Transcript Document

Chapter 21
Electrochemistry:
Chemical Change and Electrical Work
Key Points About Redox Reactions
•Oxidation (electron loss) always accompanies reduction
(electron gain).
•The oxidizing agent is reduced, and the reducing agent is
oxidized.
•The number of electrons gained by the oxidizing agent
always equals the number lost by the reducing agent.
A summary of redox terminology
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
OXIDATION
One reactant loses electrons.
Zn loses electrons.
Reducing agent is oxidized.
Zn is the reducing
agent and becomes
oxidized.
Oxidation number increases.
The oxidation number
of Zn increases from x
to +2.
REDUCTION
Other reactant gains
electrons.
Oxidizing agent is reduced.
Hydrogen ion gains
electrons.
Hydrogen ion is the oxidizing agent
and becomes reduced.
Oxidation number decreases. The oxidation number of H
decreases from +1 to 0.
Half-Reaction Method for Balancing Redox Reactions
Summary: This method divides the overall redox reaction into
oxidation and reduction half-reactions.
•Each reaction is balanced for mass (atoms) and charge.
•One or both are multiplied by some integer to make the number of
electrons gained and lost equal.
•The half-reactions are then recombined to give the balanced redox
equation.
•The separation of half-reactions reflects actual physical
separations in electrochemical cells.
Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq)
Cr3+(aq) + I2(aq)
1. Divide the reaction into half-reactions Determine the O.N.s for the species undergoing redox.
+6
-1
2Cr2O7 (aq) + I-(aq)
Cr2O72I-
Cr3+
I2
+3
0
3+
Cr (aq) + I2(aq)
Cr is going from +6 to +3
I is going from -1 to 0
2. Balance atoms and charges in each half-reaction -
14H+(aq) + Cr2O72net: +12
6e- + 14H+(aq) + Cr2O72-
2 Cr3+
net: +6
2 Cr3+
+ 7H2O(l)
Add 6e- to left.
+ 7H2O(l)
Balancing Redox Reactions in Acidic Solution
6e- + 14H+(aq) + Cr2O722 I-
2 Cr3+
I2
continued
+ 7H2O(l)
+ 2e-
Cr(+6) is the oxidizing agent and I(-1) is the reducing agent.
3. Multiply each half-reaction by an integer, if necessary 2 I-
I2 + 2e-
X3
4. Add the half-reactions together 6e- + 14H+ + Cr2O726 I14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
2 Cr3+
+ 7H2O(l)
3 I2 + 6e2Cr3+(aq) + 3I2(s) + 7H2O(l)
Do a final check on atoms and charges.
Balancing Redox Reactions in Basic Solution
Balance the reaction in acid and then add OH- so as to neutralize
the H+ ions.
14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
2Cr3+(aq) + 3I2(s) + 7H2O(l)
+ 14OH-(aq)
14H2O + Cr2O72- + 6 I-
+ 14OH-(aq)
2Cr3+ + 3I2 + 7H2O + 14OH-
Reconcile the number of water molecules.
7H2O + Cr2O72- + 6 I-
2Cr3+ + 3I2 + 14OH-
Do a final check on atoms and charges.
The redox reaction between dichromate ion and iodide ion.
Cr2O72-
I-
Cr3+ + I2
Practice 1
PROBLEM:
Balancing Redox Reactions by the Half-Reaction
Method
Permanganate ion is a strong oxidizing agent, and its deep purple
color makes it useful as an indicator in redox titrations. It reacts in
basic solution with the oxalate ion to form carbonate ion and solid
mangaese dioxide. Balance the skeleton ionic reaction that occurs
between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O42-(aq)
MnO2(s) + CO32-(aq)
General characteristics of voltaic and electrolytic cells
VOLTAIC CELL
System
Energydoes
is released
work on
from
its
spontaneous
surroundings
redox reaction
Oxidation half-reaction
X
X+ + e-
ELECTROLYTIC CELL
Surroundings(power
Energy is absorbed tosupply)
drive a
nonspontaneous
redox reaction
do work on system(cell)
Oxidation half-reaction
AA + e-
Reduction half-reaction
Y++ e- Y
Reduction half-reaction
B++ eB
Overall (cell) reaction
X + Y+
X+ + Y; DG < 0
Overall (cell) reaction
A- + B+
A + B; DG > 0
The spontaneous reaction between zinc and copper(II) ion
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A voltaic cell based on the zinc-copper reaction
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2eCu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Notation for a Voltaic Cell
components of
anode compartment
components of
cathode compartment
(oxidation half-cell)
(reduction half-cell)
phase of lower phase of higher
oxidation state oxidation state
phase of higher
oxidation state
phase of lower
oxidation state
phase boundary between half-cells
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)
Examples:
Zn(s)
Zn2+(aq) + 2e-
Cu2+(aq) + 2e-
Cu(s)
graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite
inert electrode
A voltaic cell using inactive electrodes
Oxidation half-reaction
2I-(aq)
I2(s) + 2e-
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l)
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq)
2Mn2+(aq) + 5I2(s) + 8H2O(l)
Practice 2
PROBLEM:
Diagramming Voltaic Cells
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.
Why Does a Voltaic Cell Work?
The spontaneous reaction occurs as a result of the different
abilities of materials (such as metals) to give up their electrons
and the ability of the electrons to flow through the circuit.
Ecell > 0 for a spontaneous reaction
1 Volt (V) = 1 Joule (J)/ Coulomb (C)
Voltages of Some Voltaic Cells
Voltaic Cell
Voltage (V)
Common alkaline battery
1.5
Lead-acid car battery (6 cells = 12V)
2.0
Calculator battery (mercury)
1.3
Electric eel (~5000 cells in 6-ft eel = 750V)
0.15
Nerve of giant squid (across cell membrane)
0.070
Determining an unknown E0half-cell with the standard
reference (hydrogen) electrode
Oxidation half-reaction
Zn(s) Zn2+(aq) + 2e-
Overall (cell) reaction
Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)
Reduction half-reaction
2H3O+(aq) + 2eH2(g) + 2H2O(l)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Practice 3
PROBLEM:
Calculating an Unknown E0half-cell from E0cell
A voltaic cell houses the reaction between aqueous bromine and
zinc metal:
Br2(aq) + Zn(s)
Zn2+(aq) + 2Br-(aq)
E0cell = 1.83V
Calculate E0bromine given E0zinc = 0.76V
Selected Standard Electrode Potentials (298K)
Half-Reaction
F2(g) + 2e2F-(aq)
Cl2(g) + 2e2Cl-(aq)
MnO2(g) + 4H+(aq) + 2eMn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3eNO(g) + 2H2O(l)
Ag+(aq) + eAg(s)
Fe3+(g) + eFe2+(aq)
O2(g) + 2H2O(l) + 4e4OH-(aq)
Cu2+(aq) + 2eCu(s)
2H+(aq) + 2eH2(g)
N2(g) + 5H+(aq) + 4eN2H5+(aq)
Fe2+(aq) + 2eFe(s)
2H2O(l) + 2eH2(g) + 2OH-(aq)
Na+(aq) + eNa(s)
Li+(aq) + eLi(s)
E0(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
•By convention, electrode potentials are written as reductions.
•When pairing two half-cells, you must reverse one reduction half-cell to
produce an oxidation half-cell. Reverse the sign of the potential.
•The reduction half-cell potential and the oxidation half-cell potential are
added to obtain the E0cell.
•When writing a spontaneous redox reaction, the left side (reactants)
must contain the stronger oxidizing and reducing agents.
Example:
Zn(s)
stronger
reducing agent
+
Cu2+(aq)
stronger
oxidizing agent
Zn2+(aq)
weaker
oxidizing agent
+
Cu(s)
weaker
reducing agent
Practice 4
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
PROBLEM: (a) Combine the following three half-reactions into three
spontaneous, balanced equations (A, B, and C), and calculate
E0cell for each.
(b) Rank the relative strengths of the oxidizing and reducing agents:
(1) NO3-(aq) + 4H+(aq) + 3e(2) N2(g) + 5H+(aq) + 4e(3) MnO2(s) +4H+(aq) + 2e-
NO(g) + 2H2O(l)
N2H5+(aq)
Mn2+(aq) + 2H2O(l)
E0 = 0.96V
E0 = -0.23V
E0 = 1.23V
Free Energy and Electrical Work
DG a -Ecell
-Ecell =
DG = wmax = charge x (-Ecell)
-wmax
DG = -n F Ecell
charge
In the standard state charge = n F
DG0 = -n F E0cell
n = #mols eF = Faraday constant
F = 96,485 C/mol
e-
1V = 1J/C
F = 9.65x104J/V*mol e-
DG0 = - RT ln K
E0cell = - (RT/n F) ln K
The interrelationship of DG0, E0, and K
DG0
DG0 = -nFEocell
Reaction at
standard-state
conditions
DG0
K
E0cell
<0
>1
>0
0
1
0
at equilibrium
>0
<1
<0
nonspontaneous
spontaneous
DG0 = -RT lnK
By substituting standard state
values into E0cell, we get
E0
K
cell
E0cell = -RT lnK
nF
E0cell = (0.0592V/n) log K (at 250C)
Practice 5
PROBLEM:
Calculating K and DG0 from E0cell
Lead can displace silver from solution:
Pb(s) + 2Ag+(aq)
Pb2+(aq) + 2Ag(s)
As a consequence, silver is a valuable by-product in the industrial extraction
of lead from its ore. Calculate K and DG0 at 250C for this reaction.
Practice 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+] = 0.010M
[H+] = 2.5M
PH = 0.30atm
2
Calculate Ecell at
250C.
The Effect of Concentration on Cell Potential
DG = DG0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
RT
Ecell =
E0
cell
-
ln Q
nF
•When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell
•When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell
•When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell
0.0592
Ecell =
E0
cell
n
log Q
The relation between Ecell and log Q for the zinc-copper cell
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A concentration cell based on the Cu/Cu2+ half-reaction
Oxidation half-reaction
Cu(s)
Cu2+(aq, 0.1M) + 2e-
Reduction half-reaction
Cu2+(aq, 1.0M) + 2eCu(s)
Overall (cell) reaction
Cu2+(aq,1.0M)
Cu2+(aq, 0.1M)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Practice 7
PROBLEM:
Calculating the Potential of a Concentration Cell
A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A,
electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B
dips into 4.0x10-4M AgNO3. What is the cell potential at 298K?
Which electrode has a positive charge?
The corrosion of iron
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Enhanced corrosion at sea
The effect of metal-metal contact on the corrosion of iron
faster corrosion
cathodic protection
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The use of sacrificial anodes to prevent iron corrosion
The tin-copper reaction as the basis of a voltaic and an
electrolytic cell
voltaic cell
Oxidation half-reaction
Sn(s) Sn2+(aq) + 2eReduction half-reaction
Cu2+(aq) + 2eCu(s)
Overall (cell) reaction
Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
electrolytic cell
Oxidation half-reaction
Cu(s)
Cu2+(aq) + 2eReduction half-reaction
Sn2+(aq) + 2eSn(s)
Overall (cell) reaction
Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The processes occurring during the discharge and
recharge of a lead-acid battery
VOLTAIC(discharge)
ELECTROLYTIC(recharge)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
-
Practice 8
PROBLEM:
Predicting the Electrolysis Products of a Molten
Salt Mixture
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 halfreactions and the overall cell reaction.
The electrolysis of water
Overall (cell) reaction
2H2O(l)
H2(g) + O2(g)
Oxidation half-reaction
2H2O(l) 4H+(aq) + O2(g) + 4e-
Reduction half-reaction
2H2O(l) + 4e2H2(g) + 2OH-(aq)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Practice 9
PROBLEM:
Predicting the Electrolysis Products of Aqueous
Ionic Solutions
What products form during electrolysis of aqueous solution of the
following salts: (a) KBr; (b) AgNO3; (c) MgSO4?
A summary diagram for the stoichiometry of electrolysis
MASS (g)
of substance
oxidized or
reduced
M(g/mol)
AMOUNT (MOL)
of substance
oxidized or
reduced
AMOUNT (MOL)
of electrons
transferred
balanced
half-reaction
Faraday
constant
(C/mol e-)
CHARGE (C)
time(s)
CURRENT (A)
Practice 10
Applying the Relationship Among Current, Time,
and Amount of Substance
PROBLEM:
A technician is plating a faucet with 0.86g of Cr from an electrolytic
bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the
plating, what current is needed?