electrochemistry brown, lemay ch 20 ap chemistry
TRANSCRIPT
2
20.1: Oxidation-reduction reactions“Redox”: rxns with a change in oxidation number• At least one element’s oxidation number will
increase and one will decrease• Consists of two half-reactions:
oxidation: loss of e-; oxidation number increasesreduction: gain of e-; oxidation number decreases
(“reduces”)
Oxidizing agent or oxidant: causes another substance to be oxidized; gains e-
Reducing agent or reductant: causes another substance to be reduced; loses e-
3
Ex: Determine the half-reactions.
Cr2O72- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2
(g)+6
-2 -1 +3
0
Reduction half-reaction:
Cr2O72- (aq) → Cr3+ (aq)
Oxidizing agent
Oxidation half-reaction:
Cl1- (aq) → Cl2 (g)
Reducing agent
4
20.2: Balancing Redox Reactions“Method of Half-Reactions” steps:1. Assign oxidation numbers to all species; based on
this, split the half-reactions of reduction and oxidation.
2. For each, balance all elements except H and O.3. Balance oxygen by adding H2O to the opposite
side.4. Balance hydrogen by adding H+ to the opposite
side.5. Balance charges by adding e- to side with overall
positive charge.6. Multiply each half-reaction by an integer so there
are an equal number of e- in each.7. Add the half reactions; cancel any species; check
final balance.
5
Ex: Balance the following reaction, which takes place in acidic solution:Cr2O7
2- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2 (g)
Reduction:
Cr2O72- (aq) → Cr3+ (aq)2 + 7 H2O (l)14 H+ (aq) +6 e- +
Oxidation:
Cl- (aq) → Cl2 (g)
6 Cl- (aq) → 3 Cl2 (g) + 6 e-
2 + 2 e-3 [ ]
6
6 e- + 14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7
H2O (l)
6 Cl- (aq) → 3 Cl2 (g) + 6 e-
14 H+ (aq) + Cr2O72- (aq) + 6 Cl- (aq) →
2 Cr3+ (aq) + 7 H2O (l) + 3 Cl2 (g)
•Since occurs in acidic solution, H+ appears as a reactant
•If reaction occurs in basic solution, balance using same method, but neutralize H+ with OH- as a last step.
7
Ex: In basic solution, MnO4
- (aq) + Br- (aq) → MnO2 (s) + BrO3- (aq)
+7
-2 -1 +4
+5
-2 -2
Reduction: MnO4- (aq) → MnO2 (s)
MnO4- (aq) → MnO2 (s) + 2 H2O
(l)4 H+ (aq) + MnO4
- (aq) → MnO2 (s) + 2 H2O (l)3 e- + 4 H+ (aq) + MnO4
- (aq) → MnO2 (s) + 2 H2O (l)
8
Oxidation: Br- (aq) → BrO3- (aq)
3 H2O (l) + Br- (aq) → BrO3- (aq)
3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq)
3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e-
2[3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O
(l)]
6 e- + 8 H+(aq) + 2 MnO4-(aq) → 2 MnO2(s) + 4 H2O(l)
3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e-
2 H+(aq) + 2 MnO4- (aq) + Br- (aq) →
2 MnO2 (s) + BrO3- (aq) + H2O (l)
9
Neutralize with OH-:2 H+(aq) + 2 MnO4
-(aq) + Br- (aq) + 2 OH- (aq) →2 MnO2 (s) + BrO3
- (aq) + H2O (l) + 2 OH- (aq)
2 H2O (l) + 2 MnO4-(aq) + Br- (aq) →
2 MnO2 (s) + BrO3- (aq) + H2O (l) + 2 OH- (aq)
Simplify:H2O (l) + 2 MnO4
-(aq) + Br- (aq) →
2 MnO2 (s) + BrO3- (aq) + 2 OH- (aq)
10
20.3: Voltaic (or Galvanic) Cells
• Device that utilizes energy released in a spontaneous electrochemical reaction by directing e- transfer along an external pathway, rather than directly between reactants
Electrodes: metals (or graphite) connected by an external circuitAnode: half-cell where oxidation occursCathode: half-cell where reduction occurs
Luigi Galvani(1737–1798)
Alessandro Volta(1745–1827)
11
Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)Oxidation half-cell: Zn (s) → Zn2+ (aq) + 2 e-Reduction half-cell: Cu2+ (aq) + 2 e- → Cu (s)
anode
Zn Cu
(-) (+)
salt bridge, saturated with
NaNO3 (aq)
cathodee-
flowNa1+NO3
1-
Zn2+ (aq)
Cu2+ (aq)
* Notation:Anode| Anode product|Salt |Cathode|Cathode productZn(s)|Zn2+(aq, 1 M)|NaNO3 (saturated)|Cu2+(aq, 1 M)|
Cu(s)
13
20.4: Electromotive force (emf or E)• Potential difference between cathode and
anode that causes a flow of e-.– The result of the difference in electric fields
produced at each electrode
• Also called cell potential or cell voltage, Ecell
• Units = Volt (1 V = 1 J / 1 C)
• Standard conditions (25°C, 1 atm, 1 M): Eºcell
• Eºcell is unique to each set of half-cells involved Charles-Augustin de Coulomb
(1736–1806)
14
Standard Reduction Potentials• Assigned to each reduction reactionEºcell = Eºreduction(cathode) - Eº reduction(anode)
• All Eºred are referenced to standard hydrogen electrode (S.H.E.):
2 H+ (aq, 1 M) + 2 e- → H2 (g, 1 atm)
Eºreduction = 0 V
15
Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)• Anode: Zn (s) → Zn2+ (aq) + 2 e-
– This is oxidation, but Eºreduction is recorded as the reduction reaction:
Zn2+ (aq) + 2 e- → Zn (s), where Eºreduction = - 0.76 V
• Cathode:Cu2+ (aq) + 2 e- → Cu (s) Eºreduction = + 0.34 V
Eºcell = Eºreduction (cathode) - Eºreduction (anode)= 0.34 V – (- 0.76 V)
Eºcell = +1.10 V
16
• Positive voltages = reaction is spontaneous (cell will produce voltage, G is -)
• Changing the stoichiometric coefficients of a half-cell does not affect the value of Eº, since potential is a measure of the energy per electrical charge.
• However, having more moles means the reaction will continue for a greater time.
17
Using the Standard Reduction Potential table
• When Eºred is very negative, reduction is difficult (but oxidation is easy):Li1+ (aq) + e- → Li (s) Eºred = - 3.05 V– Li (s) oxidizes easily, so it is the strongest
reducing agent; lower on the Standard Reduction Potential chart signifies the anode.
• When Eºred is very positive, reduction is easy:
F2 (g) + 2 e- → 2 F1- Eºred = + 2.87 V– F2 reduces easily, so it is the strongest oxidizing
agent; higher on the chart signifies the cathode.
• Predicting metal “activity series”: most reactive metals are near bottom
18
20.5: Spontaneity of Redox Reactions• Gibbs free energy and emf:
Go = - n F Eo
n = # of moles of e- transferred
Faraday’s constant, F= 96,500 C/mol of e-
• 1F = 1 mole of e-, which has a charge of 96,500 C• Units = (mol e-)(C/mol e-)(V) = (mol e-)(C/mol e-)
(J/C) = J
• As Eºcell increases, Gº decreases (becomes more likely to be spontaneous)
Michael Faraday(1791-1867)
19
20.6: Effect of Concentration on Ecell
The Nernst Equation:G = Go + RT ln Q and Go = - n
F Eo
(R = 8.314 J/(mol•K))
- n F E = - n F Eº + RT ln Q Q
n
RTEE ln0
F
Walther Nernst
(1864-1941)
20
• At equilibrium, G = 0
Go K Eocell Reaction
Negative > 1 Positive Spontaneous
0 1 0 Equilibrium
Positive < 1 Negative
Non-spontaneous
Kn
RTE ln0
F
Kn
E log0591.00 since T º = 298 K
Equilibrium composition measurements
Calorimetry and entropy: S0 and S0
Electrical measurements: E0 and E0
cell
3 Thermodynamic Quantities:
22
* E0cell of cells
• Model of a portion of a cell membrane shows a K+
channel (blue) and a Na+ channel (red). The area outside the cell (top) is rich in Na+ and low in K+. Inside the cell, the fluids are relatively rich in K+ and low in Na+.
23
20.8-9: Electrolytic cells• Electrical energy is used to cause a
non-spontaneous reaction to occur.• Current (I): a measure of the charge
(Q) per unit time (t)
• Units = amps or amperes (A)1 A = 1 C / 1 s
t
QI
André-Marie Ampère(1775 – 1836)
24
Ex: Na+ (l) + Cl- (l) → Na (l) + Cl2 (g)Oxidation half-cell: 2 Cl- (l) → Cl2 (g) + 2 e-
Reduction half-cell: Na+ (l) + e- → Na (l)
anode
Pt Pt(-)(+)
cathode
e- flow
NaCl (l)
Cl2 (g) Na (l)
• In a voltaic cell: anode is (-), cathode is (+).
• In an electrolytic cell: anode is (+), cathode is (-).
25
20.9: Quantitative ElectrolysisEx: A current of 0.452 A is passed
through an electrolytic cell containing molten NaCl for 1.5 hours.
• Write the electrode reactions.
• Calculate the mass of products formed at each electrodes
Oxidation: 2 Cl- (l) → Cl2 (g) + 2 e-
Reduction: Na+ (l) + e- → Na (l)
Strategy: calculate the total charge of e- transferred, and use F to convert to moles.
26
C 2441sec) A)(5400 (0.452Q
Na mol 1
g 22.99
1
Na mol 1
C 96,500
1C) (2441Na Mass
FF
g 0.58
2
22 Cl mol 1
g 70.90
2
Cl mol 1
C 96,500
1C) (2441Cl Mass
FF
g 0.90
tIQor t
QI
27
Electrolysis of Aqueous Mixtures• If the voltage on an aqueous mixture of CuCl2 and
ZnCl2 in an electrolytic cell is slowly increased, what products will form at each electrode?
o What exists in solution that could be reduced?Cu2+, Zn2+, and H2O (not Cl-)
Cu2+ + 2 e- → Cu Eºred = 0.34 V
Zn2+ + 2 e- → Zn Eºred = - 0.76 V
2 H2O + 2 e- → H2 + 2 OH- Eºred = - 0.83 V
o Since Eºred (Cu) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the cathode):Cu (s) > Zn (s) > H2 (g)
28
• If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in an electrolytic cell is slowly increased, what products will form at each electrode?
o What exists in solution that could be oxidized?Cl- and H2O (not Cu2+ and Zn2+)
2 H2O → O2 + 4 H+ + 4 e- Eºox = - 1.23 V2 Cl- → Cl2 + 2 e- Eºox = - 1.36 V
o Since Eºox (H2O) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the anode):O2 (g) > Cl2 (g)