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. Fe 3+ + V 2+ Oxidizing Reducing agent agent Fe 2+ + V 3+ (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 mole of electrons has a charge of (1.602 X 10 -19 C)(6.022 X 10 -19 C, so a 10 23 mol -1 ) = 9.649 X 10 4 C, 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, Δ
G
, 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 = -Δ
G
(14-4) Δ
G
= -work = -
E
·
q Relation between free-energy difference and electric potential difference:
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 q = n · F charge and moles:
Charge Moles C/mole (coulombs, C)
Relation between
Work =
E · q work and voltage:
Joules, J Volts, V Coulombs
Relation between free-energy difference
Δ
G = -nFE and electric potential difference:
Joules
Ohm’s law: I = E / R
Current Volts Resistance (A) (V) (ohms, Ω)
Electric power: P =
work/s =
E · I
Power J/s Volts Amperes (watts, W)
Box 13-1 Ohm’s Law, Conductance, and Molecular Wire
3
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-cells
9 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
) | CdCl 2 (
aq
) | AgCl(
s
) | Ag(
s
) The cell in Figure 14-6 is Cd(
s
) | Cd(NO 3 ) 2 (
aq
) || AgNO 3 (
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 10 6 Ω per student to 10 4 Ω 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, E
o
, for each half cell 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 = 1/2H 2 (
g
,
A
= 1) (14-11) We
arbitrarily
assign a potential of 0 to the standard hydrogen electrode at 25 o C. 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
) | H 2 (
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.
Cd 2+ + 2e = Cd(
s
) (14-12) S.H.E. || Cd + (
aq
,
A
= 1) | Cd(
s
) In this case, we observe a
negative
voltage of -0.402V.
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 (
E o
, which applies when all activities are unity) and a term showing the dependence on reagent concentrations.
Nernst Equation for a Half-Reaction
a
A +
n
e =
b
B
Nernst equation: E o R
= standard reduction potential (
A
A =
A
B = 1) = gas constant (8.314J/(K · mol) = 8.314 (V · C)/(K · mol))
T n F
= temperature (K) = number of electrons in the half-reaction = Faraday constant (9.649 A
i
= activity of species
i
X 10 4 C/mol)
The logarithmic term in the Nernst equation is the
reaction quotient,
Q
.
Q
=
A
B b /
A
A a (14-14)
Nernst equation at 25 o C:
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
E o
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
E o
.
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 E
o
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, E o
. 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
a b
=
b
log
a
:
Ag + + e = Ag(
s
) Box 13-2 shows that
neither E o nor E depends on how we write the reaction.
Box 13-3 Latimer Diagrams: How to find E
o
for a New Half Reaction
A
Latimer diagram
displays standard reduction potentials
E o
, connecting various oxidation state of an element.
11
IO 3 + 5H + +4e = HOI + 2H 2 O
E o
= +1.154 V IO 3 + 6H + + 6e = I + 3H 2 O Δ
G o
= -
nFE o When two reactions are added to give a third reaction, the sum of the individual
Δ
G o values must equal the overall value of
Δ
G o .
But, because Δ
G o
1 + Δ
G o
2 = Δ
G o
3 , we can solve for
E o
3 :
An Intuitive Way to Think About Cell Potentials
2
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 -
E o
+ = 0.222 V (14-17)
E +
=
E o
+ - 0.059 16 log[Cl ] = 0.222 – 0.059 16 log (0.033 4) = 0.309. V (14-18) Ag + + e = Ag(
s
)
E o
+ = 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
) | PbF 2 (
s
) | F (
aq
) || Cu 2+ (
aq
) | Cu(
s
) Right half-cell: Cu 2+ + 2e = Cu(
s
) Left half-cell: PbF 2 (
s
) + 2e = Pb(
s
) + 2F (14-20) Left half-cell Pb 2+ + 2e = Pb(
s
) (14-21)
The Nernst Equation Is Used in Measuring Standard Reduction Potentials
13-5 E
o
and the Equilibrium Constant
A galvanic cell produces electricity because the cell reaction is not at equilibrium.
Right electrode:
a
A +
n
e =
c
C
E o
+ Left electrode:
d
D +
n
e =
b
B
E o
-.
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 E o from K: Finding K from E o :
(at 25 o C) (at 25 o C)
Finding K for Net Reactions That Are Not Redox Reactions
FeCO 3 (
s
) + 2e Fe 2+ + 2e = Fe(
s
) + CO 3 2-
E o
+ = Fe(
s
)
E o
= -0.756 V = -0.44 V FeCO 3 (
s
) = Fe 2+ Iron(II) + CO 3 2 Carbonate
E o
= -0.756
K = K
sp = 10 (2)(-0.31
6 )/(0.059 16) = 10 – -11 (-0.44) = 0.31
6 V Half-reaction:
E o
+ Half-reaction :
E o
Net reaction:
E o
=
E o
+ -
E o
-
K
= 10
n
E ° /0.059 16
13-6 Cells as Chemical Probes
15 1.
E
quilibrium
between
the two half-cells
2.
E
quilibrium
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
) CH 3 CO 2 H = CH 3 CO 2 + H + AgCl(
s
) + e = Ag(
s
) + Cl (
aq
, 0.10 M)
E o
+ = 0.222 V 2H + (
aq
, ? M) + 2e = H 2 (
g
, 1.00 bar)
E o
= 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
E o
, 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 E
o
’
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
E o
’
(read “ E zero prime ” ).
Relation Between E
o
and E
o
’
a
A +
n
e =
b
B +
m
H +
E o
To find
E o
’ , 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 E o
’
:
All of this is called E o ’ when pH = 7
To convert [A] or [B] into F A or F B , 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.