### Balancing redox equations by sight

Reduction oxidation (redox) equations are very difficult to balance by sight. However, sometimes you will see a redox equation included in a "balance by sight" assignment. Below, you will find my efforts to explain how some of the commonly asked equations could be balanced by sight.

If you are still new to the techniques of balancing chemical equations, my advice to you would be to go work on simpler problems and then come back here when you've mastered those simpler equations.

If you decide to not follow my advice, then dive right in!

Problem #1: Cu + HNO3 ---> Cu(NO3)2 + NO2 + H2O

Solution:

1) The key is to see that the HNO3 coefficient must be twice the value of the H2O coefficient. This is because H only appears on the left in the HNO3 and appears in the right only in the H2O. Like this:

Cu + 2HNO3 ---> Cu(NO3)2 + NO2 + H2O

2) The above will not work (even though the H is now balanced). There is now no way to balance the N or the O, since there is already a 2 in front of the nitric acid, so do this:

Cu + 4HNO3 ---> Cu(NO3)2 + NO2 + 2H2O

What happened here is that the 2HNO3 & H2O combination will not work, so I went to a combination of 4HNO3 & 2H2O. Let us continue balancing, hoping the the 4 & 2 coefficients will work.

3) Now, balance the N:

Cu + 4HNO3 ---> Cu(NO3)2 + 2NO2 + 2H2O

4) Check the oxygen:

twelve on each side. It's balanced.

Comment: I saw this equation on Yahoo Answers presented this way:

H+(aq) + Cu(s) + NO3¯(aq) ---> H2O(ℓ) + NO2(g) + Cu2+(aq) + NO3¯(aq)

and the question asked to balance the net ionic equation. I recommended that it be balanced as a molecular equation and then converted to the net ionic. The final answer is this:

4H+(aq) + Cu(s) + 2NO3¯(aq) ---> 2H2O(ℓ) + 2NO2(g) + Cu2+(aq)

Problem #2: Cu + HNO3 ---> Cu(NO3)2 + NO + H2O

Solution:

1) Count up the oxygens on the right of the unbalanced to get 8. That suggests this coefficient:

Cu + 8HNO3 ---> Cu(NO3)2 + NO + H2O

Since the oxygens on the left only come in threes, 24 is the least common multiple of 3 and 8. Also, we know that we need an even coefficient in front of the nitric acid because the hydrogen on the right only comes in twos.

What we now do is continue balancing, aiming to get 24 oxygens on the right-hand side.

2) Balance the H:

Cu + 8HNO3 ---> Cu(NO3)2 + NO + 4H2O

3) Now comes a guess. We need an even number of oxygens on the right (24 on the left, so 24 needed on the right) and we can't do that with an odd coefficient in front of the NO, so the guess is to put a 2 in front of the NO and see what happens:

Cu + 8HNO3 ---> Cu(NO3)2 + 2NO + 4H2O

4) Balance the N:

Cu + 8HNO3 ---> 3Cu(NO3)2 + 2NO + 4H2O

5) Check the oxygens and they are now balanced, so finish off with the copper:

3Cu + 8HNO3 ---> 3Cu(NO3)2 + 2NO + 4H2O

Problem #3: Cu + HNO3 ---> CuNO3 + NO2 + H2O

Solution:

1) Balance the hydrogen:

Cu + 2HNO3 ---> CuNO3 + NO2 + H2O
And it's balanced. Well, that was easy!

Problem #4: Cu + HNO3 ---> CuNO3 + NO + H2O

Solution:

1) Balance the hydrogen:

Cu + 2HNO3 ---> CuNO3 + NO + H2O

The oxygen cannot be balanced because changing the NO or the H2O coefficient will immediately cause the N or the H to be out of balance.

2) Try 4 and 2 as coefficients to balance the hydrogen:

Cu + 4HNO3 ---> CuNO3 + NO + 2H2O

Now, we need to see if we can balance the nitrogen next, keeping in mind that we want to get to 12 oxygens on the right.

3) Balance the N:

Cu + 4HNO3 ---> 3CuNO3 + NO + 2H2O

The reason I picked a 3 was that was the fastest way to get me near to 12 oxygens. I could have put the 3 on the NO, but that leaves me well short of 12 oxygens. I also could have tried coefficients of 2 in fron of the CuNO3 and the NO. You may try it out to see if using 2 and 2 works.

4) Since the O is now balanced, finish with the copper:

3Cu + 4HNO3 ---> 3CuNO3 + NO + 2H2O

Problem #5: Cu + HNO3 ---> NH4NO3 + Cu(NO3)2 + H2O

Solution:

1) Count the oxygens on the right-hand side to get 10. However, the oxygens on the left-hand side only come in threes, so do this:

Cu + 10HNO3 ---> NH4NO3 + Cu(NO3)2 + H2O

2) Now, an educated guess. We need to get 10 nitrogens and let's guess that the rest of them (note that there are already two nitrogens in the NH4NO3) can be balanced by the Cu(NO3)2:

4 Cu + 10HNO3 ---> NH4NO3 + 4Cu(NO3)2 + H2O

3) Now, we need to play with the hydrogen and see what happens to the oxygen also. There are 4 hydrogen in the NH4NO3, so we try this:

4Cu + 10HNO3 ---> NH4NO3 + 4Cu(NO3)2 + 3H2O

And it's done.

Problem #6: Ag + HNO3 ---> NO + AgNO3 + H2O

Solution:

Comment: the key to balancing this equation by sight is to work with the hydrogens. The reason for that? Hydrogen is in only one place on the left and only one place on the right.

1) First attempt:

Ag + 2HNO3 ---> NO + AgNO3 + H2O

This fails because, while the nitrogen and silver are balanced, the oxygen is not. Any attempt to fix the oxygen imbalance will affect the nitrogen, silver or hydrogen balance. Trying to fix that would then throw the oxygen out of balance, if you ever did balance the oxygen.

2) Second attempt:

Ag + 4HNO3 ---> NO + AgNO3 + 2H2O

OK, so far. The next step is to balance the nitrogens, but how? Do I use coefficients of 3 and 1 on the right-hand side? Or 2 and 2? The only way is trial and error. (Also, note that there are two possible 3 and 1 possibilities.)

3) Let's try 2 and 2:

Ag + 4HNO3 ---> 2NO + 2AgNO3 + 2H2O

Do the oxygens balance? No, we have have 12 on the left, but only 10 on the right. There is no way to fix the oxygen without impacting at least one of the other elements, all of which are balanced (including the silver, which would be a trivial thing to balance with 2Ag on the left).

4) Let's try 3 and 1:

Ag + 4HNO3 ---> NO + 3AgNO3 + 2H2O

Now the oxygens balance.

5) Continue to the final answer by balancing the silver:

3Ag + 4HNO3 ---> NO + 3AgNO3 + 2H2O

Comment: this 3 and 1 is the other possible arrangement:

Ag + 4HNO3 ---> 3NO + AgNO3 + 2H2O
This one fails because of the oxygen imbalance (12 on the left and 8 on the right). I deliberately picked the correct 3 and 1 distribution (out of the two possible) to use first.

Problem #7: H3AsO3(aq) + I3¯(aq) ---> I¯(aq) + H3AsO4(aq)

Comment: on this one I use the usual redox practice of using water and hydrogen ion, even though they are not written in the equation. This is because, even though they are not written, we know by context that they are present.

This can be confusing if you are looking at these equation in a "balance equations by sight" unit. You might think I am adding things when I should not. Please trust me that, in this case, I am not. The water and the hydrogen ion are already present, just not written into the equation you are given.

Solution:

1) Balance the I:

H3AsO3(aq) + I3¯(aq) ---> 3I¯(aq) + H3AsO4(aq)

2) Balance the O

H2O + H3AsO3(aq) + I3¯(aq) ---> 3I¯(aq) + H3AsO4(aq)

3) Balance the H

H2O + H3AsO3(aq) + I3¯(aq)---> 3I¯(aq) + H3AsO4(aq) + 2H+

The final answer is balanced for atoms and charge, a requirement in the redox world.

Problem #8: NH4NO3 ---> N2 + O2 + H2O

Solution:

1) Nitrogen already balanced. Balance hydrogen:

NH4NO3 ---> N2 + O2 + 2H2O

I picked hydrogen because it is in one place on each side, whereas the oxygen is in one place on the reactant side, but two places on the product side.

2) Balance oxygen:

NH4NO3 ---> N2 + 12O2 + 2H2O

3) Clear the fraction:

2NH4NO3 ---> 2N2 + O2 + 4H2O

Problem #9: FeS2 + O2 + H2O ---> Fe(OH)3 + SO42- + H+

Solution:

1) Balance the S

FeS2 + O2 + H2O ----> Fe(OH)3 + 2SO42- + H+

2) Balance the H:

FeS2 + O2 + 2H2O ----> Fe(OH)3 + 2SO42- + H+

3) Balance the O:

FeS2 + 92O2 + 2H2O ----> Fe(OH)3 + 2SO42- + H+

4) Clear the fraction:

2FeS2 + 9O2 + 4H2O ----> 2Fe(OH)3 + 4SO42- + 2H+

Problem #10: NO2 + H2O -------> NO + HNO3

Solution:

1) Balance the H first since it is in only one place on each side of the arrow:

NO2 + H2O -------> NO + 2HNO3

2) Balance the N since it is less complicated than trying the O:

3NO2 + H2O -------> NO + 2HNO3

3) Check the oxygen to find 7 oxygen atoms on each side. It's balanced.

Problem #11: PbO2 + HCl ---> H2O + PbCl2 + Cl2

Solution:

1) The O in PbO2 and the H in HCl are going to go into the H2O, so we need two H for every one O. Let us balance the O first:

PbO2 + HCl ---> 2H2O + PbCl2 + Cl2

2) Now we need 4 H on the left:

PbO2 + 4HCl ---> 2H2O + PbCl2 + Cl2

This gives us 4 Cl on the left-hand side and we see that there are 4 Cl already present on the right-hand side. Since the Pb was already balanced, we are done. It's balanced.

Problem #12: Fe + V2O3 ---> Fe2O3 + VO

Solution:

1) Balance the Fe:

2Fe + V2O3 ---> Fe2O3 + VO

2) Balance the V:

2Fe + V2O3 ---> Fe2O3 + 2VO

3) Balance the oxygen:

It seems clear the we must try to balance the oxygen using the V2O3/VO pair. From the equation in step two, we try this:
2Fe + 2V2O3 ---> Fe2O3 + 4VO

We see that the oxygens are still not balanced, but the difference was reduced. In step two, there are 3 O on the left and 5 O on the right, for a difference of two. In the attempt just above, there are 6 O on the left and 7 O on the right, for a difference of one. So, we try this:

2Fe + 3V2O3 ---> Fe2O3 + 6VO

There are now 9 O on each side of the equation. The equation is balanced.

Problem #13: (NH4)2Cr2O7 ---> N2 + H2O + Cr2O3

Solution:

Balance the hydrogen next because it's in one place on the left and one place on the right:

(NH4)2Cr2O7 ---> N2 + 4H2O + Cr2O3

Check the oxygen: seven on the left; four plus three on the right for seven. Oxygen is balanced.

We're done!

Problem #14: N2H4 + KBrO4 ---> N2 + KBr + H2O

Solution:

That means that only hydrogen and oxygen need to be balanced. I will pick the oxygen to do first:

N2H4 + KBrO4 ---> N2 + KBr + 4H2O

I picked the oxygen because, if I picked the hydrogen, I would have put a 2 in front of the water, but then I would have to change it to a 4 in order to balance the oxygen. Now, balance the hydrogen:

2N2H4 + KBrO4 ---> N2 + KBr + 4H2O

That changes the nitrogen, so balance it:

2N2H4 + KBrO4 ---> 2N2 + KBr + 4H2O

And we're done.

Problem #15: H2O + O2 + As ---> HAsO2

Solution:

1) Pick the hydrogen to balance rather than the oxygen:

H2O + O2 + As ---> 2HAsO2

The reason for this is because, if I balanced the oxygen first and then balanced the hydrogen, I'd have to go back and re-balance the oxygen.

2) Balance the oxygen:

H2O + 32O2 + 2As ---> 2HAsO2

I also balanced the arsenic.

3) Clear the fraction:

2H2O + 3O2 + 4As ---> 4HAsO2

and it's done.

Problem #16: H2O2 + N2H4 ---> N2 + O2 + H2O

Solution:

I plan to hold the N2H4 coefficient at one and use the H2O2/H2O pair to balance the hydrogen. I will then check my oxygen situation and modify the O2 coefficient as needed.

Notice that the presence of four H in the N2H4 will make the H2O coefficient always be two greater than the H2O2 coefficient.

1) Let's make the H2O coefficient be two greater and see what happens to the H:

H2O2 + N2H4 ---> N2 + O2 + 3H2O

Whelp, that made a mess of things. The H is balanced, but the O is very messed up and there's no way I can make them balanced, except . . . .

2) I'm going to keep with my plan of attacking the H and explain why just below:

2H2O2 + N2H4 ---> N2 + O2 + 4H2O

In step one, I had two O on the left and 5 on the right, for a gap of 3. In step two, I have 4 O on the left and 6 O on the right, for a gap of two. I've closed the gap by one.

3) Will that continue to work? The short answer is yes. I'm going to proceed to a set of coefficients that balance the O and see what happens to the H:

4H2O2 + N2H4 ---> N2 + O2 + 6H2O

I increased the two coefficients by two more and that balances the oxygen. I then check my H and find that they are balanced as well.

Notice that half-way through, I abandoned my plan to focus on the H and looked more closely at the O. However, if I had continued to ignore the O and balanced the H, I would have wound up in the exact same place.

Problem #17: Cu + H2SO4 ---> CuSO4 + H2O + SO2

Solution:

1) Balance the sulfur:

Cu + 2H2SO4 ---> CuSO4 + H2O + SO2

2) Balance the hydrogen:

Cu + 2H2SO4 ---> CuSO4 + 2H2O + SO2

3) Check the oxygen:

Eight on each side. The equation is balanced!

Problem #18: Au + NaCN + O2 + H2O ---> NaAu(CN)2 + NaOH

Solution:

1) Balance the cyanide:

Au + 2NaCN + O2 + H2O ---> NaAu(CN)2 + NaOH

This also has the effect of balancing the sodium.

2) Now, my problem is with the oxygen and the hydrogen both going to the same place, that being the hydroxide. Note also that the OH of hydroxide has one O for every one H. I need to make the oxygen and water on the left combine to produce OH (meaning one O for every one H). There is a way to do that:

Au + 2NaCN + 12O2 + H2O ---> NaAu(CN)2 + 2NaOH

This, however, has the unfortunate drawback of putting the sodium out of balance.

3) The sodium can be put into balance like this:

2Au + 4NaCN + 12O2 + H2O ---> 2NaAu(CN)2 + 2NaOH

I increased the coefficient on the NaAu(CN)2 from one to two. Then, I put a four in front of the NaCN. That balanced the sodium as well as the cyanide. I then put a two in front of the Au to balance it.

4) The equation is now balanced, but with a fractional coefficient. So, I multiply through by two:

4Au + 8NaCN + O2 + 2H2O ---> 4NaAu(CN)2 + 4NaOH

Bonus Problem: C6H6 + H2O2 ---> CO2 + H2O

Solution:

1) Balance the carbon:

C6H6 + H2O2 ---> 6CO2 + H2O

Oxygen on left = 2, oxygen on right = 13; difference = 11

2) Tentative attempt to balance the hydrogen:

C6H6 + 6H2O2 ---> 6CO2 + 9H2O

Notice the the coefficient on the H2O2 must always be 3 less than the coefficient on the H2O.

Oxygen on left = 12, oxygen on right = 21; difference = 9

Comment: this next move is based on experience. Notice that there is a decreasing gap in the amount of oxygens. As I continue to increase the coefficients for the H2O2 and the H2O. Since the gap is still large, let us try this . . . .

3) Double the coefficients for the H2O2 and the H2O:

C6H6 + 12H2O2 ---> 6CO2 + 18H2O

This actually throws the hydrogen out of balance as well as not obeying our need to have the H2O2 coefficient be three less than the H2O coefficient. The oxygens are still not balanced, but the amount of oxygen on each side has been lowered from 9 to 6.

4) Fix the coefficient to obey the "three less" rule:

C6H6 + 15H2O2 ---> 6CO2 + 18H2O

This balances the equation.

Notice that, when the oxygen was out of balance by six, the hydrogen was also out of balance by six. Adding three more H2O2 added six H as well as six O.

That was a hard one!