Experiment 7: Oxidation Reduction Reactions€¦ · 7-2 Subsequently the two oppositely charged...

7-1

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:

[[Mg2+

]

]

Mg [ Cl ]

-

[ Cl ]-

Cl

Cl

transfer e-

transfer e-

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

net ionic equation: 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) Zn(NO3)2(aq) + Cu(s) Zn(s) + Cu(NO3)2(aq)

net ionic equation: 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.

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In summary then:

X loses electron(s) Y gains electron(s)

X is oxidized Y is reduced

X is the reducing agent Y is the oxidizing agent

X increases its oxidation number 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.

X Y

electron

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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.

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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+

reduced half rxn oxidized half rxn

Step 1 MnO4- Mn

2+ Cl

- Cl2

Step 2a MnO4- Mn

2+ no change Mn balanced 2Cl

- Cl2 Cl now balanced

Step 2b MnO4- Mn2+

+ 4H2O balance the 4 O’s on the left 2Cl- Cl2 no change O balanced

add 4H2O (aren’t any)

Step 2c 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 2Cl

- Cl2

the sum of the charges on left side is ( 8(+1) – 1 = +7… the sum of the charges on left side is 2(-1)=-2

on the right side its +2… add -5 ( 5e--) to balance 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) 5x (2Cl

- Cl2 + 2 e

-)

10 e- + 16 H

+ + 2MnO4

- 2Mn

2+ + 8 H2O 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

- 2Mn

2+ + 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)

16 H+ + 2 MnO4

- + 10 Cl

- 2 Mn

2+ + 5 Cl2 + 8 H2O

+ 16 OH-1

+16 OH -1

( H +1 and OH-1 make H2O) 16 H2O + + 2 MnO4- + 10 Cl

- 2 Mn

2+ + 5 Cl2 + 8 H2O +16 OH

-1

cancel H2O on opposite sides 8H2O + + 2 MnO4- + 10 Cl

- 2 Mn

2+ + 5 Cl2 + 16 OH

-1

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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.

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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


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


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,

7-8

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-


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

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4. The thiosulfate (S2O3 2-

) - bromine reaction. Add 10 drops of Br2 water to a six inch test tube

and then add 0.10 M sodium thiosulfate drop by drop until the solution goes coloress. Add 3 drops of

0.10 M AgNO3 solution. A yellow precipitate of AgBr confirms that Br- was formed. The thiosulfate

ion is oxidized to tetrathionate ion (S4O6 2-

). Record your observations and write a balanced reaction

for the equation using the half reaction method. Discard the residue in the silver residues container.

5. Hydrogen peroxide-Iron (II) reaction. Transfer a few crystals (8-12) of solid FeSO4 into a 6-inch

test tube. Dissolve the crystals in about 20 drops of water. Add 3 drops of 3% H2O2. Record any

observations. Add 5 drops of dilute ammonium hydroxide. Record your observations. The gelatinous

product is a hydrate of Fe2O3. The product from the reduction of the hydrogen peroxide in this reaction

is water. Record your observations and write a balanced equation for the reaction using the half reaction

method. The residue (rusty water and ammonia) may be washed down the sink.

6. Oxalate-permanganate reaction. Place 20 drops of 0.10 M potassium oxalate, K2C2O4, in a six inch

test tube. Add 3 drops of dilute (3 M) sulfuric acid to the tube. Add one drop of 0.10 M potassium

permangante to the solution. If it remains colored, warm gently in a hot water bath until the solution

goes colorless. After the solution goes colorless continue adding KMnO4 one drop at a time, stirring and

counting the drops until the solution turns pink or brown and remains colored. Record the total drops of

KMnO4 solution (including the first drop you used) needed to react completely with the 20 drops of

K2C2O4. The oxalate ion is oxidized to CO2 and the permanganate is reduced to almost colorless Mn 2+

ion. Write an equation for the reaction. How does your ratio of drops potassium permanganate solution

used / drops potassium oxalate solution used, compare to the coefficients of your balanced equation?

Dispose of the solution in the metal ion residues waste container.

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Oxidation-Reduction Reactions Report Name __________________

A. Colors of the free halogens in water and cyclohexane

Color of Chlorine in water ______________ in cyclohexane _______________

Color of Bromine in water ______________ in cyclohexane _______________

Color of Iodine in water _______________ in cyclohexane _______________

B. The relative strengths of halogens as oxidizing agents.

REMEMBER: K is a spectator ion!!!!

Observations: Cl2 + KBr _______________________________________________

Net ionic equation for any reaction ___________________________________________

Observations: I2 + KBr _______________________________________________

Net ionic equation for any reaction. ___________________________________________

Observations: Cl2 + KI _______________________________________________


Observations Br2 + KI_______________________________________________


Observations: Br2 + KCl _______________________________________________


Observations I2 + KCl _______________________________________________


Conclusions: Rank the half reaction couples by writing the half reactions in a table such

that the strongest reducing agent will be on the top left and the strongest oxidizing agent

will be on the bottom right. Use the periodic table to predict where fluorine should fit.

Relative strengths of Br2, Cl2, I2, and F2 and their ions as oxidizing-reducing agents.

Strongest reducing agent Weakest oxidizing agent

____________--> _______ + 2e-

____________--> ________+ 2e-

____________--> ________+ 2e-

__________ ________ + 2e-

Weakest reducing agent Strongest oxidizing agent

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C. Observations: Zn + Cu2+

____________________________________________________

Net ionic equation for any reaction ____________________________________________

Observations: Zn + Pb2+

____________________________________________________

Net ionic equation for any reaction. ____________________________________________

.Observations: Cu + Pb 2+

____________________________________________________

Net ionic equation for any reaction. ____________________________________________

Observations Cu + Zn 2+

____________________________________________________

Net ionic equation for any reaction____________________________________________

Observations Pb + Cu 2+

____________________________________________________


Observations: Pb + Zn 2+

____________________________________________________


D. Observations: Cu + Ag+

____________________________________________________


Strongest reducing agent Weakest oxidizing agent

____________--> _______ + 2e-

____________--> ________+ 2e-

____________--> ________+ 2e-

__________ ________ + 2e-

Weakest reducing agent Strongest oxidizing agent

A table like the one above can be used to predict whether a reaction will occur. For a

reaction to occur the reducing agent must be higher in the table than the oxidizing agent. Use your

table to predict whether each of the following reactions will occur. Write the expected

products or no reaction in each blank.

a. Zn + Ag+ _________________

b. Ag + Pb 2+

________________

c. Pb + Ag +

____________________

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E. Other oxidation-reduction reactions:

1. The MnO4-/ HSO3

- reaction version A

Observations: Note all the color changes you see.

Balance the reaction:

MnO4- + HSO3

- Mn

2+ + SO4

2-

2. The MnO4-/ HSO3

- reaction version B

What is the difference between this reaction and the previous reaction?

Observations:

Balance the reaction:

MnO4- + HSO3

- MnO4

2- + SO4

2-

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3. The H2O2/I - reaction

Observations:

Balance the reaction

H2O2 + I - H2O + I2

4. Reaction of S2O3 2-

ion with Br2

Observations:

Balanced net ionic equation for the reaction:

7-14

5. Reaction of Fe2+

ion with H2O2

Observations:

Balanced net ionic equation for the reaction

6. Reaction of MnO4- ion with C2O4

2- ion

Observations:

Number of drops MnO4- solution to react completely with 20 drops C2O4

-2 solution.

Balanced net ionic equation for the reaction

7-15

Post Lab

1. Balance the following oxidation-reduction equations by the half reaction

method in acidic solution

a. Cr2O7 2-

+ Cl -1 Cl2 + Cr

3+ (acidic )

b. Zn + H2SO4 Zn 2+

+ H2S (acidic)

c. I- + IO3

- I2 (acidic)

d. MnO4- + C2O4

2- Mn

2+ + CO2 (basic)

e. Cr(OH)3 + ClO3 1-

CrO4 2-

+ Cl-1

(basic)

f. Iron filings are added to FeCl3 solution.

Fe + Fe3+

→ Fe2+

g. Bismuth metal is dissolved in hot concentrated HNO3 and a brown gas is

given off.

Bi + NO3- → Bi

3+ + NO2(g)

h. A mixture of Na2S. NaClO. and NaOH solutions is warmed, giving a

suspended precipitate.

S2-

+ ClO- → S

0 + Cl

-

i. SO2 gas is bubbled into K2Cr2O7 solution (acidic).

SO2 + Cr2O72-

→ Cr3+

+ SO42-

j. CN- (aq) + MnO4

- (aq) CO2 (g) + NO (g) + MnO2 (s) (acidic)

k. CH4 (g) + CrO42-

(aq) CO2 (g) + Cr(OH)3(s) (acidic)

Convert the above balanced equation to an equation balanced in base.

l. S2O8 2-

(aq) + Cl- (aq) ClO

- (aq) + SO4

2-(aq) (acidic)

Convert the above balanced equation to an equation balanced in base.

2. Imagine that the hypothetical elements, A, B, C, and D, form the ions A2+

, B2+

, C2+

,

and D2+

, respectively. The following equations indicate reactions which can, or

cannot, occur. Use this information to write a potential series for the cations.

B2+

+ A A2+

+ B B2+

+ D N.R.

A2+

+ C C2+

+ A

______________________________

______________________________

______________________________

______________________________

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Name___________________________ Prelab report

1. Determine the oxidation number of all of the element in the substances below.

MnO2 Fe2O3 Na2O2 H2 K2Cr2O7

Mn ________ Fe _________ Na _____________ H ______ K ________

O ________ O _________ O _____________ Cr ________

O _________

2. Balance the half reaction:

NO3 -1

NO2

Is nitrogen in NO3- oxidized, or is it reduced?

3. Balance the equation below as it occurs in acidic solution.

MnO4- + Fe

2+ Fe

3+ + MnO2

In the above equation circle the oxidizing agent.

What element is being oxidized in the above reaction? What element is being reduced?

Experiment 7: Oxidation Reduction Reactions€¦ · 7-2 Subsequently the two oppositely charged...

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Transcript of Experiment 7: Oxidation Reduction Reactions€¦ · 7-2 Subsequently the two oppositely charged...