REDOX REACTIONS (Ch. 4.4)
Redox reactions involve a change in oxidation state or one or more
atoms in each substance involved
Oxidation involves loss of electrons from a species e.g.
Fe(II)
Fe(III)
(iron is oxidised)
Reduction involves gain of electrons e.g.
CO2
CO (carbon is reduced)
“Red-Ox”
(O I L - R I G)
Oxidation State (p. 134)
- the charge an atom would have if the valence electrons in a molecule
or complex were completely transferred to the most electronegative
atoms
e.g. in H-O-H
oxidation state of H is +1
oxidation state of O is -2
e.g. iodine trifluoride IF3
oxidation state of F is always -1
oxidation state of I is +3
e.g. dichromate ion, Cr2O72oxidation state of O is -2 (total of -14 for all Os)
⇒ total of +12 for metal atoms
⇒ oxidation state of Cr is +6
⇒ can write as CrVI2O72-
1
Oxidation State (p. 134)
Oxidation State (p. 134)
- the charge an atom would have if the valence electrons in a molecule
or complex were completely transferred to the most electronegative
atoms
e.g. in manganate ion, MnO4oxidation state of O is -2 (total of -8 for all oxygens)
⇒ oxidation state of Mn must be +7
⇒ can write as MnVIIO4Q: Deduce the oxidation state of the heaviest atom in each of the
following compounds/complexes
A:
a) [CoCl4]2b) H2SeO4
c) Fe2(CO3)3
2
Oxidation State (p. 134)
- the charge an atom would have if the valence electrons in a molecule
or complex were completely transferred to the most electronegative
atoms
e.g. in manganate ion, MnO4oxidation state of O is -2 (total of -8 for all oxygens)
⇒ oxidation state of Mn must be +7
⇒ can write as MnVIIO4Q: Deduce the oxidation state of the heaviest atom in each of the
following compounds/complexes
A:
a) [CoIICl4]2b) H2SeVIO4
c) FeIII2(CO3)3
Ex.
s
4.4 ee
&Q
4.48
s
& 4 4.46,
.50
Elements have oxidation state of zero
- formation of water is a redox reaction:
2H2 + O2
2H2O
oxygen is reduced to -2 oxidation state
hydrogen is oxidised to +1 oxidation state
Reaction of bromine and phosphorus:
2P + 3Br2
2PBr3
bromine is reduced to -1 oxidation state
phosphorus is oxidised to +3 oxidation state
3
2Mg
+
2Mg2+
O2
O.I.L
R.I.G.
2O2-
+
2 x 2e- transfer
a metal becomes oxidised during its combustion (from 0 to +II state)
O becomes reduced (from 0 to -II state)
‘oxidation’ now used for any reaction that involves loss of electrons
Mg
+
Mg2+
Br2
2Br-
+
2e- transfer
Mg is oxidised from 0 to +II state. Br reduced from 0 to -I state
For both reactions, oxidation of Mg is described by same ‘halfreaction’
Mg
Mg2+ +
2e-
O.I.L
R.I.G.
Half Reactions (p. 132)
Fe2O3
+
3CO
2Fe(l)
+
3CO2
6e- transfer
Fe(III)
C(II)
Fe(0)
C(IV)
Fe becomes reduced. Half equation:
2Fe3+
+
6e-
2Fe(l)
(electrons
gained)
C becomes oxidised. Half reaction:
CO
+
O2-
CO2
+
2e-
(electrons
lost)
wherever one species is oxidised, other species must also be
reduced………the reaction is red-ox
4
Oxidizing Agents and Reducing Agents (p. 132)
O.I.L
R.I.G.
-species that oxidises another (removes electrons) is called an
oxidising agent:
4Al
+
Reducing
agent
(gets oxidised)
2Al2O3
3O2
Oxidising
agent
(gets reduced)
O oxidises Al and, in doing so, gets reduced to -2 oxidation state
-species that reduces another (supplies electrons) is called the
reducing agent:
Al reduces O and, in doing so, gets oxidised to +3 ox. state
…useful oxidising agents typically have an atom in a high (electrondeficient) oxidation state e.g. KNVO3, KMnVIIO4, CrVIO3.
NO3- ions oxidise Cu metal in acidic solution - mechanism
depends on acid concentration:
O.I.L
R.I.G.
In dilute acid:
0
3Cu(s) +
Reducing
agent
+5
8H+(aq)
+ 2NO3
+2
- (aq)
3Cu2+(aq)
Oxidising
agent
+2
+ 2NO(g) + 4H2O(l)
colourless
gas
In concentrated acid:
0
+5
Cu(s) + 4H+(aq) + 2NO3- (aq)
Reducing
agent
Oxidising
agent
+2
+4
Cu2+(aq) + 2NO2(g) + 4H2O(l)
brown
gas
5
Q: When acidic potassium dichromate (K2Cr2O7) solution is mixed with
iron (II) chloride, the Fe and Cr ions that form are both in the +3
oxidation state. Which species is oxidised and which reduced?
A: 1) Determine solubilities and ions produced in solution.
-potassium dichromate (K2Cr2O7) dissolves to give K+ and Cr2O72- ions
-iron (II) chloride produces Fe2+ cations
2) Determine oxidation numbers of all metals in reactants and products:
Dichromate (Cr2O72-) contains Cr in the +6 oxidation state
Therefore:
Cr changes oxidation state from +6 to +3 (reduction)
Fe changes oxidation state from +2 to +3 (oxidation)
Therefore:
K2Cr2O7 is the oxidising agent
FeCl2 is the reducing agent
O.I.L
R.I.G.
Hydrogen Displacement Reactions (p. 138)
-most metals displace hydrogen from acids to form hydrogen
gas e.g.
2e- transfer
Sn(s) +
2HCl(aq)
SnCl2(s)
+
H2(g)
(H+ is reduced to H whilst the metal is oxidised)
-more reactive metals e.g. Mg, Zn, Al react
with steam to displace H
K(s) + H2O
-the most reactive metals e.g. Gp I metals
displace H from cold water and alcohols:
Cs(s) + CH3OH(l)
CsOCH3(aq) + H2(g)
-unreactive metals e.g. Ag, Cu, Au, Pt do not displace H from acids
6
O.I.L
R.I.G.
Metal Displacement Reactions (p. 139)
2e- transfer
Zn(s) +
Reducing
agent
Cu2+(aq)
Cu(s)
+
Zn2+(aq)
Oxidising
agent
Zn is a more strongly reducing metal than Cu;
Zn metal spontaneously displaces Cu2+ ions from solution
O.I.L
R.I.G.
7
Metal Displacement Reactions (p. 139)
Zn2+ (aq)
+
Cu (s)
O.I.L
R.I.G.
No reaction
Zn is a more strongly reducing metal than Cu
Reverse reaction won’t work:
- Cu cannot reduce Zn2+ ions to Zn metal
(Zn is a more strongly reducing than Cu)
Likewise:
H+ (aq)
+
Ag (s)
No reaction
- Ag cannot reduce protons (H+) to hydrogen gas
(H is a more strongly reducing than Ag)
- Metals displace each other based on their relative reducing power
i.e. their tendency to be oxidised to M+ ions
-we can arrange then in order of their reducing power:
The activity series
or
electrochemical series
8
Q: Could we use cadmium metal to reduce
stannous oxide to tin metal?
A: Cd lies above Sn in the
activity series
⇒ Cd more reducing than Sn
⇒ SnO can successfully be
reduced by Cd
try
Q4
.52
Q: Will concentrated
hydrochloric acid dissolve
mercury and libaerate
hydrogen gas?
A: H lies above Hg in the
activity series
⇒ H more reducing than Hg
⇒ H+ cannot be successfully
reduced by Hg
Halogen Displacement (p. 140)
-Halogens displace each other based on their relative oxidising power
i.e. their tendency to be reduced to X- ions or the -1 ox. state
The order of oxidising power is:
Cl2 (g)
0
+
2KBr (aq)
-1
F2 > Cl2 > Br2 > I2
most
oxidising
least
oxidising
(most easily reduced)
(least easily reduced)
2KCl (s)
-1
+
Br2 (l)
0
(Cl is a more strongly oxidising than Br, so bromide is oxidised to
bromide and chlorine is reduced to chloride)
O.I.L
R.I.G.
9
Halogen Displacement (p. 140)
-Halogens displace each other based on their relative oxidising power
i.e. their tendency to be reduced to X- ions or the -1 ox. state
The order of oxidising power is:
2NH4I (aq) + Br2 (g)
-1
F2 > Cl2 > Br2 > I2
2NH4Br(s)
0
+
I2(l)
0
-1
(Br is a more strongly oxidising than I, so iodide is oxidised to iodine
and bromine is reduced to bromide)
I2 (g)
+
SrCl2 (aq)
try
Q4
.44
No reaction
-1
0
(Reverse reactions won’t work:
I2 cannot oxidise bromide to elemental bromine)
O.I.L
R.I.G.
-We can use half equations to deduce the
stoichiometry of a redox reaction:
e.g. the oxidation of Al to alumina (Al2O3)
try
Q4
.44
oxidation state change
Oxidation half reaction (electrons lost):
+
3eAl
Al3+
0
+3
Reduction half reaction (electrons gained):
O2
+
4e2O2-
0
-2
In redox reaction, number of electrons gained = number lost
⇒ multiply ox. eqn by 4 and red. eqn by 3 to give 12 electrons
transferred:
4Al
4Al3+ +
12e23O2
+
12e
6O
4Al
+
3O2
4Al3+
+
6O2- (2Al2O3)
O.I.L
R.I.G.
10
Returning to a reaction we looked at earlier…..
Q: deduce the stoichiometry of the reaction between acidic potassium
dichromate (K2Cr2O7) solution and aqueous iron (II) chloride?
A: First, write imbalanced equation, identifying ox. numbers:
+6
Cr2O7
+3
+2
2-(aq)
Oxidising
agent
+
Fe2+(aq)
+
H+(aq)
Cr3+(aq)
+3
+ Fe3+(aq) + H2O
Reducing
agent
next, we must break down into oxidation and reduction half
reactions………..
Balancing Redox Equations Using Half Reactions
+6
+2
Cr2O72-(aq) + Fe2+(aq) + H+(aq)
+3
+3
Cr3+(aq) + Fe3+(aq) + H2O
Red:
H+ required on LHS to remove O from metal and form water.
Cr reduced (3 electrons gained per Cr) ⇒ put six electrons on LHS
Cr2O72-(aq) + 14H+ + 6e2Cr3+(aq) + 7H2O
Ox: Electrons lost from iron (put a single electron on RHS)
Fe3+(aq) + eFe2+(aq)
Need same number of electrons on each side ⇒ multiply ox. eqn. by 6
6Fe2+(aq)
Cr2O72-(aq) + 14H+ + 6eCr2O72-(aq) + 6Fe2+(aq) + 14H+
6Fe3+(aq) + 6e2Cr3+(aq) + 7H2O
(now add
eqns)
2Cr3+(aq) + 6Fe3+(aq) + 7H2O
11
Balancing Redox Equations Using Half Reactions
returning to another reaction we met earlier...
Q: When dilute nitric acid is poured onto copper metal, a copper (II)
salt and NO gas are formed. Write the redox half equations in order to
balance the overall stoichiometric equation.
A: First, determine what’s being oxidised and what reduced….
nitric acid (+5) is being reduced to NO (+2)…………..3 electron gain
copper metal (0) is being oxidised to copper (+2)………2 electron loss
Imbalanced half equations first:
Red:
NO3- (aq)
NO(g)
Ox:
Cu(s)
Cu2+(aq)
O.I.L
R.I.G.
Balancing Redox Equations Using Half Reactions
+5
Red:
Ox:
O.I.L
R.I.G.
+2
NO3- (aq)
Cu(s)
NO(g)
Cu2+(aq)
Red: H+ will combine with O to form water. Electrons gained (put on
LHS):
NO3- (aq) + 4H+(aq) + 3eNO(g) + 2H2O
Ox:
Cu(s)
Cu2+(aq) + 2e-
Need same number of electrons on each side
⇒ multiply ox eqn. by 3 and red. eqn by 2 to give 6 electron transfer:
(now add
eqns)
2NO3- (aq) + 8H+ (aq) + 6e3Cu(s)
2NO(g) + 4H2O
3Cu2+(aq) + 6e-
2NO3- (aq) + 8H+ (aq) + 3Cu(s)
2NO(g) + 3Cu2+(aq) + 4H2O
12
Balancing Redox Equations Using Half Reactions
Q: When potassium manganate solution is mixed with H2SO3
(sulphurous acid), the products are the hydrogen sulphate anion (HSO4-)
and the Mn2+ cation. Use half equations balanced with water and
protons, deduce the full, balanced ionic equation for the reaction.
try
at hom
e
+
A:
2MnVIIO4- (aq) + H+ (aq) + 5H2SIVO3 (aq)
2Mn2+ (aq) + 5HSVIO4- (aq) + 3H2O (l)
Balancing Redox Equations Using Half Reactions
A: Red:
Ox:
MnVIIO4- (aq)
H2SIVO3 (aq)
Mn2+(g)
HSVIO4- (aq)
now balance each half equation with protons and water:
Mn2+(aq) + 4H2O
Red: MnVIIO4- (aq) + 8H+(aq)
IV
Ox: H2S O3 (aq) + H2O
HSVIO4- (aq) + 3H+(aq)
now balance charges with electrons:
Red: MnVIIO4- (aq) + 8H+(aq) + 5eOx: H2SIVO3 (aq) + H2O
Mn2+(aq) + 4H2O
HSVIO4- (aq) + 3H+(aq) + 2e-
Need same number of electrons on each side
⇒ multiply equations so there are 10 electrons on each side:
3
Red: 2MnVIIO4- (aq) + 16H+(aq) + 10e2Mn2+(aq) + 8H2O
Ox: 5H2SIVO3 (aq) + 5H2O
5HSVIO4- (aq) + 15H+(aq) + 10e2MnVIIO4- (aq) + H+ (aq) + 5H2SIVO3 (aq)
2Mn2+ (aq) + 5HSVIO4- (aq) + 3H2O
13
Disproportionation Reactions (p. 141)
-Special type of redox process - a reaction where an element is both
oxidised and reduced
e.g. 2H2O2
-1
e.g. 3HONO
+3
2H2O + O2
-2
0
HNO3 + H2O + 2NO
+5
+2
The element must have at least three different oxidation states for
disproportionation to be possible
The S.M. must have the element in an intermediate ox. state
Q: Predict the outcome of the following redox
reactions and identify the type of reaction:
try
Q4
.56
Ag(s) + Mg(NO3)2(aq)
a) Mg + 2AgNO3(aq)
Mg is far higher in activity series than Ag so Mg reduces Ag
ions are rearranged - a displacement reaction
b) 2NO(g)
O2 + N2
Compound breaks apart - a decomposition. N is reduced whilst O is
simultaneously oxidised
c) Cl2(g) + CsF(aq)
No reaction
F2 is more oxidising than Cl2 so fluoride will not reduce chlorine gas
to chloride ion.
d) 2CuCl(aq) + 6H2O
+1
Cu[(H2O)6]Cl2(aq)
+2
+ Cu(s)
0
Copper (I) disproportionates into Cu(II) and elemental copper
14
Quantitativity & Stoichiometry in Aqueous Reactions
Q: If we mix 10-2 L of 0.35 M AgNO3 with aqueous K3PO4
(15 mL, 100 mM) a white precipitate immediately forms
a) what is the formula of the precipitate?
b) write the full ionic equation
AgNO3 (aq) + K3PO4 (aq)
Ag3PO4
Ag+ (aq) + NO3- (aq) + Ag3PO4(s)
c) what is the maximum theoretical yield (mg to 3 s.f.) of precipitate?
moles Ag = 3.5 x 10-3 mol (limiting reagent)
moles PO4 = 1.5 x 10-3 mol (excess reagent)
⇒ moles Ag3PO4 = 3.5 x 10-3 mol / 3 = 1.1667 x 10-3 mol
⇒ 0.48845 g (488 mg)
[mol. wt. (Ag3PO4) = 418.67 g/mol]
d) assuming no silver cations remain in solution, what is the final
concentration of phosphate ions after precipitation (mM to 2 s.f.)?
Moles PO43- remaining = (1.5 - 1.1667) x 10-3 mol = 3.333 x 10-4 mol
⇒ [PO43-] = 2.9 mM
Quantitativity & Stoichiometry in Aqueous Reactions
Q: If we mix 60 mL of 0.6 M HCl with Ba(OH)2 (25 mL, 650 mM)
heat is evolved but no precipitate forms. If the mixture is evaporated
a salt remains.
a) what is the formula of the residual salt and the equation of its
formation? BaCl2
Ba(OH)2 (aq) + 2HCl (aq)
BaCl2 (aq) + 2H2O (l)
b) what is the maximum theoretical yield (g to 3 s.f.) of the salt?
moles Ba = 0.01625 mol (limiting reagent)
moles Cl = 0.036 mol (excess reagent)
⇒ moles BaCl2 = 0.01625 mol ⇒ 3.38 g
c) what is the final concentration of chloride ions after mixing?
Moles Cl- remaining = 0.036 mol - (2 x 0.01625 mol) = 0.0035 mol
⇒ [Cl-] = 0.04118 ⇒ 41.2 mM
15