Balancingg Oxidation-Reduction
Reactions - neutral and acid solutions
Balancing an oxidation-reduction reaction is a little simpler
if the reaction is either neutral or acidic,
acidic and a little more
difficult if the solution is basic. If the problem doesn’t
specifically state that the reaction conditions are basic, you
should always use the simpler procedure given in this tutorial
for acid or neutral conditions.
Balancingg Oxidation-Reduction
Reactions - neutral and acid solutions
The steps to balancing oxidation-reduction (redox) reactions are:
1. Write separate equations for reduction and oxidation
h lf reactions.
half
ti
2. For each half reaction
a. First balance all elements except O and H.
b Add H2O to balance O
b.
O.
c. Add H+ to balance H.
d. Add electrons to balance net charge.
3 If necessary,
3.
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.
q
together
g
and cancel any
y
4. Add the two ½ reaction equations
common terms.
5. Double check that all species and charges balance.
E ample 1.
Example
1
Let’ss apply these steps to balance the reaction:
Let
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 1. Write separate equations for reduction and
oxidation half reactions.
In making ½ reactions you are looking for elements that are in
one oxidation state as a reactant, and a different oxidation state
as a product.
d t
The first obvious ½ reaction is:
Fe2+ = Fe3+
Here it is clear that the Fe is going from a +2
to a +3 oxidation state.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 1. Write separate equations for reduction and
oxidation half reactions.
By process of elimination, your guess for the other ½ reaction
would then be :
M O4- = Mn
MnO
M 2+
If you go through the oxidation state calculation you would
find that the Mn here goes from a +7 state to a +2 state, but I
don’t usually do this. It takes too much time, and it doesn’t
really help with balancing the equation.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 1. Write separate equations for reduction and
oxidation half reactions.
I usually don’t even bother to think about elements changing
oxidation states. I just look for the same element in different
compounds
d on the
th right
i ht an left
l ft side
id off the
th equation.
ti
Thus,
Th
I would just focus on the two main elements, Fe and Mn
and make ½ reactions that connect them as products and
reactants:
Fe2+ = Fe3+
MnO4- = Mn2+
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 2. For each half reaction:
a. First Balance all elements except O and H
b Add H2O to balance O
b.
c. Add H+ to balance H
d. Add electrons to balance net charge
Fe2+ = Fe3+
A. Fe balanced no problem
B. No O’s to balance so don’t worry- be happy
C. No H’s either - don’t worry - be happy
D (next page)
D.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 2. For each half reaction:
Part d. Add electrons to balance net charge
Fe2+ = Fe3+
Right now you have +2 charge on the left hand side of the
equation
ti andd a +3 on the
th right.
i ht To
T gett these
th
in
i balance
b l
you
always add electrons to the side with the largest positive
charge to lower the charge to match the other side of the
equation. So here you add 1 e- to the right hand side so the
net charge on both sides is +2:
Fe2+ = Fe3+ +1e-
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Fe2 = Fe3+ +1eAt this point you now know that this is your oxidation
½ reaction.
reaction
How? Well, one way to tell is to look at the oxidation states.
F iincreases it
Fe
its oxidation
id ti state
t t from
f
+2 to
t +3,
+3 andd oxidation
id ti is
i
defined as the increase in an oxidation state.
The other way is from the electrons. Whenever the electrons
are on the right had side of the equation, you have an
oxidation ½ reaction.
reaction This is the easy way because you don
don’tt
have to figure out oxidation states!
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 2 for other half reaction:
a. First Balance all elements except O and H
b. Add H2O to balance O
c Add H+ to balance H
c.
d. Add electrons to balance net charge
MnO4-6Mn2+
A. Mn balanced no problem
B. 4 O
O’ss on the left, so you need to add 4 H2O
O’ss on
the right to balance the O’s
MnO4-6Mn2+ + 4H2O
C (Next Page)
C.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 2 for other half reaction:
a. First Balance all elements except O and H
b. Add H2O to balance O
c Add H+ to balance H
c.
d. Add electrons to balance net charge
MnO4-6Mn2+ + 4H2O
C. Now you have 8 H’s on the right, so you have to
add 8 H+’ss to the left to make the H
H’ss balance:
8H+ + MnO4-6Mn2+ + 4H2O
D. (Next Page)
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 2 for other half reaction:
a. First Balance all elements except O and H
b. Add H2O to balance O
c Add H+ to balance H
c.
d. Add electrons to balance net charge
8H+ + MnO4-6Mn2+ + 4H2O
D. Now you have +8-1 = +7 charge on the left hand
side of the equation and a +22 on the right. So here
you add 5 e- to the left hand side so the net charge
on both sides is +2:
5e- + 8H+ + MnO4-6Mn2+ + 4H2O
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
55e- + 8H+ + MnO
M O4-6Mn
6M 2+ + 4H2O
At this point you now know that this is your reduction
½ reaction.
How? Again
Again, if you spent the time do determine the
oxidation state of the Mn, you would find that it is +7 in
MnO4- and +2 in Mn2+ so the oxidation date is reduced,
h
hence
the
th name reduction
d ti reaction.
ti
But the far easier way is to look where the electrons are.
Electrons on the left, reduction.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 3
St
3. If necessary multiply
lti l one or b
both
th ½ reaction
ti
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.
Here are your two ½ reactions:
Fe2+ = Fe3+ +1e&
5 - + 8H+ + MnO
5e
M O4- = Mn
M 2+ + 4H2O
There is only 1 electron in the first reaction, but
there are 5 electrons in the second reaction.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 3
St
3. If necessary multiply
lti l one or b
both
th ½ reaction
ti
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.
Fe2+ = Fe3+ +1e- & 5e- + 8H+ + MnO4- = Mn2+ + 4H2O
Since the total number of electrons in both ½ reactions
h
have
t be
to
b equal,
l we have
h
to
t multiply
lti l the
th first
fi t reaction
ti by
b
5 to turn it into the equation
(Fe2+ = Fe3+ +1e-)×5 = 5 Fe2+ = 5Fe3+ +5eSo the number of electrons in BOTH reactions is the same.
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 4
4. Add the two ½ reaction equations together
and cancel any common terms.
5 Fe
F 2+ = 5Fe
5F 3+ +5e
+5 5e- + 8H+ + MnO4- = Mn2+ + 4H2O
= 5e- + 8H+ + MnO4- + 5 Fe2+ =
Mn2+ + 4H2O + 5Fe3+ +5eThe only common terms are the 5 e- so we have
8H+ + MnO4- + 5 Fe2+6Mn2+ + 4H2O + 5Fe3+
Now fill in
N
i the
th physical
h i l forms
f
so I don’t
d ’t take
t k points
i t off.
ff
8H+(aq) + MnO4-(aq) + 5 Fe2+(aq) =
Mn2+ (aq) + 4H2O(l) + 5Fe3+(aq)
Balance the equation:
q
MnO4-(aq) + Fe2+(aq) = Fe3+(aq) + Mn2+(aq)
Step 5
5. Double check that all species and charges balance
balance.
8H+(aq)+MnO4-(aq)+5 Fe2+(aq) = Mn2+(aq)+4H2O(l)+5Fe3+(aq)
H8
=
4(2)
Mn
1
=1
O
4
=
4
Fe
5
=
5
Charge
+8
-11
5(+2) = +2
0
5(+3)
8-1+10 = +17
=
2 + 15 = +17
It all
ll balances
b l
so it mustt be
b good.
d
Example
p 2:
Cr2O7-2 + C2H5OH = Cr+3 + CO2
Step 1. Half reactions:
Cr2O7-2 = Cr+3 & C2H5OH = CO2
Step
p 2: Balancing
g ½ reactions
Cr2O7-2 = Cr+3 & C2H5OH = CO2
Atoms EXCEPT ‘O’
O and ‘H’
H
Cr2O7-2 = 2 Cr+3
&
Balancing
B
l i O
Cr2O7-2 = 2 Cr+3 + 7 H2O &
C2H5OH = 2 CO2
3 H2O + C2H5OH = 2 CO2
Balancing H
14 H++Cr2O7-2 = 2 Cr+3+7 H2O &3 H2O+C2H5OH=2 CO2+12 H+
Balancing Charge
6 e- + 14 H+ + Cr2O7-2 = 2 Cr+3 + 7 H2O
3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 e-
Step
p 3: Gettingg the same number of electrons
in both equations.
EQN 1: 6 e- + 14 H+ + Cr2O7-2 = 2 Cr+3 + 7 H2O
EQN 2: 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 eThe first equation has 6 electrons, the second has 12,
so we need to multiply the first equation by 2
EQN 1: 2×(6 e- + 14 H+ + Cr2O7-2 = 2 Cr+3 + 7 H2O)
EQN 1: 12 e- + 28 H+ + 2 Cr2O7-2 = 4 Cr+3 + 14 H2O
EQN 2: 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 e-
St 4:
Step
4 Adding
Addi equations
ti
ttogether….
th
EQN 1:12 e- + 28 H+ + 2 Cr2O7-2 = 4 Cr+3 + 14 H2O
+
EQN 22: 3 H2O + C2H5OH = 2 CO2 + 12 H+ + 12 eNet:
12 e- + 28 H+ + 2 Cr2O7-2 + 3 H2O + C2H5OH =
4 Cr+3 + 14 H2O + 2 CO2 + 12 H+ + 12 e-
St 4:
Step
4 …and
d removing
i common terms
t
Net:
12 e- + 28 H+ + 2 Cr2O7-2 + 3 H2O + C2H5OH =
4C
Cr+3 + 14 H2O + 2 CO2 + 12 H+ + 12 e-
Removing terms:
16 H+ + 2 Cr2O7-2 + C2H5OH =
4 Cr+3 + 11 H2O + 2 CO2
St 4:
Step
4 Double
D bl checking
h ki the
th balance
b l
16 H+ + 2 Cr2O7-2 + C2H5OH = 4 Cr+3 + 11 H2O + 2 CO2
H
16
6
=
11(2)
C
Cr
2(2)
=
4
O
2(7)
1
=
11
2(2)
C
2
=
2(1)
Charge
16+
2(-2)
0
= 4(+3)
0
0
It all seems to balance so it looks good!
Final Ans
Answer
er
16 H+(aq) + 2 Cr2O7-2(aq) + C2H5OH(aq) =
q) + 11 H2O ((l)) + 2 CO2(g)
4 Cr+3((aq)
(I almost forgot to put in the physical forms)
Practice Problems
Cu(s) + NO3-(aq) = Cu2+(aq) + NO(g)
I-(aq)
( ) + ClO-(aq)
( ) I3-(aq)
( ) + Cl-(aq)
( )
H3AsO4(aq) + Zn(s) = AsH3(g) + Zn2+(aq)
(Answers are on the next few pages)
Cu(s)
( ) + NO3-((aq)
q) = Cu2+((aq)
q) + NO(g)
(g)
½ rxns: Cu(s) = Cu2+(aq) & NO3-(aq) = NO(g)
Balance atoms:
Cu(s) = Cu2+(aq) &
NO3-(aq) = NO(g) + 2H2O
4H+(aq)+ NO3-(aq) = NO(g) + 2H2O
Balance charge:
Cu(s) = Cu2+(aq) + 2e3e- + 4H+(aq)+ NO3-(aq) = NO(g) + 2H2O
Get # of electrons the same:
(Cu(s) = Cu2+(aq) + 2e-)x3
2x( 3e- + 4H+(aq)
(aq)+ NO3-(aq) = NO(g) + 2H2O)
3Cu(s) = 3Cu2+(aq) + 6e6e- + 8H+(aq)+ 2NO3-(aq) = 2NO(g) + 4H2O
Adding together:
3Cu(s) + 6e- + 8H+(aq)+ 2NO3-(aq)
= 3Cu2+(aq) + 6e- + 2NO(g) + 4H2O
Cu(s)
( ) + NO3- ((aq)
q) = Cu2+((aq)
q) + NO(g)
(g)
Removing common terms:
3Cu(s) + 8H+(aq)+ 2NO3-(aq) = 3Cu2+(aq) + 2NO(g) + 4H2O
Checking
Ch
ki balance:
b l
Cu 3
H
8
N
2
O
2(3)
Charge
+8
2(
2(-1)
1)
=3
=
=
=
= 3(2+)
4(2)
2
2(1)
4(1)
Finall Answer:
Fi
A
3Cu(s) + 8H+(aq)+ 2NO3-(aq) = 3Cu2+(aq) + 2NO(g) + 4H2O(l)
I-(aq)
( ) + ClO-(aq)
( ) = I3-(aq)
( ) + Cl-(aq)
( )
½ Rxns: I-(aq) = I3-(aq) & ClO-(aq) = Cl-(aq)
Balancing atoms:
3 I-(aq) = I3-(aq) & ClO-(aq) = Cl-(aq) + H2O
2H+ + ClO-(aq)
( ) = Cl-(aq)
( ) + H2 O
Balancing Charge:
3 I-(aq) = I3-(aq) + 2e- & 2e- + 2H+ + ClO-(aq) = Cl-(aq) + H2O
No need to multiply equations because they are the same.
Adding equations:
3 I-(aq) + 2e- + 2H+ + ClO-(aq) = I3-(aq) + 2e- + Cl-(aq) + H2O
Removing common terms:
3 I-(aq) + 2H+ + ClO-(aq) = I3-(aq) + Cl-(aq) + H2O
I-(aq)
( ) + ClO-(aq)
( ) = I3-(aq)
( ) + Cl-(aq)
( )
3 I-(aq) + 2H+ + ClO-(aq) = I3-(aq) + Cl-(aq) + H2O
Double checking the balance:
I 3
= 1(3)
H
2(1)
=
Cl
1
=
O
1
Charge
3(-1)
3(
1)
2(+1)
2(
1)
-11 =
-11
-3+2-1 = -2
=
-2
2
1
1
-11
Putting
P
tti in
i physical
h i l forms:
f
3 I-(aq) + 2H+(aq) + ClO-(aq) = I3-(aq) + Cl-(aq) + H2O(l)
H3AsO
A O4(aq)
( ) + Zn(s)
Z ( ) = AsH
A H3(g)
( ) + Zn
Z 2+(aq)
( )
½ Reactions: H3AsO4(aq) = AsH3(g); Zn(s) = Zn2+(aq)
B l i
Balancing:
H3AsO4(aq) = AsH3(g) + 4H2O
8H+ + H3AsO4(aq) = AsH3(g) + 4H2O
8 e- +8H+ + H3AsO4(aq) = AsH3(g) + 4H2O
Zn(s) = Zn2+(aq) + 2 eMultiplying right equations by 4:
4 Zn(s) = 4 Zn2+(aq) + 8 eAdding equations together:
8 e- +8H+ + H3AsO
A O4(aq)
( ) + 4 Zn(s)
Z ( )=
AsH3(g) + 4H2O + 4 Zn2+(aq) + 8 eRemoving common terms:
8H+ + H3AsO4(aq) + 4 Zn(s) = AsH3(g) + 4H2O + 4 Zn2+(aq)
H3AsO
A O4(aq)
( ) + Zn(s)
Z ( ) = AsH
A H3(g)
( ) + Zn
Z 2+(aq)
( )
8H+ + H3AsO4((aq)
q) + 4 Zn(s)
( ) = AsH3(g) + 4H2O + 4 Zn2+((aq)
q)
Double checking Balance:
H8 3
As
1
O
4
Z
Zn
4
Charge
( )
8(+1)
=
3
= 1
=4(1)
4(2)
4
4
=
4(+2)
( )
Putting in physical form:
8H+(aq) +H3AsO4(aq) +4 Zn(s) =AsH
AsH3(g) +4H2O(l) +4 Zn2+(aq)