Transcript Document

Oxidation-Reduction
Larry J. Scheffler
Lincoln High School
LEO
says
GER!
Loss of Electrons = Oxidation
Gain of Electrons = Reduction
Oxidation Numbers
• Oxidation is the loss of electrons;
Reduction is the gain of electrons
• Oxidation and reduction go together.
Whenever a substance loses electrons
and another substance gains electrons
• Oxidation Numbers is a system that we
can use to keep track of electron
transfers
Oxidation Numbers
Oxidation numbers always refer to
single atoms
The oxidation number of an
uncombined element is always 0
O2, H2, Ne, Zn
The oxidation number of Hydrogen
is usually +1 Hydrides are an
exception They are -1
HCl, H2SO4
The oxidation number of Oxygen is
usually -2 Peroxides are an exception
They are –1. In OF2 oxygen is a +2
H2O, NO2, etc.
Oxidation numbers of monatomic
ions follow the charge of the ion
O2-, Zn2+
The sum of oxidation numbers is
zero for a neutral compound. It is
the charge on a polyatomic ion
LiMnO4
SO42-
Practice Assigning Oxidation
Numbers
NO2
O is -2 x 2 = -4
N must equal +4
N2O5
O is -2 x 5 = -10
N must equal +10/2 = +5
HClO3
O is -2 x 3 = -6; H is +1;
Cl must equal +5
HNO3
O is -2 x 3 = -6; H is +1;
N must equal +5
Ca(NO3)2
O is -2 x 3 = -6 x 2 = -12;
Ca is +2;
N must equal +10/2 = +5
KMnO4
O is -2 x 4 = -8; K is +1;
Mn must equal +7
Practice Assigning
Oxidation Numbers
Fe(OH)3
O is -2 x 3 = -6; Fe is +3;
H is +1 x 3 = +3
K2Cr2O7
O is -2 x 7 = -14;
K is +1 x 2 = +2;
Cr must equal +12/2 = +6
CO32-
O is -2 x 3 = -6; Charge = -2
means that C must be +4
CN-
N is -3; Charge = -1 means that
C must be +2
K3Fe(CN)6
N is -3 x 6 = -18;
C is +2 x 6 = +12;
K is +1 x 3 =+3; Fe must be +3
CH4
H is +1 x 4 = +4;
C must be -4
Using Oxidation Numbers
•
Careful examination of the oxidation
numbers of atoms in an equation allows
us to determine what is oxidized and
what is reduced in an oxidation-reduction
reaction.
Using Oxidation Numbers
• An increase in the oxidation number indicates
that an atom has lost electrons and therefore
oxidized.
• A decrease in the oxidation number indicates
that an atom has gained electrons and therefore
reduced
• Example
Zn
+ CuSO4  ZnSO4 + Cu
0
+2 +6-2
+2+6-2
0
Zn: 0  + 2  Oxidized
Cu: +2  0  Reduced
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cl2
Cu
+
KBr

KCl +
Br2
+ HNO3  Cu(NO3)2 + NO2 + H2O
HNO3 +
I2
 HIO3 +
NO2
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cl2
0
+
KBr
+1-1

KCl +
+1-1
Br increases from –1 to 0 -- oxidized
Cl decreases from 0 to –1 -- Reduced
K remains unchanged at +1
Br2
0
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
0
+1+5-2
+2 +5-2
+4 –2 +1-2
• Cu increases from 0 to +2. It is oxidized
• Only part of the N in nitric acid changes from +5
to +4. It is reduced
• The nitrogen that ends up in copper nitrate
remains unchanged
Exercise
For each of the following reactions
find the element oxidized and the
element reduced
HNO3 +
I2
 HIO3 +
NO2
1 +5 -2
0
+1+5-2
+4-2
•
N is reduced from +5 to +4. It is reduced.
•
I is increased from 0 to +5 It is oxidized
•
The hydrogen and oxygen remain unchanged.
Oxidation-Reduction
Reactions
• All RedOx reactions have one element
oxidized and one element reduced.
• The compound that supplies the electrons (is
oxidized) is the reducing agent.
• The compound that accepts the electrons (is
reduced) is the oxidizing agent.
• Occasionally, the same element may undergo
both oxidation and reduction. This is known
as an auto-oxidation reduction.
Balancing Redox Reactions
• Many chemical reactions involving
oxidations and reductions are complex and
very difficult to balance by the “guess and
check” methods we learned earlier.
• For complicated reactions, a more
systematic approach is required.
Balancing Redox Reactions
There are several basic steps
1. Assign oxidation numbers to the species in the reaction to
find which atom is oxidized and which is being reduced.
(Just learned this.)
2. Write half reactions for the oxidation and reduction
Take compounds where “action” took place, split them and write them
as individual reactions. There will be 2 half reactions.
3. Balance the atoms that change in the half reaction.
If there are atoms of an element on one side, there must be EQUAL
number of those same atoms on the other side.
Use coefficients in front of atoms/compounds. (this does NOT include
H or O)
4. Balance each half-equation for Oxygen by adding as many
H2O to the opposite side that need Oxygens.
Balancing Redox Reactions
There are several basic steps (cont.)
5. Balance each half-equation for Hydrogen by adding as
many H+ needed to the opposite side from water was
added.
6. Balance each half-equation for charge by adding as
many e- needed to get equal charges on each side of
the half reactions.
7. The number of electrons in EACH half-reaction must be
equal (and on opposite sides). Multiply any halfequations to get the # of electrons to be equal.
8. Combine the half reactions, cancelling out anything that
is the same on both sides, including electrons.
9. Check your work. Make sure that both the atoms and
charges balance
Exercise
Balancing Redox Equations 1
Cu + HNO3  Cu(NO3)2 + NO + H2O
Balancing Redox Equations 2
HNO3 + I2  HIO3 + NO2 + H2O
Metal Displacement Reactions
• The electrochemical cell potentials form the basis for
predicting which metals will react with salt solution of
other metals
• This order of reactivity of metals in single replacement
reactions is called the activity series
• The solid of more reactive metals will displace ions of
a less reactive metal from solution.
**(Higher metal will replace the ions lower )**
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Metal Displacement Reactions
**(Higher metal will replace the ions lower)**
• Oxidizing and reducing agents are not the same strength
• The relative reactivity of metals is based on potentials of half
reactions (reductions).
• Elements with very different potentials (far apart on table)
react most vigorously. More reactive the element, the
stronger the reducing agents (they are more easily oxidized).
• They show the potential difference, in volts, between the
electrodes of an electrochemical cell.
• A positive value indicates a spontaneous reaction (indicates
that the direction is positive.)
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Standard Electrode
Potentials in Aqueous
Solution at 25°C
Cathode (Reduction)
Half-Reaction
Standard Potential
E° (volts)
Li+(aq) + e- Li(s)
-3.04
 K(s)
-2.92
K+(aq)
Ca2+(aq)
+
2e-
 Ca(s)
-2.71
Mg2+(aq) + 2e-  Mg(s)
-2.38
Al3+(aq) + 3e Al(s)
-1.66
2H2O(l) + 2e-  H2(g) + 2OH-(aq)
-0.83
 Zn(s)
-0.76
Cr3+(aq) + 3e-  Cr(s)
-0.74
 Fe(s)
-0.41
Cd2+(aq) + 2e-  Cd(s)
-0.40
Ni2+(aq) + 2e-  Ni(s)
-0.23
 Sn(s)
-0.14
Pb2+(aq) + 2e-  Pb(s)
-0.13
 Fe(s)
-0.04
2H+(aq) + 2e-  H2(g)
0.00
Sn4+(aq) + 2e-  Sn2+(aq)
0.15
Cu2+(aq) + e-  Cu+(aq)
0.16
Cu2+(aq) + 2e- Cu(s)
0.34
 Cu(s)
0.52
Fe2+(aq)
Sn2+(aq)
Fe3+(aq)
+
+
+
+
Cu+(aq)
+
2e2e-
2e3e-
e-
Fe3+(aq) + e-  Fe2+(aq)
0.77
 2Hg(l)
0.80
Hg2
2+(aq)
+
2e-
Ag+(aq) + e-  Ag(s)
0.80
Hg2+(aq) + 2e-  Hg(l)
0.85
2Hg2+(aq)
+
2e-
 Hg2
2+(aq)
Ce4+(aq) + e-  Ce3+(aq)
Most reactive
-2.76
Na+(aq) + e-  Na(s)
Zn2+(aq)
Reference Point
+
e-
Decreasing
reactivity
0.90
1.44
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Electrochemical Cells
The Metal Activity
Series
(most reactive to
least reactive)
Metal
K
Na
Li
Sr
Ion
K+
Na+
Li+
Sr2+
Ca
Ca2+
Mg
Mg2+
Al
Al3+
C
included for comparison
Zn
Zn2+
Cr
Cr2+
Fe
Fe2+
Cd
Cd2+
Co
Co2+
Ni
Ni2+
Sn
Sn2+
Pb
Pb2+
H2
H+
Cu
Cu2+
Ag
Ag+
Hg
Hg2+
Au
Pt
Au3+
Pt2+
Oxidized (-0.76V)
included for comparison
Reduced (+0.34V)
21
The Activity Series
K
K+ + e-  K
-2.93
Ca
Ca2+ + 2 e- Ca
-2.87
Na
Na+ + e-  Na
-2.71
Mg
Mg2+ + 2 e-  Mg -2.37
Al
Al3+ + 3 e  - Al
-1.66
Zn
Zn2+ + 2 e-  Zn
-0.763
Fe
Fe2+ + 2 e-  Fe
-0.440
Sn
Sn2+ + 2 e-  Sn
-0.136
Pb
Pb2+ + 2 e-  Pb
-0.126
H
2 H+ + 2 e-  H2
Cu
Cu2+ + 2 e-  Cu +0.337
Ag
Ag+ + e-  Ag
0.000
+ 0.799
– Elements with highly negative
reduction potentials are not easily
reduced but they are easily
oxidized.
– Since metals react by being
oxidized, the more negative the
reduction potential, the more
reactive the element.
– Elements higher in the table (more
negative potential) can displace any
ions lower (more positive potential).
– So Zn + CuCl2  ZnCl2 + Cu
Cu + ZnCl2  No Reaction
**(Higher metal will replace the ions lower )**
The activity Series is really a reduction potential table arranged from negative to positive
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Calculating Cell Potentials From
Standard Reduction Potenials
• Calculate the cell
potential for a cell
made from silver and
zinc electrodes.
From the standard reduction table:
Zn+2 + 2 e-  Zn
- 0.763 v
Ag+ + e-  Ag
+ 0.799v
Since there must be one oxidation and one reduction, the
direction of one of two half reactions above must be reversed.
Reversing the zinc half reaction making it the oxidation would
yield a positive cell potential
Zn  Zn+2 + 2 e(+0.763 V)
Ag+ + e-  Ag
(+0.799 V)
Zn(s) + Ag+(aq)  Zn+2(aq) + Ag(s)
Cell potential = +1.562 volts
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Application Question 2
Mg2+ + 2e-  Mg
E°= - 2.37 V
Zn2+ + 2e-  Zn
E°= - 0.76 V
• Does Zn react with Mg2+?
• Does Mg react with Zn2+?
If Mg is oxidized Mg  Mg2+ + 2eCombining this with the reduction of Zn2+
Leaves an overall positive cell potential
Eo = +2.37 v
Eo = - 0.76 v
+ 1.66 v
Therefore Mg reacts with Zn2+.
Zn does not react with Mg 2+
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Electrochemical Cells
• Two types of cells
– Voltaic, Galvanic
– Electrolytic
• One cell uses chemistry to make electrical energy.
– Voltaic (battery)
• One cell uses electrical energy to make chemistry.
– Electrolytic
Voltaic Cells
In spontaneous
oxidation-reduction
(redox) reactions,
electrons are
transferred and energy
is released.
To be spontaneous, must
have a positive cell
potential (based on
Standard Potential
Table).
Voltaic Cells
• A typical VOLTAIC
cell looks like this.
• The oxidation occurs
at the anode.(OIL)
• The reduction occurs
at the cathode. (RIG)
Voltaic Cells
• Memory Device:
– Red Cat. has Paws
– An Ox
• Reduction happens at
the Cathode and is
PAWSitive (positive)
• Oxidation happens at
the Anode. (negative)
Voltaic Cells
• Electrons will not flow
unless there is a
completed circuit.
• Salt bridge is used,
usually a U-shaped
tube that contains a
salt solution, to keep
the charges balanced.
– Cations move toward
the cathode.
– Anions move toward
the anode.
Voltaic Cells
• At Anode, electrons
leave atoms, become
ions and moves into
solution (erodes
electrode).
• At Cathode, electrons
join ions (in solution)
and comes metal
atoms (plates onto
electrode)
e- are released from atom,
becomes ion
Ion comes close and gains e-,
becomes metal
Electrolytic Cells
• Electrolytic cells use electrical energy to bring about a
non-spontaneous reaction.
• An electrolytic cell is made up of three parts.
– Source of direct current (Battery)
– Two electrodes
• Cathode (negative pole) – reduction
*NOTE: no longer PAWS-itive
– The Red Cat has lost its PAWS.
• Anode (positive pole) – oxidation
– An electrolytic (ionic compound) that balances the flow of
electrons through the circuit.
• Electrons still flow from anode to cathode.
Cell
Construction
-
+
battery
power
source
eeconductive
medium
(-)
(+)
vessel
inert
electrodes
Sign or polarity of electrodes
What chemical species would be
present in a vessel of molten sodium
chloride, NaCl (l)?
Na+
Cl-
Let’s examine the electrolytic cell for molten NaCl.
Molten NaCl
Observe the reactions at the electrodes
-
battery
+
Cl2 (g) escapes
Na (l)
Cl-
Na+
Cl
(-)
electrode
half-cell
Na+ + e-  Na
NaCl (l)
Na+
-
Cl
-
(+)
Na+
electrode
half-cell
2Cl-  Cl2 + 2e-
Molten NaCl
At the microscopic level
e-
battery
+
NaCl (l)
cations
migrate
toward
(-)
electrode
cathode
Na+
+
e-
 Na
Na+
Cl-
Na+
Cl
(-)
-
Cl
-
e(+)
Na+
anions
migrate
toward
(+)
electrode
anode
2Cl-  Cl2 + 2e-
A Typical Electrolytic Cell
Molten NaCl Electrolytic Cell
cathode half-cell (-)
REDUCTION
Na+ + e-  Na
(Red Cat has lost its PAWS-itive)
anode half-cell (+)
OXIDATION
2Cl-  Cl2 + 2e-
overall cell reaction
2Na+ + 2Cl-  2Na + Cl2
Non-spontaneous reaction!
X2