CHEM 1105
1.
REDOX REACTIONS
Definition of Oxidation and Reduction
The old definition of oxidation was "addition of oxygen".
Consider the following reactions:
Na2O (2Na+ + O2-)
(a) 2Na + 4O2
NaCl (Na+ + Cl-)
(b) Na + 4Cl2
By the old definition, reaction (a) is an oxidation. But in both cases Na is converted into Na+. Hence,
in reaction (b) Na is also oxidized. In both cases, Na lost electrons in forming Na+ ions.
Oxidation and reduction occur simultaneously; when one species is oxidized, the other is reduced.
Hence we call these redox reactions. In reaction (a), Na is oxidized by losing electrons and O is
reduced by gaining electrons to form O2- ions. Similarly, in reaction (b) Cl is reduced.
New definition: Oxidation is the loss of electrons and reduction is the gain of electrons.
LEO GER is a convenient way of remembering this (Loss of Electrons is Oxidation; Gain of
Electrons is Reduction). [Another memory aid is OIL RIG].
In reaction (b), we can say that Na was oxidized by Cl and Cl was reduced by Na. Hence, we call Cl
the oxidizing agent and Na the reducing agent. In general, the species oxidized is the reducing
agent and the species reduced is the oxidizing agent.
H
H
2.
Oxidation Numbers
In reaction (b), Na and Cl were changed to NaCl in a redox reaction. In NaCl, we have Na+ ions and
Cl-ions and we say that Na+ has an oxidation number of +1 and Cl- has an oxidation number of -1.
We can consider the free elements (in this case, Na and Cl) to have an oxidation number of 0. We
can hence define oxidation as an increase in oxidation numner (Na: from 0 to +1) and reduction
as a decrease in oxidation number (Cl: from 0 to -1) In more complicated species (compounds and
polyatomic ions), we can make rules for calculating an oxidation number of an element in the ion or
compound, but we must remember that in those cases the use of oxidation numbers is only a
convenient bookkeeping device.
Oxidation Number Rules
1.
The sum of all oxidation numbers of elements in a compound is zero.
2.
The oxidation number of any free element is zero.
The oxidation number of a monatomic ion is the charge on the ion.
3.
4.
The sum of all oxidation numbers of elements in a polyatomic ion is the charge on the ion.
5.
The oxidation number of any Group 1 metal in combination is +1.
6.
The oxidation number of any Group 2 metal in combination is +2.
7.
The oxidation number of any halogen (Group 7) in combination with one other element other
than oxygen is -1.
8.
The oxidation number of hydrogen in combination is +1 unless the H is present as H-1
(hydride, as in NaH) in which case the oxidation number is -1.
9.
The oxidation number of oxygen in combination is -2 unless the O is present as peroxide in
which case the oxidation number is -1.
10.
The oxidation number of fluorine is always -1.
Exercises
Calculate the oxidation number of the underlined element in each case.
HCl
HNO3
H2SO4
Cr2O72-
CuCl2
K2XeF6
S2O32N2O3
O3
NH4+
Mg(ClO4)2
-23.
Balancing Redox Reaction Equations in Aqueous Acid
The Half-Reaction (Ion-Electron) Method
This method divides the reaction into two half-reactions; one involving the species being oxidized
and the other involving the species being reduced.
(1)
(2)
(3)
(4)
(5)
Use oxidation numbers to identify which species is oxidized and which is reduced.
Write the half-reactions. For each half-reaction, balance the element undergoing oxidation
number change, then the O (by adding H2O) and finally the H (by adding H+). Then balance
the charge by adding electrons.
Multiply one or both half-reactions by the smallest integer(s) so that they have the same
number of electrons on opposite sides.
Add the multiplied half-reactions and cancel the charges and any other species that occur on
both sides.
Check that the resulting equation is balanced for mass (atoms) and charge.
Examples
H
Simple; no other atoms needed
Fe3+ + ClFe
Solution
(1)
+
Cl2
Fe3+ is being reduced to Fe (ox. no. decreases from 3 to 0)
Cl- is being oxidized to Cl2 (ox. no. increases from -1 to 0)
H
Fe3+
Fe
3+
Fe + 3e
(2)
Cl2Cl2Cl(3)
H
H
H
H
Fe (balanced) (i)
Cl2
Cl2
Cl2 + 2e- (balanced) (ii)
H 2Fe (iii)
H
+ 6e (iv)
+ 6Cl H 2Fe + 3Cl
2Fe3+ + 6e3Cl2
6Cl-
(i) x 2 gives:
(ii) x 3 gives:
-
-
(4)
add (iii) + (iv): 2Fe3+
(5)
Check: LHS 2Fe; RHS 2Fe; LHS 6Cl; RHS 6Cl; LHS charge = 0; RHS charge = 0
EQUATION IS BALANCED
2
More complex; oxygen (from H2O) and H (from H+) needed
MnO4- + Cl-
H
Mn2+
+
Cl2
Solution
(1)
Mn in MnO4- is being reduced Mn2+(ox. no. decreases from 7 to 2)
Cl- is being oxidized to Cl2 (ox. no. increases from -1 to 0)
-3(2)
H
H
H
H
H Cl
H Cl
H Cl
MnO4Mn2+
MnO4
Mn2+ + 4H2O
Mn2+ + 4H2O
8H+ + MnO4+
Mn2+ + 4H2O (balanced) (i)
8H + MnO4 + 5e
Cl2Cl2Cl-
2
2
2
+ 2e- (balanced) (ii)
H
(3)
(i) x 2 gives: 16H+ + 2MnO4- + 10e(ii) x 5 gives:
10Cl-
(4)
add (iii) + (iv): 16H+ + 2MnO4- + 10Cl-
(5)
Check: LHS 2Mn; RHS 2Mn; LHS 10Cl; RHS 10Cl;
LHS 16H; RHS 16H; LHS 8 O; RHS 8 O
LHS charge = 4; RHS charge = 4
EQUATION IS BALANCED
H
(a)
S2O32- +
(b)
HNO3
(c)
(d)
(e)
H
I2
I-
2Mn2+ + 8H2O (iii)
5Cl2 + 10e- (iv)
H 2Mn
2+ +
5Cl2 + 8H2O
S4O62-
+
H NO + S
Cu + HNO
H Cu + NO
H IO + Cl
I + ClO
Cr O
+ CH OH H Cr
+ CH O
+
H2S
2+
3
-
2
-
3
2-
2 7
-
3
3+
3
2
Disproportionation
A disproportionation reaction is a redox reaction in which one substance is both oxidized and
reduced. In the reaction below, Cl0 (in Cl2) is being reduced to Cl- (in HCl) and is being oxidized to
Cl with an ox. no. of 1 in HOCl. Each half-reaction starts with the same reactant.
Cl2 + H2O
H
(f)
P4
(g)
MnO42-
H
PH3
H
HOCl
+
MnO4-
+
HCl
H2PO2+
MnO2