Lab 7 (S)

Question 1

Oxidation-reduction (redox) reactions

Answer 1

involve the transfer of electrons. The number of electrons lost by one entity must equal the corresponding gain of electrons by another entity.

Question 2

Oxidation-reduction (redox) reaction EX:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

Answer 2

Electrons are transferred from Zn(s) to H+(aq) within the solution. In this reaction, zinc loses electrons and is oxidized to Zn2+.

The oxidation of zinc provides electrons that are used in the reduction of the H+(aq).

Zinc is referred to as the reducing agent (provides electrons for the
reduction of H+(aq)).

Similarly, H+(aq) gains electrons and is reduced to H2(g).

The reduction of H+(aq) requires electrons that result in the oxidation of the Zn(s). H+(aq) is referred to as the oxidizing agent (attracts electrons resulting in the oxidation of Zn).

Question 3

Oxidation-reduction (redox) reaction EX:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

which is the oxidizing and reducing agent

Answer 3

Of two compounds, the stronger reducing agent is the one that has the greater tendency to lose electrons (i.e. most easily oxidized).

Experimentally, this reaction proceeds spontaneously in the direction written; therefore, Zn(s) must have a greater tendency to lose electrons than H2(g).

However, if we were to replace Zn(s) with Cu(s) in reaction (1), we would find that Cu(s) does not react with H+(aq) to give H2(g) and Cu2+(aq).

Therefore, Cu(s) is a weaker reducing agent than H2(g), or stated the other way around, H2(g) must be a stronger reducing agent than Cu(s).

Question 4

When ranking the oxidizing-reducing power of reagents

Answer 4

it is conventional to write them in terms of reduction half-reactions.

The stronger oxidizing agents (see Chemistry Data Sheet) have the
larger reduction potential and are listed in order of decreasing reduction potential.

Question 5

Cu2+(aq) + 2e– → Cu(s)(weaker reducing agent) (2)

2H+(aq) + 2e– → H2(g) (3)

Zn2+(aq) + 2e– → Zn(s)(Zn(s): stronger reducing agent than H2(g) or Cu(s) (4)

Answer 5

With this kind of ordering, an equation for the spontaneous reaction can be found by taking the
more positive reduction half-reaction equation and adding it to the reverse of the more negative
reduction half-reaction equation.

Thus, the reactions obtained by

(eq 2 plus the reverse of eq 4) giving:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) (5)

and (eq 2 plus the reverse of eq 3) giving:
Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq) (6)

are spontaneous as written.

You will determine the ordering for the reduction potentials of the
halogens in this experiment in a manner like that described above.

Question 6

An easier, and more quantitative way of ranking the reduction potentials of reagents can be
achieved by

Answer 6

_{measuring the voltage (potential difference) in a voltaic cell.}

_{The reactants are in separate compartments, and the circuit is completed by a salt bridge (filter
paper wetted with KNO3 in this experiment) and an external wire connection through a voltmeter.}

_{The salt bridge maintains charge neutrality in each half-cell and allows a current to flow, but
prevents physical mixing of the reagents in the two half-cells. Ideally, the voltmeter should draw
no current.}

_{Electrons flow from the Zn anode, which is the negative electrode, to the Cu cathode, which is the positive electrode, through the connecting wire.}

Question 7

what occurs at the cathode and what occurs at the anode?

Answer 7

Reduction always occurs at the cathode and oxidation always occurs at the anode.

Question 8

When the positive terminal of the voltmeter is connected to the cathode, and the negative terminal
is connected to the anode, a positive voltage is measured on the voltmeter.

Answer 8

The magnitude of this voltage, called the cell voltage, is a measure of the driving force, or the degree of spontaneity of the reaction.

_{For the cell shown in Fig. 1, the measured voltage is about 1.10 volts.}

_{If the 1.0 mol/L CuSO4 solution was replaced by 1.0 mol/L H2SO4, reduction of H+(aq) would replace the reduction of Cu2+(aq), and the measured voltage would be about 0.76 volts.}

_{These observations give both the relative ordering of the reducing agents as Cu < H2 < Zn, and a quantitative measure of that
ordering.}

_{Such an ordering of reducing agents in sequence of increasing strength is also known as a substitution or a displacement series.}

_{Typically, this series is organized into a table of standard
electrode reduction potentials.}

Question 9

Spontaneous Cell Reactions: reduction

Answer 9

Any two half-cells can be combined to form an electrochemical cell. Electrons given off in one half-
cell must equal those gained by the other. For the cell reaction to occur spontaneously, Ecell must be positive.

When half-cell reactions are written as reduction processes, and are ordered in a displacement series, so that the stronger reducing agent is at the bottom (electrode potential is
more negative), an equation for a spontaneous reaction will always result by taking a half-cell reaction and adding to it any oxidation half-cell reaction below it.

Question 10

Spontaneous Cell Reactions: oxidation

Answer 10

An oxidation reaction is simply
the reverse of a reduction reaction. If the direction of the reaction is reversed, then the sign on the
half-cell potential must also be reversed.

In a voltaic cell the half-cell with the more positive reduction potential will be the reduction (cathode) half-cell and the other half-cell will be the oxidation (anode) half-cell.

The cell potential will be the difference of the two standard reduction potentials where the subtraction takes into the account the reversal of the half-cell potential at the anode.

E°cell = E°red (cathode) – E°red (anode)

Question 11

Halogens

Answer 11

are nonmetallic elements of group 17 of the Periodic Table which exist as nonpolar diatomic molecules, X2. (e.g. chlorine, Cl2; bromine, Br2; iodine, I2.)

Question 12

Halides:

Answer 12

negatively charged monatomic ions of group 17 elements of the periodic table. (e.g. chloride, Cl–; bromide, Br–; iodide, I–).

Question 13

There is a rule of solubility for polar/charged and nonpolar substances in solvents that states “like dissolves like”.

Answer 13

a) Polar/charged entities are very soluble in polar solvents. (The most common polar solvent is water.) Therefore, halides (Cl–, Br–, I–) are very soluble in water and practically insoluble in nonpolar solvents. (The less polar solvent used in this experiment is dichloromethane, CH2Cl2.)

b) Nonpolar entities are slightly soluble or insoluble in water and generally become more soluble as the polarity of the solvent decreases. For this reason, halogens (Cl2, Br2, I2) are very
soluble in dichloromethane and only slightly soluble in water.

Question 14

The difference in solubility characteristics

Answer 14

allows for the separation of halogens and halides.

If an aqueous solution that contains both halogens and halides, is mixed with a less polar solvent (e.g.
dichloromethane), then the halide ions will be found in the aqueous layer and the halogens will be
found primarily in the nonpolar or less polar layer.

Question 15

Some organic solvents,

Answer 15

like dichloromethane, are not miscible with water.

Water and CH2Cl2 form two layers.

You will observe these two layers in the test tube when you perform Part II of this experiment.

The upper layer is an aqueous layer because water has a lower density than dichloromethane.

The bottom layer is the dichloromethane layer.

Question 16

Each halogen has a distinctive colour in CH2Cl2.

While,

The halide ions are not soluble in CH2Cl2 and remain in the aqueous layer.

Answer 16

Specifically, iodine is purple (pink if dilute), bromine is orange (yellow if dilute) and chlorine is pale yellow (almost colourless if dilute) in
CH2Cl2.

Halides are essentially
colourless in aqueous solution.

Question 17

Consequently, if a reaction of a halogen or halide ion takes place in
aqueous solution, and the solution is then shaken with CH2Cl2,

Answer 17

any halogen will dissolve in the
CH2Cl2 but the halide will not.

The presence of a halogen can be detected by observing the colour
of the CH2Cl2 layer.

Since each halogen has a distinct colour in CH2Cl2, comparison of the colour of the CH2Cl2 layer after reagents are allowed to react will allow for the identification of the halogen in the dichloromethane.

Question 18

if I–(aq) and Br2(aq) are allowed to react,

Answer 18

two possible results may be expected: orange CH2Cl2 layer or purple CH2Cl2 layer.

If the CH2Cl2 layer is orange, this suggests that no reaction occurred (i.e. Br2 is a reactant and remains unchanged after time for reaction has been allowed).

If the CH2Cl2 layer is purple, this suggests the formation of I2.

Since I– (aq) is a reactant, and I2 is detected as a reaction product, a redox reaction is assumed to have
occurred (2I– ⇌ I2 + 2 e–).

This procedure will be used to determine the order of substitution among the halogens, and to place them in the displacement series obtained by voltaic cell measurements.

Question 19

Experimental Method

Answer 19

set up a series of voltaic cells from which quantitative measurements of electrode potentials can be made for five metal half-cells and for I2|I–.

You will be able to determine the order of displacement for the five metals from your measurements.

By testing for the displacement of
halogens, you will qualitatively determine the order of
substitution among them.

Then, by using the results of the (I2|I–) half reaction, you will be able to relate the halogen substitution series to the metal substitution series.

Question 20

The apparatus we will use for the quantitative measurement

Answer 20

of electrode potentials of all possible combinations involving the five metals and their salt solutions.

A polystyrene plate with 24 flat-bottomed wells serves as the apparatus.

A solution of KNO3 is placed in a central well and strips of filter paper extending from the KNO3 solution to the solutions in the surrounding
wells will function as the salt bridge (once saturated with KNO3).

The 1.0 mol/L metal ion solutions
are in the surrounding wells and are in contact with a strip of the corresponding metal, which acts
as an electrode.

Each metal strip and solution forms a half-cell. The filter paper saturated with KNO3 serves as the salt bridge.

Question 21

Advance Study Assignment

1. What is the difference between an electrolytic cell and a voltaic cell?

Answer 21

In an electrolytic cell, electrical energy is used to produce a nonspontaneous chemical
change.

In a voltaic cell, a spontaneous chemical reaction is used to produce electrical energy.

Question 22

Advance Study Assignment

2. What is the purpose of the filter paper saturated with KNO3 between half-cells?

Why is a salt bridge essential to the voltaic cell?

What would happen if the KNO3 were omitted?

Answer 22

Salt bridge provides electrical contact between the half-cells without physical mixing and maintains ion charge neutrality in each cell.

Without KNO3, the potential difference between the cells cannot be measured.

Question 23

Advance Study Assignment

4. Why would an inert graphite electrode be used in the C(graphite)|(I–, I2, I3–) half-cell?

Answer 23

An inert electrode is needed because an I2(s) electrode does not conduct electricity.

Question 24

Advance Study Assignment

5. CH2Cl2 is colourless. What is the colour of I2 dissolved in CH2Cl2?

Answer 24

purple

Question 25

Advance Study Assignment

6. A student added 1 mL of CH2Cl2 into a test tube with 1 mL of KI.

Then a few crystals of a metal
salt, MNO3, was added and the mixture shaken.

a) The CH2Cl2 would be purple because I- was oxidized to form I2 and I2 is more soluble in the CH2Cl2 layer.

b) What colour would the CH2Cl2 be if the M+ does not react with the I–?

Answer 25

a) If M oxidizes the I–, what colour would the CH2Cl2 be?

b) colourless.

Why? If I- does not react, I- will remain in aqueous solution where it is a colourless species.