### 1.4 Balancing Redox Equations using Oxidation Numbers

In previous chemistry classes you learned how to balance equations. Following the Law of Conservation of Mass you learned that the number of atoms of each element must be the same on both the reactant and product side of the equation.

Many redox reactions cannot easily be balanced just by counting atoms. Consider the following net ionic equation:

Cu

_{(s)}+ Ag^{+}_{(aq)}→ Cu^{2+}_{(aq)}+ Ag_{(s)}If you simply count atoms, the equation appears to be balanced - 1 copper atom or ion on each side of the equation, and one silver. But do you see what

balanced -isn'tthe charges!If you total the electrical charge on the reactant side of the equation you find a total charge of +1, versus +2 on the product side. Charges represent gain or loss of electrons, and, like atoms, electrons are conserved during chemical reactions.There are two common techniques we can use to help us balance redox reactions - the oxidation number method and the half-reaction method. We'll look at the oxidation number method first.

## Balancing Equations using Oxidation Numbers

You may have already guessed how we will balance equations using the oxidation number method. Let's create our summary table for the copper-silver reaction:

Cu

_{(s)}+ Ag^{+}_{(aq)}→ Cu^{2+}_{(aq)}+ Ag_{(s)}

element |
initial ox no |
final ox no |
change in e ^{-} |
|||||

Cu |
0 |
→ |
+2 |
lost 2 |
||||

Ag |
+1 |
→ |
0 |
gain 1 |

We can see that the number of electrons lost by copper does not equal the number gained by silver. We need to correct that, so we will multiply Ag by 2, giving us a total of two silvers. (We'll multiply copper by one - it won't change anything but will help keep us organized):

element |
initial ox no |
final ox no |
change in e ^{-} |
balance for electrons |
||||

Cu |
0 |
→ |
+2 |
lost 2 |
× |
1 |
= |
2 |

Ag |
+1 |
→ |
0 |
gain 1 |
× |
2 |
= |
2 |

We now are balanced for electrons - two electrons will transfer, from copper to silver.

The highlighted values - our multipliers to balance electrons - will become our balancing coefficients in the equation. Our chart helps us to keep organized and see that we should put a "1" in front of copper and a "2" in front of silver. Our balanced equation:

1 Cu

_{(s)}+ 2Ag^{+}_{(aq)}→ 1 Cu^{2+}_{(aq)}+ 2Ag_{(s)}It is not necessary to put the "1" in front of copper.

Let's work through several more examples. As we go we'll learn several tricks that we'll need to use.

Balance the following reaction using the oxidation number method:

MnO

_{4}^{1-}+ Fe^{2+}+ H^{1+}→ Mn^{2+}+ Fe^{3+}+ H_{2}OThe next step is to determine oxidation numbers. In the summary table below I will only include items whose oxidation numbers change. Since the number of electrons lost must equal the number of electrons gained, we will multiply by values that give us equal numbers of electrons:

element |
initial ox no |
final ox no |
change in e ^{-} |
balance for electrons |
||||

Mn |
+7 |
→ |
+2 |
5 |
× |
1 |
= |
5 |

Fe |
+2 |
→ |
+3 |
1 |
× |
5 |
= |
5 |

Balancing our equation for electrons we get:

1 MnO

_{4}^{1-}+ 5 Fe^{2+}+ H^{1+}→ 1 Mn^{2+}+ 5 Fe^{3+}+ H_{2}O

But wait- the equation is not balanced for hydrogen and oxygen atoms! After balancing for electrons, it is still necessary to balance for all other atoms in the equation. Using inspection we see that there are 4 oxygen on the reactant side of the equation (1 MnO_{4}^{1-}), but only 1 on the product side. Put a 4 in front of H_{2}O to correct this:1 MnO

_{4}^{1-}+ 5 Fe^{2+}+ H^{1+}→ 1 Mn^{2+}+ 5 Fe^{3+}+ 4 H_{2}OWe now have 8 hydrogen on the product side (4 H

_{2}O), so we will need 8 on the reactant side as well. This gives us our final balanced equation:1 MnO

_{4}^{1-}+ 5 Fe^{2+}+ 8 H^{1+}→ 1 Mn^{2+}+ 5 Fe^{3+}+ 4 H_{2}O

This example will show us another very important trick. Balance the following equation:

NH

_{3}+ O_{2}→ NO_{2}+ H_{2}ODetermine oxidation numbers and set up a summary table - but don't finish it just yet:

element |
initial ox no |
final ox no |
change in e ^{-} |
balance for electrons |
||||||

N |
-3 |
→ |
+4 |
7 |
||||||

O |
0 |
→ |
-2 |
2 |

Before using a multiplier to get the electrons to match, notice the subscript with oxygen - O_{2}. In our summary chart we base our oxidation number changes on a single atom, but our formula tells us that we must have two oxygen. You will save some time and frustration if we take this into account at leastnow. So in our summary table we will add some columns to change our minimum number of atoms and electrons involved. Then we complete the chart: |

element |
initial ox no |
final ox no |
change in e ^{-} |
No. atoms |
No. e ^{-} |
balance for electrons |
||||||

N |
-3 |
→ |
+4 |
7 |
= |
7 |
× |
4 |
= |
28 |
||

O |
0 |
→ |
-2 |
2 |
× |
2 |
= |
4 |
× |
7 |
= |
28 |

We now have our multipliers for the balanced equation "4" for nitrogen and "7" for oxygen - but which oxygen??? The one on the reactant side or the two different compounds that contain oxygen on the product side???

Here's where our trick becomes more useful, but will require some trial and error. Since were were counting oxygen atoms in the O

_{2}molecule on the reactant side of the equation, that's where we'll use the "7". (You could make the same argument about NO_{2}, but since nitrogen's oxidation number also changed we will use nitrogen's balancing coefficient there).4 NH

_{3}+ 7 O_{2}→ 4 NO_{2}+ H_{2}OThe last step is to balance for hydrogen atoms (and finishing oxygen), which will mean placing a 6 in front of H

_{2}O:4 NH

_{3}+ 7 O_{2}→ 4 NO_{2}+ 6 H_{2}O

A fairly easy but long one. Use the trick given in the last example to help solve it. You may want to try it on your own before looking at the solution. Balance:

K

_{2}Cr_{2}O_{7}+ NaI + H_{2}SO_{4}→ Cr_{2}(SO_{4})_{3}+ I_{2}+ H_{2}O + Na_{2}SO_{4}+ K_{2}SO_{4}Your first concern is to make sure you correctly determine all oxidation numbers. You can simply your work for those tricky polyatomic ions such as SO

_{4}^{2-}if you realize that the S in SO_{4}^{2-}will always be the same as long as the SO_{4}^{2-}remains intact. Since the only place you see sulfur in this reaction is in SO_{4}^{2-}, sulfur's oxidation number is not going to change.Similarly, hydrogen and oxygen are always in compounds, so their oxidation numbers also won't change during the reaction. That narrows down the list of elements to check.

element |
initial ox no |
final ox no |
change in e ^{-} |
No. atoms |
No. e ^{-} |
balance for electrons |
||||||

Cr |
+6 |
→ |
+3 |
3 |
||||||||

I |
+1 |
→ |
0 |
1 |

Next, check for any subscripts associated with either of these two elements - we see that Cr always has a subscript of "2" (in both K

_{2}Cr_{2}O_{7}and Cr_{2}(SO_{4})_{3}), and I has a subscript in I_{2}. So we'll add that to our summary chart to get a total number of electrons transferred, andbalance.then

element |
initial ox no |
final ox no |
change in e ^{-} |
No. atoms |
No. e ^{-} |
balance for electrons |
||||||

Cr |
+6 |
→ |
+3 |
3 |
× |
2 |
= |
6 |
× |
1 |
= |
6 |

I |
+1 |
→ |
0 |
1 |
× |
2 |
= |
2 |
× |
3 |
= |
6 |

Our table now tells us to use a balancing coefficient of "1" for Cr on both sides of the equation and "3" for iodine. Since we counted the atoms in I

_{2 }(and not HI), the "3" will go in front of I_{2}:

1K_{2}Cr_{2}O_{7}+ NaI + H_{2}SO_{4}→1Cr_{2}(SO_{4})_{3}+3I_{2}+ H_{2}O + Na_{2}SO_{4}+ K_{2}SO_{4}With these numbers in place, we now balance for atoms in the remainder of the equation to get our final answer:

1 K

_{2}Cr_{2}O_{7}+ 6 NaI + 7 H_{2}SO_{4}→ 1 Cr_{2}(SO_{4})_{3}+ 3 I_{2}+ 7 H_{2}O + 3 Na_{2}SO_{4}+ 1 K_{2}SO_{4}

One more tricky one. Balance

Zn + HNO

_{3}→ Zn(NO_{3})_{2}+ NO_{2}+ H_{2}ODetermine oxidation numbers and create your summary chart:

element |
initial ox no |
final ox no |
change in e ^{-} |
No. atoms |
No. e ^{-} |
balance for electrons |
||||||

Zn |
0 |
→ |
+2 |
2 |
||||||||

N |
+5 |
→ |
+4 |
1 |

The main thing to notice is that N appears in two separate products - Zn(NO

_{3})_{2}and NO_{2}. Should we consider the subscript for nitrogen from Zn(NO_{3})_{2}? In this case no, because this compound also contains Zn, the oxidized element. Also, the oxidation number for nitrogen does not change from HNO_{3}to Zn(NO_{3})_{2}.

element |
initial ox no |
final ox no |
change in e ^{-} |
balance for electrons |
||||

Zn |
0 |
→ |
+2 |
2 |
× |
1 |
= |
2 |

N |
+5 |
→ |
+4 |
1 |
× |
2 |
= |
2 |

We now get our balancing coefficients from our summary table. A "1" will be placed in front of Zn, but which N should we use for the "2"? If you put it in front of both HNO

_{3}and NO_{2}you'll find you cannot balance for nitrogen atoms. Since the oxidation number for nitrogen changed in becoming NO_{2}, we will try it there first. Some trial-and-error may be required:

1Zn + HNO_{3}→1Zn(NO_{3})_{2}+2NO_{2}+ H_{2}OWith the 2 in place in front of NO

_{2}, we can now balance the rest of the equation for atoms. Doing so gives us the final answer:1 Zn + 4 HNO

_{3}→ 1 Zn(NO_{3})_{2}+ 2 NO_{2}+ 2 H_{2}O

Balancing by oxidation number can be easy or difficult, depending on the equation you are given to balance. If you sometimes struggle with the more difficult examples, don't worry - you do get better with practice. Focus first on solving the simpler equations.

Be sure to complete the practice questions (Set 3, Question 1) before moving on to the next section.