Post on 18-Apr-2018
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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
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,
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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
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Record your observations and then balance the equation using the half-reaction method.
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 _______________________________________________
Net ionic equation for any reaction ___________________________________________
Observations Br2 + KI_______________________________________________
Net ionic equation for any reaction. ___________________________________________
Observations: Br2 + KCl _______________________________________________
Net ionic equation for any reaction. ___________________________________________
Observations I2 + KCl _______________________________________________
Net ionic equation for any reaction ___________________________________________
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+
____________________________________________________
Net ionic equation for any reaction ____________________________________________
Observations: Pb + Zn 2+
____________________________________________________
Net ionic equation for any reaction ____________________________________________
D. Observations: Cu + Ag+
____________________________________________________
Net ionic equation for any reaction ____________________________________________
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:
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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
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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?