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Saturday, April 25, 2009 http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method

Redox reactions - in which charge is transferred between reagents - can be balanced by inspection, although it can be extremely difficult. But let's start with a very simple example:

Cu + Fe3+ -> Cu2+ + Fe2+

At first sight the equation seems to be already balanced, but if we check charges - it is not. There is +3 charge on the left, and +4 charge on the right, so we have to do something about it. Using the inspection method we should select the most complicated molecule first. Let's say it is Fe3+:

Cu + 1Fe3+ -> Cu2+ + Fe2+

If so, we can already add 1 in front of Fe2+:

Cu + 1Fe3+ -> Cu2+ + 1Fe2+

Trying to balance copper will lead us nowhere (atoms are balanced, but charges will be left unbalanced as they were from the beginning), so let's look at the charge - it can be balanced just like atoms. We have +3 on the left and +2 on the right - so we need additional +1 on the right side. Do you remember that we can use fractions at this stage? Half Cu2+ will do:

Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+

What is wrong now is the left side of equation - too much copper at the moment. Once again, half will do:

1/2Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+

Final touch - multiply everything by 2 to remove fractions:

Cu + 2Fe3+ -> Cu2+ + 2Fe2+

Atoms - balanced (one of copper and two of iron on both sides). Charge - balanced (+6 on both sides). We have just balanced the redox equation by using the inspection method. But in the case of more difficult ones - like this for example:

FeSO4 + KMnO4 +H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O

this approach is a waste of time. Much easier and faster ways of balancing redox reactions are oxidation numbers method and half reaction method.

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When balancing redox reactions we have always - apart from all the rules pertaining to balancing chemical equations - additional information about electrons moving. In the case of oxidation numbers method we assume electrons are transferred between atoms (which is only an approximation), in the case of half reactions method we assume there are two systems exchanging electrons (which is much closer to reality, although what we observe may be a multistep process with numerous intermediate reagents).

Let's try to use half reactions method to balance reaction equation already mentioned in the balancing by inspection section:

FeSO4 + KMnO4 +H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O

First of all, let's concentrate on what is really important - on the net ionic reaction:

Fe2+ + MnO4- + H+ -> Fe3+ + Mn2+ + H2O

All other ions (K+ and SO42-) are only spectators and we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method.

At first glance you can see that in the reaction iron gets oxidized and permanganate gets reduced:

Fe2+ -> Fe3+

MnO4- -> Mn2+

We will start equation balancing with balancing these half reactions - using electrons to balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. To balance charge we will add one electron on the right side:

Fe2+ -> Fe3+ + e-

We can use electrons safely, as the final step of balancing will be electron cancellation.

Now we have to balance permanganate reduction half reaction:

MnO4- -> Mn2+

Before balancing charges we have to balance atoms. What to do with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can add H+ on the left and H2O on the right:

MnO4- + H+ -> Mn2+ + H2O

Using simple balancing by inspection we will add two coefficients to balance atoms:

MnO4- + 8H+ -> Mn2+ + 4H2O

Once the atoms are balanced it is time to balance charge with electrons - there is +7 charge on the left side of the equation and +2 on the right side. To balance charges we have to add 5 electrons on the left:

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

At the moment we have two balanced half reactions. Here comes the final trick - we add these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to use just numbers of electrons, although if often means you will have to find lowest coefficients later). To have five electrons in both equations we have to multiply first equation by 5:

5Fe2+ -> 5Fe3+ + 5e-

and when we add both half reactions we will get

MnO4- + 8H+ + 5Fe2+ + 5e- -> Mn2+ + 4H2O + 5Fe3+ + 5e-

Canceling out the electrons:

MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+

And the reaction is balanced.

Balancing hydrogen and oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can add H+ on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH- and water to do the trick, for example half reaction:

ClO- -> Cl-

is not balanced, but once we add water and OH-:

ClO- + H2O -> Cl- + 2OH-

we have not only balanced atoms but we are also ready to balance charge by adding two electrons on the left:

ClO- + H2O + 2e- -> Cl- + 2OH-

and the half reaction is ready to be used. General rule says that if the reaction takes place in acidic conditions we use water and H+ to balance oxygens, and if the reaction takes place in basic conditions - we use OH- and water. Don't worry if it looks like the reaction produces H+ in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.

Also note that in really hard cases, when you have no idea what is really going on in the solution, you may look for the half reactions (and amount of electrons consumed/produced) in the standard potentials tables.

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Before we will get to explanation very important disclaimer: oxidation numbers don't exist. They were invented to help in charge accounting needed when balancing redox reaction equations, but they don't refer to any real life chemical concept.

The general idea behind the oxidation numbers (ON) method for balancing chemical equations is that electrons are transferred between charged atoms. These charges - assigned to atoms - are called oxidation numbers, just to remind you that they don't reflect real structure of the reagents.

There are several simple rules used for assigning oxidation numbers to every atom present in any compound:

* First of all, charged mono atomic ion has oxidation number equal to its charge. Thus Na+ has oxidation number +1, Fe3+ has oxidation number +3, F- has oxidation number of -1 and S2- has oxidation number of -2.

* Second rule says that the oxidation number of a free element is always 0. Thus oxidation number of Cu is 0, oxidation number of O in O2 is 0, the same holds for S in S8 and so on.

* Oxygen in almost all compounds has oxidation number -2.

* Hydrogen in almost all compounds has oxidation number +1.

* Some elements usually have the same oxidation number in their compounds:

o alkali metals - Li, Na, K, Rb, Cs - oxidation numbers are +1

o alkaline earth metals - Be, Mg, Ca, Sr, Ba - oxidation numbers are +2

o halogens (except when they form compounds with oxygen or one another) - oxidation numbers are -1 (always true for fluorine)

* Last rule says that the charge of the ion or molecule equals sum of oxidation numbers of all atoms.

There are some exceptions to the rules 2 and 3 - for example oxygen in peroxides has oxidation number of -1, and hydrogen in hydrides has oxidation number -1 too, but these are very rare cases.

Before we will try to balance any equations let's use above rules to assign oxidation numbers to atoms in several substances.

For example - what is oxidation number of sulfur in SO2? Particle is not charged, so oxidation number of sulfur must equal sum of oxidation numbers of oxygens, but with the opposite sign. Oxygen oxidation number is -2, there are two oxygens - that gives -4 together, so sulfur must have ON=+4.

What is oxidation number of atoms in CrO42-? Oxygen is -2 and there are 4 oxygens - that gives overall of -8, ion has charge of -2, so central atom must have ON=+6.

How do we use oxidation numbers for balancing? First of all, we have to understand that oxidation means increase of oxidation number, while reduction means decrease of oxidation number. In both cases change of oxidation number is due to electrons lost (oxidation) or gained (reduction). We calculate oxidation numbers for all atoms present in the reaction equation (note that it is not that hard as it sounds, as for most atoms oxidation numbers will not change) and we look for a ratio that makes the number of electrons lost equal to the number of electrons gained. That gives us additional information needed for reaction balancing.

Let's try with following reaction:

KIO3 + KI + H2SO4 -> K2SO4 + H2O + I2

First of all - we don't need any spectators here, as they are only making things look more difficult then they are in reality. Quick glance tells us that the net ionic reaction is

IO3- + I- + H+ -> H2O + I2

Looks like IO3- is oxidizing agent here and I- is reducting agent. I- has oxidation number of -1, iodine in IO3- has oxidation number of +5. On the right side in I2 both iodine atoms have oxidation number 0. It means that iodine in IO3- must gain 5 electrons. These electrons come from I- - one for every I- ion. Assuming (just like we do in the inspection method) that IO3- is the most complicated molecule and it's coefficient is 1 we will need five I- for the redox process to complete:

1IO3- + 5I- + H+ -> H2O + I2

Now that the ration between oxidizer and reducing agent is known we use simple techniques we know from the inspection method to balance remaining elements. There are six atoms of iodine on the left, so we need three I2 molecules to balance iodine:

1IO3- + 5I- + H+ -> H2O + 3I2

And the final, trivial step is balancing oxygen, hydrogen and water:

IO3- + 5I- + 6H+ -> 3H2O + 3I2

Other case we can try is oxidation of Mn2+ with NaBiO3 in acidic conditions:

Mn2+ + BiO3- + H+ -> MnO4- + Bi3+ + H2O

Using methods for oxidation numbers calculation we can easily check that manganese is oxidized from +2 to +7 (freeing five electrons) and bismuth is reduced from +5 to +3 (accepting two electrons). To balance electrons transferred we can put coefficients 2 and 5 on the left side of reaction equation:

2Mn2+ + 5BiO3- + H+ -> MnO4- + Bi3+ + H2O

Rest can be balanced by inspection and is not difficult to do, yielding:

2Mn2+ + 5BiO3- + 14H+ -> 2MnO4- + 5Bi3+ + 7H2O

Now the same equation can be also easily balanced as a full (non net-ionic) version:

4MnSO4 + 10NaBiO3 + 14H2SO4 -> 4NaMnO4 + 5Bi2(SO4)3 + 14H2O + 3Na2SO4

Apart from the three already described methods, there is also a general method, often less user friendly - but thanks to its systematic approach perfect for use in computer programs. That's the method EBAS - our chemical reaction equation balancer - uses.

Let's try an algebraic method for

H3BO3 -> H4B6O11 + H2O

It can be rather easily balanced by inspection, but let's try a more systematic approach.

What does 'balanced' mean? It means that for every element, there are the same number of atoms on both sides of the reaction equation. Our reaction has three coefficients A, B and C:

AH3BO3 -> BH4B6O11 + CH2O

From a mathematical point of view 'balanced' means that for every element we can write an equation, comparing the number of atoms on both sides of the reaction equation. An example for boron:

A = 6×B + 0×C

where coefficients 1, 6 and 0 are taken from the compound formulae. We can write similar equations for all elements - hydrogen:

3×A = 4×B + 2×C

and oxygen:

3×A = 11×B + C

As there are no free terms in this set of equations, it has a trivial solution (A = B = C = 0) which we are not interested in. We have three equations, and three unknowns. Such equation set is not a thing that you may want to solve manually, although when balancing chemical equations in most cases it can be done relatively easy, as most equations don't contain all unknowns. In this case we have very simple equation A = 6×B that we can use to substitute 6×B for A in the second and third equation to get:

18×B = 4×B + 2×C

18×B = 11×B + C

After some rearranging:

7×B = C

7×B = C

Both equations are identical. In algebra it usually means that the set of equations doesn't have a unique solution, but in the case of chemical equations we have one additional information - all coefficients must be integer and they must be the smallest ones. To find them we can assume one of the coefficients to be 1:

B = 1

If so

A = 6

C = 7

and indeed

6H3BO3 -> H4B6O11 + 7H2O

is the balanced reaction equation.

This first example doesn't look convincing - why do we have to solve set of equations when the reaction equation can be easily balanced by other means? Good point - but what if the reaction can be not easily balanced?

P2I4 + P4 + H2O -> PH4I + H3PO4

Try for a moment. Looks easy but soon gets surprisingly hard and the coefficients become pretty high, which makes you wonder if you have not made some mistake (*see some balancing hints at the bottom of that page). What about general, algebraic method?

We need five coefficients, and there are only four equations (one for each element present) - but it shouldn't bother us, as we know that we have additional information that works as an additional equation.

AP2I4 + BP4 + CH2O -> DPH4I + EH3PO4

Setting up equations:

P: 2×A + 4×B = D + E

I: 4×A = D

H: 2×C = 4×D + 3×E

O: C = 4×E

Balances for iodine and oxygen make this set look much easier then expected. C = 4×E and D = 4×A are substitutions that we are about to use to reduce number of unknowns:

1st equation: 2×A + 4×B = 4×A + E

3rd equation: 8×E = 4×D + 3×E

So we have now after some canceling:

4×B = 2×A + E

4×A = D

5×E = 4×D

C = 4×E

For someone fluent in mathematics it is obvious that we have already finished - it is now enough to assume that one of the variables equals 1 to calculate values of all others. Assuming A = 1 and simply substituting calculated values we have:

A = 1

B = 13/10

C = 64/5

D = 4

E = 16/5

These are hardly integer, but all we have to do is to find the smallest common denominator to have a list of integer coefficients in numerators. In this case the smallest common denominator is 10, so if we multiply all numbers by 10 we get:

A = 10

B = 13

C = 128

D = 40

E = 32

And you may check that these are the correct coefficients. Imagine finding them by inspection method!

It's also easy to use the algebraic method to balance redox reaction with charged species. Let's try it for

ACr2O72- + BH+ + CFe2+ -> DCr3+ + EH2O + FFe3+

What equations do we have? Four balances of atoms:

Cr: 2A = D

O: 7A = E

H: B = 2E

Fe: C = F

But that's not enough to balance equation - we have six coefficients and four equations. For algebraic method we can have one equation less than variables - so we need a fifth equation. So far we have not accounted for charge balance, and that will be our last equation needed to balance reaction:

-2A + B + 2C = 3D + 3F

(note that sign of coefficients in the last equation depends on the charge sign). We are ready to solve. First of all, we know that

E = 7A

so

B = 14A

Let's get rid of B and D in the charge balance equation:

-2A + 14A + 2C = 6A + 3F

Sorting, and using C=F:

C = 6A

That's almost ready. Let's put A = 1 and calculate all other coefficients simply using already known values:

C = 6

D = 2

E = 7

B = 14

F = 6

and balanced equation takes form:

Cr2O72- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+

That's all. Not that hard.

Probably the most important characteristics of the algebraic method is that - contrary to the inspection method - is guaranteed to give you an answer. If the reaction can be balanced, you will find coefficients. If the reaction can't be balanced - you will find it out seeing that there are more unknowns than independent equations (remember - one more is not a problem), or that equations are contradictory. With inspection method you will never have a proof that the equation can't be balanced.

There is an additional reason for this method to be important. Computers are very effective in solving sets of simultaneous equations, for example using method called Gauss elimination. What is difficult for humans is a perfect task for the number cruncher built in every processor. That's why EBAS is able to balance the so-called Blakley equation (20 unknowns in 19 equations) in a blink of an eye on a 10 year old PC.

Note that there are - although rare - cases, when reaction occurring in reality cannot be balanced with the algebraic approach. Some of these cases (together with explanation) are described in "failures" section.

*Balancing hints for P2I4 + P4 + H2O -> PH4I + H3PO4 reaction:

Contrary to what balancing by inspection rules say, start with balancing oxygen from phosphoric acid, then balance hydrogen, iodine, and finally phosphorus. Leaving oxygen and hydrogen to the end is asking for troubles.

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