Transcript Electrochemistry Power Point

Chapter 20
Electrochemistry
Electrochemistry
Electrochemistry is the branch of chemistry
that deals with the interconversion of
electrical energy and chemical energy.
Electrochemistry processes are redox
(Oxidation-Reduction) reactions in which
the energy released by a spontaneous
reaction is converted to electricity or in
which electrical energy is used to cause a
nonspontaneous reaction to occur.
Oxidation-Reduction Reactions Redox
Oxidation-Reduction Reactions are considered
electron-transfer reactions.
Also known as Redox Reactions.
Remember: LEO goes GER
Loss of Electrons means Oxidized & Gain of Electrons
means Reduced
If you say an element is Oxidized, then it is called a
Reducing agent because it donates electrons to
another element.
If you say an element is Reduced, then it is called an
Oxidizing agent because it accepts electrons from
another element.
Let’s take a look at the formation of calcium oxide (CaO)
from calcium and oxygen:
2Ca(s) + O2(g) -> 2CaO(s)
For convenience, we think of this process as 2 steps, one
involving the loss of electrons and one involving the gain of
electrons. This separate steps are called a half-reaction,
which explicitly shows the electrons involved in a redox
reaction.
Loss of electrons: 2Ca -> 2Ca2+ + 4eGain of electrons: O2 + 4e- -> 2O2The sum of the half-reactions gives the overall reaction:
2Ca + O2 + 4e- -> 2Ca2+ + 2O2- + 4eWe cancel the electrons that appear on both sides to get:
2Ca + O2 -> 2Ca2+ + 2O2Finally, the Ca2+ and the O2- ions combine to form CaO:
2Ca2+ + 2O2- -> 2CaO
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Rules for Assigning Oxidation States
1.
2.
3.
4.
5.
6.
7.
The oxidation state of an atom in an element is 0.
In a neutral molecule, the sum of the oxidation numbers of all the
atoms must equal 0.
The oxidation state of a monatomic ion is the same as its charge.
In its compound, fluorine is always assigned an oxidation sate of -1.
Oxygen is usually assigned an oxidation of -2 in its covalent
compounds, such as CO, CO2, SO2. Exceptions to this rule includes
peroxides (compounds containing the O22- group), where each oxygen
is assigned an oxidation state of -1, as in hydrogen peroxide (H2O2),
and OF2 in which oxygen is assigned a +2 oxidation state.
In its covalent compounds with nonmetals, hydrogen is assigned an
oxidation state of +1. For example HCl. When hydrogen is bonded to
a nonmetal in a binary compound, it is assigned an oxidation state of 1. For example LiH.
For an ion, the sum of the oxidation states must equal the charge of
the ion. For example, the sum of the oxidation states must equal -2 in
CO32-.
Half-Reaction Method for Balancing
Redox Reactions in Acidic Solutions
1. Write separate equations for the oxidation and reduction
half-reactions and show the oxidation state of all
elements.
2. For each half-reaction,
a. Balance all of the elements except hydrogen
and oxygen.
b. Balance oxygen using H2O.
c. Balance hydrogen using H+.
d. Balance the charge using electrons.
3. If necessary, multiply one or both balanced half-reactions
by an integer to equalize the number of electrons
transferred in the two half-reactions.
4. Add the half-reactions, and cancel identical species.
5. Check that the elements and charges are balanced.
Half-Reaction Method for Balancing
Redox Reactions in Basic Solutions
1.
2.
3.
4.
5.
6.
7.
Write separate equations for the oxidation and reduction half-reactions
and show the oxidation state of all elements.
For each half-reaction,
a. Balance all of the elements except hydrogen and oxygen.
b. Balance oxygen using H2O.
c. Balance hydrogen using H+.
d. Balance the charge using electrons.
If necessary, multiply one or both balanced half-reactions by an integer
to equalize the number of electrons transferred in the two half-reactions.
Add the half-reactions, and cancel identical species.
To both sides of the equation obtained, add a number of OH- ions that is
equal to the number of H+ ions. (We want to eliminate the H+ by
forming H2O.)
Form H2O on the side containing both H+ and OH- ions, and eliminate
the number of H2O molecules that appear on both sides of the equation.
Check that the elements and charges are balanced.
Redox Reactions
Cu(s) + 2NO3-(aq) + 4H+ -> 2NO2(g) + Cu2+(aq) + 2H2O(l)
2Ag+(aq) + Cu(s) -> 2Ag(s) + Cu2+ (aq)
2NO(g) + O2(g) -> 2NO2(g)
2Fe(s) + 3Cl2(g) -> 2FeCl3(s)
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Electrochemical Cells
If we place a piece of Zinc metal into a CuSO4 solution, Zn is
oxidized to Zn2+ ions while Cu2+ ions are reduced to metallic
copper:
Zn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s)
The electrons are transferred directly from the reducing agent (Zn) to
the oxidizing agent (Cu2+) in solution. However, if we physically
separate the oxidizing agent from the reducing agent, the transfer
of electrons can take place via an external conducting medium (a
metal wire). As the reaction progresses, it sets up a constant flow
of electrons and hence generates electricity (it produces electrical
work such as driving an electric motor).
An electrochemical cell is the experimental apparatus for
generating electricity through the use of a spontaneous redox
reaction. (An electrochemical cell is sometimes referred to as a
galvanic cell or voltaic cell, after the scientists Luigi Galvani and
Allessandro Volta, who constructed early versions of this device.)
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External
Wire that
electrons
travel on.
Anode – Where oxidation (loss of
electrons) occurs.
Cathode – Where reduction (gain of
electrons) occurs.
Salt Bridge – Contains inert electrolyte
that allows cations to move toward cathode
and anions toward anode so build up of (+)
and (–) charges in the solutions does not
occur and stop the reaction.
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This Particular Galvanic / Voltaic Cell is Known as the Daniell Cell
When the concentrations of the Cu2+ and Zn2+ ions are
both 1.0 M, we find that the voltage or emf of the Daniell
cell is 1.10 V at 25 oC
Galvanic/Voltaic Cells
An electric current flows from the anode to the cathode because there is a
difference in electrical potential energy between the electrodes. This is
similar to the flow of gas from a high-pressure region to a low-pressure
region.
Experimentally, the difference in electrical potential between the anode and
the cathode is measured by a voltmeter and the reading (in volts) is
called cell voltage. Two other terms, electromotive force or emf (E)
and cell potential are also used to denote cell voltage.
The conventional notation for representing electrochemical cells is the cell
diagram. For the previous cell we used, if we assume the concentrations
of Zn2+ and Cu2+ ions are 1 M, the cell diagram is:
Zn(s) Zn2+ (1 M)
Cu2+ (1 M) Cu(s)
The single vertical line represents a phase boundary. For example, the
zinc electrode is a solid and the Zn2+ ions (from ZnSO4) are in solution.
Thus we draw a line between Zn and Zn2+ to show the phase boundary.
The double vertical line denotes the salt bridge.
You must always write the anode to the left of the double lines, and the
cathode on the right of the double lines to show the movement of
electrons from anode to cathode.
Standard Reduction Potentials
When the concentrations of the Cu2+ and Zn2+ ions are both
1.0 M, we find that the voltage or emf of the Daniell cell is
1.10 V at 25 oC.
This voltage must be related directly to the redox reactions,
but how?
The measured emf of the cell is the sum of the electrical
potentials at the Zn and Cu electrodes. If we know one of
these electrode potentials, we could obtain the other from
subtraction (from 1.10 V).
It is impossible to measure the potential of just a single
electrode, but if we set the potential value of a particular
electrode to zero, we can use it to determine the relative
potentials of other electrodes.
The hydrogen electrode, shown in next slide, serves as the
reference for this purpose.
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The hydrogen electrode, shown in the previous slide, serves as the
reference for setting the potential value of a particular electrode
to zero so that we can determine the relative potentials of other
electrodes.
Hydrogen gas is bubbled into a hydrochloric acid solution at 25
oC. The platinum electrode has two functions. First, it
provides a surface on which the dissociation of hydrogen
molecules can take place:
H2 g 2H+ + 2eSecond, it serves as an electrical conductor to the external circuit.
Under standard state conditions (when the pressure of H2 is 1 atm
and the concentration of HCl solution is 1 M), the potential for
the reduction of H+ at 25 oC is taken to be exactly zero:
2H+ (1 M) + 2e- g H2 (1 atm)
Eo = 0 V
We can use hydrogen electrode potential = 0 to figure out the
other potentials of other kinds of electrodes.
We can use the hydrogen potential = 0 to figure out the
previous Daniell Cell voltage = 1.1 Volts, by breaking it
up into two separate parts. For the Daniell Cell:
Zn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s)
If we use hydrogen as the middle point, the cell diagram is:
Zn(s) Zn2+ (1 M)
H+ (1 M) H2(1 atm)
Pt (s)
(The platinum electrode provides the surface on which the
reduction can take place.)
The half reactions are:
Anode (oxidation):
Zn(s) g Zn2+ (1 M) + 2eCathode (reduction): 2H+ (1 M) + 2e- g H2 (1 atm)
Overall: Zn(s) + 2H+ (1 M) g Zn2+ (1 M) + H2 (1 atm)
The standard emf of any cell, Eocell, is:
Eocell = Eocathode - Eoanode
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The emf for this cell is =
0.76 V at 25 oC
The emf for this cell is =
0.34 V at 25 oC
The standard emf of any cell, Eocell, is:
Eocell = Eocathode – Eoanode
Therefore, for the first half reaction of Zn and H2:
Eocell = EoH+/H2 – EoZn2+/Zn
0.76 V = 0 - EoZn2+/Zn Therefore: (EoZn2+/Zn = -0.76 V)
We can also find the standard electrode potential for copper
with Hydrogen, where the cell diagram is:
Pt (s) H2(1 atm) H+ (1 M) Cu2+ (1 M) Cu(s)
The half reactions are:
Anode (oxidation):
H2 (1 atm) g 2H+ (1 M) + 2eCathode (reduction):
Cu2+ (1 M) + 2e- g Cu(s)
Overall: H2 (1 atm) + Cu2+ (1 M) g 2H+ (1 M) + Cu(s)
Therefore, for the first half reaction of Cu and H2:
Eocell = EoCu2+/Cu – EoH+/H2
0.34 V = EoCu2+/Cu – 0 Therefore: EoCu2+/Cu = 0.34 V
For the Daniell Cell (Cu and Zn) we can now write
that the combination of both will be:
Eocell = Eocathode – Eoanode
Eocell = EoCu2+/Cu - EoZn2+/Zn
Eocell = 0.34 V – (-0.76 V)
Eocell = 1.10 V
This is an example of how we use hydrogen as a
reference point to find the standard electrode
potential of any galvantic cell.
Free Energy and Chemical Equilibrium
(This is from Ch 19.7) If we start a reaction in solution
with all the reactants in their standard states, 1 M
concentrations, then as soon as the reaction starts, the
standard-state conditions will no longer exist.
Under these conditions, (not standard-state), we must use
DG rather than DGo. The relationship between the two
is:
DG = DGo + RT ln Q
(R is the gas law constant, 8.314 J / K x mol, T is the
absolute temperature of the reaction, and Q is the
reaction quotient derived from the equilibrium
expression.)
Free Energy and Chemical Equilibrium
Let’s look at 2 different types of cases for
DG = DGo + RT ln Q:
1) If DGo is a large (-) value, the RT ln Q term will not
become (+) enough to match the DGo term until a
significant amount of product has formed.
2) If DGo is a large (+) value, the RT ln Q term will be more
negative than DGo is (+) only as long as very little
product formation has occurred and the concentration of
the reactant is high relative to that of the product.
At equilibrium, by definition, DG = 0 and Q = K. Therefore,
we get:
0 = DGo + RT ln K
DGo = -RT ln K
Spontaneity of Redox Reactions
To relate cell potential, Eocell to thermodynamic quantities such as DGo
and K, we can change chemical energy into electrical energy by the
following equation:
DGo = -nFEocell
Where n = the moles of electrons that pass through the circuit, F =
faraday = 96,500 J / V x mol and Eocell is the cell potential and is
spontaneous when it is (+).
Remember that from previous slides:
DGo = -RT ln K
Where R is the gas law constant 8.314 J/K x mol, T is the temperature
in Kelvin and K is the equilibrium constant of the reaction.
Therefore:
nFEocell = RT ln K
For a spontaneous reaction, K > 1 , Eocell = (+) and DGo = (-)
For a reaction at equilibrium, K = 1 , Eocell = (0) and DGo = (0)
For a non-spontaneous reaction, K < 1 , Eocell = (-) and DGo = (+)
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The Effect of Concentration on Cell emf
So far we have been talking about redox reactions with reactants and
products in their standard states. But this is difficult to find. There
does exist a relationship between the emf of a cell and the
concentration of reactants and products when they are not in their
standard states.
The Nernst Equation
Consider a redox reaction:
aA + bB g cC + dD
DG = DGo + RT ln Q
Because DG = -nFE and DGo = -nFEo, we can get:
-nFE = -nFEo + RT ln Q
Which is the Nernst Equation: E = Eo - (RT/nF) ln Q
This can also be written as:
E = Eo - (0.0257 V/n) ln Q
Corrosion
Corrosion is the term usually applied to the deterioration of metals by
an electrochemical process.
The most familiar example is the formation of rust from iron. Oxygen
gas, water and acid in the atmosphere must be present for the
formation of rust. Iron serves as the anode:
Fe(s) g Fe2+(aq) + 2eThe electron given up by the
iron is used to reduce the
oxygen in the atmosphere
into water (along with the
presence of acid in the atmosphere).
O2(g) + 4H+(aq) + 4e- g 2H2O(l)
The overall redox rxn is:
2Fe(s) + O2(g) + 4H+(aq) g
2Fe2+(aq) + 2H2O(l)
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The Fe2+ ions formed at the anode are further oxidized by oxygen into Rust:
4Fe2+(aq) + O2(g) + (4+2x)H2O(l) g 2Fe2O3.xH2O(s) + 8H+(aq)
Other metals such
as copper and
silver also
corrode.
Electrolysis
In contrast to the spontaneous redox reactions, which result in the
coversion of chemical energy into electrical energy, electrolysis is
the process in which electrical energy is used to cause a
nonspontaneous chemical reaction to occur.
An electrolytic cell is an apparatus for carrying out electrolysis.
The same principles underlie electrolysis and the processes that
take place in electrochemical cells.
Electrolysis of Molten Sodium Chloride
In its molten state, sodium chloride, an ionic compound, can be
electrolyzed to form sodium metal and chlorine. The diagram on
the next slide is known as the Downs cell, which is used for largescale electrolysis of NaCl.
The electrolytic cell contains a pair of electrodes connected to the
battery. The battery serves as an “electron pump”, driving
electrons to the cathode, where reduction occurs, and withdrawing
electrons from the anode, where oxidation occurs.
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Theoretical estimates show that the Eo value for the overall process is about –4V, which
means that this is a nonspontaneous process. Therefore, a minimum of 4 V must be
supplied by the battery to carry out the reaction.
(a) Is a practical example of the electrolytic Downs cell – you can actually see the way
liquid Na and gaseous Cl2 is harnessed.
(b) Is a simplified example of the electrolytic Downs cell.
Electrolysis of Water
We all know that water does not spontaneously decompose into Hydrogen and
Oxygen gas under normal standard room conditions. This is because the
standard free-energy change for the reaction is large and positive.
But, this reaction can be induced in a cell like the one shown on the next slide.
This electrolytic cell consists of a pair of electrodes made of a nonreactive
metal, such as platinum, immersed in water. When the electrodes are
connected to the battery, nothing happens because there is not enough ions in
pure water to carry much of an electric current.
The reaction occurs readily in a 0.1 M H2SO4 solution because there are a
sufficient number of ions to conduct electricity. Immediately, gas bubbles
begin to appear at both electrodes.
The process at the anode is:
2H2O(l) g O2(g) + 4H+(aq) + 4eThe process at the cathode is:
H+(aq) + e- g ½ H2(g)
The overall reaction is given by:
Anode (oxidation): 2H2O(l) g O2(g) + 4H+(aq) + 4eCathode (reduction):
4(H+(aq) + e- g ½ H2(g))
.
Overall:
2H2O(l) g 2H2(g) + O2(g)
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Quantitative Aspect of Electrolysis
The quantitative treatment of electrolysis was developed primarily by Faraday. He
observed that the mass of product formed (or reactant consumed) at an electrode
is proportional to both the amount of electricity transferred at the electrode and
the molar mass of the substance in question.
For example, in the electrolysis of molten NaCl, the cathode reaction tells us that
one Na atom is produced when one Na+ ion accepts an electron from the
electrode. To reduce 1 mole of Na+ ions, we must supply Avogadro’s number of
electrons to the cathode. On the other hand, the stoichiometry of the anode
reaction shows that oxidation of two Cl- ions yields one chlorine molecule.
Therefore, the formation of 1 mole of Cl2 results in the transfer of 2 moles of
electrons from the Cl- ions to the anode.
In an electrolysis experiment, we generally measure the current (in amperes, A)
that passes through an electrolytic cell in a given period of time. The
relationship between charge (in coulombs, C) and current is:
1C = 1A x 1 second
A coulomb is the quantity of electrical charge passing any point in the circuit in 1
second when the current is 1 ampere.
Calculating the quantity of a substance produced in electrolysis
Example problem: Suppose a current of 0.452 A is passed through the cell for 1.50 hrs.
How much product will be formed at the anode and at the cathode in a molten CaCl2
electrolytic cell?
1. First step is to determine which species will be oxidized at the anode and which species
will be reduced at the cathode:
Anode (oxidation): 2Cl-(l) g Cl2(g) + 2eCathode (reduction): Ca2+(l) + 2e- gCa(l)
Overall: Ca2+(l) + 2Cl-(l) gCa(l) + Cl2(g)
2.
Find the charge (the number of electrons) that passes through the electrolytic cell.
The quantity of calcium metal and chlorine gas formed will depend on the charge =
current x time:
? C = 0.452 A x 1.50 hrs x 3600 s x 1 C = 2.44 x 103 C
1hr
1A x s
3.
The mass can be calculated from the Coulomb to electron to mole to gram connection.
1 mole of e-s = 96,500 C and 2 mole of e-s are required to reduce 1 mole of Ca2+ ions,
the mass of Ca metal formed would then be:
? g Ca = 2.44 x 103 C x 1 mol e-s x 1 mol Ca x 40.08 g Ca = 0.507 g Ca
96,500 C 2 mol e-s
1 mol Ca
? g Cl2 = 2.44 x 103 C x 1 mol e-s x 1 mol Cl2 x 70.90 g Cl2 = 0.896 g Cl2
96,500 C 2 mol e-s
1 mol Cl2