Download 9. Oxidation and reduction

9. Oxidation and
reduction
9.1 Introduction
• Oxidation = loss of one or more electrons from a
substance
• Reduction = gain of one or more electrons
0
0
2Mg (s) + O2 (g)
2Mg2+ + 4e-
2Mg
2
O + 4e
-
2O
2-
2+
2-
2MgO (s)
Oxidation half-reaction (loses e-)
Reduction half-reaction (gaines e-)
http://www.youtube.com/watch?v=Mx5JJWI2aaw
9.2 Redox reactions
• Redox reaction: a chemical reaction where electrons are
transferred from one element to another.
• Oxidizing agent —electron acceptor; the species is reduced.
• Reducing agent—electron donor; the species is oxidized.
• Oxidation and reduction always occur together:
Oxidation numbers
• The oxidation number of an atom in a compound is a
measure of the electron possession it has relative to
the atom in a pure element.
• The oxidation number makes it possible to keep track
of the relative electron density in a compound and
how it changes during a reaction.
Rules for determining oxidation numbers
1) Free elements (uncombined state) have an oxidation number
of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2) In monatomic ions, the oxidation number is equal to the charge
on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3) For simple ionic compounds with two elements, the oxidation
number of each element is equal to the charge of the ion.
e.g. MgO:
Mg2+ O2(Mg = +2; O = -2)
NaCl:
Na+ Cl(Na = +1; Cl = -1)
4) Covalent compounds are assumed to be ionic with the more
electronegative element forming the negative ion.
e.g.
H2O ( H = +1; O = -2)
HCl
( H = +1; Cl = -1)
5) The sum of all the oxidation numbers in a compound is equal
to zero.
e.g. Al2O3 [2 x (+3) + 3 x (-2)] = 0
8) The oxidation number of oxygen is always –2, except in peroxides
(H2O2 and O22- ), when it is –1 or in fluorine oxide (F2O) , when it is +2.
9) The oxidation number of hydrogen is +1 except in certain metal
hydrides (e.g. NaH), when it is –1.
10) alkali metals: +1
earth alkali metals: +2
ammonium:
+1
aluminium:
+3
phosphate:
-3
nitrate, hydroxide: -1
halides: -1
oxide, sulfide: -2
phosphide, nitride: -3
carbonate, sulfate: -2
• Some elements, especially transition metals and certain nonmetals,
have variable oxidation states!
Ex. 1 Assign oxidation numbers.
a) NO3
b) Cr2O7
c) NaH
e) KMnO4
f) Al2S3
g) of nitrogen in HNO3, N2H4, NO2+, N2, NO2-,
NO2, N2O4, N2O, NO
Oxidation numbers in names of compounds
●
Ex 3. Write the name of the following compounds:
FeO
Fe2O3
CuO
PbO2
SnCl4
AgBr
HNO3
• Ex. 3 Identify the substance oxidized / reduced and the
oxidizing / reducing agent in the following reaction:
3 SO2 + 2 HNO3 + 2 H2O →
3 H2SO4 + 2 NO
●
●
The higher the positive number, the more the atom has
lost control over electrons:
→ increase in oxidation number = oxidation
The higher the negative number, the more the atom has
gained electron control:
→ decrease in oxidation number = reduction
The same element can be both oxidized and reduced in
the same reaction. This is called disproportionation:
3 Cl2 + 6 OH− → 5 Cl− + ClO3 - + 3 H2O
2 Sn2+ → Sn + Sn4+
Balancing Redox Reactions
Ex. The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid
solution.
1) Write the unbalanced equation for the reaction in
ionic form.
Fe2+ + Cr2O72-
Fe3+ + Cr3+
2) Assign oxidation numbers to see what is being
oxidized and what reduced.
3) The redox reaction is then divided into two halfreactions (half-equations), which are first balanced
separately and then combined.
4) Balance the atoms other than O and H in each halfreaction.
Cr2O722Cr3+
Fe2+
Fe3+
5) For reactions in acid, add H2O to balance O atoms
and H+ to balance H atoms.
Cr2O722Cr3+ + 7H2O
14H+ + Cr2O72-
2Cr3+ + 7H2O
6) Add electrons to one side of each half-reaction to
balance the charges on the half-reaction.
6e- + 14H+ + Cr2O72Fe2+
2Cr3+ + 7H2O
Fe3+ + 1e-
7) If necessary, equalize the number of electrons in the
two half-reactions by multiplying the half-reactions by
appropriate coefficients.
6e- + 14H+ + Cr2O72-
6Fe2+ →
6Fe3+ + 6e-
→
2Cr3+ + 7H2O
8. Add the two half-reactions together, cancelling out
anything that is the same on both sides. The number of
electrons on both sides must cancel.
Oxidation:
Reduction:
6Fe2+
→
6Fe3+ +
6e- + 14H+ + Cr2O72-
14H+ + Cr2O72- + 6Fe2+
→
6e-
→ 2Cr3+ + 7H2O
6Fe3+ + 2Cr3+ + 7H2O
9. Verify that the number of atoms and the charges are
balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
19.1
●
Write the overall redox equation for the
reaction where manganate (VII) oxidizes a
solution of iron (II) ions to iron (III) ions.
9.3 Reactivity
●
●
●
As some substances are better at oxidizing than others,
they can be placed in order of their oxidizing ability.
The reactivity series was initially determined
experimentally by comparing the reactivity of metals.
The most reactive metals give up electrons readily and
can displace the ions of another metal in aqueous
solution.
Use the reactivity series to predict whether reactions will
occur between the following elements:
9.3 Reactivity series
●
Substances can be placed in order of reactivity:
• Metals high in the series can displace hydrogen ions from
cold water to form hydrogen gas and metal hydroxides.
For example:
2 Na (s) + 2H2O (l) → 2Na+ + 2OH- (aq) + H2 (g)
• All metals above hydrogen in the series react with acids
to form hydrogen gas and a metal salt. For example:
Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)
Metals below hydrogen in the series do not displace
hydrogen, but may still react with acids:
http://www.youtube.com/watch?v=A4XITC225uk
http://www.youtube.com/watch?v=pvhDkqXa3CQ
Cu (s) + 4 HNO3 (conc) → Cu(NO3)2 (aq) + 2 NO2(g) + 2
H2O (l)
9.4 Voltaic cells
• A voltaic cell (or galvanic cell) uses spontaneous redox
reactions to generate electrical energy.
• The oxidizing and reducing agents can be separated into
two half-cells and connected by an external wire.
• The electrons of the half-reactions flow via the wire from
the oxidizing half-cell to the reducing half-cell and a useful
current is obtained.
• The salt bridge (e.g. piece of filter paper soaked in
aqueous KCl) enables the flow of ions, thus completing
the electrical circuit.
ANODE:
Negative electrode
Oxidation
Cathode:
Positive electrode
Reduction
9.5 Electrolytic cells
●
●
●
●
Electricity from an external power supply is used to
bring about a non-spontaneous redox reaction.
The source of electric power (electrons) is a battery or a
DC power source.
The power source pushes the electrons towards the
cathode where they enter the electrolyte.
The ions in the electrolyte migrate to the electrodes.
●
●
Positive ions (=cations) are attracted to the negative
electrode (cathode).
Negative ions (=anions) are attracted to the positive
electrode (anode).
ttp://www.pearsonhotlinks.co.uk/9780435994402.aspx