Redox Equations
q
under Basic
Conditions
Basic
B
i conditions
diti
means that
th t you have
h
more OH- andd very
little H+ in your solution. In fact, you have so little H+ that
it can’t appear in the equation because it doesn’t exist!
The way you will balance redox reactions under basic
conditions is exactly the same way you balanced equations
under neutral and acidic solution, the only difference is that
we will include one additional step to remove the H+ in our
equation
ti andd replace
l
it with
ith OH-.
Redox Equations
q
under Basic
Conditions
The steps to balancing oxidation-reduction
oxidation reduction (redox) reactions under
basic conditions are:
1. Write separate equations for reduction and oxidation half reactions.
2 For each half reaction
2.
a. First balance all elements except O and H.
b. Add H2O to balance O.
c. Add H+ to balance H.
d. Add electrons to balance net charge.
3. If necessary, multiply one or both ½ reaction equations by integers
so the total number of electrons used in one reaction is equal to the
total number of electrons furnished by the other reaction.
4. Add the two ½ reaction equations together and cancel any
common terms.
5 Add OH- to
5.
t BOTH sides
id off the
th equation
ti to
t change
h
any H+’s
’ iinto
t H2O.
O
6. Double check that all species and charges balance.
E ample 1.
Example
1
Let’s try this example
Balance the equation:
NO2-(aq) + Al(s) = NH3(g) + AlO2-(aq)
under
nder basic conditions
conditions.
NO2-(aq)
( ) + Al(s)
Al( ) = NH3(g)
( ) + AlO2-(aq)
( )
½ Reactions:
NO2-(aq) = NH3(g) Al(s) = AlO2-(aq)
Balancing:
NO2-(aq) = NH3(g) + 2H2O
7H+ + NO2-(aq) = NH3(g) + 2H2O
6e- + 7H+ + NO2-((aq)
q) = NH3(g) + 2H2O
2H2O + Al(s) = AlO2-(aq)
2H2O + Al(s) = AlO2-(aq) + 4H+
2H2O + Al(s) = AlO2-(aq) + 4H+ + 3eMultiplying
l l
the
h secondd equation by
b 2 so we have
h
6 electrons:
l
4H2O + 2 Al(s) = 2 AlO2-(aq) + 8H+ + 6e-
NO2-(aq)
( ) + Al(s)
Al( ) = NH3(g)
( ) + AlO2-(aq)
( )
Adding equations together:
6e- + 7H+ + NO2-(aq) = NH3(g) + 2H2O
+
4H2O + 2 Al(s) = 2 AlO2-(aq) + 8H+ + 6e6e- + 7H+ + NO2-(aq) + 4H2O + 2 Al(s) =
NH3(g) + 2H2O + 2 AlO2-(aq) + 8H+ + 6eRemoving common terms:
NO2-(aq) +2 H2O +2Al(s) = NH3(g) +2AlO2-(aq) +1H+
Now to our added rule for basic solutions (next page)
NO2-(aq)
( ) + Al(s)
Al( ) = NH3(g)
( ) + AlO2-(aq)
( )
NO2-((aq)
q) +2 H2O +2Al(s)
( ) = NH3(g) + 2 AlO2-((aq)
q) + 1 H+
Now to our added rule for basic solutions:
5 Add OH- to BOTH sides of the equation to change any H+’s
5.
s into H2O
To apply this rule find the number of H+’s in the equation.
In this case there is 1 H+ on the right hand side of the equation,
equation
so we add 1 OH- to BOTH sides of the equation so it remains
balanced.
The trick is that on the side with the H+, the H+ and the OH- can
pp
from the
be combined to make H2O,, so the H+ disappears
equation! (Next page)
NO2-(aq)
( ) + Al(s)
Al( ) = NH3(g)
( ) + AlO2-(aq)
( )
NO2-((aq)
q) +2 H2O +2 Al(s)
( ) = NH3(g) + 2 AlO2-((aq)
q) + 1 H+
+ OH+ OHRewriting:
g
NO2-(aq) +2 H2O +2Al(s) + OH- =
NH3(g) +2AlO2-(aq) +1 H++ OHCombining H+ and OH-:
NO2-(aq) +2 H2O +2Al(s) + OH- = NH3(g) +2AlO2-(aq) +1H2O
NO2-(aq)
( ) + Al(s)
Al( ) = NH3(g)
( ) + AlO2-(aq)
( )
Checkingg to remove common terms a second time:
NO2-(aq) +2H2O +2Al(s) + OH- = NH3(g) +2AlO2-(aq) +1H2O
Becomes:
NO2-(aq) + H2O +2Al(s) + OH- = NH3(g) +2AlO2-(aq)
Double checking our balance:
NO2-(aq) + H2O +2Al(s) +OH- = NH3(g) +2AlO2-(aq)
N
1
=1
O
2
1
1 =
2(2)
( )
H
2
1 =
3
Al
2
=
2
Charge
-1
-1 =
2(-1)
NO2-(aq)
( ) + Al(s)
Al( ) = NH3(g)
( ) + AlO2-(aq)
( )
Adding physical forms we get our final answer:
NO2-(aq) + H2O(l) + 2Al(s) + OH-(aq) = NH3(g) +2 AlO2-(aq)
E ample 2.
Example
2
Let’s try this example
Balance the equation:
Al(s) + MnO4-(aq) = MnO2(s) + Al(OH)4-(aq)
under basic conditions.
Al( ) + MnO
Al(s)
M O4-(aq)
( ) = MnO
M O2(s)
( ) + Al(OH)4-(aq)
( )
½ Reactions:
Al(s) = Al(OH)4-(aq)
MnO4-(aq) = MnO2(s)
Balancing:
4 H2O + Al(s) = Al(OH)4-(aq)
4 H2O + Al(s) = Al(OH)4-(aq) + 4H+
4 H2O + Al(
Al(s)) = Al(OH)4-(aq)
( ) + 4H+ + 3e
3 MnO
M
O4-(aq)
( ) = MnO
M O2(s)
( ) + 2 H2 O
4H+ +MnO4-(aq) = MnO2(s) + 2 H2O
3e- + 4H+ +MnO4-(aq) = MnO2(s) + 2 H2O
Al( ) + MnO
Al(s)
M O4-(aq)
( ) = MnO
M O2(s)
( ) + Al(OH)4-(aq)
( )
Both equations
q
have the same # off electrons,
so we don’t have to multiply anything!
Adding equations together:
4 H2O + Al(s) = Al(OH)4-(aq) + 4H+ + 3e+
3 - + 4H+ +MnO
3e
M O4-(aq)
( ) = MnO
M O2(s)
( ) + 2 H2 O
4 H2O + Al(s)
( ) + 3e- + 4H+ +MnO4-((aq)
q) =
Al(OH)4-(aq) + 4H+ + 3e- + MnO2(s) + 2 H2O
Removing common terms:
2 H2O + Al(s) +MnO4-(aq) =Al(OH)4-(aq) + MnO2(s)
Al( ) + MnO
Al(s)
M O4-(aq)
( ) = MnO
M O2(s)
( ) + Al(OH)4-(aq)
( )
2 H2O + Al(s)
( ) +MnO4-((aq)
q) = Al(OH)
( )4-((aq)
q) + MnO2((s))
How about that. There aren’t any H+’s so I don’t have to
do anything! This happens every once in a while
while.
Double checking overall balance:
2 H2O + Al(
Al(s)) +MnO
M O4-(aq)
( ) = Al(OH)4-(aq)
( ) + MnO
M O2(s)
()
H
2(2)
=
4
O
2
4
=
4
2
Al
1
= 1
Mn
1
1
Charge
-1
= -1
Al( ) + MnO
Al(s)
M O4-(aq)
( ) = MnO
M O2(s)
( ) + Al(OH)4-(aq)
( )
Putting in physical forms to get our final answer:
2 H2O(l) + Al(s) +MnO4-(aq) = Al(OH)4-(aq) + MnO2(s)
Practice problems
Here are a couple of problems for you to try.
Balance the followingg two reactions under BASIC conditions:
CN-(aq) + MnO4-(aq) = CNO-(aq) + MnO2(s)
Cl2(g) = Cl-(aq) + OCl-(aq)
Practice problems
CN-(aq) + MnO4-(aq) = CNO-(aq) + MnO2(s)
Answer: 3 CN-(aq) + 2 MnO4-(aq) + 1 H2O(l) =
3 CNO-(aq) + 2 MnO2(s) + 2 OH- (aq)
Cl2(g) = Cl-(aq) + OCl-(aq)
Answer:
2 Cl2(g) + 4OH- (aq) = 2 Cl-(aq) + 2 OCl-(aq) + 2 H2O (l)
If you didn’t get these answers, look over my notes on the next
few pages, otherwise you can exit from this tutorial now.
CN-(aq)
( ) + MnO
M O4-(aq)
( ) = CNO-(aq)
( ) + MnO
M O2(s)
()
½ Reactions:
CN-(aq) = CNO-(aq)
MnO4-(aq) = MnO2(s)
Balancing:
H2O + CN-((aq)
q) = CNO-((aq)
q)
H2O + CN-(aq) = CNO-(aq) + 2H+
H2O + CN-(aq) = CNO-(aq) + 2H+ +2eMnO4-(aq) = MnO2(s) + 2H2O
4H+ + MnO4-(aq) = MnO2(s) + 2H2O
3e- + 4H+ + MnO4-(aq) = MnO2(s) + 2H2O
CN-(aq)
( ) + MnO
M O4-(aq)
( ) = CNO-(aq)
( ) + MnO
M O2(s)
()
½ Reactions:
H2O + CN-(aq) = CNO-(aq) + 2H+ +2e3e- + 4H+ + MnO4-(aq) = MnO2(s) + 2H2O
So we need to multiply the first reaction by 3 and the
second by 2!
3x(H2O + CN-(aq) = CNO-(aq) + 2H+ +2e- )
2x( 3e- + 4H+ + MnO4-(aq) = MnO2(s) + 2H2O)
So we have:
3 H2O + 3 CN-(aq) = 3 CNO-(aq) + 6H+ + 6 e6e- + 8 H+ + 2 MnO4-(aq) = 2 MnO2(s) + 4 H2O
CN-(aq)
( ) + MnO
M O4-(aq)
( ) = CNO-(aq)
( ) + MnO
M O2(s)
()
Adding together:
3 H2O + 3 CN-(aq) = 3 CNO-(aq) + 6H+ + 6 e+
6e- + 8 H+ + 2 MnO4-(aq) = 2 MnO2(s) + 4 H2O
We get:
3 H2O + 3 CN-(aq) + 6e- + 8 H+ + 2 MnO4-(aq) =
3 CNO-(aq) + 6H+ + 6 e- +2 MnO2(s) + 4 H2O
Removing common terms:
3 CN-(aq) + 2 H+ + 2 MnO4-(aq) =
3 CNO-(aq) +2 MnO2(s) + 1 H2O
CN-(aq)
( ) + MnO
M O4-(aq)
( ) = CNO-(aq)
( ) + MnO
M O2(s)
()
Last equation:
3 CN-(aq) + 2 H+ + 2 MnO4-(aq) =
3 CNO-(aq) +2 MnO2(s) + 1 H2O
Adding 2OH- to both sides:
3 CN-(aq) + 2 H+ + 2 MnO4-(aq) + 2OH- =
3 CNO-(aq) +2 MnO2(s) + 1 H2O + 2OHCombining the H+ and OH- on the left side to get H2O:
3 CN-(aq) + 2 MnO4-(aq) + 2H2O =
3 CNO-(aq) +2 MnO2(s) + 1 H2O + 2OH-
CN-(aq)
( ) + MnO
M O4-(aq)
( ) = CNO-(aq)
( ) + MnO
M O2(s)
()
Last equation:
3 CN-(aq) + 2 MnO4-(aq) + 2H2O =
3 CNO-(aq) +2 MnO2(s) + 1 H2O + 2OH-
Removing common terms a second time:
3 CN-(aq) + 2 MnO4-(aq) + 1 H2O =
3 CNO-(aq) +2 MnO2(s) + 2OH-
CN-(aq)
( ) + MnO
M O4-(aq)
( ) = CNO-(aq)
( ) + MnO
M O2(s)
()
Checking the balance:
3 CN-(aq) + 2 MnO4-(aq) + 1H2O = 3 CNO-(aq) +2 MnO2(s) + 2OHC
3
=3
N
3
=3
Mn
2
=
2
O
2(4)
1 =
3
2(2)
2(1)
H
2 =
2
Charge
3(-1)
2(-1)
= 3(-1)
2(-1)
Putting
P
i in
i the
h physical
h i l forms
f
to get our final
fi l equation:
i
3 CN-(aq) + 2 MnO4-(aq) + 1 H2O(l) =
q) + 2 MnO2((s)) + 2 OH- ((aq)
q)
3 CNO-((aq)
Cl2(g)
( ) = Cl-(aq)
( ) + OCl-(aq)
( )
½ Reactions:
Cl2(g) = Cl-((aq)
q)
Cl2(g) = OCl-(aq)
(Note: This is a tricky one,
one the same compound on the left is
undergoing BOTH oxidation and reductions at the same time.
This is called a disproportionation reaction, and you will see
reactions
i
like
lik this
hi every once in
i a while.)
hil )
Cl2(g)
( ) = Cl-(aq)
( ) + OCl-(aq)
( )
Balancing:
Cl2(g)
( ) = 2 Cl-(aq)
( )
2e- + Cl2(g) = 2 Cl-(aq)
Cl2(g) = 2 OCl-(aq)
2 H2O + Cl2(g) = 2 OCl-(aq)
2 H2O + Cl2(g) = 2 OCl-(aq) + 4 H+
2 H2O + Cl2(g) = 2 OCl-(aq) + 4 H+ + 2 eBoth equations have 2 electrons so we can combine without
a multiplication:
2e- + Cl2(g) + 2 H2O + Cl2(g) =
2 Cl-(aq)
( ) + 2 OCl-(aq)
( ) + 4H+ + 2 e-
Cl2(g)
( ) =Cl
Cl-(aq)
( ) + OCl-(aq)
( )
Last equation:
2 - + Cl2(g)
2e
( ) + 2 H2O + Cl2(g)
( )=
2 Cl-(aq) + 2 OCl-(aq) + 4H+ + 2 eRemoving
R
i common terms andd combing
bi Cl2:
2 Cl2(g) + 2 H2O = 2 Cl-(aq) + 2 OCl-(aq) + 4H+
Adding 4OHAddi
4OH to both
b h sides
id off the
h equation:
i
2Cl2(g) + 2H2O + 4OH- = 2 Cl-(aq) +2OCl-(aq) + 4H+ + 4OHCombing the H+ and OH- on the right side of the equation:
2Cl2(g) + 2H2O + 4OH- = 2 Cl-(aq) +2OCl-(aq) + 4H2O
Removing common terms:
2Cl2(g) + 4OH- = 2 Cl-(aq) + 2 OCl-(aq) + 2 H2O
Cl2(g)
( ) = Cl-(aq)
( ) + OCl-(aq)
( )
Last equation:
2 Cl2(g)
( ) + 4OH- = 2 Cl-(aq)
( ) + 2 OCl-(aq)
( ) + 2 H2 O
Checkingg balance:
2 Cl2(g) + 4OH- = 2 Cl-(aq) + 2 OCl-(aq) + 2 H2O
Cl
2(2)
= 2
2
O
4(1) =
2(1)
2(1)
H
4(1) =
2(2)
Charge
4(-1) = 2(-1)
2(-1)
Putting in physical forms for final answer:
g + 4OH- (aq)
q = 2 Cl-(aq)
q + 2 OCl-(aq)
q + 2 H2O (l)
2 Cl2(g)