# how do you balance equations

im confused on balancing equations ..some are easy but some  just dont make sense

Posted Thu 14th June, 2012 @ 17:40 by Onyeka

To balance an equation, you first look at the oxidation states of the elements before and after, but disregard oxygen and hydrogen. This is easily shown by writing out half equationsof the elements changing oxidation states. By doing this you can identify the loss/gain of electrons. from that you can decipher how many electrons are needed on which side. Then you add H+ ions until u have used up all the oxygen to make water.

this probably makes no sense so i will illustrate with a fairly difficult example, the reduction of a dichromate ion (VI) to Cr3+ using a solution of Fe2+

1) write half equations for both of these reactions

Fe2+ --> Fe3+  +  e-

a) balancing this half equation requires steps of its own, this is what we know so far.

(Cr2O7)2-                    --> 2Cr3+

b) lets add electrons to allow for the change in oxidation state. in this example, chromium is going from +6 to +3. meaning that each chromium atom will gain 3 electrons. and since we have 2, 6 will be gained overall

(Cr2O7)2-  +  6e-          --> 2Cr3+

c) now lets even up the charges with some H+ ions. this in practice is done by acidifying the solution. if you add up the charges, you have -8 on the left, and +6 on the right. (2 of the 3+)

so from this you need to make the left = to the right in terms of charges. so you can and 14H+ to the left and the charges become equal

notice that this is exactly how many hydrogens you need to balance out the left hand side for water... you make 7H2O.

(Cr2O7)2-  +  14H+  + 6e-  --> 2Cr3+ + 7H2O

2) now make the 2 half equations = by balancing the electrons. in this case you would multiply the top 1 by 6 so that 6 electrons are created. After that you add the two half equations together b putting all the left hand side things on the left, and the same for the right

(Cr2O7)2- + 6Fe2+  +  6e- + 14H+  --> 2Cr3+  +  6Fe3+  + 6e-  + 7H20

3) cancel aything that appears on both sides, and you have the full equation for the reaction balanced in terms of electrons and everything

(Cr2O7)2-  +  6Fe2+  +  14H+  -->  2Cr3+  +  6Fe3+  + 7H2O

_________________________________________________________________________________________

balancing equations when its for electrons is also easy.

Fe2+ --> Fe3+  + e-

Zn2+   + 2e-  -->  Zn(s)

make the electrons balanc by doubling iron then add them

2Fe2+  +  Zn2+  -->  2Fe3+  +   Zn(s)

Answered Thu 14th June, 2012 @ 19:37 by melvin

my teachers just said that nothing can "disappear" so both side have to be equal all the time. they also said leave hydrogen and oxygen the last elements to be balanced

Answered Thu 14th June, 2012 @ 22:54 by fboco

To put it simply. The sides must have the same number of atoms.

Lets say for example H + O2 ------> O2 +H20 (I dont think this is really a proper equation but its just an example)

As you can see there are only 2 oxygen atoms in the reactants whereas there are 3 in the products.

Now if we put a 2 in front of the oxygen atoms in the reactants we will have 4 in total

H + 2O2 ---------> H2O + O2

Now we must balance it out so we have 4 oxygen atoms in the products

H +202 -----------> 2H20 + O2

You get the idea. In the end you would end up with something like this:

4H + 202 ------------> 2H20 + O2

In both reactants and products we now have 4 hydrogen atoms, and 4 oxygen atoms.

Answered Fri 15th June, 2012 @ 11:39 by Tezz