Download Balancing Redox Reactions

Balancing Redox
Reactions
Chem 12
Application of oxidation numbers:


Oxidation = an increase
in oxidation number
Reduction = a decrease
in oxidation number
Using oxidation numbers to identify redox
reactions:

Example: Determine if the following reaction is a
redox reaction:
-4+1
0 -2+1-1 +1-1
CH (g) + Cl (g) CH Cl(g) + HCl(g)
4


2
3
Oxidation number of C increases from -4 to -2
Oxidation number of Cl decreases from 0 to -1

Yes, redox rxn!
TO BALANCE A REDOX REACTION…
Zn +
+
Cu

2+
Zn
+ Cu
Zn 
+
Cu  Cu
2+
Zn
BREAK IT INTO HALF-REACTIONS!!!
THEN BALANCE EACH REACTION…
Zn +
+
Cu

2+
Zn
+ Cu
Zn 
+
+
Cu  Cu
2+
Zn
2e
…FOR BOTH MASS AND CHARGE
THEN BALANCE EACH REACTION…
Zn +
+
Cu

2+
Zn
+ Cu
Zn 
+
+
1e + Cu  Cu
2+
Zn
2e
…FOR BOTH MASS AND CHARGE
EQUILIZE ELECTRON TRANSFER…
Zn +
+
Cu

2+
Zn
+ Cu
Zn 
+
+
2(1e + Cu  Cu)
2+
Zn
2e
AND ADD THE REACTIONS TOGETHER
Zn +
+
Cu

2+
Zn
Zn 
+
+
2(1e + Cu  Cu)
2+
Zn
+
+ Cu
Zn +
+
2Cu

2e
2+
Zn
+ 2Cu
2C2O4
+
3+
Fe
 CO2 +
2+
Fe
C2O42-  CO2
Fe3+  Fe2+
BREAK IT INTO HALF-REACTIONS!!!
THEN BALANCE EACH REACTION…
2C2O4
+
3+
Fe
 CO2 +
2+
Fe
C2O42-  2CO2
Fe3+  Fe2+
…FOR BOTH MASS AND CHARGE
THEN BALANCE EACH REACTION…
2C2O4
+
3+
Fe
 CO2 +
2+
Fe
C2O42-  2CO2 + 2eFe3+  Fe2+
…FOR BOTH MASS AND CHARGE
THEN BALANCE EACH REACTION…
2C2O4
+
3+
Fe
 CO2 +
2+
Fe
C2O42-  2CO2 + 2e1e- + Fe3+  Fe2+
…FOR BOTH MASS AND CHARGE
EQUILIZE ELECTRON TRANSFER…
2C2O4
+
3+
Fe
 CO2 +
2+
Fe
C2O42-  2CO2 + 2e2(1e- + Fe3+  Fe2+)
AND ADD THE REACTIONS TOGETHER
2C2O4
+
+
3+
Fe
 CO2 +
2+
Fe
C2O42-  2CO2 + 2e2(1e- + Fe3+  Fe2+)
C2O42- + 2Fe3+  2CO2 + 2Fe2+
BALANCING IN ACIDIC SOLUTION
(the steps…..)
1.
2.
3.
4.
5.
6.
7.
8.
Assign oxidation charges to all atoms in equation
Break into half reactions
Balance each half reaction for all atoms (except O
and H)
Add H2O to balance O (because in an acidic
solution)
Add H+ ions to balance H.
Add electrons to reduction reaction to balance
charges
Equalize electron transfer (make e- lost = egained)
Add reactions!
BALANCING IN BASIC SOLUTION
(the steps…..)
Steps 1-5 same as acidic!
6.
Adjust for basic conditions by adding to both sides
the same number of OH- ions as H+ already there.
7.
Simplify each half reaction by combining H+ and
OH- to make water
8.
Cancel any water present on both sides of the half
reactions
9.
Add electrons to reduction reaction to balance
charges
10. Equalize electron transfer (make e- lost = egained)
11. Add reactions!
Examples:
1.
2.
Write a balanced half reaction that shows
the reduction of permanganate ions to
maganese (II) ions in acidic conditions.
Write a balanced half reaction that shows
the oxidation of S2O32- to SO32- in basic
conditions
Day 1: Try it:



Pg 728 # 13 and section review # 2
Pg 732 # 19, 20
Pg 734 # 23, 24
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2-

3+
Cr
+
4+
Ni

23+
Cr2O7  Cr
2+
Ni
4+
Ni
BREAK IT INTO HALF-REACTIONS!!!
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2-

3+
Cr
+
4+
Ni

+
23+
Cr2O7  Cr
2+
Ni
4+
Ni
2e
THEN BALANCE EACH REACTION…
…FOR BOTH MASS AND CHARGE
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2-

3+
Cr
+

+
23+
Cr2O7  2Cr
2+
Ni
4+
Ni
2e
4+
Ni
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2+
Ni
Cr2O7
2-
2-

3+
Cr
+
4+
Ni

+
3+
 2Cr + 7H2O
4+
Ni
2e
WE CAN BALANCE OXYGEN WITH
WATER MOLECULES BECAUSE THE
REACTION HAPPENS IN SOLUTION
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2+
Ni
2-


4+
Ni
3+
Cr
+
+
4+
Ni
2e
14H+ + Cr2O72-  2Cr3+ + 7H2O
WE CAN BALANCE HYDROGEN
WITH H+ IONS SINCE IT IS AN ACIDIC
SOLUTION
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2+
Ni
2-


4+
Ni
3+
Cr
+
+
4+
Ni
2e
6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2+
3(Ni
2-


4+
Ni
3+
Cr
+
+
4+
Ni
2e )
6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O
EQUILIZE ELECTRON TRANSFER…
BALANCING IN ACIDIC SOLUTION
2+
Ni
+ Cr2O7
2+
3(Ni
+
2-


4+
Ni
3+
Cr
+
+
4+
Ni
2e )
6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O
14H+ + Cr2O72- + 3Ni2+  3Ni4+ + 2Cr3+ + 7H2O
AND ADD THE REACTIONS TOGETHER
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
+
Ag
+ NO
Ag 
NO3  NO
+
Ag
SEE IF YOU CAN PREDICT EACH STEP
BEFORE ADVANCING THE SLIDE
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
+
Ag
Ag 
+
NO3  NO
+
Ag
+ NO
1e
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
+
Ag
+ NO
Ag 
+
NO3  NO + 2H2O
+
Ag
1e
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
+
Ag
+ NO
Ag 
+
+
4H + NO3  NO + 2H2O
+
Ag
1e
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
Ag 
+
Ag
+
Ag
+
+ NO
1e
3e- + 4H+ + NO3-  NO + 2H2O
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
3(Ag 
+
Ag
+
Ag
+
+ NO
1e )
3e- + 4H+ + NO3-  NO + 2H2O
BALANCING IN ACIDIC SOLUTION
Ag + NO3 
-
3(Ag 
+
Ag
+
Ag
+
+ NO
1e )
- + 4H+ + NO -  NO + 2H O
3e
3
2
+
4H+ + NO3- + 3Ag  3Ag+ + NO + 2H2O
BALANCING IN BASIC SOLUTION
To balance a redox reaction in a
basic solution,
where hydroxide (OH-) ions outnumber
hydrogen (H+) ions,
FIRST BALANCE IT AS IF IT WERE IN
ACIDIC SOLUTION
BALANCING IN BASIC SOLUTION
In the previous example…
4H+ + NO3- + 3Ag  3Ag+ + NO + 2H2O
… is balanced in acidic solution
BALANCING IN BASIC SOLUTION
To balance in basic solution…
4H+ + NO3- + 3Ag  3Ag+ + NO + 2H2O
4OH4OH… add an amount of hydroxide
equal to the amount of H+ to
each side of the reaction
BALANCING IN BASIC SOLUTION
To balance in basic solution…
4H+ + NO3- + 3Ag  3Ag+ + NO + 2H2O
4OH4OH…Combine the H+ and OH- to
make water
BALANCING IN BASIC SOLUTION
To balance in basic solution…
4H2O + NO3- + 3Ag  3Ag+ + NO + 2H2O
4OH-
…Combine the H+ and OH- to
make water
BALANCING IN BASIC SOLUTION
To balance in basic solution…
2
4H2O + NO3- + 3Ag  3Ag+ + NO + 2H2O
4OH…since there is water on both
sides, we can cancel some out
BALANCING IN BASIC SOLUTION
To balance in basic solution…
2H2O + NO3- + 3Ag  3Ag+ + NO + 4OH-
… and rewrite the final balanced
equation
ASSIGN OXIDATION NUMBERS TO EACH ELEMENT IN THE REACTION
0
Zn +
+1
+
Cu
The oxidation number
of zinc increases from
zero to positive two
by losing two
electrons.
ZINC IS OXIDIZED

+2
2+
Zn
0
+ Cu
The oxidation number
of copper decreases
from positive one to
zero by gaining one
electron.
COPPER IS
REDUCED
Try it:

Page 739 # 27(a, b, c), 28
ASSIGN OXIDATION NUMBERS TO EACH ELEMENT IN THE REACTION
+1
+7 -2
+2
+2
+3
+1 -2
H+ + MnO4- + Fe2+  Mn2+ + Fe3+ + H2O
The oxidation number
of Mn increases from
+7 to +2 by gaining
five electrons.
MANGANESE IS
REDUCED
The oxidation number
of Fe increases from
+2 to +3 by losing
one electron.
IRON IS OXIDIZED
+1
+7 -2
+2
+2
+3
+1 -2
H+ + MnO4- + Fe2+  Mn2+ + Fe3+ + H2O
MANGANESE IS REDUCED
IRON IS OXIDIZED
The compound that contains the element being
reduced is called the OXIDIZING AGENT
The compound that contains the element being
oxidized is called the REDUCING AGENT