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Problems Chapter 21:
2,9,10,11,13,17,22,29,31,38,40,44,46,50,53,58,62,64,65,70,
72,73,82,85,87
Wonderful job with the presentations I was very impressed with
all of them.
Exam 3 March 17
Comprehensive Final Exam: March 24 7:30AM - 9:30 C114
This is an electrochemical cell, also called a voltaic
or galvanic cell. Chemists name parts
electrochemical cell names.
Electron
Flow e- ------->
Anode (-):
Oxidation
always
occurs here
-
Anode
Electrolyte
Salt bridge with non-reacting
but conductive salt to replace
lost electrons and reduced
cations
+
Cathode
Electrolyte
Cathode (+):
Reduction
always
occurs here.
Example Problem: Making a Voltaic Cell Diagram
Diagram, show balanced equations, and write the notation for a
voltaic cell that consists of one half-cell with a Cr solid electrode
in a Cr(NO3)3 solution, another half-cell with an Ag solid
electrode in an AgNO3 solution, connected by a KNO3 salt
bridge. What would the cell potential be under standard state
conditions?
−0.74 = Cr 3+ + 2 e− −−→ Cr(s)
Example Problem: Making a Voltaic Cell Diagram
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
oxidation half-reaction:
Cr(s)
Cr3+(aq) + 3e-
reduction half-reaction:
Ag+(aq) + e-
Ag(s)
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
Galvanic cells can also be made using “non-active
electrode”, or electrodes that don’t undergo reaction
directly in the cell.
The non-active
electrodes are
typically carbon
or platinum.
reduction half-reaction:
MnO4-(aq) + 8H+(aq) + 5eoxidation half-reaction:
2I-(aq)
I2(s) + 2e-
Mn2+(aq) + 4H2O(l)
overall (cell) reaction:
2MnO4-(aq) + 16H+(aq) + 10I-(aq)
2Mn2+(aq) + 5I2(s) + 8H2O(l)
We have and use standard cell notation for nonactive electrodes too as shown below.
graphite | I-(aq) | I2(s)
|| H (aq), MnO -(aq) | Mn
+
4
2+(aq)
| graphite
inert electrode
inert electrode
anode--oxidation occurs
cathode--reduction occurs
Salt Bridge
The electric potential difference (electromotive force
(emf)) is caused by differences in electron donating
tendencies of metals and ions. This tendency is
tabulated in a table of reduction potentials.
•
In order to tell what
predict what is
oxidized and what
species is reduced you
must refer to a table of
reduction potentials or
be given the reaction
that occurs.
The purpose of a galvanic cell is to convert
electrochemical potential energy to work! We need
to relate electric potential E in volts to energy as
they have different units.
1. The SI unit for electric
potential is the Volt and
for charge it is the
Coulomb.
By definition a 1 V
potential across two
electrodes can provide 1 J
of energy for each
Coulomb of charge that is
the moves between the
electrodes.
1 Joule
1 Volt =
Coulomb
The cell potential that is generated between a redox
pair (called E˚cell) depends on four main factors
which include:
(1) molarity of solutions
in each half-cell
(2) the metals and ions in
each half-cell (redox
potential)
(3) temperature of the
solution.
We must adopt a “yardstick” or a single half-cell from
which we measure every other half-cell. It’s called the
standard hydrogen electrode or SHE.
The electrochemical potential (or emf) can do work
just like a gravitational potential can do work. We
say the system moves from higher chemical
potential to lower chemical potential.
Strong oxidant + Strong reductant
Chemical Potential
Weak Oxidant + Weak Reductant
Reactants ==> Products
An analogy of “water pressure” to electrochemical
potential difference in a spontaneous redox system.
It is caused by the different abilities of metals to
give up electrons (strong to weak).
Zn(s) + Cu2+(aq)
Chemical
Potential
Zn2+(aq)
+ Cu(s)
Eoxidized
Ereduced
−0.74 = Cr 3+ + 2 e− −−→ Cr(s)
The Standard Hydrogen Electrode consists of an
inert Pt wire electrode placed in a tube containing 1M
HCl solution. H2 gas is bubbled through the tube into
the acid and connected to another arbitrary half-cell.
The reduction half-reaction
is defined as zero!
2e- + 2H+ (1 M)
H2 (1 atm)
E˚cell = 0 V
This electrode is our zero
point or “sea-level”
Standard cell potential with the SHE are all done under
“standard conditions” which means 1 M for electrolyte
concentrations, any gases are at 1 atm, temperature at
298K.
Standard
Conditions
--1 Molar
Electrolytes
--All gases (if any)
1 atmosphere
--Temperature =
298K = 25˚C
The other part of
the galvanic cell is
any metal and its
electrolyte that
we choose.
Suppose we assemble the following
electrochemical cell and ask what the cell
potential would be?
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Zn(s)
Zn2+ (1 M) + 2e E0oxidized = 0.76 V
2H+ (1 M) + 2e-
H2
E0reduced = 0.0 V
E0cell = 0.76 V
The Standard cell potential for all half-cells
combinations under standard conditions is given
by a simple equation shown below:
E˚cell =
E˚cell =
E˚red(cathode) standard reduction
potential of the
substance reduced
-
E˚red(anode)
standard reduction
potential of the
substance oxidized
Notice that both numbers will be the magnitude and sign
of given in the standard reduction table.
What is the standard emf of an electrochemical cell
made of a Cd electrode in a 1.0 M Cd(NO3)2 solution
and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e-
Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e-
Cr (s)
E0 = -0.74 V
What will be oxidized and what will be reduced?
How can we tell?
What is the standard emf of an electrochemical cell made
of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr
electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e-
Cd (s) E0 = -0.40 V Cd is the stronger oxidizer
Cr3+ (aq) + 3e-
Cr (s)
Anode (oxidation):
Cd will oxidize Cr
E0 = -0.74 V
Cr3+ (1 M) + 3e- x 2
Cr (s)
Cathode (reduction): 2e- + Cd2+ (1 M)
2Cr (s) + 3Cd2+ (1 M)
3Cd (s) + 2Cr3+ (1 M)
0
0 = E0
Ecell
cathode - Eanode
0 = -0.40 – (-0.74)
Ecell
0 = 0.34 V
Ecell
Cd (s) x 3