7.3 – BALANCING REDOX REACTIONS USING OXIDATION NUMBERS UNIT 7 – REDOX REACTIONS & ELECTROCHEMISTRY

A NEW LOOK AT BALANCING EQUATIONS 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. According to that, this equation is balanced: 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 isn't balanced - the charges! Charges must also be balanced in a reaction so we must use another method to balance this redox reaction.

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 1 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 nochange in e - Cu0→+2lost 2 Ag+1→0gain 1

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): Ehe element initial ox no final ox no change in e - balance for electrons Cu0→+2lost 2× 1=2 Ag+1→0gain 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)

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 2 Balance the following reaction using the oxidation number method: MnO Fe 2+ + H 1+ → Mn 2+ + Fe 3+ + H 2 O The 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: H element initial ox no final ox no change in e - balance for electrons Mn+7→+25× 1=5 Fe+2→+31× 5=5

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 2 Balancing our equation for electrons we get: 1 MnO Fe 2+ + H 1+ → 1 Mn 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 Fe 2+ + H 1+ → 1 Mn Fe H 2 O We 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 Fe H 1+ → 1 Mn Fe H 2 O

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 3 Balance the following equation: NH 3 + O 2 → NO 2 + H 2 O Determine 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→+47 O0→-22

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 3 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 at least two oxygen. You will save some time and frustration if we take this into account now. 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: Yep element initial ox no final ox no change in e - No. atoms No. e - balance for electrons N-3→+47=7× 4=28 O0→-22 × 2=4× 7=28

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 3 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 we 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 O 2 → 4 NO 2 + H 2 O The 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 O 2 → 4 NO H 2 O

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 4 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 simplify 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.

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 4 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→+33 I+1→01

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 4 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, and then balance. Water element initial ox no final ox no change in e - No. atoms No. e - balance for electrons Cr+6→+33×2=6×1=6 I+1→01×2=2×3=6

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 4 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 : 1 K 2 Cr 2 O 7 + NaI + H 2 SO 4 → 1 Cr 2 (SO 4 ) I 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 NaI + 7 H 2 SO 4 → 1 Cr 2 (SO 4 ) I H 2 O + 3 Na 2 SO K 2 SO 4

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 5 Balance Zn + HNO 3 → Zn(NO 3 ) 2 + NO 2 + H 2 O Determine oxidation numbers and create your summary chart: element initial ox no final ox nochange in e - No. atoms No. e - balance for electrons Zn0→+22 N+5→+41

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 5 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 Yep element initial ox no final ox nochange in e - balance for electrons Zn0→+22× 1=2 N+5→+41× 2=2

BALANCING EQUATIONS USING OXIDATION NUMBERS – EXAMPLE 5 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: 1 Zn + HNO 3 → 1 Zn(NO 3 ) NO 2 + H 2 O With 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 ) NO H 2 O