Transcript electrochem ppt

Lecture PowerPoint
Chemistry
The Molecular Nature of
Matter and Change
Sixth Edition
Martin S. Silberberg
21-1
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 21
Electrochemistry:
Chemical Change and Electrical Work
21-2
Electrochemistry:
Chemical Change and Electrical Work
21.1 Redox Reactions and Electrochemical Cells
21.2 Voltaic Cells: Using Spontaneous Reactions to
Generate Electrical Energy
21.3 Cell Potential: Output of a Voltaic Cell
21.4 Free Energy and Electrical Work
21.5 Electrochemical Processes in Batteries
21.6 Corrosion: An Environmental Voltaic Cell
21.7 Electrolytic Cells: Using Electrical Energy to Drive
Nonspontaneous Reactions
21-3
Overview of Redox Reactions
Oxidation is the loss of electrons and reduction is the
gain of electrons. These processes occur simultaneously.
Oxidation results in an increase in O.N. while reduction
results in a decrease in O.N.
The oxidizing agent takes electrons from the substance
being oxidized. The oxidizing agent is therefore reduced.
The reducing agent takes electrons from the substance
being oxidized. The reducing agent is therefore oxidized.
21-4
Figure 21.1 A summary of redox terminology, as applied to the
reaction of zinc with hydrogen ion.
0
Zn(s) +
21-5
+1
2H+(aq)
→
+2
Zn2+(aq)
0
+ H2(g)
Half-Reaction Method for
Balancing Redox Reactions
The half-reaction method divides a redox reaction into
its oxidation and reduction half-reactions.
- This reflects their physical separation in electrochemical cells.
This method does not require assigning O.N.s.
The half-reaction method is easier to apply to reactions in
acidic or basic solutions.
21-6
Steps in the Half-Reaction Method
• Divide the skeleton reaction into two half-reactions, each
of which contains the oxidized and reduced forms of one
of the species.
• Balance the atoms and charges in each half-reaction.
– First balance atoms other than O and H, then O, then H.
– Charge is balanced by adding electrons (e-) to the left side in
the reduction half-reaction and to the right side in the
oxidation half-reaction.
• If necessary, multiply one or both half-reactions by an
integer so that
– number of e- gained in reduction = number of e- lost in oxidation
• Add the balanced half-reactions, and include states of
matter.
21-7
Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s)
Step 1: Divide the reaction into half-reactions.
Cr2O72- → Cr3+
I→ I2
Step 2: Balance the atoms and charges in each half-reaction.
For the Cr2O72-/Cr3+ half-reaction:
Balance atoms other than O and H:
Cr2O72- → 2Cr3+
Balance O atoms by adding H2O molecules:
Cr2O72- → 2Cr3+ + 7H2O
21-8
Balance H atoms by adding H+ ions:
14H+ + Cr2O72- → 2Cr3+ + 7H2O
Balance charges by adding electrons:
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
This is the reduction half-reaction. Cr2O72- is reduced, and is the
oxidizing agent. The O.N. of Cr decreases from +6 to +3.
For the I-/I2 half-reaction:
Balance atoms other than O and H:
2I- → I2
There are no O or H atoms, so we balance charges by adding electrons:
2I- → I2 + 2eThis is the oxidation half-reaction. I- is oxidized, and is the reducing
agent. The O.N. of I increases from -1 to 0.
21-9
Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
The reduction half-reaction shows that 6e- are gained; the oxidation
half-reaction shows only 2e- being lost and must be multiplied by 3:
3(2I- → I2 + 2e-)
6I- → 3I2 + 6e-
Step 4: Add the half-reactions, canceling substances that appear on
both sides, and include states of matter. Electrons must always cancel.
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
6I- → 3I2 + 6e6I-(aq) + 14H+(aq) + Cr2O72-(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)
21-10
Balancing Redox Reactions in Basic Solution
An acidic solution contains H+ ions and H2O. We use H+
ions to balance H atoms.
A basic solution contains OH- ions and H2O. To balance
H atoms, we proceed as if in acidic solution, and then
add one OH- ion to both sides of the equation.
For every OH- ion and H+ ion that appear on the same
side of the equation we form an H2O molecule.
Excess H2O molecules are canceled in the final step,
when we cancel electrons and other common species.
21-11
Sample Problem 21.1
Balancing a Redox Reaction in Basic
Solution
PROBLEM: Permanganate ion reacts in basic solution with oxalate
ion to form carbonate ion and solid manganese dioxide.
Balance the skeleton ionic equation for the reaction
between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) [basic solution]
PLAN: We follow the numbered steps as described in the text, and
proceed through step 4 as if this reaction occurs in acidic
solution. Then we add the appropriate number of OH- ions
and cancel excess H2O molecules.
SOLUTION:
Step 1: Divide the reaction into half-reactions.
MnO4- → MnO2
C2O42- → CO32-
21-12
Sample Problem 21.1
Step 2: Balance the atoms and charges in each half-reaction.
Balance atoms other than O and H:
MnO4- → MnO2
C2O42- → 2CO32-
Balance O atoms by adding H2O molecules:
MnO4- → MnO2 + 2H2O
2H2O + C2O42- → 2CO32-
Balance H atoms by adding H+ ions:
4H+ + MnO4- → MnO2 + 2H2O
2H2O + C2O42- → 2CO32- + 4H+
Balance charges by adding electrons:
3e- + 4H+ + MnO4- → MnO2 + 2H2O 2H2O + C2O42- → 2CO32- + 4H+ + 2e[reduction]
[oxidation]
21-13
Sample Problem 21.1
Step 3: Multiply each half-reaction, if necessary, by an integer so that
the number of e- lost in the oxidation equals the number of e- gained
in the reduction.
x2
6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
x3
6H2O + 3C2O42- → 6CO32- + 12H+ + 6e-
Step 4: Add the half-reactions, canceling substances that appear on
both sides.
6e- + 8H+ + 2MnO4- → 2MnO2 + 4H2O
2 6H2O + 3C2O42- → 6CO32- +412H+ + 6e2MnO4- + 2H2O + 3C2O42- → 2MnO2 + 6CO32- + 4H+
21-14
Sample Problem 21.1
Basic. Add OH- to both sides of the equation to neutralize H+, and
cancel H2O.
2MnO4- + 2H2O + 3C2O42- + 4OH- → 2MnO2 + 6CO32- + [4H+ + 4OH-]
2MnO4- + 2H2O + 3C2O42- + 4OH- → 2MnO2 + 6CO32- + 2 4H2O
Including states of matter gives the final balanced equation:
2MnO4-(aq) + 3C2O42-(aq) + 4OH-(aq) → 2MnO2(s) + 6CO32-(aq) + 2H2O(l)
21-15
Electrochemical Cells
A voltaic cell uses a spontaneous redox reaction
(DG < 0) to generate electrical energy.
- The system does work on the surroundings.
A electrolytic cell uses electrical energy to drive a
nonspontaneous reaction (DG > 0).
- The surroundings do work on the system.
Both types of cell are constructed using two electrodes
placed in an electrolyte solution.
The anode is the electrode at which oxidation occurs.
The cathode is the electrode at which reduction occurs.
21-16
Figure 21.2 General characteristics of (A) voltaic and (B) electrolytic
cells.
21-17
Spontaneous Redox Reactions
A strip of zinc metal in a solution of Cu2+ ions will react
spontaneously:
Cu2+(aq) + 2e- → Cu(s)
Zn(s) → Zn2+(aq) + 2e-
[reduction]
[oxidation]
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)
Zn is oxidized, and loses electrons to Cu2+.
Although e- are being transferred, electrical energy is not
generated because the reacting substances are in the
same container.
21-18
Figure 21.3 The spontaneous reaction between zinc and
copper(II) ion.
21-19
Construction of a Voltaic Cell
Each half-reaction takes place in its own half-cell, so
that the reactions are physically separate.
Each half-cell consists of an electrode in an electrolyte
solution.
The half-cells are connected by the external circuit.
A salt bridge completes the electrical circuit.
21-20
Figure 21.4A
A voltaic cell based on the zinc-copper reaction.
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
21-21
Operation of the Voltaic Cell
Oxidation (loss of e-) occurs at the anode, which is
therefore the source of e-.
Zn(s) → Zn2+(aq) + 2eOver time, the Zn(s) anode decreases in mass and the
[Zn2+] in the electrolyte solution increases.
Reduction (gain of e-) occurs at the cathode, where the eare used up.
Cu2+(aq) + 2e- → Cu(s)
Over time, the [Cu2+] in this half-cell decreases and the
mass of the Cu(s) cathode increases.
21-22
Figure 21.4B
A voltaic cell based on the zinc-copper reaction.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s)
After several hours, the
Zn anode weighs less as
Zn is oxidized to Zn2+.
The Cu cathode gains
mass over time as Cu2+
ions are reduced to Cu.
21-23
Charges of the Electrodes
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
The anode produces e- by the oxidation of Zn(s). The
anode is the negative electrode in a voltaic cell.
Electrons flow through the external wire from the anode
to the cathode, where they are used to reduce Cu2+ ions.
The cathode is the positive electrode in a voltaic cell.
21-24
The Salt Bridge
The salt bridge completes the electrical circuit and allows
ions to flow through both half-cells.
As Zn is oxidized at the anode, Zn2+ ions are formed and
enter the solution.
Cu2+ ions leave solution to be reduced at the cathode.
The salt bridge maintains electrical neutrality by allowing
excess Zn2+ ions to enter from the anode, and excess
negative ions to enter from the cathode.
A salt bridge contains nonreacting cations and anions,
often K+ and NO3-, dissolved in a gel.
21-25
Flow of Charge in a Voltaic Cell
Electrons flow through the wire from anode to cathode.
Zn(s) → Zn2+(aq) + 2eZn2+
Cations move through the salt
bridge from the anode solution
to the cathode solution.
Cu2+(aq) + 2e- → Cu(s)
SO42Anions move through the salt
bridge from the cathode solution
to the anode solution.
By convention, a voltaic cell is shown with the anode on
the left and the cathode on the right.
21-26
Active and Inactive Electrodes
An active electrode is an active component in its halfcell and is a reactant or product in the overall reaction.
An inactive electrode provides a surface for the reaction
and completes the circuit. It does not participate actively
in the overall reaction.
- Inactive electrodes are necessary when none of the reaction
components can be used as an electrode.
Inactive electrodes are usually unreactive substances such
as graphite or platinum.
21-27
Figure 21.5
A voltaic cell using inactive electrodes.
Oxidation half-reaction
2I-(aq) → I2(s) + 2e-
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq) → 2Mn2+(aq) + 5I2(s) + 8H2O(l)
21-28
Notation for a Voltaic Cell
The components of each half-cell are written in the same
order as in their half-reactions.
The anode components
are written on the left.
The cathode components
are written on the right.
Zn(s)│Zn2+(aq)║Cu2+(aq) │Cu(s)
The single line shows a phase
boundary between the
components of a half-cell.
The double line shows that the halfcells are physically separated.
If needed, concentrations of dissolved components
are given in parentheses. (If not stated, it is assumed
that they are 1 M.)
21-29
Notation for a Voltaic Cell
graphite I-(aq)│I2(s)║MnO4-(aq), H+(aq), Mn2+(aq) │graphite
The inert electrode is specified.
A comma is used to show components
that are in the same phase.
21-30
Sample Problem 21.2
Describing a Voltaic Cell with Diagram
and Notation
PROBLEM: Draw a diagram, show balanced equations, and write the
notation for a voltaic cell that consists of one half-cell with
a Cr bar in a Cr(NO3)3 solution, another half-cell with an
Ag bar in an AgNO3 solution, and a KNO3 salt bridge.
Measurement indicates that the Cr electrode is negative
relative to the Ag electrode.
PLAN: From the given contents of the half-cells, we write the halfreactions. To determine which is the anode compartment
(oxidation) and which is the cathode (reduction), we note the
relative electrode charges. Electrons are released into the
anode during oxidation, so it has a negative charge. Since Cr
is negative, it must be the anode, and Ag is the cathode.
21-31
Sample Problem 21.2
SOLUTION:
The half-reactions are:
Ag+(aq) + e- → Ag(s)
[reduction; cathode]
Cr(s) → Cr3+(aq) + 3e- [oxidation; anode]
The balanced overall equation is:
3Ag+ + Cr(s) → 3Ag(s) + Cr3+(aq)
The cell notation is given by:
Cr(s)│Cr3+(aq)║Ag+(aq)│Ag(s)
The cell diagram shows the anode on
the left and the cathode on the right.
21-32
Electrical Potential and the Voltaic Cell
When the switch is closed and no reaction is occurring,
each half-cell is in an equilibrium state:
Zn(s)
Cu(s)
Zn2+(aq) + 2eCu2+(aq) + 2e-
(in Zn metal)
(in Cu metal)
Zn is a stronger reducing agent than Cu, so the position of
the Zn equilibrium lies farther to the right.
Zn has a higher electrical potential than Cu. When the
switch is closed, e- flow from Zn to Cu to equalize the
difference in electrical potential
The spontaneous reaction occurs as a result of the different
abilities of these metals to give up their electrons.
21-33
Cell Potential
A voltaic cell converts the DG of a spontaneous redox
reaction into the kinetic energy of electrons.
The cell potential (Ecell) of a voltaic cell depends on the
difference in electrical potential between the two
electrodes.
Cell potential is also called the voltage of the cell or the
electromotive forces (emf).
Ecell > 0 for a spontaneous process.
21-34
Table 21.1 Voltages of Some Voltaic Cells
Voltaic Cell
21-35
Voltage (V)
Common alkaline flashlight battery
1.5
Lead-acid car battery (6 cells ≈ 12 V)
2.1
Calculator battery (mercury)
1.3
Lithium-ion laptop battery
3.7
Electric eel (~5000 cells in 6-ft eel
= 750 V)
0.15
Nerve of giant squid
(across cell membrane)
0.070
Figure 21.6
Measuring the standard cell potential of a zinccopper cell.
The standard cell potential is designated E°cell and is
measured at a specified temperature with no current
flowing and all components in their standard states.
21-36
Standard Electrode Potentials
The standard electrode potential (E°half-cell) is the potential
of a given half-reaction when all components are in their
standard states.
By convention, all standard electrode potentials refer to
the half-reaction written as a reduction.
The standard cell potential depends on the difference
between the abilities of the two electrodes to act as
reducing agents.
E°cell = E°cathode (reduction) - E°anode (oxidation)
21-37
The Standard Hydrogen Electrode
Half-cell potentials are measured relative to a standard
reference half-cell.
The standard hydrogen electrode has a standard
electrode potential defined as zero (E°reference = 0.00 V).
This standard electrode consists of a Pt electrode with H2
gas at 1 atm bubbling through it. The Pt electrode is
immersed in 1 M strong acid.
2H+(aq; 1 M) + 2e-
21-38
H2(g; 1 atm)
E°ref = 0.00V
Figure 21.7
Determining an unknown E°half-cell with the standard
reference (hydrogen) electrode.
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e−
Reduction half-reaction
2H3O+(aq) + 2e- → H2(g) + 2H2O(l)
Overall (cell) reaction
Zn(s) + 2H3O+(aq) → Zn2+(aq) + H2(g) + 2H2O(l)
21-39
Sample Problem 21.3
Calculating an Unknown E°half-cell from E°cell
PROBLEM: A voltaic cell houses the reaction between aqueous bromine
and zinc metal:
Br2(aq) + Zn(s) → Zn2+(aq) + 2Br-(aq)
E°cell = 1.83 V.
Calculate E°bromine, given that E°zInc = -0.76 V
PLAN: E°cell is positive, so the reaction is spontaneous as
written. By dividing the reaction into half-reactions, we
see that Br2 is reduced and Zn is oxidized; thus, the zinc
half-cell contains the anode. We can use the equation for
E°cell to calculate E°bromine.
SOLUTION:
Br2(aq) + 2e- → 2Br-(aq)
Zn(s) → Zn2+(aq) + 2e-
21-40
[reduction; cathode]
[oxidation; anode]
E°zinc = -0.76 V
Sample Problem 21.3
E°cell = E°cathode − E°anode
1.83 = E°bromine – (-0.76)
1.83 + 0.76 = E°bromine
E°bromine = 1.07 V
21-41
Comparing E°half-cell values
Standard electrode potentials refer to the half-reaction as
a reduction.
E° values therefore reflect the ability of the reactant to act
as an oxidizing agent.
The more positive the E° value, the more readily the
reactant will act as an oxidizing agent.
The more negative the E° value, the more readily the
product will act as a reducing agent.
21-42
Table 21.2 Selected Standard Electrode Potentials (298 K)
21-43
F2(g) + 2e−
2F−(aq)
Cl2(g) + 2e−
2Cl−(aq)
MnO2(g) + 4H+(aq) + 2e−
Mn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3e−
NO(g) + 2H2O(l)
Ag+(aq) + e−
Ag(s)
Fe3+(g) + e−
Fe2+(aq)
O2(g) + 2H2O(l) + 4e−
4OH−(aq)
Cu2+(aq) + 2e−
Cu(s)
2H+(aq) + 2e−
H2(g)
N2(g) + 5H+(aq) + 4e−
N2H5+(aq)
Fe2+(aq) + 2e−
Fe(s)
2H2O(l) + 2e−
H2(g) + 2OH−(aq)
Na+(aq) + e−
Na(s)
Li+(aq) + e−
Li(s)
E°(V)
+2.87
+1.36
+1.23
+0.96
+0.80
+0.77
+0.40
+0.34
0.00
−0.23
−0.44
−0.83
−2.71
−3.05
strength of reducing agent
strength of oxidizing agent
Half-Reaction
Writing Spontaneous Redox Reactions
Each half-reaction contains both a reducing agent and an
oxidizing agent.
The stronger oxidizing and reducing agents react
spontaneously to form the weaker ones.
A spontaneous redox reaction (E°cell > 0) will occur
between an oxidizing agent and any reducing agent that
lies below it in the emf series (i.e., one that has a less
positive value for E°).
The oxidizing agent is the reactant from the half-reaction
with the more positive E°half-cell.
21-44
Using half-reactions to write a spontaneous redox reaction:
Sn2+(aq) + 2e- → Sn(s) E°tin = -0.14 V
Ag+(aq) + e- → Ag(s) E°silver = 0.80 V
Step 1: Reverse one of the half-reactions into an oxidation step
so that the difference between the E° values will be positive.
Here the Ag+/Ag half-reaction has the more positive E° value, so it
must be the reduction. This half-reaction remains as written.
We reverse the Sn2+/Sn half-reaction, but we do not reverse the sign:
Sn(s) → Sn2+(aq) + 2e- E°tin = -0.14 V
21-45
Step 2: Multiply the half-reactions if necessary so that the number
of e- lost is equal to the number or e- gained.
2Ag+(aq) + 2e- → 2Ag(s)
E°silver = 0.80 V
Note that we multiply the equation but not the value for E°.
Step 3: Add the reactions together, cancelling common species.
Calculate E°cell = E°cathode - E°anode.
Sn(s) → Sn2+(aq) + 2e2Ag+(aq) + 2e- → 2Ag(s)
E°tin = -0.14 V
E°silver = 0.80 V
Sn(s) + 2Ag+(aq) → 2Ag(s) + Sn2+(aq)
E°cell = 0.94 V
E°cell = E°silver – E°tin = 0.80 – (-0.14) = 0.94 V
21-46
Sample Problem 21.4
Writing Spontaneous Redox Reactions and
Ranking Oxidizing and Reducing Agents by
Strength
PROBLEM: (a) Combine the following three half-reactions into three
balanced equations for spontaneous reactions (A, B,
and C), and calculate E°cell for each.
(b) Rank the relative strengths of the oxidizing and reducing
agents.
(1) NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O(l) E° = 0.96 V
(2) N2(g) + 5H+(aq) + 4e- → N2H5+(aq)
E° = -0.23 V
(3) MnO2(s) +4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l) E° = 1.23 V
PLAN: To write the redox equations, we combine the possible
pairs of half-reactions. In each case the half-reaction with
the less positive value for E° will be reversed. We make elost equal to e- gained, add the half-reactions and calculate
E°cell. We can then rank the relative strengths of the
oxidizing and reducing agents by comparing E° values.
21-47
Sample Problem 21.4
SOLUTION: (a)
For (1) and (2), equation (2) has the smaller, less positive E° value:
(1) NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2H2O(l)
(2)
N2H5+(aq) → N2(g) + 5H+(aq) + 4e-
E° = 0.96 V
E° = -0.23 V
We multiply equation (1) by 4 and equation (2) by 3:
(1) 4NO3-(aq) + 16H+(aq) + 12e- → 4NO(g) + 8H2O(l) E° = 0.96 V
(2)
3N2H5+(aq) → 3N2(g) + 15H+(aq) + 12e- E° = -0.23 V
(A) 4NO3-(aq) + 3N2H5+(aq) + H+(aq) → 3N2(g) + 4NO(g) + 8H2O(l)
E°cell = 0.96 V – (-0.23 V) = 1.19 V
21-48
Sample Problem 21.4
For (1) and (3), equation (1) has the smaller, less positive E° value:
(1) NO(g) + 2H2O(l) → NO3-(aq) + 4H+(aq) + 3e-
E° = 0.96 V
(3) MnO2(s) +4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l) E° = 1.23 V
We multiply equation (1) by 2 and equation (3) by 3:
(1) 2NO(g) + 4H2O(l) → 2NO3-(aq) + 8H+(aq) + 6e-
E° = 0.96 V
(3) 3MnO2(s) +12H+(aq) + 6e- → 3Mn2+(aq) + 6H2O(l) E° = 1.23 V
(B) 3MnO2(s) + 4H+(aq) + 2NO(g) → 2Mn2+(aq) + 2NO3-(aq) + 2H2O(l)
E°cell = 1.23 V – (0.96 V) = 0.27 V
21-49
Sample Problem 21.4
For (2) and (3), equation (2) has the smaller, less positive E° value:
(2)
N2H5+(aq) → N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)
We multiply equation (3) by 2:
(2)
N2H5+(aq) → N2(g) + 5H+(aq) + 4e(3) 2MnO2(s) +8H+(aq) + 4e- → 2Mn2+(aq) + 4H2O(l)
E° = -0.23 V
E° = 1.23 V
E° = -0.23 V
E° = 1.23 V
(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq) → N2(g) + 2Mn2+(aq) + 4H2O(l)
E°cell = 1.23 V – (-0.23 V) = 1.46 V
21-50
Sample Problem 21.4:
(b) We first rank the oxidizing and reducing agents within each
equation, then we can compare E°cell values.
Equation (A)
Oxidizing agents: NO3- > N2
Reducing Agents: N2H5+ > NO
Equation (B)
Oxidizing agents: MnO2 > NO3Reducing Agents: N2H5+ > NO
Equation (C)
Oxidizing agents: MnO2 > N2
Reducing Agents: N2H5+ > Mn2+
Comparing the relative strengths from the E°cell values:
Oxidizing agents:
Reducing agents:
21-51
MnO2 > NO3- > N2
N2H5+ > NO > Mn2+
The Activity Series of the Metals
Metals that can displace H2 from acid are metals that
are stronger reducing agents than H2.
2H+(aq) + 2e- → H2(g)
Fe(s)
→ Fe2+ + 2e-
E° = 0.00V
E° = -0.44 V
Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g) E°cell = 0.44 V
The lower the metal is in the list of half-cell potentials,
the more positive its E°half-cell and the stronger it is as a
reducing agent.
The larger (more positive) the E°half-cell of a metal, the
more active a metal it is.
21-52
The Activity Series of the Metals
Metals that cannot displace H2 from acid are metals that
are weaker reducing agents than H2.
2H+(aq) + 2e- → H2(g)
2Ag(s)
→ 2Ag+ + 2e-
E° = 0.00V
E° = 0.80 V
2Ag(s) + 2H+(aq) → 2Ag+(aq) + H2(g) E°cell = -0.80 V
The higher the metal is in the list of half-cell potentials,
the smaller its E°half-cell and the weaker it is as a
reducing agent.
The smaller (more negative) the E°half-cell of a metal, the
less active a metal it is.
21-53
The Activity Series of the Metals
Metals that can displace H2 from water are metals
whose half-reactions lie below that of H2O:
2H2O(l) + 2e- → H2(g) + 2OH-(aq)
2Na(s)
→ 2Na+(aq) + 2e-
E° = -0.42 V
E° = -2.17 V
2Na(s) + 2H2O(l) → 2Na+(aq) + H2(g) + 2OH-(aq) E°cell = 2.29 V
21-54
The Activity Series of the Metals
We can also predict whether one metal can displace
another from solution. Any metal that is lower in the list
of electrode potentials (i.e., has a larger E° value) will
reduce the ion of a metal higher up the list.
Zn(s) → Zn2+(aq) + 2eFe2+(aq) + 2e- → Fe(s)
E° = -0.76V
E° = -0.44V
Zn (s) + Fe2+(aq) → Zn2+(aq) + Fe(s) E°cell = 0.32 V
21-55
Figure 21.8
The reaction of calcium in water.
Oxidation half-reaction
Ca(s) → Ca2+(aq) + 2e-
Reduction half-reaction
2H2O(l) + 2e- → H2(g) + 2OH-(aq)
Overall (cell) reaction
Ca(s) + 2H2O(l) → Ca2+(aq) + H2(g) + 2OH-(aq)
21-56
Figure 21.9
A dental “voltaic cell.”
Biting down with a filled tooth on a scrap of aluminum foil will cause
pain. The foil acts as an active anode (E°aluminum = -1.66 V), saliva as
the electrolyte, and the filling as an inactive cathode as O2 is reduced
to H2O.
21-57
Free Energy and Electrical Work
For a spontaneous redox reaction, DG < 0 and Ecell > 0.
n = mol of e- transferred
F is the Faraday constant
= 9.65x104 J/V·mol e-
DG = -nFEcell
Under standard conditions, DG° = -nFE°cell and
RT ln K
E°cell =
nF
or
0.0592 V log K
E°cell =
n
for T = 298.15 K
21-58
The interrelationship of DG°, E°cell, and K.
Figure 21.10
Reaction Parameters at the Standard State
DG°
E°cell
K
E°cell =
21-59
RT ln K
nF
DG°
K
E°cell
Reaction at standardstate conditions
<0
>1
>0
spontaneous
0
1
0
at equilibrium
>0
<1
<0
nonspontaneous
Sample Problem 21.5
Calculating K and DG° from E°cell
PROBLEM: Lead can displace silver from solution, and silver occurs in
trace amounts in some ores of lead.
Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s)
As a consequence, silver is a valuable byproduct in the
industrial extraction of lead from its ore. Calculate K and
DG° at 298.15 K for this reaction.
PLAN: We divide the spontaneous redox reaction into the half-reactions
and use values from Appendix D to calculate E°cell. From this we
can find K and DG°.
SOLUTION:
Writing the half-reactions with their E° values:
(1) Ag+(aq) + e- → Ag(s)
(2) Pb2+(aq) + 2e- → Pb(s)
21-60
E° = 0.80 V
E° = -0.13 V
Sample Problem 21.5
We need to reverse equation (2) and multiply equation (1) by 2:
(1) 2Ag+(aq) + 2e- → 2Ag(s)
E° = 0.80 V
(2) Pb(s)
→ Pb2+(aq) + 2e- E° = -0.13 V
2Ag+(aq) + Pb(s) → 2Ag(s) + Pb2+(aq) Ecell = 0.80 – (-0.13) = 0.93 V
E°cell =
RT ln K
0.0592 V log K = 0.93 V
=
nF
2
log K =
0.93 V x 2
= 31.42
0.0592 V
K = 2.6x1031
DG° = -nFE°cell = - 2 mol e x 96.5 kJ x 0.93 V
mol rxn
V·mol e-
= -1.8x102 kJ/mol rxn
21-61
Cell Potential and Concentration
Nernst Equation
Ecell = E°cell - RT ln Q
nF
• When Q < 1, [reactant] > [product], ln Q < 0, so Ecell > E°cell
• When Q = 1, [reactant] = [product], ln Q = 0, so Ecell = E°cell
• When Q > 1, [reactant] < [product], ln Q > 0, so Ecell < E°cell
We can simplify the equation as before for T = 298.15 K:
Ecell = E°cell - 0.0592 V log Q
n
21-62
Sample Problem 21.6
Using the Nernst Equation to Calculate Ecell
PROBLEM: In a test of a new reference electrode, a chemist constructs
a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+
half-cell under the following conditions:
[Zn2+] = 0.010 M
[H+] = 2.5 M
PH2 = 0.30 atm
Calculate Ecell at 298 K.
PLAN: To apply the Nernst equation and determine Ecell, we must
know E°cell and Q. We write the equation for the spontaneous
reaction and calculate E°cell from standard electrode
potentials. We must convert the given pressure to molarity in
order to have consistent units.
SOLUTION:
(1) 2H+(aq) + 2e- → H2(g)
(2) Zn(s)
→ Zn2+(aq) + 2e-
E° = 0.00 V
E° = -0.76 V
2H+(aq) + Zn(s) → H2(g) + Zn2+(aq) E°cell = 0.00 – (-0.76) = 0.76 V
21-63
Sample Problem 21.6
Converting pressure to molarity:
n
V
=
P
0.30 atm
=
= 1.2x10-2 M
RT
0.0821 atm·L x 298.15 K
mol·K
[H2][Zn2+] = 0.012 x 0.010
Q=
= 1.9x10-5
(2.5)2
[H+]2
Solving for Ecell at 25°C (298.15 K), with n = 2:
Ecell = E°cell - 0.0592 V log Q
n
= 1.10 V -
21-64
0.0592 V
log(1.9x10-5)
2
= 0.76 – (-0.14 V) = 0.90 V
Figure 21.11A
The relation between Ecell and log Q for the zinccopper cell.
If the reaction starts with [Zn2+] < [Cu2+] (Q < 1), Ecell is higher than the
standard cell potential.
As the reaction proceeds, [Zn2+] decreases and [Cu2+] increases, so
Ecell drops. Eventually the system reaches equilibrium and the cell can
no longer do work.
21-65
Figure 21.11B
The relation between Ecell and log Q for the zinccopper cell.
A summary of the changes in Ecell as any voltaic cell operates.
21-66
Concentration Cells
A concentration cell exploits the effect of concentration
changes on cell potential.
The cell has the same half-reaction in both cell
compartments, but with different concentrations of
electrolyte:
Cu(s) → Cu2+(aq; 0.10 M) + 2eCu2+(aq; 1.0 M) → Cu(s)
[anode; oxidation]
[cathode; reduction]
Cu2+(aq; 1.0 M) → Cu2+(aq; 0.10 M)
As long as the concentrations of the solutions are
different, the cell potential is > 0 and the cell can do work.
21-67
Figure 21.12 A concentration cell based on the Cu/Cu2+ half-reaction.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Oxidation half-reaction
Cu(s) → Cu2+(aq, 0.1 M) + 2eReduction half-reaction
Cu2+(aq, 1.0 M) + 2e- → Cu(s)
Overall (cell) reaction
Cu2+(aq,1.0 M) → Cu2+(aq, 0.1 M)
21-68
Ecell > 0 as long as the half-cell
concentrations are different.
The cell is no longer able to do work
once the concentrations are equal.
Sample Problem 21.7
Calculating the Potential of a Concentration
Cell
PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells. In
half-cell A, the electrolyte is 0.0100 M AgNO3; in half-cell
B, it is 4.0x10-4 M AgNO3. What is the cell potential at
298.15 K?
PLAN: The standard half-cell potentials are identical, so E°cell is
zero, and we find Ecell from the Nernst equation. Half-cell A
has a higher [Ag+], so Ag+ ions are reduced and plate out on
electrode A, which is therefore the cathode. In half-cell B, Ag
atoms of the electrode are oxidized and Ag+ ions enter the
solution. Electrode B is thus the anode. As for all voltaic cells,
the cathode is positive and the anode is negative.
21-69
Sample Problem 21.7
SOLUTION: The [Ag+] decreases in half-cell A and increases in halfcell B, so the spontaneous reaction is:
Ag+(aq; 0.010 M) [half-cell A] → Ag+(aq; 4.0x10-4 M) [half-cell B]
[Ag+]dil
0.0592
V
Ecell = E°cell log
[Ag+]conc
1
= 0.0 V -
21-70
4.0x10-4
0.0592 log
0.010
= 0.0828 V
Figure 21.13
Laboratory measurement of pH.
The operation of a pH meter illustrates an important application of
concentration cells. The glass electrode monitors the [H+] of the
solution relative to its own fixed internal [H+].
An older style of pH meter
includes two electrodes.
21-71
Modern pH meters use a
combination electrode.
Table 21.3 Some Ions Measured with Ion-Specific Electrodes
21-72
Species Detected
Typical Sample
NH3/NH4+
Industrial wastewater, seawater
CO2/HCO3-
Blood, groundwater
F-
Drinking water, urine, soil, industrial
stack gases
Br-
Grain, plant tissue
I-
Milk, pharmaceuticals
NO3-
Soil, fertilizer, drinking water
K+
Blood serum, soil, wine
H+
Laboratory solutions, soil, natural
waters
Figure 21.14
Minimocroanalysis.
A microelectrode records electrical impulses of a single neuron in a
monkey’s visual cortex. The electrical potential of a nerve cell is due to
the difference in concentration of [Na+] and [K+] ions inside and outside
the cell.
21-73
Electrochemical Processes in Batteries
A battery consists of self-contained voltaic cells arranged
in series, so their individual voltages are added.
A primary battery cannot be recharged. The battery is
“dead” when the cell reaction has reached equilibrium.
A secondary battery is rechargeable. Once it has run
down, electrical energy is supplied to reverse the cell
reaction and form more reactant.
21-74
Figure 21.15
Alkaline battery.
Anode (oxidation):
Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2eCathode (reduction): MnO2(s) + 2H2O(l) + 2e- → Mn(OH)2(s) + 2OH-(aq)
Overall (cell) reaction:
Zn(s) + MnO2(s) + H2O(l) → ZnO(s) + Mn(OH)2(s) Ecell = 1.5 V
21-75
Figure 21.16
Silver button battery.
Anode (oxidation):
Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2eCathode (reduction): Ag2O(s) + H2O(l) + 2e- → 2Ag(s) + 2OH-(aq)
Overall (cell) reaction:
Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)
Ecell = 1.6 V
The mercury battery uses HgO as the oxidizing agent instead of
Ag2O and has cell potential of 1.3 V.
21-76
Figure 21.17
Lithium battery.
The primary lithium battery is widely used
in watches, implanted medical devices,
and remote-control devices.
Anode (oxidation):
3.5Li(s) → 3.5Li+ + 3.5eCathode (reduction):
AgV2O5.5 + 3.5Li- + 3.5e- → Li3.5V2O5.5
Overall (cell) reaction:
AgV2O5.5 + 3.5Li(s) → Li3.5V2O5.5
21-77
Figure 21.18
Lead-acid battery.
The lead-acid car battery is a secondary battery and is rechargeable.
21-78
The reactions in a lead-acid battery:
The cell generates electrical energy when it discharges as a voltaic cell.
Anode (oxidation): Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2eCathode (reduction):
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- → PbSO4(s) + 2H2O(l)
Overall (cell) reaction (discharge):
PbO2(s) + Pb(s) + H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Ecell = 2.1 V
Overall (cell) reaction (recharge):
2PbSO4(s) + 2H2O(l) → PbO2(s) + Pb(s) + H2SO4(aq)
21-79
Figure 21.19
Nickel-metal hydride battery
Anode (oxidation):
MH(s) + OH-(aq) → M(s) + H2O(l) + eCathode (reduction): NiO(OH)(s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq)
Overall (cell) reaction:
MH(s) + NiO(OH)(s) → M(s) + Ni(OH)2(s) Ecell = 1.4 V
21-80
Figure 21.20
Lithium-ion battery.
Anode (oxidation):
LixC6(s) → xLi+ + xe- + C6(s)
Cathode (reduction):
Li1-xMn2O4(s) + xLi+ + xe- → LiMn2O4(s)
Overall (cell) reaction:
LixC6(s) + Li1-xMn2O4(s) → LiMn2O4(s)
Ecell = 3.7 V
The secondary (rechargeable) lithium-ion battery is used to power laptop
computers, cell phones, and camcorders.
21-81
Fuel Cells
In a fuel cell, also called a flow cell, reactants enter the
cell and products leave, generating electricity through
controlled combustion.
Reaction rates are lower in fuel cells than in other
batteries, so an electrocatalyst is used to decrease
the activation energy.
21-82
Figure 21.21
Hydrogen fuel cell.
Anode (oxidation):
2H2(g) → 4H+(aq) + 4eCathode (reduction): O2(g) + 4H+(aq) + 4e- → 2H2O(g)
Overall (cell) reaction:
2H2(g) + O2(g) → 2H2O(g) Ecell = 1.2 V
21-83
Corrosion: an Environmental Voltaic Cell
Corrosion is the process whereby metals are oxidized to
their oxides and sulfides.
The rusting of iron is a common form of corrosion.
- Rust is not a direct product of the reaction between Fe and
O2, but arises through a complex electrochemical process.
- Rusting requires moisture, and occurs more quickly at low pH, in
ionic solutions, and when the iron is in contact with a less active
metal.
21-84
The Rusting of Iron
The loss of iron:
Fe(s) → Fe2+(aq) + 2e- [anodic region; oxidation]
O2(g) + 4H+(aq) + 4e- → 2H2O(l)
[cathodic region; reduction]
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) [overall]
The rusting process:
Overall reaction:
2Fe2+(aq) + ½O2(g) + (2 + n)H2O(l) → Fe2O3·nH2O(s) + 4H+(aq)
H+ ions are consumed in the first step, so lowering the pH increases the
overall rate of the process. H+ ions act as a catalyst, since they are
regenerated in the second part of the process.
21-85
Figure 21.22
21-86
The corrosion of iron.
Figure 21.23
Enhanced corrosion at sea.
The high ion concentration of seawater enhances the corrosion of
iron in hulls and anchors.
21-87
Figure 21.24
The effect of metal-metal contact on the corrosion
of iron.
Fe in contact with Cu corrodes
faster.
21-88
Fe in contact with Zn does not
corrode. The process is known
as cathodic protection.
Figure 21.25 The use of sacrificial anodes to prevent iron corrosion.
In cathodic protection, an active metal, such as zinc, magnesium, or
aluminum, acts as the anode and is sacrificed instead of the iron.
21-89
Electrolytic Cells
An electrolytic cell uses electrical energy from an
external source to drive a nonspontaneous redox
reaction.
Cu(s) → Cu2+(aq) + 2eSn2+(aq) + 2e- → Sn(s)
Cu(s) + Sn2+(aq) → Cu2+(aq) + Sn(s)
[anode; oxidation]
[cathode; reduction]
E°cell = -0.48 V and ΔG° = 93 kJ
As with a voltaic cell, oxidation occurs at the anode and
reduction takes place at the cathode.
An external source supplies the cathode with electrons,
which is negative, and removes then from the anode,
which is positive. Electrons flow from cathode to anode.
21-90
Figure 21.26
The tin-copper reaction as the basis of a voltaic and
an electrolytic cell.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sn(s) → Sn2+(aq) + 2eCu2+(aq) + 2e- → Cu(s)
Cu(s) → Cu2+(aq) + 2eSn2+(aq) + 2e- → Sn(s)
Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq) Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq)
voltaic cell
21-91
electrolytic cell
Figure 21.27
The processes occurring during the discharge and
recharge of a lead-acid battery.
VOLTAIC (discharge)
Switch
ELECTROLYTIC (recharge)
21-92
Table 21.4 Comparison of Voltaic and Electrolytic Cells
Electrode
Cell Type
DG
Ecell
Name
Process
Sign
Voltaic
<0
>0
Anode
Oxidation
-
Voltaic
<0
>0
Cathode
Reduction
+
Electrolytic
>0
<0
Anode
Oxidation
+
Electrolytic
>0
<0
Cathode
Reduction
-
21-93
Products of Electrolysis
Electrolysis is the splitting (lysing) of a substance by the
input of electrical energy.
During electrolysis of a pure, molten salt, the cation will
be reduced and the anion will be oxidized.
During electrolysis of a mixture of molten salts
- the more easily oxidized species (stronger reducing agent) reacts at
the anode, and
- the more easily reduced species (stronger oxidizing agent) reacts at
the cathode.
21-94
Sample Problem 21.8
Predicting the Electrolysis Products of a
Molten Salt Mixture
PROBLEM: A chemical engineer melts a naturally occurring mixture of
NaBr and MgCl2 and decomposes it in an electrolytic cell.
Predict the substance formed at each electrode, and write
balanced half-reactions and the overall cell reaction.
PLAN: We need to determine which metal and nonmetal will form more
easily at the electrodes. We list the ions as oxidizing or reducing
agents.
If a metal holds its electrons more tightly than another, it has a
higher ionization energy (IE). Its cation will gain electrons more
easily, and it will be the stronger oxidizing agent.
If a nonmetal holds its electrons less tightly than another, it has a
lower electronegativity (EN). Its anion will lose electrons more
easily, and it will be the reducing agent.
21-95
Sample Problem 21.8
SOLUTION:
Possible oxidizing agents: Na+, Mg2+
Possible reducing agents: Br-, ClMg is to the right of Na in Period 3. IE increases from left to right across
the period, so Mg has the higher IE and gives up its electrons less
easily. The Mg2+ ion has a greater attraction for e- than the Na+ ion.
Mg2+(l) + 2e- → Mg(l)
[cathode; reduction]
Br is below Cl in Group 7A. EN decreases down the group, so Br
accepts e- less readily than Cl. The Br- ion will lose its e- more easily, so
it is more easily oxidized.
2Br-(l) → Br2(g) + 2eThe overall cell reaction is:
21-96
[anode; oxidation]
Mg2+(l) + 2Br-(l) → Mg(l) + Br2(g)
Figure 21.28
The electrolysis of water.
Overall (cell) reaction
2H2O(l) → H2(g) + O2(g)
Oxidation half-reaction
2H2O(l) → 4H+(aq) + O2(g) + 4e-
21-97
Reduction half-reaction
2H2O(l) + 4e- → 2H2(g) + 2OH-(aq)
Electrolysis of Aqueous Salt Solutions
When an aqueous salt solution is electrolyzed
- The strongest oxidizing agent (most positive electrode potential) is
reduced, and
- The strongest reducing agent (most negative electrode potential) is
oxidized.
Overvoltage is the additional voltage needed (above
that predicted by E° values) to produce gases at metal
electrodes.
Overvoltage needs to be taken into account when
predicting the products of electrolysis for aqueous
solutions.
Overvoltage is 0.4 – 0.6 V for H2(g) or O2(g).
21-98
Summary of the Electrolysis of Aqueous Salt Solutions
• Cations of less active metals (Au, Ag, Cu, Cr,
Pt, Cd) are reduced to the metal.
• Cations of more active metals are not reduced.
H2O is reduced instead.
• Anions that are oxidized, because of
overvoltage from O2 formation, include the
halides, except for F-.
• Anions that are not oxidized include F- and
common oxoanions. H2O is oxidized instead.
21-99
Sample Problem 21.9
Predicting the Electrolysis Products of
Aqueous Salt Solutions
PROBLEM: What products form at which electrode during electrolysis of
aqueous solution of the following salts?
(a) KBr (b) AgNO3 (c) MgSO4
PLAN: We identify the reacting ions and compare their electrode
potentials with those of water, taking the 0.4 – 0.6 V overvoltage
into account. The reduction half-reaction with the less negative
E° occurs at the cathode, while the oxidation half-reaction with
the less positive E° occurs at the anode.
SOLUTION:
(a) KBr
K+(aq) + e- → K(s)
E° = -2.93
2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V
Despite the overvoltage, which makes E for the reduction of water
between -0.8 and -1.0 V, H2O is still easier to reduce than K+, so
H2(g) forms at the cathode.
21-
Sample Problem 21.9
2Br-(aq) → Br2(l) + 2e2H2O(l) → O2(g) + 4H+(aq) + 4e-
E° = 1.07 V
E° = 0.82 V
The overvoltage makes E for the oxidation of water between 1.2
and 1.4 V. Br- is therefore easier to oxidize than water, so Br2(g)
forms at the anode.
(b) AgNO3
Ag+(aq) + e- → Ag(s)
E° = 0.80 V
2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V
As the cation of an inactive metal, Ag+ is a better oxidizing agent
than H2O, so Ag(s) forms at the cathode.
NO3- cannot be oxidized, because N is already in its highest (+5)
oxidation state. Thus O2(g) forms at the anode:
2H2O(l) → O2(g) + 4H+(aq) + 4e-
21-
Sample Problem 21.9
(c) MgSO4
Mg2+(aq) + 2e- → Mg(s)
E° = -2.37 V
2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V
Mg2+ is a much weaker oxidizing agent than H2O, so H2(g) forms at
the cathode.
SO42- cannot be oxidized, because S is already in its highest (+6)
oxidation state. Thus O2(g) forms at the anode:
2H2O(l) → O2(g) + 4H+(aq) + 4e-
21-
Stoichiometry of Electrolysis
Faraday’s law of electrolysis states that the amount of
substance produced at each electrode is directly
proportional to the quantity of charge flowing through
the cell.
The current flowing through the cell is the amount of
charge per unit time. Current is measured in amperes.
Current x time = charge
21-
Figure 21.29
A summary diagram for the stoichiometry of
electrolysis.
MASS (g)
of substance
oxidized or
reduced
CURRENT
(A)
M (g/mol)
AMOUNT (mol)
of substance
oxidized or
reduced
21-
balanced
half-reaction
time (s)
AMOUNT (mol)
of electrons
transferred
CHARGE
(C)
Faraday
constant
(C/mol e-)
Sample Problem 21.10 Applying the Relationship Among Current,
Time, and Amount of Substance
PROBLEM: A technician plates a faucet with 0.86 g of Cr metal by
electrolysis of aqueous Cr2(SO4)3. If 12.5 min is allowed for
the plating, what current is needed?
PLAN: To find the current, we divide charge by time, so we need to find
the charge. We write the half-reaction for Cr3+ reduction to get
the amount (mol) of e- transferred per mole of Cr. We convert
mass of Cr needed to amount (mol) of Cr. We can then use the
Faraday constant to find charge and current.
mass (g) of Cr needed
divide by M
mol of Cr
3 mol e- = 1 mol Cr
mol e- transferred
Charge (C)
1 mol e- = 9.65x104 C
21-
current (A)
divide by time in s
Sample Problem 21.10
SOLUTION:
Cr3+(aq) + 3e- → Cr(s)
0.86 g Cr x
1 mol Cr x 3 mol e- = 0.050 mol e52.00 g Cr
1 mol Cr
4
Charge (C) = 0.050 mol e- x 9.65x10 C = 4.8x103 C
1 mol e-
charge (C)
4.8x103 C x
=
Current (A) =
time (s)
12.5 min
21-
1 min
60 s
= 6.4 C/s = 6.4 A
Chemical Connections
Figure B21.1 The mitochondrion
Mitochondria are subcellular particles outside the cell nucleus that
control the electron-transport chain, an essential part of energy
production in living organisms.
21-
Chemical Connections
Figure B21.2 The main energy-yielding steps in the electrontransport chain (ETC).
21-
Chemical Connections
Figure B21.3 Coupling electron transport to proton transport to
ATP synthesis.
21-