Download Redox Reactions and Electrochemistry

Redox Reactions and
Electrochemistry
Problem Set
Chapter 5: 21-26, Chapter 21: 15-17, 32, 34, 43, 53, 72, 74
Oxidation/Reduction & Electrochemistry
Oxidation – a reaction in which a substance gains
oxygen atoms
(e.g. the oxidation of a hydrocarbon)
H
R
C
H
H
OH
O2
R
C
H
O
O2
H
R
O
O2
C
H
O2
C
R
OH
O
C
O + H2O
(i.e. this is equivalent to the combustion of a hydrocarbon)
Reduction - a reaction in which a substance loses
oxygen atoms
Oxidation/Reduction & Electrochemistry
• A different type of “redox” (i.e. reduction plus oxidation)
reaction does not involve gain or loss of oxygen.
• Tarnishing of silver is a redox reaction that produces Ag2S.
• This spontaneous reaction can be reversed with a coupled
reaction (add 80 g of baking soda and 80 g of table salt per
litre of near boiling water in an aluminum pan to a depth
covering the silver object).
3 Ag2S + 2 Al + 6 H2O → 6 Ag + 2 Al(OH)3 + 3 H2S
1
Oxidation/Reduction & Electrochemistry
Let’s look at a simple case of this type of redox reaction:
Cu+2/Zn
Cu /Zn+2
Cu+2(aq) + Zn(s)
Cu(s) + Zn+2(aq)
Redox Reactions
Cu+2(aq) + Zn(s)
Cu(s) + Zn+2(aq)
The reaction can be represented by two half-reactions
in which electrons are either gained or lost and the
“oxidation state” of elements changes :
Cu+2(aq) + 2eZn(s)
Cu(s)
Zn+2(aq) + 2e-
oxidation state of Cu +2
0
oxidation state of Zn
+2
0
Reduction – a process in which electrons are gained.
(The oxidation state of an element decreases and
electrons appear on the left side of the half-reaction.)
Oxidation – a process in which electrons are lost.
(The oxidation state of an element increases and
electrons appear on the right side of the half-reaction.)
Oxidation States - Review
1) Oxidation state of an atom in a free element is 0.
2) Total of the oxidation states of atoms in a molecule or ion
is equal to the total charge on the molecule or ion.
3) Group 1A and Group 2A metals have oxidation state of +1
and +2 respectively.
4) F always has an oxidation state of –1. Cl also has
oxidation state of –1 unless it is bonded to oxygen or
fluorine.
5) H almost always has an oxidation state of +1.
6) O has oxidation state of –2 (unless bonded to itself or F).
7) When bound to metals, group 7A, 6A and 5A elements
have oxidation states of -1, -2, -3 respectively.
2
Electronegativities-Review
Example
What is the formal oxidation state of P in:
+1
-2
H3PO4
: 3 x (+1) + ? + 4 x (-2) = 0 (charge on molecule)
oxidation state of P = +5
?
H2PO4-
: 2 x (+1) + (O.S. of P) + 4 x (-2) = -1 (charge on ion)
oxidation state of P = +5
HPO42-
: oxidation state of P = +5
If the oxidation state of elements do not change in a
reaction, it is NOT a redox reaction!
H3PO4(aq) + 3 OH-(aq)
3H2O + PO43-(aq) acid/base reaction
Balancing Redox Reactions
Half-reaction method (not the same as method in textbook).
1) Identify species in which the oxidation state of an
element is changing. Write the skeleton half-reactions
including balancing of the redox atoms if necessary.
2) Identify oxidation state on both sides of equation for
elements that have a change in oxidation state.
3) Add appropriate number of electrons to either left or
right to balance oxidation states of redox atom(s).
4) Balance changes on left and right side of equation by
adding H+ (if in acidic solution) or OH- (if in basic
solution).
5) Add appropriate number of H2O’s to left or right side of
equation to balance atoms in the half-reaction.
3
Balancing Redox Reactions, cont’d
At this point, both half-reactions should be
balanced. The next step is to combine the two halfreactions to form an overall equation.
6) Multiply through each half-reactions by appropriate
coefficients to match electrons in each half-reaction.
(i.e. number of electrons lost by the oxidized species
must equal the number gained by the reduced one)
7) Add half-reactions and cancel electrons and other
common species on left and right sides of the
equation.
8) Check Reaction! It should be balanced in terms of
oxidation states, charge and atoms.
IF NOT, YOU HAVE MADE A MISTAKE!
Examples
Determining sulfite in wastewater.
Sulfite is reacted with permanganate to produce sulfate
and Mn(II) ion in acidic solution. Balance the redox
reaction.
SO32- + MnO4-
+4
SO42- + Mn2+
skeleton reaction
+6
SO32-
SO42-
identify oxidation states
SO32-
SO42- + 2e-
balance O.S. with electrons
SO32-
SO42- + 2e- + 2H+
balance charges with H+
H2O + SO32-
SO42- + 2e- + 2H+
balance atoms with H2O
Cont’d, Mn half-reaction
+7
+2
MnO4MnO4 +
-
Mn2+
5e-
identify oxidation states
Mn2+
8 H+ + MnO4- + 5e8 H+ + MnO4- + 5e-
balance O.S. with electrons
Mn2+
balance charges with H+
Mn2+ + 4H2O balance atoms with H2O
Balanced Half-Reactions
H2O + SO328
H+ +
MnO4 +
-
SO42- + 2e- + 2H+
5e-
Mn2+
+ 4H2O
x5
x2
to balance e-’s
4
Balancing full equation
Balanced full reaction:
5H2O + 5SO32-
5SO42- + 10e- + 10H+
6
3
16 H+ + 2MnO4- + 10e-
2Mn2+ + 8H2O
5SO32- + 2MnO4- + 6 H+
5SO42- + 2Mn2+ + 3H2O
Check atom balance. OK
Try example 5.7 using this approach, use OH- to
balance charge in basic solution. Much easier.
This method forces you to know oxidation states.
Another Example
Write the half reaction for Cr2O72- Æ Cr+3
+6
Cr2O72-(aq)
Cr+3(aq)
Cr2O72-(aq)
2
Cr+3
Cr2O72-(aq)
2 Cr+3(aq)
Cr2O72-(aq)
+3
+
2(3e-)
Cr2O72-(aq) + 6e14
H+
+
Cr2O72-(aq)
2
Cr+3
skeleton (in acidic solution),
(aq)
(aq)
2 Cr+3(aq)
+
6e-
14 H+ + Cr2O72-(aq) + 6e-
2
Cr+3
balance redox atoms
determine O.S. of redox atoms
balance O.S. with e-’s
”
“
+
(aq) balance charges with H
2 Cr+3(aq) + 7H2O balance atoms with H2O
Extra Practise Balancing Redox
Reactions (solutions on web site)
#1) Cl2 → ClO- + Cl- in basic solution
#2) I- + IO3- → I2 in acidic solution
#3) H2O2(aq) → O2(g)
(in either acidic or basic solution)
5
Disproportionation Reactions
A disproportionation reaction occurs when an element in
a substance is both oxidized and reduced.
Example:
H
O O
H
-1
-2
2 H2O2(aq)
Hydrogen peroxide: antiseptic
agent, O2 acts as germicide
0
2 H2O(l) + O2(aq)
Ox/Red Agents, cont’d
Oxidizing Agent – a chemical substance that oxidizes
(removes electrons from) other substances in a
chemical reaction. In the process of oxidizing
something, the oxidant becomes reduced; it’s
oxidation state decreases.
Reducing Agent – a chemical substance that reduces
(loses electrons to) other substances. In the process
of reducing, the reductant becomes oxidized; it’s
oxidation state increases.
Oxidizing and Reducing Agents
Removes
electrons
Oxidation States of Nitrogen
Loses
electrons
(best oxidizing agent)
(best reducing agent)
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Oxidizing Agents
O2 – Probably the most common and most important
oxidant known to us. Ubiquitous.
Organic Oxidation Schemes (Example: methane)
-4
-2
O2
CH4
0
H3C OH
methane
methanol
(alkane)
(alcohol)
O2
H2C O
O2
+2
O
HC
OH
O2
+4
O C O
formaldehyde
(aldehyde)
formic acid
(carboxylic acid)
carbon dioxide
(inorganic carbon)
Other oxidizing agents
Oxides in their highest oxidation state are frequently
strong oxidizing agents.
+6
+5
NO3
+4
HNO3
+2
NO2
NO
strong oxidizing agents
+7
+5
HClO4
ClO3-
+1
0
N2 O
N2
weaker oxidizing agents
+3
+1
HClO2
HOCl
HNO3 and HClO4 are oxidizing acids.
Non-oxidizing acids – HCl, HBr, HI, acids for which
the only possible reduction half-reaction is:
2H+(aq) + 2e-
H2(g)
Oxidizing Agents, cont’d
HNO3 is a much stronger oxidizing agent than H+.
Metals that dissolve in dilute
H+ to produce H2
Metals that will not
dissolve
Li, Na, K (1A metals)
Mg, Ca (2A metals)
Al, Zn
Fe, Sn, Pb
Cu,
Ag,
Au,
Hg
Practice problem – Cu will dissolve in HNO3 producing
Cu+2 in solution and the brown gas NO2. Write a
balanced equation for this process.
7
Electrochemistry
(a) Cu(s) / Ag+(aq)
(b) Cu(s) / Zn2+(aq)
Cu2+(aq)/ Ag(s)
No reaction!
Spontaneous!
Not spontaneous!
(∆G < 0)
Ag+(aq) + eCu(s)
(∆G > 0)
Ag(s)
Cu2+(aq) + 2 e-
reduction
oxidation
Electrochemical Cells
Flow of electrons
(current) can do work.
We can connect halfreactions in separate
containers through
an electrical circuit.
This will produce a
current (electron
flow) and voltage
according to the
spontaneity of the
reactions.
Atomic view of a
Voltaic (galvanic) cell
Salt bridge
(e.g. KNO3)
maintains
neutrality
Anode – oxidation
Cathode - reduction
8
Cell Diagrams
• anode (oxidation) is placed at left side of diagram
• cathode (reduction) is placed on right side of diagram
• boundary line, |, indicates a boundary between different
phase (i.e. solution|solid)
• a double boundary line || indicates a boundary (i.e. salt
bridge) between the two half-cell compartments
anode
(oxidation)
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
half-cell
salt bridge
cathode
(reduction)
half-cell
Voltages and Current
anode
cathode
-
+
Electromotive Force (EMF) - The voltage difference
between two solutions provides a measure of the
driving force of the electron transfer reaction.
Standard Electrode Potentials
In electronics and electricity theory, a voltage is a
measurement of the potential to do electrical work
measured between two points in a circuit. Absolute
measurements of potential (voltage) at a single point
are meaningless, UNLESS, they are measured against
some known reference.
In electricity, that reference is known as “ground”.
In electrochemistry, that reference is the standard
hydrogen electrode (SHE).
A Standard Electrode potential, Eo, measures the
tendency for the reduction process to occur at an
electrode, when all species have unit activity
(substances in solution are ~ 1.0 M or, if gases, are at
1 bar {~1 atm} pressure).
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Standard Hydrogen Electrode (SHE)
aH2 = 1.0 ~ PH2 = 1.0 bar ~ 1.0 atm
aH3O+ = aH+ = 1.0 ~ [H+] = 1M
|| H+(aq) (1M) | H2(g) (1 atm) | Pt
2H+(aq) + 2e-
H2(g) Eo = 0.00 V
frequently written as: EoH+(aq)/H2(g)
Easy to reduce,
hard to oxidize
(good oxidizing
agents)
↑
↓
Hard to reduce,
easy to oxidize
(good reducing
agents)
Standard Electrode (reduction) Potentials
The potential of an electrochemical cell under
standard conditions may be calculated by
Eo cell = Eocathode – Eoanode
where the Eo’s are standard reduction potentials taken
from a table.
The cathode is the electrode at which reduction
occurs (electrons on left side of equation, oxidation
state decreasing).
The anode is the electrode at which oxidation occurs
(electrons on right side of equation, oxidation state
increasing).
Also for a spontaneous reaction, Eo cell > 0, as we will
see shortly.
10
Example 21-2
A new battery system currently under study for possible
use in electric vehicles is the ZnCl2 battery.
Reaction: Zn(s) + Cl2(g)
ZnCl2(aq)
What is the standard potential of the cell, Eo.
Zn(s)
EoZn2+/Zn = -0.763V
Zn2+(aq) + 2eoxidation potential
Zn(s)
Zn2+(aq) + 2e-
Cl2(g) + 2e-
EoZn/Zn2+ = - EoZn2+/Zn = -(-0.763V) = +0.763 V
2Cl-(aq)
Zn(s) + Cl2(g)
EoCl2/Cl- =
ZnCl2(aq)
Eo
+ 1.358 V
cell =
+ 2.121 V
OR
Eocell = Eocathode – Eoanode = 1.358 – (- 0.763)V = 2.121 V
Spontaneous change in a Cell
Previously, it was said Ecell > 0 for a spontaneous
reaction. Where did this come from?
Electrical work:
Welectrical = Q V
Q = charge, V = voltage
If Q in coloumbs, V in volts, W in joules
Related:
P = iV
{P = power
i = current (charge/time), V = voltage}
If i - coloumbs/sec (Amp), V -volts, P- joules/sec = watts
In an electrochemical cell,
Q=nxF
n = moles of electrons
F = charge/mole of electrons = Faraday
F = 96485 C/ mole of electrons
V = Ecell
Spontaneous change, cont’d
Welectrical = Q V = nFEcell
This applies to a reversible process (implying that the
reaction is carried out slowly enough that the system
maintains equilibrium). Previously it was argued that the
amount of work we can extract from a chemical process
is equal –∆G (pg 796, Petrucci “Are You Wondering”
box).
∆G = – Welectrical
∆G = – nFEcell
∆Go = – nFEocell
If Eocell > 0, ∆Go < 0 and the reaction is spontaneous
If Eocell < 0, ∆Go > 0 and reaction is nonspontaneous
11
Behavior of Metals
Previously we said that experimental evidence shows the
following:
Metals that dissolve in
Metals that will
dilute H+ to produce H2
not dissolve
Li, Na, K (1A metals)
Mg, Ca (2A metals)
Al, Zn
Fe, Sn, Pb
Cu,
Ag,
Au,
Hg
Now we can better understand this:
Mn+(aq) + n e-
M(s)
2 H+(aq) + 2 e-
oxidation
H2(g)
reduction
Eo = 0.00V
Eocell = Eocathode – Eoanode = 0 – EoM+/M
If EoM+/M < 0, Eocell > 0, the process is spontaneous.
If EoM+/M > 0, a stronger oxidizing agent than H+ is required
(i.e. HNO3, HClO4) .
Nernst Equation
Previously, we talked about standard
electrode potentials in which everything
was in its standard state.
Very rarely are things in standard state!
∆G = ∆Go + RT ln Q
R = gas constant
T = temperature (K)
Q = reaction quotient
-nFEcell = -nFEocell + RT ln Q
Ecell = Eocell – RT/nF ln Q = Eocell – RT/ (2.303 nF) log Q
E
cell
o
= Ecell
−
0.0592
logQ,
n
for T = 25o C
Nernst Equation
Applications of the Nernst Equation
1) Draw the condensed
cell diagram for the
voltaic cell pictured at
right.
2) Calculate the value
of Ecell.
1) Pt|Fe2+(0.1M), Fe3+ (0.2M) || Ag+ (1.0M)|Ag(s)
2) The cell is in nonstandard conditions so we need to
apply the Nernst equation - we will need to find Eocell , n,
and Q.
12
cont’d
What is Q ?
From table of Standard Reduction Potentials:
Fe3+ + eAg+
+
e-
Fe2+
EoFe3+/Fe2+ = 0.771 V
Ag
EoAg+/Ag = 0.800 V
Overall reaction:
Cathode:
Ag+ + e-
Ag(s)
Anode:
Fe2+
Fe3+
after we combine
+
Ag+(aq) + Fe2+(aq)
half-reactions, n = 1
e-
Fe3+(aq) + Ag(s)
Q=
What is Eocell ?
[Fe 3+ ]
[Fe 2+ ][Ag+ ]
Eocell = Eocathode – Eoanode = EoAg+/Ag – EoFe3+/Fe2+
= 0.800V – 0.771V = 0.029V
Cont’d
0.0592
E cell = E −
log Q
n
0.0592
[Fe3+ ]
= 0.029 V log
1
[Fe2+ ][Ag+ ]
o
cell
= 0.029 V - 0.0592 log
(0.20M)
= 0.011V
(0.10M)(1.0M)
Example A: Calculate Ecell for the following cell
Al|Al3+(0.36M) || Sn4+ (0.086 M), Sn2+ (0.54 M) |Pt
Al3+ + 3e-
Al
Sn4+ + 2e-
Sn2+
Cathode:
Sn4+
Anode:
EoAl3+/Al = -1.676 V
EoSn4+/Sn2+ = 0.154 V
+
Al
2eAl3+
Sn2+ x 3
+ 3e-
x2
cont’d
3 Sn4+(aq) + 2 Al(s)
Q=
3 Sn2+(aq) + 2 Al3+ (aq)
after we combine
half-reactions, n = 6
[Sn2 + ]3 [Al3 + ]2
[Sn4 + ]3
Eocell = Eocathode – Eoanode = EoSn4+/Sn2+ – EoAl3+/Al
= 0.154 – (-1.676)V = 1.830V
0.0592
log Q
n
0.0592
[Sn2+ ]3 [Al3+ ]2
= 1.830 V log
6
[Sn4 + ]3
E
cell
= Eocell −
= 1.830 V -
0.0592
(0.54M)3 (0.36M)2
= 1.815V
log
6
(0.086)3
13
Change in Ecell with Conditions
Zn(s) + Cu2+(aq)
o
E cell = E cell
−
Slope = -59/2 mV per decade
change in log{[Zn2+]/[Cu2+]}
Zn2+(aq) + Cu (s)
[Zn 2+ ]
0.0592
0.0592
log
logQ = 1.103V −
2
n
[Cu 2+ ]
If we let cell reaction proceed,
reaction shifts to right, [Zn2+]
increases, [Cu2+] decreases and Ecell
decreases. When does it stop?
It stops at equilibrium, Ecell = 0.00V
0.0592
Keq !!
0.0 = Eocell −
logQ eq
n
K eq = 10
 nEocell 


 0.0592 
We can calculate Keq
from Eo values! For
above reaction,
Keq = 1.5 x1037
Concentration Cells
Both half-cells are the same chemical system, just
different concentrations. The driving force (i.e. the
EMF) is provided by the difference in concentrations.
Pt|H2 (g, 1.0 atm)| H+ (x M) || H+ (1 M) |H2(g,1.0 atm)|Pt(s)
Concentration Cell
Cathode:
2 H+ (1M) + 2e-
Anode:
H2(g)
H2(g)
2 H+ (xM) + 2e-
Eocell = Eocathode – Eoanode = 0.00 – 0.00 = 0.00V
E cell = Eocell −
=−
[H+ ]2
0.0592
0.0592
logQ = 0.00V −
log + 2 anode
n
2
[H ] cathode
0.0592
X2
 0.0592 
log 2 = − 
 ( 2logX) = −0.0592logX
2
1
 2 
Since pH = -log X,
Ecell = 0.0592 pH
This concentration cell behaves as a pH meter! Other
concentration cells can be used to measure unknown
concentrations of other species (i.e. potentiometry).
14
Determination of Ksp (see Example 21-10)
From measured Ecell,
determine Ksp.
Solution:
Set up Nernst equation
with Ag+ (xM) at anode,
0.1M at cathode.
Solve Nernst equation to
get x.
x = [Ag+] = S, [I-] = S
Ksp = S2
Ag(s) | Ag+ (sat. AgI) || Ag+ (0.1M) |Ag(s)
Electrolysis
The use of an externally applied voltage to force an
electrochemical reaction, even if it is naturally
nonspontaneous.
Spontaneous!
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
What about the reverse process?
Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)
Eocell = + 1.10 V
Nonspontaneous!
Eocell = - 1.10 V
But if we apply a potential > 1.10 V across the cell, we
overcome the natural negative voltage, thus providing
the driving force to make the reaction proceed.
Current is in opposite direction of voltaic, or galvanic,
cell.
Galvanic and Electrolytic Cells
electron flow reversed
External energy
(voltage) source
Galvanic Cell
Electrolytic Cell
Regardless of the cell type, anode and cathode always
defined by the process: oxidation at the anode,
reduction at the cathode.
15
Zn/Cu2+ electrolysis example continued...
The amount of current that flows in the electrolytic cell
tells us how much Zn has been produced or how much
Cu2+ has dissolved.
Faraday’s Law of Electrolysis:
The number of moles of product formed in an electrolysis
cell by an electric current is chemically equivalent to the
number of moles of electrons supplied.
or
charge
(coulombs)
Note: 1 A = 1C/s
time
(seconds)
Q = nF = it
current
(amperes)
moles of
electrons
Faradays constant = 96485 C /mole of e-
Example 21-12
Electrodeposition of Cu can be used to determine Cu2+
content of sample.
Cathode: Cu2+ + 2 eAnode: 2H2O
Cu(s)
O2(g) + 4H+(aq) + 4e-
What mass of Cu is deposited in 1 hr if current = 1.62A?
Solution: Find moles of electrons, then find moles of
Cu, then find mass of Cu.
Mole of e- = 1.62 A (C/s) x 3600 sec x 1/(96485 C/mole e-)
Mole of Cu = mole e- x 1 mole Cu / 2 mole eMass Cu = moles Cu x 63.456 g Cu/mole Cu
Answer = 1.92 g of Cu deposited in 1 hour
Cont’d
Example B: How long will it take to produce 2.62 L of O2(g)
at 26.2oC and 738 mmHg at a Pt anode with a constant
current of 2.13A?
Solution: Find moles of O2, then find moles of electrons,
then find charge, then find time. 738
atm × 2.62L
PV
760
Mole of O2:
n=
=
RT 0.08206Latmmol −1K −1 × 299.35K
(recall anode reaction: 2H2O
O2(g) + 4H+(aq) + 4e-)
Mole of electrons = moles of O2 x 4 mole electrons/ mole O2
Charge = moles of electrons x F (C/mole of electrons)
Time = Charge (C)/Current (C/s)
Answer = 18829 sec = 5.23 hr = 5 hr & 14min
16
Chlor-Alkali Process
Electrolysis of NaCl solutions
Cl2(g) + 2 eanode
-EoCl2/Cl- = -1.358 V
2 Cl-(aq)
2 H2 O + 2 e -
H2(g) + 2 OH-(aq) cathode EoH2O/H2 = -0.828 V
2 H2O + 2 Cl-(aq)
Cl2(g) + H2(g) + 2OH-(aq)
Eocell = -2.19 V
- Cl2 produced at anode
- H2 and NaOH(aq) produced at cathode
- membrane allows Na+ movement
- 11% NaOH and 15% NaCl is
concentrated, NaCl crystallized and
removed
- final product, 50% NaOH (1% NaCl
impurity), Cl2, H2
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