资源简介

(共79张PPT)
Chapter 19
Electrochemistry
Electrochemistry
19
19.2 Galvanic Cells
19.3 Standard Reduction Potentials
19.4 Spontaneity of Redox Reactions Under Standard-State Conditions
19.5 Spontaneity of Redox Reactions Under Conditions Other Than Standard State
The Nernst Equation
Concentration Cells
19.6 Batteries
Dry Cells and Alkaline Batteries
Lead Storage Batteries
Lithium-ion Batteries
Fuel Cells
19.7 Electrolysis
Electrolysis of Molten Sodium Chloride
Electrolysis of Water
Electrolysis of an Aqueous Sodium Chloride Solution
Quantitative Applications of Electrolysis
19.8 Corrosion
Static Electricity Happens Everywhere
Lightning
Triboelectricity 摩擦电
Atoms bind electrons slightly differently.
If you rub any two neutral different materials together, some electrons may transfer from one to the other resulting in oppositely charged objects.
The amount of charge is generally small - a nanocoulomb or so.
Tribo means rubbing
Triboelectricity
Triboelectric Series
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Galvanic Cells
19.2
When zinc metal is placed in a copper(II) solution, Zn is oxidized and Cu2+ ions are reduced.
Galvanic Cells
The experimental apparatus for generating electricity through the use of a spontaneous reaction is called a galvanic cell.
Galvanic Cells
Electric current flows from anode to cathode because there is a difference in electrical potential energy between the electrodes.
The electrical potential is measured by a voltmeter and is called the cell potential (Ecell ).
Cell notation:
Zn(s) │ Zn2+ (1 M) ││ Cu2+ (1 M) │ Cu(s)
anode
cathode
Notation for a Voltaic Cell
components of anode compartment
(oxidation half-cell)
components of cathode compartment
(reduction half-cell)
phase of lower oxidation state
phase of higher oxidation state
phase of higher oxidation state
phase of lower oxidation state
phase boundary between half-cells
Examples:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)
Zn(s) Zn2+(aq) + 2e-
Cu2+(aq) + 2e- Cu(s)
graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq), Mn2+(aq) | graphite
inert electrode
Standard Reduction Potentials
When the concentrations of Zn2+ and Cu2+ are 1 molar at 25°C, the cell voltage is 1.10 V.
19.3
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Voltages of Some Voltaic Cells
Voltaic Cell
Voltage (V)
Common alkaline battery
Lead-acid car battery (6 cells = 12V)
Calculator battery (mercury)
Electric eel (~5000 cells in 6-ft eel = 750V)
Nerve of giant squid (across cell membrane)
2.0
1.5
1.3
0.15
0.070
Metal Surface
Solution
Electrode Potential 电极电位
Potential Difference here and forms a Bilayer 双电层
Electrode potential depends on the properties of the metal, temperature, solution characteristics and concentration.
Agar-agar 琼脂 + KNO3
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)
A voltaic cell using inactive electrodes.
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5e-
Mn2+(aq) + 4H2O(l)
Oxidation half-reaction
2I-(aq) I2(s) + 2e-
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)
Standard Reduction Potentials
The hydrogen electrode is used as the arbitrary standard in measuring standard cell potentials.
H2 → 2H+ + 2e– E° = 0 V
Standard Reduction Potentials
Zn(s) │ Zn2+ (1 M) ││ H+ (1 M) │ H2(g) │ Pt(s)
Zn(s) → Zn2+ (1 M) + 2e–
2H+(1 M) + 2e– → H2 (1 atm)
Zn(s) + 2H+(1 M) → Zn2+ (1 M) H2 (1 atm)
Anode (oxidation)
Cathode (reduction)
Overall:
Standard Reduction Potentials
Zn(s) │ Zn2+ (1 M) ││ H+ (1 M) │ H2(g) │ Pt(s)
Zn(s) → Zn2+ (1 M) + 2e–
2H+(1 M) + 2e– → H2 (1 atm)
Zn(s) + 2H+(1 M) → Zn2+ (1 M) + H2 (1 atm)
Anode (oxidation)
Cathode (reduction)
Overall:
E°cell = E°cathode – E°anode
E°cell = E°H+ /H2 – E°Zn2+/Zn
0.76 V = 0 – E°Zn2+/Zn
E°Zn2+/Zn = –0.76 V
Standard Reduction Potentials
Pt(s) │ H2(g) │H+ (1 M) ││Cu2+ (1 M) │Cu(s)
H2(1 atm) → 2H+ (1 M) + 2e–
Cu2+ (1 M) + 2e– → Cu (s)
H2(1 atm) + Cu2+ (1 M) → 2H+ (1 M) + Cu(s)
Anode (oxidation)
Cathode (reduction)
Overall:
Standard Reduction Potentials
Pt(s) │ H2(g) │H+ (1 M) ││Cu2+ (1 M) │Cu(s)
H2(1 atm) → 2H+ (1 M) + 2e–
Cu2+ (1 M) + 2e– → Cu (s)
H2(1 atm) + Cu2+ (1 M) → 2H+ (1 M) + Cu(s)
Anode (oxidation)
Cathode (reduction)
Overall:
E°cell = E°cathode – E°anode
E°cell = E°Cu2+/Cu – E°H+/H2
0.34 V = 0 – E°H+/H2
E°H+/H2 = –0.34 V
Standard Reduction Potentials
Cu2+ (1 M) + 2e– → Cu (s)
Zn(s) + Cu2+ (1 M) → Zn2+ (1 M) + Cu(s)
Anode (oxidation)
Cathode (reduction)
Overall:
E°cell = E°cathode – E°anode
E°cell = E°Cu2+ /Cu – E°Zn2+/Zn
E°cell = 0.34 V –(–0.76 V)
E°cell = 1.10 V
Zn(s) │ Zn2+ (1 M) ││ Cu2+ (1 M) │ Cu(s)
Zn(s) → Zn2+ (1 M) + 2e–
Standard Reduction Potentials
Standard Reduction Potentials at 25°C (See Table 19.1)
Half-Reaction E°(V)
Cl2 (l) + 2e– → 2Cl–(aq) 1.36
Br2 (l) + 2e– → 2Br–(aq) 1.07
Ag+ (aq) + e– → Ag(s) 0.80
Cu2+ (aq) + 2e– → Cu(s) 0.34
2H+(aq) + 2e– → H2(g) 0.00
Pb2+ (aq) + 2e– → Pb(s) –0.13
Cd2+ (aq) + 2e– → Cd(s) –0.40
Zn2+ (aq) + 2e– → Zn(s) –0.76
Mn2+ (aq) + 2e– → Mn(s) –1.66
Standard Reduction Potentials
2Ag+ (1 M) + 2e– → 2Ag(s)
Zn(s) + 2Ag+ (1 M) → Zn2+ (1 M) + 2Ag(s)
Anode (oxidation)
Cathode (reduction)
Overall:
Zn(s) │ Zn2+ (1 M) ││ Ag+ (1 M) │ Ag(s)
Zn(s) → Zn2+ (1 M) + 2e–
Half-Reaction E°(V)
Ag+ (aq) + e– → Ag(s) 0.80
Cu2+ (aq) + 2e– → Cu(s) 0.34
2H+(aq) + 2e– → H2(g) 0.00
Zn2+ (aq) + 2e– → Zn(s) –0.76
Mn2+ (aq) + 2e– → Mn(s) –1.66
E°cell = E°Ag+/Ag – E°Zn2+/Zn = 0.80 V – (–0.76 V) = 1.56 V
Determine the overall cell reaction and E°cell at 25°C of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Pb electrode in a 1.0 M Pb(NO3)2 solution.
Solution
Step 1: Use a standard reduction potential table to determine which electrode is the cathode and which is the anode.
Pb/Pb2+ is the cathode; Cd/Cd2+ is the anode
Standard Reduction Potentials
Half-Reaction E°(V)
Ag+ (aq) + e– → Ag(s) 0.80
Cu2+ (aq) + 2e– → Cu(s) 0.34
2H+(aq) + 2e– → H2(g) 0.00
Pb2+ (aq) + 2e– → Pb(s) –0.13
Cd2+ (aq) + 2e– → Cd(s) –0.40
Standard Reduction Potentials
Determine the overall cell reaction and E°cell at 25°C of a galvanic cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Pb electrode in a 1.0 M Pb(NO3)2 solution.
Solution
Step 2: Use the equation below to calculate E°cell.
Half-Reaction E°(V)
Pb2+ (aq) + 2e– → Pb(s) –0.13
Cd2+ (aq) + 2e– → Cd(s) –0.40
E°cell = E°cathode – E°anode
E°cell = E°Pb2+/Pb – E°Cd2+/Cd
E°cell = –0.13 V – (–0.40 V) = 0.27 V
Worked Example 19.2
Strategy Use the tabulated values of E° to determine which electrode is the cathode and which is the anode, combine cathode and anode half-cell reactions to get the overall cell reaction, and use E°cell = E°cathode – E°anode to calculate E°cell.
A galvanic cell consists of an Mg electrode in a 1.0 M Mg(NO3)2 solution and a Cd electrode in a 1.0 M Cd(NO3)2 solution. Determine the overall cell reaction, and calculate the standard cell potential at 25°C.
Solution The half-cell reactions and their standard reduction potentials are
Mg2+ + 2e- → Mg E° = –2.37 V
Cd2+ + 2e- → Cd E° = –0.40 V
Because the Cd half-cell reaction has the greater (less negative) standard reduction potential, it will occur as the reduction. The Mg half-cell reaction will occur as the oxidation. Therefore, E°cathode = –0.40 V and E°anode = –2.37 V.
Worked Example 19.2 (cont.)
Solution Adding the two half-cell reactions together gives the overall cell reaction:
The standard cell potential is
Mg → Mg2+ + 2e-
Cd2+ + 2e- → Cd
Overall: Mg + Cd2+ → Mg2+ + Cd
E°cell = E°cathode – E°anode
= E°Cd2+/Cd – E°Mg2+/Mg
= (–0.40 V) – (–2.37 V)
= 1.97 V
Think About It If you ever calculate a negative voltage for a galvanic cell potential, you have done something wrong–check your work. Under standard-state conditions, the overall cell reaction will proceed in the direction that gives a positive E°cell.
Worked Example 19.3
Strategy In each case, write the equation from the redox reaction that might take place and use E° values to determine whether or not the proposed reaction will actually occur.
Predict what redox reaction will take place, if any, when molecular bromine (Br2) is added to (a) a 1-M solution of NaI and (b) a 1-M solution of NaCl. (Assume a temperature of 25°C.)
Solution From Table 19.1:
Br2(l) + 2e- → 2Br-(aq) E° = 1.07 V
I2(l) + 2e- → 2I-(aq) E° = 0.53 V
Cl2(l) + 2e- → 2Cl-(aq) E° = 1.36 V
Worked Example 19.3 (cont.)
Solution (a) If a redox reaction is to occur, it will be oxidation of I- ions by Br2:
Br2(l) + 2I-(aq) → 2Br-(aq) + I2(s)
Because the reduction potential of Br2 is greater than that of I2, Br2 will be reduced to Br- and I- will be oxidized to I2. Thus, the preceding reaction will occur.
(b) In this case, the proposed reaction is the reduction of Br2 by Cl- ions:
Br2(l) + 2Cl-(aq) → 2Br-(aq) + Cl2(g)
Because the reduction potential of Br2 is greater than that of I2, Br2 will be reduced to Br- and I- will be oxidized to I2. Thus, the preceding reaction will occur.
Think About It We can use E°cell = E°cathode – E°anode and treat problems of this type like galvanic cell problems. Write the proposed redox reaction, and identify the “cathode” and the “anode.” If the calculated E°cell is positive, the reaction
will occur. If the calculated E°cell is negative, the reaction will not occur.
Spontaneity of Redox Reactions Under Standard-State Conditions
E cell is related to the thermodynamic quantities ΔG° and K.
n is the number of moles electrons
F is the Faraday constant (96,500 C/mol e–)
R is the gas constant (8.314 J/mol K)
19.4
ΔG° = –nFE°cell
Spontaneity of Redox Reactions Under Standard-State Conditions
By converting to the base-10 logarithm of K, we get
E°cell =
0.0592V
n
log K (at 25°C)
Spontaneity of Redox Reactions Under Standard-State Conditions
Calculate ΔG° and K for the following reaction at 25°C:
3Mg(s) + 2Al3+(aq) 3Mg2+(aq) + 2Al(s)
Solution
Step 1: Use E° to calculate E°cell.
Half-Reaction E°(V)
Al3+ (aq) + 3e– → Al(s) –1.66
Mg2+ (aq) + 2e– → Mg(s) –2.37
E°cell = E°Al3+/Al – E°Mg2+/Mg
E°cell = –1.66 V – (–2.37 V) = 0.71 V
Spontaneity of Redox Reactions Under Standard-State Conditions
Calculate ΔG° and K for the following reaction at 25°C:
3Mg(s) + 2Al3+(aq) 3Mg2+(aq) + 2Al(s)
Solution
Step 2: Use the equation below to calculate ΔG° :
ΔG° = –(6 e–)(96500 J/V·mol e– )(0.71 V)
ΔG° = –411090 J or –411 kJ
ΔG° = –nFE°cell
Spontaneity of Redox Reactions Under Standard-State Conditions
Calculate ΔG° and K for the following reaction at 25°C:
3Mg(s) + 2Al3+(aq) 3Mg2+(aq) + 2Al(s)
Solution
Step 3: Use the equation below to calculate K :
K = 8 x 10165
Worked Example 19.4
Strategy Use E° values from Table 19.1 to calculate E° for the reaction, and then use ΔG° = –nFE°cell to calculate the standard free-energy change.
Calculate the standard free-energy change for the following reaction at 25°C:
2Au(s) + 3Ca2+(1.0 M) 2Au3+(1.0 M) + 3Ca(s)
Solution The half-cell reactions are
From Table 19.1, E°Ca2+/Ca = –2.87 V and E°Au3+/Au = 1.50 V.
Cathode (reduction): 3Ca2+(aq) + 6e- → Ca(s)
Anode (oxidation): 2Au(s) → 2Au3+(aq) + 6e-
E°cell = E°cathode – E°anode
= E°Ca2+/Ca – E°Au3+/Au
= –2.87 V – 1.50 V
= –4.37 V
Worked Example 19.4 (cont.)
Solution Next, substitute this value of E° into ΔG° = –nFE°cell to obtain ΔG°. The overall reaction show that n = 6, so
Think About It The large positive value of ΔG° indicates that reactants are favored at equilibrium, which is consistent with the fact that E° for the reaction is negative.
ΔG° = –nFE°cell
= –(6 mol e-)(96,500 J/V mol e-)(–4.37 V)
= 2.53×106 J/mol
= 2.53×103 kJ/mol
Worked Example 19.5
Strategy Use E° values from Table 19.1 to calculate E° for the reaction, and then calculate the equilibrium constant using E°cell = (0.0592 V/n)log K (rearranged to solve for K.)
Calculate the equilibrium constant for the following reaction at 25°C:
Sn(s) + 2Cu+(aq) Sn2+(aq) + 2Cu+(aq)
Solution The half-cell reactions are
From Table 19.1, E°Cu2+/Cu+ = 0.15 V and E°Sn2+/Sn = – 0.14 V.
Cathode (reduction): 2Cu2+(aq) + 2e- → 2Cu+(aq)
Anode (oxidation): Sn(s) → Sn2+(aq) + 2e-
E°cell = E°cathode – E°anode
= E°Cu2+/Cu+ – E°Sn2+/Sn
= 0.15 V – (–0.14 V)
= 0.29 V
Worked Example 19.5 (cont.)
Solution Solving E°cell = (0.0592 V/n)log K for K gives
Think About It A positive standard cell potential corresponds to a large equilibrium constant.
K = 10
= 10
= 6×109
nE°/0.0592 V
(2)(0.29 V)/0.0592 V
Spontaneity of Redox Reactions Under Conditions Other Than Standard State
Under non-standard conditions, cell potential is calculated with the Nernst equation:
When Q > 1, E < E°
When Q < 1, E > E°
19.5
Spontaneity of Redox Reactions Under Conditions Other Than Standard State
Will the following reaction occur spontaneously at 298 K if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M
Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s)
Solution
Step 1: Calculate E° using data from a standard reduction potential table.
Half-Reaction E°(V)
Cd2+ (aq) + 2e– → Cd(s) –0.40
Fe2+ (aq) + 2e– → Fe(s) –0.44
E°cell = E°Fe2+/Fe – E°Cd2+/Cd
E°cell = –0.44 V – (–0.40 V) = –0.04 V
Spontaneity of Redox Reactions Under Conditions Other Than Standard State
Will the following reaction occur spontaneously at 298 K if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M
Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s)
Solution
Step 2: Use the equation below to calculate E. If E is positive, the reaction remains spontaneous.
Worked Example 19.6
Strategy Use E° values from Table 19.1 to determine E° for the reaction, and use E = E° – (0.0592 V/n)log Q to calculate E. If E is positive, the reaction will occur spontaneously.
Predict whether the following reaction will occur spontaneously as written at 298 K:
Co(s) + Fe2+(aq) → Co2+(aq) + Fe(s)
assuming [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
Solution The half-cell reactions are
Cathode (reduction): Fe2+(aq) + 2e- → Fe(s)
Anode (oxidation): Co(s) → Co2+(aq) + 2e-
Worked Example 19.6 (cont.)
Solution
The reaction quotient, Q, for the reaction is [Co2+][Fe2+]. Therefore, Q = (0.15/0.68) = 0.22.
The negative E value indicates that the reaction is not spontaneous as written under the conditions describe.
E°cell = E°cathode – E°anode
= E°Fe2+/Fe – E°Co2+/Co
= –0.44 V – (–0.28 V)
= –0.16 V
E = E° –
0.0592 V
n
log Q
= –0.16 V –
0.0592 V
2
log 0.22
= –0.14 V
Worked Example 19.6 (cont.)
Think About It For this reaction to be spontaneous as written, the ratio of [Fe2+] to [Co2+] would have to be enormous. We can determine the required ratio by first setting E equal to zero:
For E to be positive, therefore, the ratio of [Fe2+] to [Co2+], the reciprocal of Q, would have to be greater than 3×105 to 1.
0 V = –0.16 V –
0.0592 V
n
log Q
– = log Q
(0.16 V)(2)
0.0592 V
log Q = –5.4
Q = 10–5.4 = = 4×10-6
[Co2+]
[Fe2+]
Spontaneity of Redox Reactions Under Conditions Other Than Standard State
A concentration cell is one with two half-cells containing the same components, but differing ion concentrations.
Zn(s) │ Zn2+ (0.10 M) ││ Zn2+ (1.0 M) │ Zn(s)
Spontaneity of Redox Reactions Under Conditions Other Than Standard State
A concentration cell is one with two half-cells containing the same components, but differing ion concentrations.
Zn(s) │ Zn2+ (0.10 M) ││ Zn2+ (1.0 M) │ Zn(s)
Zn2+ (1.0 M) + 2e– → Zn (s)
Zn2+ (1.0 M) → Zn2+ (0.10 M)
Anode (oxidation)
Cathode (reduction)
Overall:
Zn(s) → Zn2+ (0.10 M) + 2e–
Worked Example 19.7
Strategy Use E = E° – (0.0592 V/n)log Q to solve for the unknown concentration of silver ion. The half-cell with the higher Ag+ (1.0 M AgNO3) concentration will be the cathode; the half-cell with the lower, unknown Ag+ concentration (saturated AgCN solution) will be the anode. The overall reaction is Ag+(0.10 M) → Ag+(x M).
An electrochemical cell is constructed for the purpose of determining Ksp of silver cyanide (AgCN) at 25°C. One half-cell consists of a silver electrode in a 1.00-M solution of silver nitrate. The other half-cell consists of a silver electrode in a saturated solution of silver cyanide. The cell potential is measured and found to be 0.470 V. Determine the concentration of silver ion in the saturated silver cyanide solution and the value of Ksp for AgCN.
Worked Example 19.7 (cont.)
Solution Because this is a concentration cell, E°cell = 0 V. The reaction quotient, Q, is (x M)/(1.00 M); and the value of n is 1.
Therefore, [Ag+] = 1.15×10-8 M and Ksp for AgCN = x2 = 1.3×10-16.
Ecell = E°cell –
0.0592 V
n
log Q
0.470 V = 0 V –
0.0592 V
1
log
x
1.00
–7.939 =
log
x
1.00
10–7.939 = 1.15×10-8 =
x
1.00
x = 1.15×10-8
Think About It Remember that in a saturated solution of a salt that dissociates into two ions, the ion concentrations are equal to each other and each ion concentration is equal to the square root of Ksp.
Batteries
Dry Cell Batteries
No fluid component
Zinc container (anode)
Graphite cathode
1.5 V
19.6
2 NH4+(aq) + 2MnO2(s) + 2e– → Mn2O3(s) + 2NH3(aq) + H2O(l)
Zn(s) + 2NH4+(aq) + 2MnO2(s) → Zn2+(aq)+ Mn2O3(s) + 2NH3(aq) + H2O(l)
Anode
Cathode
Overall:
Zn(s) → Zn2+ (aq) + 2e–
Batteries
Alkaline Batteries
Basic medium
2MnO2(s) + 2H2O(l) + 2e– → 2MnO(OH)(s) + 2OH–(aq)
Zn(s) + 2H2O(l) + 2MnO2(s) → Zn(OH)2(s)+ 2MnO(OH)(s)
Anode
Cathode
Overall:
Zn(s) + 2OH–(aq) → Zn(OH)2(s) + 2e–
Batteries
Lead Storage
Batteries
6 cells
2 V per cell
12 V total
Rechargeable
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l)
Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq) → 2PbSO4(s) + 2H2O(l)
Anode
Cathode
Overall:
Pb(s) + 2SO42–(aq) → PbSO4(s) + 2e–
Batteries
Lithium ion batteries
Cell potential: 3.4 V
Rechargeable (many times)
Li+(aq) + CoO2 + e– → LiCoO2(s)
Li+(s) + CoO2 → LiCoO2(s)
Anode
Cathode
Overall:
Li(s) → Li+ + e–
Batteries
Fuel Cells
Cell potential: 1.23 V
~70% efficient
O2 (aq) + 2H2O(l) + 4e– → 4OH–(aq)
2H2(g) + O2(g) → 2H2O(l)
Anode
Cathode
Overall:
2H2(g) + 4 OH–(aq) → 4H2O(l) + 4e–
Electrolysis
The use of electric energy to drive a non-spontaneous chemical reaction is called electrolysis.
An electrolytic cell is used.
19.7
Cell Type Chemical Reaction Electric Energy
Galvanic Spontaneous Produced
Electrolytic Nonspontaneous Consumed
Energy is absorbed to drive a nonspontaneous redox reaction
General characteristics of voltaic and electrolytic cells.
VOLTAIC CELL
ELECTROLYTIC CELL
Energy is released from spontaneous redox reaction
Reduction half-reaction
Y++ e- Y
Oxidation half-reaction
X X+ + e-
System does work on its surroundings
Reduction half-reaction
B++ e- B
Oxidation half-reaction
A- A + e-
Surroundings(power supply)
do work on system(cell)
Overall (cell) reaction
X + Y+ X+ + Y; DG < 0
Overall (cell) reaction
A- + B+ A + B; DG > 0
Comparison of Voltaic and Electrolytic Cells
Cell Type
DG
Ecell
Electrode
Name
Process
Sign
Voltaic
Voltaic
Electrolytic
Electrolytic
< 0
< 0
> 0
> 0
> 0
> 0
< 0
< 0
Anode
Anode
Cathode
Cathode
Oxidation
Oxidation
Reduction
Reduction
-
-
+
+
Electrolysis
Electrolysis of molten sodium chloride; E°cell = – 4 V
2Na+(l) + 2e– → 2Na(l)
2Na+(l) + 2Cl–(l) → 2Na(l) + Cl2(g)
Anode
Cathode
Overall:
2Cl–(l) → Cl2(g) + 2e–
Electrolysis
Electrolysis of water; ΔG° = 474.4 kJ/mol
4H+(aq) + 4e– → 2H2(g)
2H2O(l) → O2(g) + 2H2(g)
Anode
Cathode
Overall:
2H2O(l) → O2(g) + 4H+(aq) + 4e–
Electrolysis
Faraday developed the quantitative treatment of electrolysis.
The quantity of chlorine gas formed depends on the charge.
Suppose 0.452 A is passed through the cell in 1.50 hr:
Ca2+(aq) + 2e– → Ca(l)
Anode
Cathode
Overall:
2Cl–(l) → Cl2(g) + 2e–
Ca2+(aq) + 2Cl–(l) → Ca(l) + Cl2(g)
Electrolysis
A constant current of 0.912 A is passed through an electrolytic cell containing molten MgCl2 for 18 h. What mass of Mg is produced
Solution
Step 1: Use the current and time to determine charge:
Step 2: Use charge to determine mass
Mg2+(l) + 2e– → Mg(l)
Worked Example 19.8
Strategy As shown in Figure 19.12, we can use current and time to determine charge. We can then convert charge to moles of electrons, and use the balanced half-reactions to determine how many moles of product form at each electrode. Finally, we can convert moles to volume.
A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reaction, and calculate the volume of gases generated at STP.
Solution The half-cell reactions for the electrolysis of water are
Remember that STP for gases means 273 K and 1 atm.
Cathode: 4H+(aq) + 4e- → 2H2(g)
Anode: 2H2O(l) → O2(g) + 4H+(aq) + 4e-
Overall: 2H2O(l) → O2(g) + 2H2(g)
Worked Example 19.8 (cont.)
Solution
At the anode:
The volume of 0.0874 mol O2 at STP is given by
Similarly, for hydrogen we write
= 3.375×104 C
coulombs = (1.26 A)(7.44 h)
3600 s
1 h
1 C
1 A s
= 0.0874 mol O2
moles of O2 = (3.375×104 C)
1 mol e-
96,500 C
1 mol O2
4 mol e-
V =
nRT
P
=
(0.0874 mol)(0.08206 L atm/mol K)(273.15 K)
1 atm
= 1.96 L O2
= 0.175 mol H2
moles of H2 = (3.375×104 C)
1 mol e-
96,500 C
1 mol H2
2 mol e-
Worked Example 19.8 (cont.)
Solution The volume of 0.175 mol H2 at STP is given by
V =
nRT
P
=
(0.175 mol)(0.08206 L atm/mol K)(273.15 K)
1 atm
= 3.92 L H2
Think About It The volume of H2 is twice that of O2 (see Figure 19.11), which is what we would expect based on Avogadro’s law (at the same temperature and pressure, volume is directly proportional to number of moles of gas; V α n.)
Corrosion
The term corrosion generally refers to the deterioration of a metal by an electrochemical process.
19.8
Latimer Diagram
A Latimer diagram provides us a concise way of presenting a great deal of information about the various oxidation states of the elements.
The arrow connecting ClO4- and ClO3- represents the half-reaction
Use Latimer diagram to determine the reduction potential for half-reactions between non-adjacent species
remember that the relation between free energy and Eo is
Go = -nF Eo.
When add two half-reactions where the electrons do not cancel, the potential of the resultant reaction is given by
As an example, to go from HClO to Cl- the potential would be given by
Eo = (1.63+1.36)/2 = 1.50V
Predicting the Stability of a Particular Form of an Element in Solution
If the potential going to the right is more positive than that going to the left, the species is unstable and will undergo, slowly or quickly, a disproportionation 歧化 reaction.
ClO2 will disproportionate to HClO and ClO3-
Pourbaix or eh-pH diagrams
the thermodynamically stable form of an element as a function of potential and pH
The Pourbaix diagram is a type of predominance diagram -- it shows the predominate form in an element will exist under a given set of environmental conditions.
give a visual representation of the oxidizing and reducing abilities of the major stable compounds of an element and are used frequently in geochemical, environmental and corrosion applications.
Kinetics is not incorporated.
Pourbaix Diagram for Manganese
How to Read a Pourbaix Diagram
Vertical lines separate species that are in acid-base equilibrium.
Non vertical lines separate species related by redox equilibria.
Horizontal lines separate species in redox equilibria not involving hydrogen or hydroxide ions.
Diagonal boundaries separate species in redox equilibria in which hydroxide or hydrogen ions are involved.
Dashed lines enclose the practical region of stability of the water solvent to oxidation or reduction.
What You Can Learn From a Pourbaix Diagram
Any point on the diagram will give the thermodynamically most stable (and theoretically most abundant) form of that element at a given potential and pH condition.
Strong oxidizing agents and oxidizing conditions are found only at the top of Pourbaix diagrams. Strong oxidizing agents have lower boundaries that are also high on the diagram. Permanganate is an oxidizing agent over all pH ranges. It is very strongly oxidizing at low pH.
Reducing agents and reducing conditions are found at the bottom of a diagram and not elsewhere. Strong reducing agents have low upper boundaries on the diagram. Manganese metal is a reducing agent over all pH ranges and is strongest in basic conditions.
When the predominance area for a given oxidation state disappears completely above or below a given pH and the element is in an intermediate oxidation state, the element will undergo disproportionation MnO42- tends to disproportionate.
A species that ranges from the top to the bottom of the diagram at a given pH will have no oxidizing or reducing properties at that pH.
Key Concepts
19
Balancing Redox Reactions
Galvanic Cells
Standard Reduction Potentials
Spontaneity of Redox Reactions Under Standard State Conditions
The Nernst Equation
Concentration Cells
Dry Cells and Alkaline Batteries
Lead Storage Batteries
Lithium-ion Batteries
Fuel Cells
Electrolysis of Molten Sodium Chloride
Electrolysis of Water
Quantitative Applications of Electrolysis
Corrosion

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