Experiment 7: Oxidation Reduction Reactions

[Pages:16]Experiment 7: Oxidation-Reduction Reactions

PURPOSE Become familiar with the concepts of oxidation and reduction and how these reactions occur. Carry out several such reactions and learn to recognize when oxidation-reduction is occurring. Develop an understanding of the relative strengths of oxidizing and reducing agents and rank oxidation-reduction couples in a series. Use an oxidation-reduction series to determine if a reaction will occur spontaneously. Balance oxidation reduction reactions by the half-reaction method.

BACKGROUND In a previous experiment, "Double Displacement Reactions", you became familiar with several reactions in which ions exchange partners in order to form new compounds. A very large number of chemical reactions are of this variety. An equally large number of reactions do not fall into this category but instead involve the exchange of electrons to form new compounds. Reactions involving exchange of electrons are called oxidation-reduction reactions.

An example of such a reaction is the reaction of magnesium metal and chlorine gas.

Mg (S) + C12 (g) MgCl2 (S)

At first glance, the electron exchange may not be apparent. However, when you examine the above reaction in more detail you can see it. It actually consists of two parts that occur simultaneously.

(1) Mg Mg2+ + 2e- Each magnesium atom loses 2 electrons to form a magnesium 2+ cation. (2) 2e- + Cl2 2 Cl- Each chlorine molecule gains 2 electrons to form 2 Cl- anions.

The reaction can be pictured as below:

transfer e-

Mg

Cl [[Mg2+] [ Cl ]-

Cl transfer e-

]

[ Cl ]-

7-1

Subsequently the two oppositely charged ionic species then interact to form magnesium chloride crystals.

An oxidation-reduction reaction may be thought of as a competition between two substances for electrons. Consider the two reactions below, which are the reverse of each other:

Reaction (1) net ionic equation:

Cu(NO3)2(aq) + Zn(s) Cu(s) + Zn(NO3)2(aq) Cu2+(aq) + Zn(s) Cu(s) + Zn2+ (aq)

reduction half-reaction:

Cu2+ (aq) + 2 e- Cu(s)

oxidation half-reaction:

Zn(s) Zn2+ (aq) + 2 e-

oxidizing agent = Cu2+ reducing agent = Zn

Reaction (2) net ionic equation:

Zn(NO3)2(aq) + Cu(s) Zn(s) + Cu(NO3)2(aq) Zn2+(aq) + Cu(s) Zn(s) + Cu2+(aq)

reduction half-reaction:

Zn2+ (aq) + 2 e- Zn(s)

oxidation half-reaction:

Cu(s) Cu2+(aq) + 2 e-

oxidizing agent = Zn2+ reducing agent = Cu

Reaction (1) will occur spontaneously and Reaction (2) will not if Cu2+ is a stronger oxidizing agent (i.e. likes to be reduced or you could say likes to acquire electrons) more than Zn2+. Conversely, reaction (2) will occur and (1) will not if Zn2+ is a stronger oxidizing agent (i.e. like to be reduced or you could say likes to acquire electrons more than Cu2+ ).

In an oxidation-reduction reaction, the species that gains electrons is reduced and the substance that loses electrons is oxidized. This may seem odd at first, but remember electrons have a negative charge so gaining an electron will lower the oxidation number of the atom receiving the electron. In the above reaction magnesium is oxidized and chlorine is reduced.

The substance that gains electrons in the reaction (i.e. is reduced) can also be called the oxidizing agent. In the Mg + Cl2 reaction, Cl2(g) is the oxidizing agent because it is reduced and in the process it oxidizes (removes electrons from) the magnesium metal. Similarly, the substance that loses electrons in a reaction (i.e. is oxidized), can be called the reducing agent. In the above reaction magnesium metal is a reducing reagent.

7-2

In summary then:

electron

X Y

X loses electron(s) X is oxidized X is the reducing agent X increases its oxidation number

Y gains electron(s) Y is reduced

Y is the oxidizing agent Y decreases its oxidation number

To understand re-dox reactions, you must be able to determine the oxidation number of elements using the rules below.

Rules for Assigning Oxidation Numbers

Always start at the top of the list and work down.

1. The total sum of the oxidation numbers of all the atoms in a molecule must add up to the charge on

the molecule.

Example:

the

oxidation

number

must

sum

to

zero

in

H2O,

and

CaSO4,

and

-3

in

PO4

-3 .

2. An uncombined element has an oxidation number of zero. (for example, Li (solid), Mg (solid), H2 ) 3.The oxidation number of a monoatomic ion = charge of the monatomic ion. Examples: Oxidation number of S2- is -2, Oxidation number of Al3+ is +3

4. When combined with other elements, Group IA elements are always +1. (for example, Li in LiCl or Na in Na2S)

4. When combined with other elements, the oxidation number of all Group 2A metals = +2 ( for example, Mg in MgF2, Ca in CaO)

5. F has an oxidation number of ?1.

6. H has an oxidation number of +1.

7. O has an oxidation number of ?2.

7-3

Rules higher in the list take precedence over those lower in the list.

PROBLEM: Determine the oxidation number (O.N.) of each element in these compounds: (a) Water (b) sulfate ion (c) Hydrogen peroxide

(a) H2O Rule (1) Charges sum to zero...... Rule (6)... H is +1.... Rule (7) .... O is -2 (b) SO4-2 Rule (1) Charges sum to -2... Rule (7)... O is -2 (there are four of them

however)... Therefore... S must be +6 Check S+6 + O-2(4) = -2 (c) H2O2 Rule (1) Charges sum to zero.......... Rule (6)... H is +1..... Rule (7) .... O is -2

HOWEVER this can not be since if H is +1 and O is -2 the sum will not be zero H+1 (2) + O-2 (2) =-2!

IF there is a conflict in the rules, ignore the rule lowest on the list... therefore H2O2 Rule (1) Charges sum to zero.......... Rule (6)... H is +1 and O has to be -1. When oxygen is -1, it is called a peroxide. You will know when this happens by applying the above rules.

A very powerful method for balancing oxidation reduction equations is the half-reaction method. The method consists of the following very systematic steps. If the steps are not followed exactly in this order, you will not be able to balance the reaction correctly. The steps are

1. Identify which atom is oxidized and which atom is being reduced. Split the reaction into two halves ? one for the reduction and one for the oxidation process 2. Balance the two half reactions separately by:

a. Balance elements other than O and H. b. Balance O by adding H2O c. Balance H by adding H+ ions d. Balance the charge by adding electrons (e-) 3. Looking at the two balanced half reactions... multiply them by some numbers so that both will have the same number of electrons. 4. Add the two half reactions together. Collect like terms and cancel terms that appear on both sides of the equation. 5. If in acidic conditions, you are done. If in basic conditions, add OH -1 to both sides to combine with the H + to form H2O. 6. Check to make sure the atoms and charges are balanced.

7-4

Example: Balance the equation:

MnO4- + Cl-1 Cl2 + Mn 2+

Looks like Cl-1 is oxidized -1 to 0 and Mn is reduced from 7+ to 2+

Step 1 Step 2a Step 2b

Step 2c

reduced half rxn

MnO4-

Mn 2+

oxidized half rxn Cl- Cl2

MnO4- Mn2+

no change Mn balanced

2Cl- Cl2 Cl now balanced

MnO4- Mn2+ + 4H2O balance the 4 O's on the left add 4H2O

2Cl- Cl2 no change O balanced (aren't any)

8H+ + MnO4- Mn 2+ + 4H2O balance the 8H's on the right 2Cl- Cl2 no change H balanced

add 8H+

(aren't any)

Step 2d

8H+ + MnO4- Mn 2+ + 4H2O

the sum of the charges on left side is ( 8(+1) ? 1 = +7... on the right side its +2... add -5 ( 5e--) to balance

2Cl- Cl2 the sum of the charges on left side is 2(-1)=-2

on the right side its 0... add -2 ( 2e-) to balance

5e- + 8H+ + MnO4- Mn 2+ + 4 H2O

2Cl- Cl2 + 2e-

Step 3 Multiplying the left equation by 2 and the night equation by 5, so both have 10 e-

2 x(5 e- + 8H+ + MnO4- Mn 2+ + 4 H2O) 10 e- + 16 H+ + 2MnO4- 2Mn 2+ + 8 H2O

5x (2Cl- Cl2 + 2 e-) 10Cl- 5 Cl2 +10e-

Step 4 (adding the two equations together and combining like terms, and cancel same terms on opposite sides)

10e- + 16H+ + 2MnO4- 2Mn 2+ + 8H2O 10Cl- 5 Cl2 +10e-

Answer in acidic condition 16H+ + 2 MnO4- + 10Cl- 2Mn2+ + 5Cl2 + 8H2O

Step 5 In basic conditions only.... add OH-1 to both sides to get rid of H+ (bases mostly contain OH-1 and H2O so they can appear in your answer but H+1 can not)

( H +1 and OH-1 make H2O)

16 H+ + 2 MnO4- + 10 Cl- 2 Mn2+ + 5 Cl2 + 8 H2O

+ 16 OH-1

+16 OH -1

16 H2O + + 2 MnO4- + 10 Cl- 2 Mn2+ + 5 Cl2 + 8 H2O +16 OH -1

cancel H2O on opposite sides 8H2O + + 2 MnO4- + 10 Cl- 2 Mn2+ + 5 Cl2 + 16 OH -1

7-5

Oxidation-Reduction Reactions Procedure

SAFETY CAUTION:

Parts A and B should be done in the hood CYCLOHEXANE: EXTREMELY FLAMMABLE LIQUID AND VAPOR. VAPOR MAY CAUSE FLASH FIRE. HARMFUL OR FATAL IF SWALLOWED. HARMFUL IF INHALED. CAUSES IRRITATION TO SKIN, EYES, AND RESPIRATORY TRACT. CHLORINE WATER: CORROSIVE. CAUSES EYE AND SKIN BURNS. CAUSES DIGESTIVE AND RESPIRATORY TRACT BURNS. BROMINE WATER: CORROSIVE. CAUSES EYE AND SKINBURNS. CAUSES DIGESTIVE AND RESPIRATORY TRACT BURNS. IODINE WATER: POISON! CAUSES SEVERE IRRITATION OR BURNS TO EVERY AREA OF CONTRACT. MAY BE FATAL IF SWALLOWED OR INHALED. VAPORS CAUSE SEVERE IRRITATION TO SKIN, EYES, AND RESPIRATORY TRACT. OXIDIZER. MAY CAUSE ALLERGIC SKIN OR RESPIRATORY REACTION. POTASSIUM BROMIDE: HARMFUL IF SWALLOWED OR INHALED. MAY CAUSE IRRITATION TO SKIN, EYES, AND RESPIRATORY TRACT. POTASSIUM IODIDE: MAY CAUSE IRRITATION TO SKIN, EYES, AND RESPIRATORY TRACT.

A. Determine the colors of free halogens in water and in cyclohexane In the second part of the lab we will using the color of solutions of cyclohexane in order to determine

whether a reaction is occurring or not. In order to do this, we must first determine the color that the elements we are working with (Cl2, Br2, and I2 ) are in cyclohexane.

1. Obtain 3 small test tubes. Put 20 drops of Cl2 water into the first tube. Put 20 drops of Br2 water into the second tube Put 20 drops of I2 water into the third tube.

Observe and record the color of each of the above aqueous solutions.

2. Add 10 drops of cyclohexane to each of the above solutions. Notice where the layer of cylohexane forms. It should be the thinner layer since you use less of it. Notice that the non-polar cyclohexane does not dissolve in the polar water.

3. Stopper each tube and, holding the stopper in place, shake each tube vigorously for 30 seconds.

4. Observe and record the color of the cyclohexane layer in each tube. You will be using cyclohexane to identify which halogen is present after a reaction takes place in the mixtures below.

5. Discard the above solutions in the cyclohexane residue waste container

B. Relative strength of halogens as oxidizing agents.

1.

The reaction of Br-1 (from KBr) with Cl2 and I2

a. Obtain 2 small test tubes.

b. Put 15 drops 0.1 M KBr into both tubes.

c. Add 10 drops of chlorine water (Cl2) the first tube and 10 drops of iodine water (I2) to the

second.

7-6

d. Add 10 drops of cyclohexane to each tube, stopper, and holding the stopper in place, shake vigorously for about 30 seconds. Bases on the color you see, what halogen is present in each tube? Write a balanced net ionic equation any the reaction that must have occurred to form the halogen present. If no reaction occurred write "No Reaction" in the blank on the data sheet.

e. Discard the above solutions in the cyclohexane residues waste container

2.

The reaction of I-1 (from KI) with Cl2 and Br2

a. Obtain 2 small test tubes.

b. Put 15 drops 0.1 M KI into both tubes.

c. Add 10 drops of chlorine (Cl2) water the first tube and 10 drops of bromine water (Br2) to the

second.

d. Add 10 drops of cyclohexane to each tube, stopper, and holding the stopper in place, shake

vigorously for about 30 seconds. Based on the color of the cyclohexane layer, what halogen is present

in each tube? Write a balanced net ionic equation any the reaction that must have occurred to form the

halogen present. If no reaction occurred write "No Reaction" in the blank on the data sheet.

e. Discard the above solutions in the cyclohexane residue waste container

3.

The reaction of Cl-1 (from KCl) with I2 and Br2

a. Obtain 2 small test tubes.

b. Put 15 drops 0.1 M KCl into both tubes.

c. Add 10 drops of iodine (I2) water the first tube and 10 drops of bromine water (Br2) to the

second.

d. Add 10 drops of cyclohexane to each tube, stopper, and, holding the stopper in place, shake

vigorously for about 30 seconds. Based on the color of the cyclohexane layer, what halogen is present

in each tube? Write a balanced net ionic equation any the reaction that must have occurred to form the

halogen present. If no reaction occurred write "No Reaction" in the blank on the data sheet.

e. Discard the above solutions in the cyclohexane residue waste container.

4. On your report page, arrange the reduction couples 2Cl- Cl2 + 2e-, 2I- I2 + 2e-, and 2Br- Br2 + 2e- into a series with the strongest reducing agent on the top left and the strongest oxidizing agent on the bottom right. This can be accomplished by looking at which combinations reacted and which did not react, and keeping in mind that elements that are easily reduced are strong oxidizing agents. Where would you predict fluorine would fit in this series? Hint: Look at the periodic table. Add it to your series in its proper position.

C. Relative strengths of copper, lead, and zinc as reducing agents. In this part of the experiment, we will investigate the behavior of small pieces of copper, lead, and

zinc by dropping a small piece of shiny metal (you may need to scuff its surface with sandpaper) into a solution of the other two metal ions.

Safety Caution:

COPPER (II) NITRATE Solution: STRONG OXIDIZER. HARMFUL IF SWALLOWED. CAUSES IRRITATION TO SKIN, EYES AND RESPIRATORY TRACT. ZINC NITRATE Solution: CAUSES IRRITATION. HARMFUL IF SWALLOWED.STRONG OXIDIZER. SILVER NITRATE: WARNING! CAUSES SEVERE EYE IRRITATION. HARMFUL IF SWALLOWED OR INHALED. CAUSES IRRITATION TO SKIN AND RESPIRATORY TRACT.AFFECTS EYES, SKIN AND RESPIRATORY TRACT. FURTHERMORE, SILVER NITRATE WILL STAIN YOUR SKIN AND CLOTHING

1. Place a piece of zinc into about 20 drops of 0.10 M copper (II) nitrate solution and another piece of zinc into 20 drops of lead (II) nitrate solution. Examine each reaction mixture and record your observations on the Report Sheet. Look for a reaction on the surface of the metal. Some reactions may be slow so let the mixtures sit before taking your observations. If you conclude from your observations that a reaction has occurred, write its net ionic equation. If no reaction occurs, do not write an equation,

7-7

write N.R. Keep in mind that the solutions are all nitrates, but the nitrate ions are not involved in the reactions (i.e. are spectator ions), and that if there is a reaction, it will involve the oxidation of the solid metal and the reduction of the metal ion from the solution. Discard solutions in the metal ion residues waste container.

2. Repeat the above experiments with the following combinations a. a piece of solid Cu + 20 drops 0.1 M lead (II) nitrate solution b. a piece of solid Cu + 20 drops 0.1 M zinc nitrate solution c. a piece of solid Pb + 20 drops 0.1 M copper (II) nitrate solution d. a piece of solid Pb + 20 drops 0.1 M zinc nitrate solution.

Again record your observations and write the net ionic equations for any reactions that happened. If no reaction occurs, do not write an equation, write N.R. Discard solutions in the metal ion residues waste container. Discard metal pieces in the waste basket.

Summarize your results in a table like you did in the halogen experiment. On your report page, arrange the reduction couples: Cu Cu +2 + 2e-, Zn Zn +2 + 2e-, and Pb Pb+2 + 2e- into a series with the strongest reducing agent on the top left and the strongest oxidizing agent on the bottom right. This can be accomplished by looking at which combination reacted and which did not react, and keeping in might that elements that are easily reduced are strong oxidizing agents.

D. The Ag Ag+ + e- couple. Place a small piece of copper into about 15 drops of 0.10 M silver nitrate solution. Write a net ionic equation for the reaction that occurs. Add silver to its proper place in the table containing Cu, Zn, and Pb. Discard solution into the silver residues waste container. Discard metal into the waste basket.

E. Other oxidation-reduction reactions. In this part of this part of the lab we will perform more complex redox reactions which must be balanced using the half reaction method. You will again notice color changes which will help you determine the reaction that occurred.

1. The MnO4-/ HSO3- reaction (version A) Place 3 mL of KMnO4 solution into a large test tube. Add 2 mL of 1.0 M H2SO4. Place a white piece of paper behind the test tube and slowly add a few milliliters of 0.01 M NaHSO3 while stirring the solution with a glass rod. Note all the color changes you see. The unbalanced reaction you just did is

MnO4- + HSO3 - Mn 2+ + SO4 2-

Record your observations and then balance the equation using the half-reaction method.

2. The MnO4-/HSO3 ? reaction (version B) In another large test tube, place 3 mL of KMnO4 solution and add 2 mL of 1.0 M NaOH. Place a white piece of paper behind the test tube and slowly add a few milliliters of 0.01 MNaHSO3 while stirring the solution with a glass rod. Note all the color changes you see. The unbalanced reaction you just did is

MnO4- + HSO3 - MnO4 2- + SO4 2-

Record your observations and then balance the equation using the half-reaction method.

3. The H2O2/I - reaction Place 3 mL of potassium iodide solution (0.1 M) in a large test tube. Add 1 drop of concentrated sulfuric acid. Add hydrogen peroxide (3 % solution) slowly while stirring until you notice a color change is produced. The unbalanced reaction you just did is

H2O2 + I - H2O + I2

7-8

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download