First, assign oxidation numbers. +1 +6 -2 +2 +1 +3 +3 Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3 Note: If you do not know how to assign oxidation numbers, visit this link: http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm Next, decide what is being oxidized and what is being reduced. Cr is being reduced because its oxidation number is decreasing. Fe is being oxidized because its oxidation number is increasing. Next, start oxidation half reaction. Fe^2+ ------> Fe(OH)3 Notice how it includes the oxidized reactant and the oxidized product. Next, balance oxidation half reaction. In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O. Also, make sure the metals are balanced prior to anything else. 3H2O + Fe^2+ -------> Fe(OH)3 Now, notice how we have created yet another imbalance. We now have too much hydrogen on the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+ ions. 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ This is the somewhat tricky part. Looking at the original equation, we see OH\'s everywhere. This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how many to add is to look at how many H^+ ions you added. The number of OH^- ions you add should be equal to the H^+ ions you added. 3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^- Notice how 3OH^- is added on both sides. HOH = H2O. This is an important concept in balancing redox reactions. 3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O Since we have the same amount of H2O on both sides, we can cancel them out completely. 3OH^- + Fe^2+ -------> Fe(OH)3 Notice the charges on each side of the equation. We need to balance them somehow. We can do this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are going to have to add 1 e- to the right. 3OH^- + Fe^2+ ----> Fe(OH)3 + e- This is our balanced oxidation half reaction!!!! Next, start the reduction half reaction. Cr2O7^2- ==> Cr(OH)3 Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion. Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to be less crowded. In the end, it won\'t matter because Na, in this situation, is purely superfluous. However, make sure to add Na back in at the end. Next, balance the reduction half reaction. In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O. Also, make sure the metals are balanced prior to anything else. Cr2O7^2- -----> 2Cr(OH)3 + H2O Now, notice how we have created yet another imbalance. We now have too much hydrogen on the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+ ions. 8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O This is the somewhat tricky part. Looking at the original equatio.

1. First, assign oxidation numbers.
+1 +6 -2 +2 +1 +3 +3
Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3
Note: If you do not know how to assign oxidation numbers, visit this link:
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add

2. should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O

3. Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.

4. ( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3
Note: If you do not know how to assign oxidation numbers, visit this link:
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.

5. Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O

6. Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how

7. many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-

8. 18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.

9. In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.

10. 3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-

11. Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.

12. Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids

13. indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+

14. This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
3OH^- + Fe^2+ -------> Fe(OH)3

15. Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.

16. 8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1

17. Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.

18. 8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.

19. ( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.

20. Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Solution
First, assign oxidation numbers.
+1 +6 -2 +2 +1 +3 +3
Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3
Note: If you do not know how to assign oxidation numbers, visit this link:
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3

21. Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3

22. Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.

23. 8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.

24. 10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Na2Cr2O7 + Fe^2+ ------- Na^+ + Fe(OH)3 + Cr(OH)3
Note: If you do not know how to assign oxidation numbers, visit this link:
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.

25. 3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!

26. Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.

27. 6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe

28. 2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Next, decide what is being oxidized and what is being reduced.
Cr is being reduced because its oxidation number is decreasing.
Fe is being oxidized because its oxidation number is increasing.
Next, start oxidation half reaction.
Fe^2+ ------> Fe(OH)3
Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+

29. This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.

30. In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't

31. want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Fe^2+ ------> Fe(OH)3

32. Notice how it includes the oxidized reactant and the oxidized product.
Next, balance oxidation half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Notice how 3OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3

33. 3H2O + Fe^2+ -------> Fe(OH)3
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the left. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add

34. should be equal to the H^+ ions you added.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H^+ + 3OH^-
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
HOH = H2O. This is an important concept in balancing redox reactions.
3OH^- + 3H2O + Fe^2+ -------> Fe(OH)3 + 3H2O
Since we have the same amount of H2O on both sides, we can cancel them out completely.
3OH^- + Fe^2+ -------> Fe(OH)3
3OH^- + Fe^2+ -------> Fe(OH)3
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
Notice the charges on each side of the equation. We need to balance them somehow. We can do
this with electrons. We have a -1 charge on the left and a neutral charge on the right, so we are
going to have to add 1 e- to the right.
3OH^- + Fe^2+ ----> Fe(OH)3 + e-
This is our balanced oxidation half reaction!!!!
Next, start the reduction half reaction.

35. Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.
7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-

36. This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na

37. Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
Cr2O7^2- ==> Cr(OH)3
Notice the 2- charge on the Cr2O7, this is because it is a polyatomic ion.
Note: Na was taken out of the equation, however, it can be used in the equation. I just like it to
be less crowded. In the end, it won't matter because Na, in this situation, is purely superfluous.
However, make sure to add Na back in at the end.
Next, balance the reduction half reaction.
In order to balance oxygen in an aqueous solution, we can assume that there is free floating H2O.
Also, make sure the metals are balanced prior to anything else.
Cr2O7^2- -----> 2Cr(OH)3 + H2O
Now, notice how we have created yet another imbalance. We now have too much hydrogen on
the right. In order to solve this since it is aqueous, we can also assume there are free floating H^+
ions.
8H^+ + Cr2O7^2- ------> 2Cr(OH)3 + H2O
This is the somewhat tricky part. Looking at the original equation, we see OH's everywhere.
This is an indicator that we have a basic solution. (OH indicates basic and HCl or other acids
indicate acidic). In basic solutions, we must add OH^- ions. An easy way to remember how
many to add is to look at how many H^+ ions you added. The number of OH^- ions you add
should be equal to the H^+ ions you added.
8H^+ 8OH^- + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Notice how 2OH^- is added on both sides.
HOH = H2O. This is an important concept in balancing redox reactions.
8H2O + Cr2O7^2- ------> 2Cr(OH)3 + H2O + 8OH^-
Since we have more H2O's on the left, we can simply subtract the H2O's from the right out of
the equation.

38. 7H2O + Cr2O7^2- --------> 2Cr(OH)3 + 8OH^-
Notice the charges on each side of the equation. There's a -2 charge on the left and a -8 charge
on the right. We need to add 6e- on the left to fix this.
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
This is our balanced reduction half reaction!!!
Now that we have our balanced half reactions, we must combine them to form the final balanced
equation.
The first step in doing this is to make the electrons in the oxidation half reaction equal the
electrons in the reduction half reaction. We do this so they can cancel out, because we don't
want electrons in our final equation.
How do we make the electrons equal each other? We simply multiply. In most cases you will just
multiply each half reaction by the number of electrons in the other half reaction, but in this case,
there is a simpler solution.
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6 +
( 3OH^- + Fe^2+ ----> Fe(OH)3 + e-)*6
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3

39. This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.
2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
6e- + 7H2O + Cr2O7^2- ------> 2Cr(OH)3 + 8OH^-
18OH^- + 6Fe^2+ +7H2O + Cr2O7^2- -------> 6Fe(OH)3 + 2Cr(OH)3 + 8OH^-1
Now, this is where we need to be extra careful and make sure everything is added and multiplied
correctly. Also, make sure to add Na back in.
Simplify.
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
This will be our final balanced equation.
Next, check to make sure equation is truly balanced.
Add up all the elements and see if they check off.
24 O = 24 O
24 H = 24 H
6 Fe = 6 Fe
2 Cr = 2 Cr
2 Na = 2 Na
Now, check the charge.

40. 2+ = 2+
Since everything checks off this equation is correct:
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3
10 OH^- + 6Fe^2+ +7H2O + Na2Cr2O7 ------> 2Na^+ + 6Fe(OH)3 + 2Cr(OH)3