the steel
from corrosion?
Corrosion is oxidation. Zinc has a greater tendency to
oxidize than iron, and will corrode instead of the steel
(which is mainly iron). Zinc acts as a sacrificial anode.
How does the
activity series
compare with the
table of reduction
potentials?
Where in the table
are the strongest
oxidizing agents?
Reducing agents?
Reduction Potentials - Example
Calculate the standard emf for the galvanic (voltaic) cell
in which acidified permanganate ion reacts with the
iron(II) ion to form the iron(III) ion and the manganese(II)
ion. Write the balanced equation for the reaction.
1. Write and balance the two half-reactions:
Fe2+(aq)  Fe3+(aq) + e8H+(aq) + MnO4-(aq) + 5e-  Mn2+(aq) +4H2O
oxidation
reduction
2. From the table of standard reduction potentials, we find
E°(Fe3+(aq) + e-  Fe2+(aq)) = 0.77V
E°(8H+(aq) + MnO4-(aq) + 5e-  Mn2+(aq) +4H2O) = 1.51V
3. E°cell = 1.51 – 0.77 = 0.74 V
4. The equation for the reaction is
5Fe2+(aq) + 8H+(aq) + MnO4-(aq)  5Fe3+(aq) + Mn2+(aq) +4H2O
Note that the equation stoichiometry does not influence the cell emf.
Gibbs Energy of an Electrochemical
Cell
The change in Gibbs Energy ΔG is the
maximum non-PV work* that can be obtained
from a chemical reaction at constant T and P:
ΔG = wmax
For an electrochemical cell, wmax = -nFE,
n = number of electrons in reaction
F = Faraday constant = 96500 J/V-mol
E = cell EMF
*or it is the minimum energy that must be
supplied to make the reaction happen.
Gibbs Energy of an Electrochemical
Cell
For an electrochemical cell, the maximum work
that can be obtained from a chemical reaction
at a temperature T is:
ΔG = -nFE
Under standard conditions, the relationship is
ΔG° = -nFE°
But we already know that
ΔG = ΔG° + RT ln Q
The Nernst Equation
The equation for Gibbs energy can be written in
terms of the emf of the electrochemical cell:
ΔG = ΔG° + RT ln Q
-nFE = -nFE° + RT ln Q
Dividing by nF gives the Nernst
E = E° - RT ln Q
nF
Equation:
The Nernst Equation
The Nernst equation lets us find the emf of an
electrochemical cell under nonstandard
conditions.
E = E° - RT ln Q
nF
At 25°C, the Nernst equation may be
note the change
written
from ln to log
E = E° - 0.0592 V log Q
n
The Nernst Equation - Example
E = E° - 0.0592 V log Q
n
What is the emf of the following cell at 298 K when [Ni2+] =
3.00 M and [Zn2+] = 0.100 M?
Zn(s) + Ni2+(aq)  Zn2+(aq) + Ni(s)
Q = [Zn2+]
[Ni2+]
= 0.0333 M
E = 0.48 - 0.0592 log .0333 = 0.48 + 0.0437 = 0.52V
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