Oxidation-reduction (redox) reactions
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Reactions in which there are changes in oxidation state
(oxidation number) between reactants and products
2 MnO4- + 10 Br- + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O
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One reactant must be oxidized (lose electrons, the reductant or
reducing agent) and another must be reduced (gain electrons,
the oxidant or oxidizing agent)
– In some instances a single reactant will be both oxidized and
reduced (disproportionation)
– in other cases two species containing the same element in
different oxidation states will combine to give a single product
with an intermediate oxidation state (comproportionation).
Oxidation-reduction (redox) reactions, cont’d
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Redox reactions may be separated into oxidation and reduction
“half-reactions”.
MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O
2 Br- → Br2 + 2 e2 MnO4- + 10 Br- + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O
– When half-reaction reagents can be isolated into separate
compartments connected by a conduit for ion migration,
electron flow through an external circuit can be utilized to
perform work (basis for batteries and electrochemical cells).
– The potential (Eo) of an electrochemical cell is the sum of the
potentials of the reduction and oxidation half-reactions
– The potential for a cell or half-reaction is related to the free
energy change for the redox reaction through the relationship
ΔGo = -nöEo where n is the number of electrons transferred
and ö is the Faraday (9.65 x 104 J V-1 mol-1)
An electrochemical cell
anode
(oxidation)
cathode
(reduction)
Zn(s) = Zn2+ + 2 e- Eo = -(-0.76) V
Cu2+ + 2 e- = Cu(s) Eo = 0.34 V
Zn(s) + Cu2+ = Zn2+ + Cu(s) Eo (cell) = (0.76 + 0.34) =1.10 V
Zn Zn2+ Cu2+ Cu
ΔGo = -nöEo
Galvanic vs electrolytic cells
For a Galvanic Cell, a spontaneous reaction takes place.
ΔE > 0 (positive)
ΔG < 0 (negative)
A positive voltage means a rxn is
thermodynamically favored and
thus reactants are “unstable”
Zn Zn2+ Cu2+ Cu
For the Galvanic cell, the cell performs electrical work on
the surroundings (acts as a battery)
For an Electrolytic Cell, a non-spontaneous reaction takes
place.
Cu Cu2+ Zn2+ Zn
ΔE < 0 (negative)
ΔG > 0 (positive)
A negative voltage means a rxn
is not thermodynamically
favored and thus reactants are
“stable”
For the electrolytic cell, electrical work is performed on the
system
Conventions relative to electrode potentials
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All half-reactions are written as reduction reactions
Potentials of half-reactions are relative to H+(aq) + e- = ½
H2(g) with Eo = 0.00 V
Potentials are recorded for standard conditions: 1.0 M conc.
for soluble reactants/products, 100 kPa (1 atm) for gases,
pure solids
Cell potentials are obtained by adding potentials for halfreactions
– Sign of potential changes when half-reaction is written
as oxidation
Changes in half-reaction or cell potentials resulting from nonstandard conditions can be determined using the Nernst equation
E = Eo -
[products]
RT
ln
nF
ö [reactants]
ö = 9.65 x 104 C mol-1
R = 8.31 V C mol-1 K-1
Half-Reaction
E° (volts)
0.15
Sn4+(aq) + 2e- = Sn2+(aq)
0.16
Cu2+(aq) + e- = Cu+(aq)
2+
Cu (aq) + 2e = Cu(s)
0.34
0.52
Cu+(aq) + e- = Cu(s)
I2(s) + 2e- = 2I-(aq)
0.54
O2(g) + 2H+(aq) + 2e- =
H2O2(aq)
0.70
0.77
Fe3+(aq) + e- = Fe2+(aq)
+
Ag (aq) + e = Ag(s)
0.80
NO3-(aq) + 4H+(aq) + 3e- =
NO(g) + 2 H2O(liq)
0.96
1.07
Br2(liq) + 2e- = 2 Br-(aq)
O2(g) + 4 H+(aq)+ 4e- = 2 H2O(liq)1.23
Cr2O72-(aq) + 14 H+(aq) + 6e- =
2 Cr3+(aq) + 7 H2O(liq)
1.33
1.36
Cl2(g) + 2e- = 2 Cl-(aq)
Ce4+(aq) + e- = Ce3+(aq)
1.44
MnO4-(aq) + 8 H+(aq) + 5e- =
Mn2+(aq) + 4 H2O(liq)
1.51
H2O2(aq) + 2 H+(aq) + 2e- =
2 H2O(liq)
1.78
1.82
Co3+(aq) + e- = Co2+(aq)
F2(g) + 2e- = 2 F-(aq)
2.87
Most oxidizing
Most reducing
Half-Reaction
E° (volts)
Li+(aq) + e- = Li(s)
-3.04
-2.92
K+(aq) + e- = K(s)
2+
Ca (aq) + 2e = Ca(s)
-2.76
-2.71
Na+(aq) + e- = Na(s)
Mg2+(aq) + 2e- = Mg(s)
-2.38
-1.66
Al3+(aq) + 3e- = Al(s)
2 H2O(liq) + 2e- =
-0.83
H2(g) + 2 OH-(aq)
2+
Zn (aq) + 2e = Zn(s)
-0.76
-0.74
Cr3+(aq) + 3e- = Cr(s)
Fe2+(aq) + 2e- = Fe(s)
-0.41
-0.40
Cd2+(aq) + 2e- = Cd(s)
Ni2+(aq) + 2e- = Ni(s)
-0.23
+
0.00
2 H (aq) + 2e = H2(g)
Most oxidizing
Most reducing
Some electrode potentials
Getting the terminology straight
A + B = A+ + BA is the reductant (reducing agent); gives up electrons, i.e., A = A+ + eB is the oxidant (oxidizing agent); gains electrons, i.e., B + e- = BA is oxidized by B
B is reduced by A
A+ is the oxidation product
B- is the reduction product
For this reaction to be favorable as written entries in a table of
electrode (half-reaction) potentials would appear as follows:
more reducing
A+ + e- = A
B + e- = B-
more oxidizing
The reduction potential for A+/A will be less positive than that for B/B-
Rules for assigning oxidation state
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The sum of the oxidation states for all atoms in a species must
equal the charge on the species.
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The oxidation state of an atom in an elemental form is zero.
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The most electronegative element is assigned a negative
oxidation state in compounds of hydrogen.
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In compounds such as HO-OH, H2N-NH2, etc. the E-E bond
does not contribute to the oxidation state of either atom.
However, in compounds such as SSO32- (S2O32-, thiosulfate)
assignment of a filled valence shell to the terminal sulfur gives
it an oxidation state of -2.
The oxidation state of a monatomic ion is equal to its charge.
Terminal atoms (or groups of atoms) are assigned an oxidation
state consistent with the charge in their monatomic anionic
form, i.e., filled valence shell.
Balancing oxidation-reduction (redox) reactions
HNO3(aq) + Cu(s) = NO2(g) + Cu2+(aq)
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Write separate half-reactions for oxidation and reduction
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Balance each half-reaction for mass utilizing water and protons as
necessary, i.e.,
– Get rid of oxygen as water
– Get rid of hydrogen as H+
– Get hydrogen from H+
– Get oxygen from water
Balance each half-reaction for charge by adding electrons to the right
(oxidation) or left (reduction) as required
Multiply half-reactions by the appropriate factors such that the electrons
lost in oxidation equal the electrons gained in reduction
Add the two half-reactions; if the same reagent appears as both reactant
and product, subtract the smaller amount from both sides of the equation
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Balancing oxidation-reduction (redox) reactions
HClO = Cl2 + ClO3-
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Write separate half-reactions for oxidation and reduction
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Balance each half-reaction for mass utilizing water and protons as
necessary, i.e.,
– Get rid of oxygen as water
– Get rid of hydrogen as H+
– Get hydrogen from H+
– Get oxygen from water
Balance each half-reaction for charge by adding electrons to the right
(oxidation) or left (reduction) as required
Multiply half-reactions by the appropriate factors such that the electrons
lost in oxidation equal the electrons gained in reduction
Add the two half-reactions; if the same reagent appears as both reactant
and product, subtract the smaller amount from both sides of the equation
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Balancing oxidation-reduction (redox) reactions
O
P4 = PH3 + H2PO- (basic solution)
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Write separate half-reactions for oxidation and reduction
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Balance each half-reaction for mass utilizing water and protons as
necessary, i.e.,
– Get rid of oxygen as water
– Get rid of hydrogen as H+
– Get hydrogen from H+
– Get oxygen from water
Balance each half-reaction for charge by adding electrons to the right
(oxidation) or left (reduction) as required
Multiply half-reactions by the appropriate factors such that the electrons
lost in oxidation equal the electrons gained in reduction
Add the two half-reactions; if the same reagent appears as both reactant
and product, subtract the smaller amount from both sides of the equation
If the reaction is conducted in basic medium, add OH- to both sides to
eliminate any H+ (H+ + OH- = H2O)
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Latimer diagrams – a way to simplify data
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Latimer diagrams summarize several electrode potentials
(half-reactions) in abbreviated form for a given element
Half-reactions are written in sequence, as reductions, with the
most highly oxidized species to the left
Only the species involving the element are shown; other
species must be added using rules for balancing halfreactions
When potential of a half-reaction is more positive than the
one to the left the species will disproportionate
Separate diagrams are required for acidic and basic
conditions
Examples of Latimer diagrams
Acidic solution
Acidic solution
Basic solution
Examples and interpretation of Latimer diagrams
1.49
Acidic solution
MnO4- + 8 H+ + 5 e- = Mn2+ + 4 H2O
Eº = 1.49
Basic solution
MnO4- + 4 H+ + 3 e- = MnO2 + 2 H2O
Eº = 0.59
Examples and interpretation of Latimer diagrams
Acidic solution
HClO2 + 2 e- + 2 H+ = HClO + H2O
Eº = 1.674
HClO2 + H2O = ClO3- + 2 H+ + 2 e-
Eº = -1.181
2 HClO2 = H+ + ClO3- + HClO
Eº = 0.493