Download Red-ox reactions Electochemistry

24/11/2014
Oxidation number
Red-ox reactions
Electrochemistry
Lecture 9
• The oxidation number of an atom
(oxidation state) represents the
number of electrons lost, gained or
unequally shared by an atom.
• It could be zero, positive or negative.
• If the oxidation number is zero it
means that the atom has the same
number of electrons as in the free
neutral atom.
Rules for assigning
oxidation number
• Positive oxidation number means the
atom has fewer electrons assigned
to it than in neutral atom.
• Negative oxidation number means
the atom has more electrons
assigned to it than in the neutral
atom.
• All elements in their free state have an
oxidation number of zero
• H is +1 (except metal hydrides where is -1,
e.g. NaH)
• O is -2 (except in peroxides where is -1 and
OF2, where is +2)
• The metallic element in an ionic compound
has a positive oxidation number
1
24/11/2014
Rules ….
• In covalent compounds the negative
oxidation number is assigned to the most
electronegative atom.
• The algebraic sum of the oxidation
numbers of the elements in a compound is
zero.
• The algebraic sum of the oxidation
numbers of the elements in a poliatomic
ion is equal to the charge of the ion.
Ion/
atom
H+
Ca2+
Fe3+
P
O2Cl-
Oxidation Molecule
number/s
+1
+2
+3
0
-2
-1
Oxidation
numbers
SO4-2
CH4
NH3
H2O
H3PO4
Na2S2O3
Multiple oxidation
numbers:
• A) N2; N2O; NO; N2O3; NO2; N2O5
0 +1
+3
+5
B) P; P2O3; P2O5
0 +3
+5
C) Cl ; Cl2; Cl2O; Cl2O3; HClO3; HClO4
-1; 0; +1;
+3
+5
+7
2
24/11/2014
Quiz
• Find the oxidation number of:
• Br (a) or S (b) in the following
compounds;
a) HBr; HBrO2; HBrO3; HBrO4
b) H2S; H2SO4; Na2S2O3
Oxidation –reduction
reaction(redox)
• Redox is a chemical process in
which the oxidation number of an
element (elements) is changed.
• This process may involve the
complete transfer of electrons to
form ionic bond or only a partial
transfer or shift of electrons to form
covalent bonds.
Oxidation
Reduction
• This process occurs whenever
the oxidation number of an
element increases as a result of
losing electrons.
Br0 – 5e →Br+5
Cu0 - 2e → Cu2+
S-2 – 8e → S+6
• Reduction occurs whenever the
oxidation number of an element
decreases as a result of gaining
electrons:
Fe3+ + e→ Fe2+
Mn+7 + 5e → Mn+2
Cl+5 + 6e → Cl-1
3
24/11/2014
• The loss of electrons from an atom or
molecule is called oxidation, and the gain
of electrons is reduction.
• Oxidation and reduction occur
simultaneously in a chemical
reaction; one cannot take place
without the other.
This can be easily remembered through
the use of mnemonic devices. Two of the
most popular are: "OIL RIG" (Oxidation Is
Loss, Reduction Is Gain)
Oxidizing agent vs. reducing
agent
• The substance that causes an
increase in the oxidation state of
another substance is called an
oxidizing agent.
• The substance that causes a
decrease in the oxidation state of
another substance is called a
reducing agent.
http://www.landfood.ubc.ca/soil200/interaction/orgmatter_air.htm
4
24/11/2014
Zn + 2HCl(aq)→ ZnCl2 + H2
O.n. 0
+1;-1
+2; -1;
0
Zn0 is oxidized, so it is a reducing
agent
H+1 is reduced, so it is an oxidizing
agent.
http://www.emc.maricopa.edu/faculty/farabee/biobk/biobookenzym.html
Quiz
• In the following reactions,
identify what is oxidized, and
what is reduced, oxidizing and
Ballancing red-ox
reactions
• The total increase in oxidation
numbers must equal the total
decrease in oxidation numbers.
reducing agent :
• A) Cl2(g)+ 2NaBr →2NaCl + Br2
• B) 2PbO(s)→ 2Pb(s) + O2(g)
5
24/11/2014
All balanced equations must satisfy
two criteria.
• 1. There must be mass balance.
That is, the same number of atoms of each
kind must appear in reactants and products.
• 2. There must be charge balance.
The sums of actual charges on the left and
right sides of the equation must equal each
• In a balanced formula unit equation,
the total charge on each side will be
equal to zero.
• In a balanced net ionic equation,
the total charge on each side might not
be zero, but it still must be equal on the
two sides of the equation.
other.
Check if the reaction is
well ballanced
• 1.
H2O2 + H+ + MnO4-→ O2 + Mn2+ + H2O
2.
Mg + 2H+ →Mg2+ + H2
3.
ClO2+ OH- → ClO2- + ClO3- + H2O
THE HALF-REACTION METHOD
1. Write as much of the overall unbalanced
equation as possible, omitting spectator ions.
2. Construct unbalanced oxidation and reduction
half-reactions (these are usually incomplete as well
as unbalanced). Show complete formulas for
polyatomic ions and molecules.
3. Balance by inspection all elements in each halfreaction, except H and O.
6
24/11/2014
4. Balance the charge in each half-reaction by
adding electrons as “products” or “reactants.”
5. Balance the electron transfer by multiplying
the balanced half-reactions by appropriate
Example
• I2 + S2O3-2→ 2I- + S4O6-2
integers.
6. Add the resulting half-reactions and
eliminate any common terms.
I2→ I(reduction half reaction)
I2→ 2 II2 + 2e→ 2 I-(balanced reduction half reaction)
S2O3-2→ S4O6-2 (oxidation half reaction)
2S2O3-2→ S4O6-2
2S2O3-2→ S4O6-2 – 2e (balanced oxidation half
reaction)
Adding H+; OH- or H2O
In acidic solution: we add only H + or H2O
(not OH- in acidic solution)
In basic solution: we add only OH - or H2O
(not H+ in alkalic solution)
I2 + 2S2O3-2 +2e→ 2I- + S4O6-2 – 2e
Full red-ox reaction
7
24/11/2014
The following chart shows how to balance
hydrogen and oxygen
In acidic solution:
 To balance O add H2O and/ then
 to balance H add H+
In basic solution
To balance O:
For each O needed :
 add two OH- to side needing O
 and add one H2O to other side
To balance H:
 Add one H2O to side needing H
 and add one OH- to other side
Change in oxidation method
1. Write as much of the overall unbalanced equation
as possible.
2. Assign oxidation numbers to find the elements
that undergo changes in oxidation numbers.
3. a) Draw a bracket to connect atoms of the
element that is oxidised. Show the increase in
oxidation number per atom. Draw a bracket to
connect atoms of the element that is reduced.
Show the decrease in oxidation number per atom.
b) determined the factors that will make the total
increase and decrease in oxidation numbers
equal.
• 4. Insert coefficients into the
equation to make the total
increase and decrease in
oxidation number equal.
• 5. Balance the other atoms by
inspection.
8
24/11/2014
Quiz
• Balance the following red-ox
reaction with both methods:
S+ HNO3→ H2SO4 + NO
S0 → S+6 reduction half reaction
S0→ S+6 + 6e
N+5 +3e → N+2 oxidation half reaction
N+5 + 3e → N+2
• Number of electrons lost: 6
Number of electrons gained: 3 =
The second mode
0
- 6e
S0 + 2N+5→ S+6 + 2N+3
equation
+2
S+ HNO3→ H2SO4 + NO
this part should doubled
S0 + 2HNO3 →H2SO4 + 2NO –
+6
+5
3e
balanced
Oxidation
Reduction
S0 → S+6 + 6e
N+5 + 3e→N+2
The numbers of electrons from oxidation is 6.
The numbers of electrons from reduction is 3.
So the numbers of electrons from reduction should be
twiced.
9
24/11/2014
Quiz
1 S+ 2HNO3→ 1 H2SO4 + 2NO
• Ag + HNO3→ AgNO3 + NO + H2O
• Balance the following equations. For each
equation tell what is oxidized, what is
reduced, what is the oxidizing
• agent, and what is the reducing agent.
• (a) Zn(s) + HClO4(aq) → Zn(ClO4)2(aq) H2(g)
• (b) K(s) + H2O(l) → KOH(aq) + H2(g)
heat
• (c) NaClO3(s)
NaCl(s) + O2(g)
Redox titration
• KI + K2CrO4 + H2SO4 → I2 +
Cr2(SO4)3 + K2SO4 + H2O
• H2O2 + KMnO4 + H2SO4→ O2 + K2SO4
+ MnSO4 + H2O
10
24/11/2014
• In reaction of some amount Fe2+
cation with permanganate ions
(KMnO4; M=0.02 mole/L) in acidic
solution 20 ml of the titrant was
used.
Solution
• Start from chemical equation:
Fe2+ + MnO4- + H+ → Fe3+ + Mn2+ + H2O
Balance it.
• How many moles (grams) of iron ions
were present in the sample?
5Fe2+ + MnO4- + 8 H+ → 5Fe3+ +
+Mn2+ + 4H2O
5 moles of Fe2+ will react with 1
mole of MnO4-
• Step 3. calculate number of KMnO4
moles used:
n = M . V = 0.02 mol/l . 0.02 l
(conversion of 20 ml into l)
n= 4 . 10-4 moles of KMnO4 used
n
for Fe
= 5 . 4. 10-4 = 2 .10-3 moles of Fe2+
11
24/11/2014
An activity series place metals in order
from most active to least active
• Na<Mg<Al<Zn<Fe<Cd<Co<Ni<Pb< H
active
A more active metal will displace a
less active metal from solution.
A more active metal:
(i) loses electrons more easily
(ii) is the stronger reductant
• H<Cu<Hg<Ag<Au
•
noble
(iii) is the weaker oxidant
(iv) has a lower standard reduction
potential
The less active metal is less likely lose
electrons
(ii) is the weaker reductant
(iii) is the stronger oxidant
(iv) has a higher standard reduction
potential
Electrochemistry
• Electrochemistry deals with the
chemical changes produced by electric
current and with the production of
electricity by chemical reactions.
http://www.ausetute.com.au/redox.html
12
24/11/2014
Many metals are purified or are plated onto
jewelry by electrochemical methods.
Digital watches, automobile starters,
• Batteries serve as power sources for
all types of gadgets
• The energy in a battery comes from
calculators, and pacemakers are just a few
a spontaneous redox reaction where
devices that depend on electrochemically
the electron transfer is forced to
produced power.
take place through a wire
Corrosion of metals is an electrochemical
process.
• The apparatus that provides
electricity in this way is called a
galvanic or voltaic cell
• Cell reactions are obtained by
adding the half-reactions
A galvanic cell. The cell consists of two half-cells where the
oxidation and reduction half-reactions take place. The salt
bridge is required for electrical neutrality. The overall cell
reaction is: 2Ag+(aq)+Cu(s) 2Ag(s)+Cu2+ (aq)
• Half-reactions are balanced
using the ion-electron method
13
24/11/2014
The electrodes are assigned the name
anode or cathode.
Reduction (electron gaining) occurs at the
cathode.
Electrons appear as reactants in this halfreaction.
Oxidation (electron lossing) occurs at the
anode.
Electrons appear as products in this halfreaction.
• There are two types of
electrical conduction in a
galvanic cell
• Metallic conduction occurs when
electrons move through the wires
• Electrolytic conduction occurs
through the liquid by movement of
The movement of ions through the salt
bridge and in solution is required for
charge neutrality.
Cations move in the general direction of
the cathode.
Anions move in the general direction of
the anode.
ions, not electrons
14
24/11/2014
• The anode has negative polarity
Quiz
because the electrons left behind by
the Cu2+ ions give it a slightly
negative charge.
Among the following cations find
• The cathode has positive polarity
those from cathode and anode part
because of the Ag+ ions “joining” the
(remember about activity series):
electrode give it a slightly positive
Cu2+; Ag+, Zn2+, Ca2+,
charge.
For convenience, a standard cell notation
has been developed by chemists.
Cathode
Cu2+
Ag+
Cu2+
Ag+
Anode
Ca2+
Cu2+
Zn2+
Ca2+
Anode half-cell is specified on the left.
Cathode half-cell is specified on the right.
Phase boundaries are indicated using “|”.
The salt bridge separates the anode
and cathode and is indicated using “||”
15
24/11/2014
• The cell diagram for the coppersilver galvanic cell is
Galvanic cells can push
electrons through a wire.
The magnitude of this ability
Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)
(anode)
(cathode)
is expressed as a potential.
The maximum potential a
given cell can generate is
called the cell potential, Ecell
• The tendency for a species to
• The standard cell potential,
E0cell, is the cell potential
measured at 298 K (25oC) with
all ion concentration 1.00 M
gain electrons and be reduced
is its reduction potential.
• When measured at standard
condition, it is called the
standard reduction potential, E0.
16
24/11/2014
• When two half-cells are connected:
• The one with the larger reduction
• The difference in the two
standard reduction potentials
gives the standard cell potential
potential will acquire electrons and
undergo reduction.
• The half-cell with the lower reduction
potential will give up electrons and
 standard reduction   standard reduction 

 

o
Ecell
  potential of the    potential of the 
 substance reduced   substance oxidized 

 

undergo oxidation.
Calculate the standard potential of cells
formed from:
Cu│Cu2+(aq) and Zn2+ (aq)│Zn
E0 = +0.34; E0= -0.76 V
Cu│Cu2+ - substance reduced (oxidizing
agent; cathode);
The galvanic cell formula is as follows:
Zn│Zn2+(aq) ││ Cu2+(aq) │Cu
Solution
E0cell = (+0.34V) – (-0.76V) = 1.12V
Zn2+│Zn – substance oxidized (reducing
agent; anode)
17
24/11/2014
Determine oxidized and reducing agent
for half cells presented below. Write the
formula for galvanic cell. Calculate the
standard potential of cells formed from:
1. Ag│Ag+ and Na+│Na
E0 = +0.80; E0= -2.71 V
2. Fe│Fe2+ and Mg2+│Mg
E0 = -0.44; E0= -2.37V
It is not possible to measure
the reduction potential of an
isolated half-cell
The standard hydrogen
electrode (SHE).
• A reference electrode, called
the standard hydrogen
electrode (SHE), has been
assigned the potential of
exactly 0 V
18
24/11/2014
• Using a hydrogen half-cell, other
reduction potentials can be
measured
A galvanic cell
comprised of
copper and
hydrogen half-cells.
The reaction is
Cu2+(aq)+H2(g)
Cu(s)+2H+(aq)
Cell notation:
Pt(s), H2(g)|H+(aq)||Cu2+(aq)|Cu(s)
Galvanic series
• The galvanic series (or electropotential
series) determines the nobility of metals
and semi-metals.
• It is the order of metal and its cation pairs
according to increase of their
• For any pair of reduction
potentials ordered most reactive
to least reactive:
• The higher (more positive) reduction
potential will occur as a reduction
• The lower (more negative) reduction
potential will be reversed and occur
as an oxidation
standard reduction potentials, E0.
19
24/11/2014
Cathode (Reduction)
Half-Reaction
Standard Potential
E0 (volts)
Li+(aq) + e- -> Li(s)
-3.04
K+(aq) + e- -> K(s)
-2.92
Ca2+(aq)
-2.76
+
2e-
-> Ca(s)
2H+(aq) + 2e- -> H2(g)
0.00
Cu2+(aq) + e- -> Cu+(aq)
0.16
AgCl(s) + e- -> Ag(s) + Cl-(aq)
0.22
Cu2+(aq) + 2e- -> Cu(s)
0.34
• In a galvanic cell, the calculated
cell potential for the spontaneous
reaction is always positive.
• If the calculated cell potential is
negative, the cell is spontaneous in
the reverse direction.
20
24/11/2014
The free energy change (ΔG)
•The maximum useful work that
can be obtained from a reaction is:
for a system can also be used
to predict if a reaction is
 G  maximum work
spontaneous.
Free energy changes and cell
potentials are related.
ΔG0 = - n F E0 cell
or
ΔG = - n F Ecell
n = number of moles of electrons
transferred
F = Faraday constant = 96,485 C/mol e
E0cell = standard cell potential in volts
ΔG
Ecell
Reaction
<0
>0
Spontaneous
0
0
At equilibrium
>0
<0
Non-spontaneous
21
24/11/2014
Calculate the ΔGo for the following
reaction:
Zn(s) + Cu2+(aq) ↔ Cu(S) + Zn2+(aq)
Step 1: Break the redox reaction into oxidation
and reduction half-reactions.
The two equilibria which are set
up in the half cells are:
Cu2+ +2 e- → Cu (reduction)
Zn→ Zn2++ 2 e- (oxidation)
The negative sign of the zinc E° value shows that it releases electrons more readily
than hydrogen does.
The positive sign of the copper E° shows that it releases electrons less
readily than hydrogen.
Step 2: Find the E0cell of the cell.
From the Table of standard reduction potentials
Cu2+ + 2e → Cu; E0 = 0.342V
Zn2+ + 2 e- → Zn; E0 = -0.762 V
E0cell = E0reduction - E0oxidation
E0cell = 0.342 V – (- 0.762) V
E0cell = 1.104V
Step 3: Find ΔG°.
There are 2 moles of electrons transferred in the
reaction for every mole of reactant, therefore
n=2.
Another important conversion is 1 volt = 1
Joule/Coulomb
ΔG° = -nFE0cell
ΔG° = -(2 mol)(9.65 x 104 C/mol)(1.104J/C)
ΔG° = - 213072 J or - 213.07 kJ
22
24/11/2014
Self-checking quiz
Self-checking quiz
Calculate the ΔGo for the following
reaction:
2Hg+ + Cd→2Hg0 + Cd+2
Calculate the ΔGo for the following
reaction:
Ni2+ + Cd0→Ni0 + Cd+2
E0 for Hg+│Hg cell is + 0.615V
E0 for Ni2+│Ni cell is - 0.25 V
E0 for Cd2+│Cd cell is -0.403V
E0 for Cd2+│Cd cell is -0.403V
The cell consists of CdSO4, Cd, Hg2SO4,
Hg is named Weston Cell.
The cell which consists from Cd (anode)
and NiO(OH) (catode) forms gumstick
batteries
Lithium – ion battery, LIB
The positive electrode half reaction
(with charging being forwards) is:
The negative electrode half-reaction is:
Disassembled Ni–Cd battery from cordless drill.
1: outer metal casing (also negative terminal)
2: separator (between electrodes)
3: positive electrode
4: negative electrode with current collector (metal
grid, connected to metal casing).
Everything is rolled.
They are based on lithium cobalt oxide (LiCoO2), which
offers high energy density, but have well-known safety
concerns, especially when damaged
23
24/11/2014
Li-ion batteries provide light weight, high energy
density power sources. They are used in :
Portable devices: mobile phones, smartphones,
laptops and tablets, digital cameras and many
others.
Electric vehicles: Because of their light weight
Li-ion batteries are used for energy storage for
many electric vehicles, electric wheelchairs,
from radio-controlled models and model aircrafts
• Equating:
G  nFEcell
(nonstanda rd conditions)
G  nFE
(standard conditions)
o
o
cell
• The cell potential can be
related to the equilibrium
constant K
o
G o   RT ln K  nFEcell
thus
o
Ecell

o
RT
ln K or K  e nFEcell / RT
nF
to the Mars Curiosity rover.
Cell potentials depend on
concentrations
G  G 0  RT ln Q thus
o
 nFEcell  nFEcell
 RT ln Q or
o
Ecell  Ecell

RT
ln Q
nF
• The last expression is a form of the Nernst
equation which relates ion concentrations
to the cell potential
• Use molar concentrations (M) for ions and
partial pressures of gases in atmospheres
when calculating Q
Example:
In a certain zinc-copper cell:
Zn(s)+Cu2+(aq)Zn2+(aq)+Cu(s)
the ion concentrations are:
[Cu2+]=0.0100 M and [Zn2+]=1.0 M.
What is the cell potential at 298 K?
o
o
o
Ecell
 ECu
 0.34 V  (0.76) V  1.10 V
2  E
Zn2
o
Ecell  Ecell

RT [ Zn 2 ]
ln
 1.04 V
2 F [Cu 2 ]
For this two electron change at 298 K
24
24/11/2014
Self-check quiz
Find the cell potential of a galvanic cell based on
the following reduction half-reactions at 25 °C
Cd2+ + 2 e- → Cd
Pb2+ + 2 e- → Pb
The Nernst equation could be also
expressed as:
E0 = -0.403 V
E0 = -0.126 V
where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M.
For this reaction to be galvanic, the cadmium reaction
must be the oxidation reaction.
Cd → Cd2+ + 2 e- E0 = +0.403 V
Pb2+ + 2 e- → Pb E0 = -0.126 V
25
24/11/2014
• Galvanic cells, commonly called
batteries, can be classified as
either primary or secondary cells
• Primary cells are not designed to be
recharged
• Secondary cells are able to be recharged
Primary cells
Secondary cells
A rechargeable battery, storage battery, or accumulator is a
type of electrical battery. It comprises one or more electrical
cells, and is a type of energy accumulator. It is known as a
secondary cell because its electrochemical reactions are
electrically reversible.
26
24/11/2014
(−) Zn | NH4Cl | MnO2,C (+).
Electricity in nature
• A battery is usually a collection
of cells connected in series
• When connected in series, the
voltage of each cell is added to
provide the total voltage of the
battery
Battery in medicine
Pacemaker
Tool for hearing improvement
27
24/11/2014
Electrochemistry in
medicine.
Electrocardiography (ECG or EKG from
Electroencephalography (EEG) is the recording of electrical
activity along the scalp
transthoracic (across the thorax or chest)
Greeek: kardia, meaning heart) is a
interpretation of the electrical activity of the
heart over a period of time, as detected by
electrodes attached to the surface of the skin
and recorded by a device external to the body
Epileptic spike and wave discharges monitored with EEG
Voltage gained ion channels
Voltage-gated ion channels for cations (sodium,
calcium, and potassium channels) and anions
(chloride channels) are described together with
electrophysiology, pharmacology and other
medical sciences.
28