6.1.3 redox reactions. oxidation numbers identify and indicate which element is oxidized and which...

6.1.3 REDOX REACTIONS

• Oxidation numbers identify and indicate which element is oxidized and which is reduced. Here's an example - the reaction between sodium metal and chlorine gas:

• It is often useful to write the oxidation number for every element above the element in the equation. Thus for our reaction we have:

• balancing coefficients in the equation  do not affect the value of the oxidation numbers.

2 Na + Cl2 → 2 NaCl

0 0      +1   -1

2 Na + Cl2 → 2 NaCl

element

initialox no   final

ox nochange in

electrons (e-)oxidized or

reduced

Na 0 → +1 lost 1 e- oxidized

Cl 0 → -1 gain 1 e reduced

EXAMPLE

2 Mg + O2 → 2 MgO

Add in the oxidation numbers• which are oxidized and which are reduced?

An increase in oxidation number indicates oxidationA decrease in oxidation number indicates reduction

Reducing agentthe substance that is oxidized.It allows another element to be reduced.Oxidizing agentthe substance that is reduced.It allows another element to be oxidized.

N2 + 2H2 → 2 NH3

EXAMPLE

elementinitialox no

 finalox no

e- oxidized orreduced

Agent

N 0 → -3 gain 3e- reduced oxidizing agent: N2

H 0 → +1 lose 1e- oxidized reducing agent: H2

ASSIGNMENTS

• Practice 6.1.3• Assignment 6.1.3

BALANCING REDOX REACTIONS

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

Cu(s) + Ag+(aq) → Cu2+

(aq) + Ag(s)

• If you simply count atoms, the equation is balanced but the charges aren’t balanced! Charges represent gain or loss of electrons, and, like atoms, electrons are conserved during chemical reactions.

1. BALANCING EQUATIONS USING OXIDATION NUMBERS

• Cu(s) + Ag+(aq) → Cu2+

(aq) + Ag(s

• 1 Cu(s) + 2Ag+(aq) → 1 Cu2+

(aq) + 2Ag(s)

element initialox no   final

ox no change in e-

balance for electrons

Cu 0 → +2 lost 2 ×  1 = 2

Ag +1 → 0 gain 1 ×  2 = 2

EXAMPLE

• MnO41- + Fe2+ + H1+ → Mn2+ + Fe3+ + H2O

• 1 MnO41- + 5 Fe2+ + H1+ → 1 Mn2++ 5 Fe3+ + H2O

• H and O are not balanced

• 1 MnO41- + 5 Fe2+ + 8 H1+ → 1 Mn2+ + 5 Fe3+ + 4

H2O

element initialox no   final

ox no change in e-

balance for electrons

Mn +7 → +2 5 ×  1 = 5

Fe +2 → +3 1 ×  5 = 5

EXAMPLE

• NH3 + O2 → NO2 + H2O

• Before using a multiplier to get the electrons to match, notice the subscript with oxygen - O2. 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:

element initialox no   final

ox nochange in

e- balance for electrons

N -3 → +4 7            

O 0 → -2 2            

element

initialox no   final

ox no change in e-

No. atom

s  No.

e-

balance for electrons

N -3 → +4 7 x 1 = 7 ×  4 = 28

O 0 → -2 2 ×  2 = 4 ×  7 = 28

• Since we were counting oxygen atoms in the O2 molecule on the reactant side of the equation, that's where we'll use the "7". Since nitrogen's oxidation number also changed we will use nitrogen's balancing coefficient there

4 NH3 + 7 O2 → 4 NO2 + H2O

• The last step is to balance for hydrogen atoms (and finishing oxygen), which will mean placing a 6 in front of H2O:

4 NH3 + 7 O2 → 4 NO2 + 6 H2O

• K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4

• Your first concern is to make sure you correctly determine all oxidation numbers. Since the only place you see sulfur in this reaction is in SO4

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.

• K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4

• Next, check for any subscripts associated with either of these two elements - we see that Cr always has a subscript of "2" (in both K2Cr2O7 and Cr2(SO4)3), and I has a subscript in I2. So we'll add that to our summary chart to get a total number of electrons transferred, and then balance

element initialox no   final

ox no change in e-

No. atom

s  No.

e-

balance for electrons

Cr +6 → +3 3                

I +1 → 0 1                

• 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 I2 (and not HI), the "3" will go in front of I2:

1 K2Cr2O7 + NaI + H2SO4 → 1 Cr2(SO4)3 + 3 I2 + H2O + Na2SO4 + K2SO4

• With these numbers in place, we now balance for atoms in the remainder of the equation to get our final answer:

1 K2Cr2O7 + 6 NaI + 7 H2SO4 → 1 Cr2(SO4)3 + 3 I2 + 7 H2O + 3 Na2SO4 + 1 K2SO4

element initialox 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

ONE MORE EXAMPLE

• One more tricky one. BalanceZn + HNO3 → Zn(NO3)2 + NO2 + H2O

• Determine oxidation numbers and create your summary chart:

• The main thing to notice is that N appears in two separate products - Zn(NO3)2 and NO2. Should we consider the subscript for nitrogen from Zn(NO3)2? In this case no, because this compound also contains Zn, the oxidized element. Also, the oxidation number for nitrogen

does not change from HNO3 to Zn(NO3)2 .

element initialox no   final

ox no change in e-

No. atom

s  No.

e-

balance for electrons

Zn 0 → +2 2                

N +5 → +4 1                

• 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 HNO3 and NO2 you'll find you cannot balance for nitrogen atoms. Since the oxidation number for nitrogen changed in becoming NO2, we will try it there first. Some trial-and-error may be required:

1 Zn + HNO3 → 1 Zn(NO3)2 + 2 NO2 + H2O

• With the 2 in place in front of NO2, we can now balance the rest of the equation for atoms. Doing so gives us the final answer:

1 Zn + 4 HNO3 → 1 Zn(NO3)2 + 2 NO2 + 2 H2O

PRACTICE PROBLEMS

• Practice problems 6.1.5- question #1

6.1.5- HALF REACTIONS

• Another way to balance redox reactions is by the half-reaction method. This technique involves breaking an equation into the oxidation reaction and the reduction reaction.

The general technique involves the following:• The overall equation is broken down into two half-reactions. If

there are any spectator ions, they are removed from the equations.

• Each half-reaction is balanced separately -atoms and then charge. Electrons are added to one side of the equation to balance charge.

• Next the two equations are compared to make sure electrons lost equal electrons gained. One will be an oxidation reaction, the other will be a reduction reaction.

• Finally the two half-reactions are added together, and any spectator ions that were removed are placed back into the equation

• Mg(s) + Cl2 (g) → MgCl2 (s)

• In this reaction, Mg is oxidized and Cl is reduced (Mg changes from 0 to +2; Cl changes from 0 to -1)• Balance the two reactions for atoms.

• Next balance the equations for charge by adding electrons

Mg → Mg+2 Cl2 → 2 Cl-

Mg → Mg+2 + 2 e- Cl2 + 2 e- → 2 Cl-

oxidation reduction

• Finally, add the two equations together: Mg + Cl2 → Mg+2 + 2 Cl-

• and reform any compounds broken apart in the earlier steps:  Mg + Cl2 → MgCl2• We see that the original equation was already

balanced, not just for atoms but for electrons as well

• Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag(s)

• Identify the elements undergoing oxidation (Cu) and reduction (Ag). The nitrate group (NO3) is a spectator ion which we will not include in our half-reactions.

• After balancing for atoms and for charge, we see that the two equations do not have the same number of electrons - there are 2 in the copper reaction but only one in the silver reaction. Multiply everything in the silver reaction by 2, then we will add the equations together:

Cu → Cu+2 + 2 e- Ag+ + 1 e- → Ag

oxidation reduction

Step 1 Step 2 Step 3

Write the balancedhalf-reactions

Balance electron

sAdd half-reactions

Cu → Cu+2 + 2 e-   Cu → Cu+2 + 2 e-

Ag + 1 e- → Ag- × 2 2 Ag+ + 2e- → 2 Ag

Add equations together Cu + 2 Ag+ → Cu+2 + 2 Ag

Reform compound/return spectator ions

Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

• Here is a reaction occurring in an acid solution.MnO4

- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O

• In this example, spectator ions have already been removed. Even though hydrogen and oxygen do not undergo changes in oxidation number they are not spectators and we need to work with them in our half-reactions.

• We determine that Mn undergoes reduction (+7 to +2) while Fe undergoes oxidation (+2 to +3). The iron half-reaction is straight forward, but the manganese reaction is more complex - we must include hydrogen and oxygen in its half-reaction:

• To balance the manganese half-reaction - first balance for Mn and O atoms. Next balance the H atoms, and finally add enough electrons to balance the charge on both sides of the equation.

Fe2+→ Fe +3 + 1e- MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O

oxidation reduction

Step 1 Step 2 Step 3

Write the balancedhalf-reactions Add half-reactions

Fe2+→ Fe +3 + 1e- × 5 5Fe2+→ 5Fe +3 + 5e-

MnO4- + 8H+ + 5e- → Mn2+ +

4H2OMnO4

- + 8 H+ + 5e- → Mn2+ + 4 H2O

Add equations togetherMnO4

- + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4H2O

• HNO3 + Cu + H+ → NO2 + Cu2+ + H2O

• Determine what is oxidized, what is reduced, and write the two balanced half-reactions (Step 1)• Balance for electrons lost = electrons gained

(Step 2)• Add equations together

Step 1 Step 2 Step 3

Write the balancedhalf-reactions Add half-reactions

Cu → Cu+2 + 2e- Cu → Cu+2 + 2e-

HNO3 + H+ + 1 e- → NO2 + H2O

× 22HNO3 + 2H+ + 2e- → 2NO2 +

2H2O

Add equations together2HNO3 + Cu + 2H+ → 2NO2 +

Cu2+ + 2H2O

ASSIGNMENT

• Finish practice problems• Assignment 6.1.5- hand-in• When balancing redox reactions, either the oxidation

number method or the half-reaction method may be used. Often you'll find that one method works best for some equations, while the other method is more suited for other reactions. Or you may find one method just easier to use. The practice exercises and assignments tell you which method to use for a reaction, but as you get get more experience you'll be able to make your own decision as to which method to use.

• Writing half-reactions, however, is a skill you will need for our final topic in this course - Electrochemistry - so be sure you can write balanced half-reactions.