Redox
Examine:
Zn + CuCl2  ZnCl2 + Cu
CuCl2 = ionic compound with copper charged +2
Cu = element with copper charged 0
Zn = element with zinc charged 0
ZnCl2 = ionic compound with zinc charged +2
There has been an exchange of charge. Since electrons are the only mobile charged
particles, there must have been and exchange of electrons. This is called redox or
reduction & oxidation.
Oxidation = loss of electrons
Reduction = gain of electrons
LEO goes GER
Oxidation #’s = charge
1) Elements = 0
O2 = 0
Fe = 0
2) Monatomic ions = charge
Mg+2 = +2
Cl-1 = -1
3) H = +1 almost always
O = -2 almost always
PO4-3
ClO-1
NO2
Cr2O7-2
Na2O2
NaH
(exception is metal hydrides NaH, eg, H = -1)
(exception is peroxides, H2O2, eg, O = -1)
-3 = P + 4(-2)
-1 = Cl + (-2)
0 = N + 2(-2)
-2 = 2 Cr + 7(-2)
0 = 2(+1) + 2(O)
0 = (+1) + H
P = +5
Cl = +1
N = +4
Cr = +6
O = -1 peroxide
H = -1 metal hydride
Balancing Redox in Neutral
Must make sure that electrons that are lost are the same as electrons being gained.
Cu+2 + Zn  Zn+2 + Cu
------------------------------------Zn
 Zn+2 + 2 eCu+2 + 2 e-  Cu
Electrons being lost are equal to the
electrons being gained so the equation
is balanced
Example
Ag+1 + Cu  Cu+2 + Ag
------------------------------------Cu
 Cu+2 + 2 eAg+1 + 1 e-  Ag
Balanced for mass but not for charge!
2 electrons being lost but
only 1 being gained
Need to make them the same so:
Cu
 Cu+2 + 2 e2 Ag+1 + 2 e-  2 Ag
2 Ag+1 + Cu  Cu+2 + 2 Ag
Rules for balancing in neutral solution
1)
2)
3)
4)
5)
determine oxidation numbers
split into ½ reactions
balance for mass
add e-‘s where they are required
recombine by making e-‘s equal
By multiplying the bottom equation
by 2, we’ve made the electrons the
same
Example
 Cr+3 + I-1
Cr+2 + I2
------------------------------------Cr+2
 Cr+3 + 1 eI2 + 2 e
 2 I-1
Balanced for charge but not for mass!
1 electron being lost and
now 2 being gained?
Need to make them the same so:
2 Cr+2
I2 + 2 e-
 2 Cr+3 + 2 e 2 I-1
2 Cr+2 + I2
 2 Cr+3 + 2 I-1
By multiplying the top equation by 2,
we’ve made the electrons the same
Balancing in Acidic Solution
Often a solution will not be neutral. It may be acidic, in which case there are plenty of
H+ ions and H2O to help balance things out.
Rules for balancing in an acidic solution:
1) determine oxidation #’s
2) split into ½ reactions
3) balance for mass that is NOT Hydrogen & Oxygen
4) add e-‘s where they are required
5) balance for O by adding H2O to the side that needs O
6) balance for H by adding H+ ions to the side that needs H
7) recombine by making e-‘s the same
Example:
 Mn+2 + CO2
MnO4-1 + H2C2O4
--------------------------------------------------------------- Mn+2 + 4 H2O)
2 (8 H+ + MnO4-1 + 5 e
5 (H2C2O4
 2 CO2 + 2 H+ + 2 e)
6 H+ + 2 MnO4-1 + 5 H2C2O4  2 Mn+2 + 10 CO2 + 8 H2O
Example:
Cu + NO3-1
 Cu+2 + NO2
---------------------------------------------------------Cu
 Cu+2 + 2 e
-1
+
2( NO3 + e + 2 H
 NO2 + H2O)
Cu + 2 NO3-1 + 4 H+
 Cu+2 + 2 NO2 + 2 H2O
Example:
I-1 + Br2
 IO3-1 + Br-1
---------------------------------------------------------I-1 + 6 e + 3 H2O
 IO3-1 + 6 H+
3(Br2
 2 Br-1 + 2 e)
I-1 + 3 Br2 + 3 H2O
 IO3-1 + 6 Br-1 + 6 H+
Balancing in Basic Solution
You can do it in a base, too.
Rules for balancing in a basic solution:
1) determine oxidation #’s
2) split into ½ reactions
3) balance for mass that is NOT Hydrogen & Oxygen
4) add e-‘s where they are required
5) charge by adding OH-1 where you need negative charges
6) add ½ as many H2O as the difference between OH’s between left and right
7) recombine by making sure the electrons are the same
Example
Bi(OH)3 + SnO2-2  Bi + SnO3-2
--------------------------------------------- Bi + 3 OH-1]
2[ Bi(OH)3 + 3 e
3[ SnO2-2 + 2 OH-1  SnO3-2 + 2 e + H2O]
2 Bi(OH)3 + 3 SnO2-2  2 Bi + 3 SnO3-2 + 3 H2O
Example
 Cr(OH)3 + ClO3-1
Cr + ClO4-1
-------------------------------------------------2 [ Cr + 3 OH-1
 Cr(OH)3 + 3 e]
3 [ ClO4-1 + 2 e + H2O ClO3-1 + 2 OH-1]
2 Cr + 3 ClO4-1 + 3 H2O  2 Cr(OH)3 + 3 ClO3-1
Example
Co+2 + OCl-1
 Co(OH)3 + Cl-1
------------------------------------------------------2 [ Co+2 + 3 OH-1
 Co(OH)3 + e]
OCl-1 +2 e + H2O  Cl-1 + 2 OH-1
2 Co+2 + OCl-1 + 4 OH-1 + H2O
 2 Co(OH)3 + Cl-1