Transcript Quantitative Chemical Analysis 7e

Chapter 13
Fundamentals of Electrochemistry
13-1 Basic Concepts
A redox reaction involves transfer of electrons from one species to
another. A species is said to be oxidized when it loses electrons. It is
reduced when it gains electrons.
An oxidizing agent, also called an oxidant, takes electrons from
another substance and becomes reduced. A reducing agent, also called
a reductant, gives electrons to another substance and is oxidized in the
process.
Fe3+ + V2+  Fe2+ + V3+
Oxidizing Reducing
agent
agent
(14-1)
Chemistry and Electricity
Electric Charge
Electric charge, q, is measured in coulombs (C).
The magnitude of the charge of a single electron is 1.602 X 10-19C, so a
mole of electrons has a charge of (1.602 X 10-19C)(6.022 X 1023mol-1) =
9.649 X 104C, which is called the Faraday constant, F.
Relation between
charge and moles:
Coulombs
mol e-
Coulombs
mol e-
Electric Current
The quantity of charge flowing each second through a circuit is called the
current. The unit of current is the ampere, abbreviated A.
In Figure 14-1, we encountered a
Pt electrode, which conducts
electrons into or out of a
chemical species in the redox
reaction.
Platinum is a common inert
electrode.
Voltage, Work, and Free Energy
The difference in electric potential, E, between two points is the work needed
(or that can be done) when moving an electric charge from one point to the
other. Potential difference is measured in volts (V).
Relation between
work and voltage:
Work
Joules
Volts
Coulombs
Work has the dimensions of energy, whose units are joules (J). One joule of
energy is gained or lost when 1 coulomb of charge moves between points whose
potentials differ by 1 volt.
The free-energy change, ΔG, for a chemical reaction conducted reversibly at
constant temperature and pressure equals the maximum possible electrical work
that can be done by the reaction on its surroundings:
Work done on surroundings
=
-ΔG
ΔG = -work = -E · q
Relation between free-energy difference
and electric potential difference:
(14-4)
Ohm’s Law
Ohm’s law states that current, I, is directly proportional to the potential
difference (voltage) across a circuit and inversely proportional to the resistance,
R, of the circuit.
Ohm’s law:
Units of resistance are ohms, assigned the Greek symbol Ω (omega).
Power
Power, P, is the work done per unit time. The SI unit of power is J/s, better known
as the watt (W).
P = work/s = (E · q)/s = E · (q/s)
(14-7)
Because q/s is the current, I, we can write
P=E · I
(14-8)
The energy appears as heat in the resistor.
Relation between
charge and moles:
Relation between
work and voltage:
q
=
n ·
Charge
Moles C/mole
(coulombs, C)
Work
=
Joules, J
E
Ohm’s law:
=
-nFE
Joules
Current
(A)
Power
(watts, W)
q
ΔG
I
P =
·
Volts, V Coulombs
Relation between free-energy difference
and electric potential difference:
Electric power:
F
= E
Volts
(V)
work/s =
J/s
/
R
Resistance
(ohms, Ω)
E ·
I
Volts Amperes
Box 13-1 Ohm’s Law, Conductance, and Molecular Wire3
13-2 Galvanic Cells
A galvanic cell (also called a voltaic cell) uses a spontaneous chemical reaction
to generate electricity.
A Cell in Action
Reduction:
Oxidation:
Net reaction:
The net reaction is composed of a reduction and an oxidation, each of which is
called a half-reaction.
Chemists define the electrode at which reduction occurs as the cathode. The
anode is the electrode at which oxidation occurs.
Salt Bridge
Cathode:
Anode:
Net reaction:
We can separate the reactants into two half-cells9 if we connect the two halves
with a salt bridge, as shown in figure 14-6.
Line Notation
| phase boundary
|| salt bridge
The cell in Figure 14-4 is represented by the line diagram
Cd(s) | CdCl2(aq) | AgCl(s) | Ag(s)
The cell in Figure 14-6 is
Cd(s) | Cd(NO3)2(aq) || AgNO3(aq) | Ag(s)
Demonstration 13-1 The Human Salt Bridge
A salt bridge is an ionic medium with a semipermeable barrier on each end.
Challenge One hundred eighty students at Virginia Tech made a salt bridge by
holding hands.8 Their resistance was lowered from 106Ω per student to 104Ω per
student by wetting everyone’s hands. Can your class beat this record?
13-3 Standard Potentials
To predict the voltage that will be observed when different half-cells are
connected to each other, the standard reduction potential, Eo, for each halfcell is measured by an experiment shown in an idealized form in Figure 14-7.
The half-reaction of interest in this diagram is
Ag+ + e- = Ag(s)
(14-10)
which occurs in the half-cell at the right connected to the positive terminal of the
potentiometer. Standard means that the activities of all species are unity.
The left half-cell, connected to the negative terminal of the potentiometer, is
called the standard hydrogen electrode (S.H.E.).
H+(aq, A = 1) + e- = 1/2H2(g, A = 1)
(14-11)
We arbitrarily assign a potential of 0 to the standard hydrogen electrode at
25oC. The voltage measured by the meter in Figure 14-7 can therefore be
assigned to Reaction 14-10, which occurs in the right half-cell.
We can arbitrarily assign a potential to Reaction 14-11 because it serves as a
reference point from which we can measure other half-cell potentials.
Pt(s) | H2 (g, A =1 ) | H+(aq, A = 1) || Ag+(aq, A = 1) | Ag(s)
or
S.H.E. || Ag+(aq, A = 1) | Ag(s)
By convention, the left-hand electrode (Pt) is attached to the negative
(reference) terminal of the potentiometer and the right-hand electrode is
attached to the positive terminal. A standard reduction potential is really a
potential difference between the potential of the reaction of interest and the
potential of S.H.E, which we have arbitrarily set to 0.
Cd2+ + 2e- = Cd(s)
S.H.E. || Cd+(aq, A = 1) | Cd(s)
In this case, we observe a negative voltage of -0.402V.
(14-12)
13-4 Nernst Equation
The net driving force for a reaction is expressed by the Nernst equation, whose
two terms include the driving force under standard conditions (Eo, which applies
when all activities are unity) and a term showing the dependence on reagent
concentrations.
Nernst Equation for a Half-Reaction
aA + ne- = bB
Nernst equation:
Eo = standard reduction potential (AA = AB = 1)
R = gas constant (8.314J/(K · mol) = 8.314 (V · C)/(K · mol))
T = temperature (K)
n = number of electrons in the half-reaction
F = Faraday constant (9.649 X 104 C/mol)
Ai = activity of species i
The logarithmic term in the Nernst equation is the reaction quotient, Q.
Q = ABb/AAa
Nernst equation at 25oC:
(14-14)
Nernst Equation for a Complete Reaction
Nernst equation for a complete cell:
The potential of each half-reaction (written as a reduction) is governed by a
Nernst equation like Equation 14-13, and the voltage for the complete reaction
is the difference between the two half-cell potential.
Step 1 Write reduction half-reactions for both half-cells and find Eo for each in
Appendix H. Multiply the half-reactions as necessary so that they both contain
the same number of electrons. When you multiply a reaction, you do not
multiply Eo.
Step 2 Write a Nernst equation for the fight half-cell, which is attached to the
positive terminal of the potentiometer. This is E+.
Step 3 Write a Nernst equation for the left half-cell, which is attached to the
negative terminal of the potentiometer. This is E-.
Step 4 Find the net cell voltage by subtraction: E = E+ - E-.
Step 5 To write a balanced net cell reaction, subtract the left half-reaction from
the fight half-reaction. (Subtraction is equivalent to reversing the left-half
reaction and adding.)
Box 13-2 Eo and the Cell Voltage Do Not Depend on How you
Write the Cell Reaction
Multiplying a half-reaction by any number does not change the standard
reduction potential, Eo. The potential difference between two points is the work
done per coulomb of charge carried through that potential difference (E =
work/q).
Multiplying a half-reaction by any number does not change the half-cell
potential, E.
The two expressions are equal because log ab = b log a:
Ag+ + e- = Ag(s)
Box 13-2 shows that neither Eo nor E depends on how we write the reaction.
Box 13-3 Latimer Diagrams: How to find Eo for a New HalfReaction
A Latimer diagram displays standard reduction potentials Eo, connecting
various oxidation state of an element.11
IO3- + 5H+ +4e- = HOI + 2H2O
Eo = +1.154 V
IO3- + 6H+ + 6e- = I- + 3H2O
ΔGo = -nFEo
When two reactions are added to give a third reaction, the sum of the individual
ΔGo values must equal the overall value of ΔGo.
But, because ΔGo1 + ΔGo2 = ΔGo3, we can solve for Eo3:
An Intuitive Way to Think About Cell Potentials2
Figure 14-8 and note that electrons always flow toward a more positive potential.
Different Descriptions of the Same Reaction
AgCl(s) + e- = Ag(s) + Cl-
Eo+ = 0.222 V
(14-17)
E+ = Eo+ - 0.059 16 log[Cl-] = 0.222 – 0.059 16 log (0.033 4) = 0.309. V
(14-18)
Ag+ + e- = Ag(s) Eo+ = 0.799 V
(14-19)
This description is just as valid as the previous one.
(for AgCl)
Advice for Finding Relevant Half-Reactions
To do this, look in the cell for an element in two oxidation states.
Pb(s) | PbF2(s) | F-(aq) || Cu2+(aq) | Cu(s)
Right half-cell:
Cu2+ + 2e- = Cu(s)
Left half-cell:
PbF2(s) + 2e- = Pb(s) + 2F-
(14-20)
Pb2+ + 2e- = Pb(s)
(14-21)
Left half-cell
The Nernst Equation Is Used in Measuring Standard
Reduction Potentials
13-5 Eo and the Equilibrium Constant
A galvanic cell produces electricity because the cell reaction is not at equilibrium.
Right electrode:
aA + ne- = cC Eo+
Left electrode:
dD + ne- = bB Eo-.
Box 14-4 Concentrations in the Operating Cell
Cell voltage is measured under conditions of negligible current flow.
The meter measures the voltage of the cell without affecting concentrations in
the cell.
Finding
Eo
from K:
Finding K from Eo:
(at 25oC)
(at 25oC)
Finding K for Net Reactions That Are Not Redox Reactions
FeCO3(s) + 2e- = Fe(s) + CO32Fe2+ + 2e- = Fe(s)
FeCO3(s) = Fe2+ + CO32-
Eo+ = -0.756 V
Eo- = -0.44 V
Eo = -0.756 – (-0.44) = 0.316 V
Iron(II)
Carbonate
K = Ksp = 10(2)(-0.316)/(0.059 16) = 10-11
Half-reaction:
Eo+
Half-reaction :
Eo-
Net reaction:
Eo = Eo+ - Eo-
-
K = 10nE°/0.059 16
13-6 Cells as Chemical Probes15
1.
Equilibrium between the two half-cells
2.
Equilibrium within each half-cell
We say that equilibrium between the two half-cells has not been established.
We allow half-cells to stand long enough to come to chemical equilibrium
within each half-cell.
AgCl(s) = Ag+(aq) + Cl-(aq)
CH3CO2H = CH3CO2- + H+
AgCl(s) + e- = Ag(s) + Cl-(aq, 0.10 M)
2H+(aq, ? M) + 2e- = H2(g, 1.00 bar)
Eo+ = 0.222 V
Eo- = 0
The measured voltage therefore allows us to find [H+] in the left half-cell:
The cell in Figure 14-9 acts as a probe to measure [H+] in the left half-cell.
Survival Tips
Step 1 Write the two half-reactions and their standard potentials.
If you choose a half-reaction for which you cannot find Eo, then
find another way to write the reaction.
Step 2 Write a Nernst equation for the net reaction and put in all
the known quantities. If all is well, there will be only one
unknown in the equation.
Step 3 Solve for the unknown concentration and use that
concentration to solve the chemical equilibrium problem that was
originally posed.
13-7 Biochemists Use Eo’
Whenever H+ appears in a redox reaction, or whenever reactants or
products are acids or bases, reduction potentials are pH dependent.
The standard potential for a redox reaction is defined for a galvanic
cell in which all activities are unity. The formal potential is the
reduction potential that applies under a specified set of conditions
(including pH, ionic strength, and concentration of complexing
agents). Biochemists call the formal potential at pH 7 Eo’ (read “E
zero prime”).
Relation Between Eo and Eo’
aA + ne- = bB + mH+
Eo
To find Eo’, we rearrange the Nernst equation to a form in
which the log term contains only the formal concentrations of
A and B raised to the powers a and b, respectively.
Recipe for Eo’:
All of this is called Eo’
when pH = 7
To convert [A] or [B] into FA or FB, we use fractional composition
equations (Section 10-5), which relate the formal (that is, total)
concentration of all forms of an acid or a base to its concentration
in a particular form:
Monoprotic system:
Diprotic system:
Curve a in Figure 14-11
shows how the calculated
formal potential for
Reaction 14-32 depends on
pH.