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APPARATUS AND CHEMICALS:
Sodium thiosulfate solution Thermometer Bunsen burner
Hydrochloric acid Measuring cylinder Piece of paper
Distilled water Conical flask Wire gauze
THEORY:
On the basis of experiments you've performed, you probably have already noticed that reactions
occur at varying speeds. There is an entire spectrum of reaction speeds, ranging from very slow
to extremely fast. For example, the rusting of iron is reasonably slow, whereas the
decomposition of TNT is extremely fast. The branch of chemistry that is concerned with the
rates of reactions is called chemical kinetics.
Experiments show that rates of reactions in solution depend upon:
- The nature of the reactants
- The concentration of the reactants
CEAC 104
GENERAL CHEMISTRY
Experiment 4
Effect of Concentration and Temperature on Rate of
Reaction (Dissappearing Cross)
Purpose: To observe the effect of concentration and temperature upon the rate of
the reaction of sodium thiosulfate with hydrochloric acid.
- The temperature
- Catalysis.
Before a reaction can occur, the reactants must come into direct contact via collisions of the
reacting particles. However, even then, the reacting particles (ions or molecules) must collide
with sufficient energy to result in a reaction; if they do not, their collisions are ineffective and
analogous to collisions of billiard balls. With these considerations in mind, we can
quantitatively explain how the various factors influence the rates of reactions.
Concentration:
Changing the concentration of a solute in solution alters the number of particles per unit volume.
The more particles present in a given volume, the greater the probability of them colliding.
Hence, increasing the concentration of a solute in solution increases the number of collisions
per unit time and therefore, increases the rate of reaction.
Temperature:
Since temperature is a measure of the average kinetic energy of a substance, an increase in
temperature increases the kinetic energy of the reactant particles. The results in an
increase in the velocity of the particles and therefore, increases the number of collisions
between them in a given period of time. Thus, the rate of reaction increases. Also, an increase
in kinetic energy results in a greater proportion of the collisions having the required energy for
reaction.
Catalyst:
Catalysts, in some cases, are believed to increase reaction rates by bringing particles into close
just a position in the correct geometrical arrangement for reaction to occur. In other instances,
catalysts offer an alternative route to the reaction, one that requires less energetic collisions
between reactant particles. If less energy is required for a successful collision, a larger
percentage of the collisions will have the required energy, and the reaction will occur faster.
Actually, the catalyst may take an active part in the reaction, but at the end of the reaction, the
catalyst can be recovered chemically unchanged.
Let’s examine now precisely what is meant by the expression rate of reaction.
Order of Reaction Defined
Consider the hypothetical reaction:
A + B → C + D [1]
The rate of reaction is measured by observing the rate of disappearance of the reactants A or B,
or the rate of appearance of the products C or D. The species observed is a matter of
convenience. For example if A, B, and D are colorless and C is colored, you could conveniently
volume. The rate of reaction can be measured by timing how long it takes for the solution to
become cloudy and the precipitation of sulfur. In other words the time taken for a certain
quantity of sulphur to form and cause the ‘X’ mark to disappear is used to determine the rate
of reaction. The rate of this reaction directly proportional with the inverse of the time taken
for a formation of precipitation of sulfur.
SAFETY PRECAUTIONS
Wear your eye protection.
Do not inhale any fumes.
Sulfur dioxide is toxic and corrosive. Dispose of the solution immediately after the
experiment following your teacher's instructions.
Wash your hands when finished.
PROCEDURE:
A.Effect of concentration
- Using a pencil, draw a cross in the middle of the filter paper.
- Place 20 mL of the 0,1 M sodium thiosulfate solution into a conical flask.
- Add 20 mL of 1 M hydrochloric acid to the flask, while starting the stop clock at the
same time.
- Swirl the flask and place it on a paper marked with a cross.
- Record the time taken for the cross to disappear.
- Repeat the experiment using 10,12, 14, 16 and 18 mL of sodium thiosulfate solution
respectively. In each case, add water to make the volume up to 20 mL and mix before
adding HCl.
- Record the results in data sheet.
B.Effect of temperature
Place 20 mL of 0.05 M sodium thiosulfate solution into a conical flask.
Warm or cool the flask gently until the temperature is about 20
0
C.
- Add 20 mL of 1 M HCl, starting a stop clock at the same time, before proceeding.
- Without delay, swirl the flask, place it on a paper marked with a cross, and record the
exact temperature of the contents of the flask.
- Record the time taken for the cross to disappear.
- Repeat the experiment, heating or cooling the thiosulfate solution to temperatures of
approximately 10
0
C, 30
0
C, 40
0
C, 50
0
C and 60
0
C respectively (before adding the HCl).
- Record the results in data sheet.
QUESTIONS
- What is the meaning of directly proportionality of two quantities?
- What is the effect of increasing the concentration on the reaction time and reaction
rate?
- What is the effect of raising the temperature on the reaction time?
- What is the effect of raising the temperature on the reaction rate? Suggest two factors
responsible for the result observed.
- Suggest a reason why it is not recommended to carry out the experiment at
temperatures higher than about 60
0
C.
- Which is the limiting reactant in the temperature experiment.
B. Effect of Temperature
1. Record your results.
Volume of
1 M HCl
(mL)
Volume of
0. 05 M
sodium
thiosulfate
solution (mL)
T (˚C) Reaction time
(s)
1/time (s
- 1
2. Draw a graph of 1/time against temperature and reaction time against temperature
using excel and write your comments about the graph. ( Hint: 1/time for this reaction
is the measure of reaction rate.)
APPARATUS AND CHEMICALS:
distilled water test tubes
sodium thiocyanate (NaSCN) graduated cylinder
ferric nitrate (Fe(NO 3
3
) white paper
ruler 250 - mL beaker
THEORY:
Most of the chemical reactions occur so as they approach a state of chemical
equilibrium. The equilibbrium state can be characterized by specifying its equilibrium
constant, i.e., by indicating the numerical value of the mass-action expression. In this
Purpose: Given the equation for a chemical equilibrium, predict and explain, the direction of
a shift in the position of an equilibrium caused by a change in the concentration of the species
on the basis of LeChatelier’s principle and finally to find the value of an equilibrium constant,
K
eq
, experimentally.
CEAC 104
GENERAL CHEMISTRY
Experiment 5
Chemical Equilibrium
PROCEDURE:
- Clean six 15-cm test tubes with distilled water and let them drain.
- To each of these test tubes add 5 mL of 0.0020 M NaSCN.
- To the fırst test tubes add 5 mL of 0.20 M Fe(NO 3
3
. This tube will serve as the
standard.
- For the other test tubes proceed as follows:
Add 10 mL of 0.20 M Fe(NO 3
3
to a graduated cylinder. Add 15 mL distilled water
so that you have a 25 mL of diluted solution. Stir thoroughly to mix. Take 5 mL from
this solution and pour it into the second test tube.
- Discard 10 mL of the diluted solution in the graduated cylinder and add 15 mL of
distilled water and again you complete it to 25 mL. Stir thoroughly. Take 5 mL from
this solution and pour it into the third test tube.
Figure 5.1 How to make a comparison of color.
- Again discard 10 mL of the solution in the graduated cylinder and add 15 mL
distilled water. Continue this procedure until you prepare six of the test tubes.
- Now the problem is to determine the concentration of FeSCN
2+
in each test tube
relative to the standard in test tube 1. Compare the color intensity in test tube 1 with that
in each of the other test tubes (see Figure 5.1). To do it, take two tubes to be compared,
hold them side-by-side and wrap a strip of white paper around both. Look down through
the solutions as shown in Figure. If color intensities appear identical, measure the
heights of the solutions in the two tubes being compared. If not, take test tube 1 and
pour out some of the standard into a clean beaker (you may need to pour some back)
until the color intensities appear identical. Do this comparison for all five tubes.
CALCULATIONS:
In calculating initial concentrations, assume that each of Fe(NO 3
3
and NaSCN
are completely dissociated. Remember also that mixing two solutions dilutes both of
them. In ca1culating equilibrium concentrations, assume that all the initial SCN
has
been converted to FeSCN
2+
in test tube 1, for the other test tubes; calculate FeSCN
2+
from the ratio of heights in the color comparison. Equilibrium concentrations of Fe
3+
and SCN
are obtained by subtracting FeSCN
2+
formed from initial Fe
3+
and SCN
each of test tube 2 to 6 calculate the value of K. Decide which of these values is most
reliable.
QUESTIONS:
- In this experiment you examine the equilibrium
Fe
3+
+ SCN
↔ FeSCN
2+
The following equilibria are somewhat competing with the above equilibrium
Fe
3+
+ SCN
↔ Fe(SCN)
2+
and Fe(SCN)
2+
+ 2SCN
↔ Fe(SCN) 3
a. To allow you to ignore these equilibria, how must their equi1ibrium constants be in
comparison with that of the equilibrium being studied?
b. Given that all ions are color1ess except for FeSCN
2+
, what effect should competing
equilibria have on the value of K determined in this experiment? Explain.
How reasonable was your assumption that all the SCN
in test tube 1 was
converted to FeSCN
2+
? (Calculate the percent of SCN
converted to FeSCN
2+
using
your best K value.)
- Why are the values of K determined for test tubes 3, 4 and 5 probably more
reliable than those determined for tubes 2 or 6?
- In your own words, give the rational behind the procedure and methodology in
this experiment.
Apparatus and Chemicals:
Copper strips or wire Cotton Ring stand, iron ring and wire
KNO 3
Agar-agar Zinc strips or wire
ZnSO 4
solution Wires CuSO 4
solution
DC voltmeter or potentiometer 250 - mL beaker HCl solution
Glass U-tubes Emery cloth 2 sets clips
Thermometer Clamps
THEORY:
Electrochemistry is that area of chemistry that deals with the relations between
chemical changes and electrical energy. It is primarily concerned with oxidation-
reduction phenomena. Chemical reactions can be used to produce electrical energy in
cells that are referred to as voltaic , or galvanic, cells. Electrical energy, on teh other
Purpose: To examine the correlation between the reactions of metals and their ions (half cells), and
to measure the voltages produced at various concentrations when two half cells are combined to form
electrochemical (voltaic) cells. The voltage of the redox reactions will be calculated theoretically via
Nernst equation.
CEAC 104
GENERAL CHEMISTRY
Experiment 6
Electrochemical Cells and Thermodynamics
hand, can be used to bring about chemical changes in what are termed electrolytic cells.
In this experiment you will investigate some of the properties of voltaic cells.
Oxidation-reduction reactions are those that involve the transfer of electrons from
one substance to another. The substance that loses electrons is said to be oxidized, while
the one gaining electrons is reduced. Thus if a piece of zinc metal were immersed into a
solution containing copper (II) ions, zinc would be oxidized by copper (II) ions. Zinc
loses electrons and is oxidized, and the copper (II) ions gain electrons anda re reduced.
We can conveniently Express these processes by the following two half-reactions,
which add to give the overall reaction:
Oxidation: 𝑍𝑛(𝑠) → 𝑍𝑛
2 +
−
Reduction: 𝐶𝑢
2 +
−
______________________________________ [1]
Net reaction: 𝑍𝑛(𝑠) + 𝐶𝑢
2 +
2 +
In principle, any spontaneous redox reaction can be used to produce electrical
energy- that is, the reaction can be used as the basis of a voltaic cell. The trick is to
seperate the two half reactions so that electrons will flow through an external circuit. A
voltaic cell that is based upon the reaction in Equation [1] and that uses a salt bridge is
shown in Figure 6.1.
The corresponding half-cell reactions are as follows:
2 +
−
𝑜𝑥
0
−
2
𝑟𝑒𝑑
0
The Standard cell emf of this cell is 0.76 V (that is, E
0
cell
= 0.76 V). Because the
Standard reduction potential of H
is 0.000 V, it is possible to calculate the Standard
oxidation potential, E
0
ox
, of Zn:
𝑐𝑒𝑙𝑙
0
𝑟𝑒𝑑
0
𝑜𝑥
0
𝑜𝑥
0
[2]
Thus the Standard oxidation potential of 0.76 V can be assigned to Zn. By
measuring other Standard-cell emf values, we can establish a series of Standard
potentials for other half-reactions.
It is important to note that the half cell potential for any oxidation is equal in magnitude
but opposite in sign to that of the reverse reduction. Hence,
2 +
−
𝑟𝑒𝑑
0
It is customary today to tabulate half-cell potentials as Standard reduction potentials
and also to refer to them as Standard electrode potentials.
EXAMPLE 1:
The cell in Figure 5.1 may be represented by the following notation:
2 +
2 +
The double bar represents the salt bridge. Given that E
0
cell
for this cell is 1.10 V and that
E
ox
is 0.76 V for zinc (see Equation [2]), find the Standard electrode potential, E
0
red
, for
the reduction of copper (𝐶𝑢
2 +
−
Solution :
𝑐𝑒𝑙𝑙
0
𝑟𝑒𝑑
0
𝑜𝑥
0
𝑟𝑒𝑑
0
𝑟𝑒𝑑
0
𝑟𝑒𝑑
0
The free-energy change, 𝛥G, associated with a chemical reaction is a measure of the
driving force or spontaneity of the process. If the free-energy change of a process is
negative, the reaction will ocur spontaneously in the direction indicated by the chemical
equation.
The cell potential of a redox process is related to the free-energy change as follows:
In this equation, F; is Faraday's constant, the electrical charge on 1 mol of electrons:
1F = 96. 500 C/mol e
−
= 96500 J/V mol e
−
and n represents the number of moles of electrons transferred in the reaction. For the case
when both reactants and products are in their standard states, Equation [3] takes the
following form:
w
max
= ΔG = −nFE
0
[ 4 ]
The maximum amount of work that can be obtained from a galvanic cell is equal to the
free energy change, ΔG, for the process.
The standard free-energy change of a chemical reaction is also related to the equilibrium
constant for the reaction as follows:
∆G° = −RT In K [ 5 ]
where R is the gas-law constant (8.314 J/K mol) and T is the temperature in Kelvin.
Consequently, E° is also related to the equilibrium constant. From Equations [4] and [5]
it follows that
−nFE° = −RTlnK
E° =
RT
nF
InK [ 6 ]
When T = 298 K, In K is converted to log K, and the appropriate values of R and 9; are
substituted, Equation [6] becomes
0
𝑙𝑜𝑔𝐾 [ 7 ]
We can see from this relation that the larger K is, the larger the standard-cell potential
will be.
In practice, most voltaic cells are not likely to be operating under standard-state
conditions. It is possible, however, to calculate the cell emf, E, under non-standard-state
conditions from a knowledge of E°, temperature, and concentrations of reactants and
products:
E = E
0
n
logQ [ 8 ]
You can see that small changes in concentrations have small effects on the cell emf.
A list of the properties of electrochemical cells and some definitions of related terms are
given in Table 6.1.
TABLE 6 .1 Summary of Properties of Electrochemical Cells and Some
Definitions
Voltaic cells: E > 0 , ΔG < 0: reaction spontaneous, K large (greater than 1)
Electrolytic cells: E < 0, ΔG > 0; reaction nonspontaneous, K small (less than 1)
Anode electrode at which oxidation occurs
Cathode electrode at which reduction occurs
Oxidizing agent-species accepting electrons to become reduced
Reducing agent-species donating electrons to become oxidized
Chemists have developed a shorthand notation for electrochemical cells, as seen in
Example 1. The notation for the Cu-Zn cell that explicitly shows concentrations is as
follows:
Zn | Zn
2+
(xM) || Cu
2+
(yM) | Cu
Anode Cathode
(oxidation) (reduction)
In this notation, the anode (oxidation half-cell) is written on the left and the cathode
(reduction half-cell) is written on the right.
Your objective in this experiment is to construct a set of three electrochemical cells and to
measure their cell potentials. With a knowledge of two half-cell potentials and the cell
potentials obtained from your measurements, you will calculate the other half-cell
potentials and the equilibrium constants for the reactions. By measuring the cell potential
as a function of temperature, you may also determine the thermodynamic constants, ΔG,
ΔH, and ΔS, for the reactions. This can be done with the aid of Equation [9]:
∆𝐺 = ∆𝐻 − 𝑇∆𝑆 [ 9 ]
ΔG may be obtained directly from measurements of the cell potential using the
relationship
A plot of ΔG versus temperature in degrees Kelvin will give - ΔS as the slope and AH as
the intercept. A more accurate measure of ΔH can be obtained, however, by substituting
ΔG and ΔS back into Equation [9] and calculating ΔH.
REVIEW QUESTIONS:
Before beginning this experiment in the laboratory, you should be able to answer the
following questions:
- Define the following: faraday, salt bridge, anode, cathode, voltaic cell, electrolytic
cell.
- Write a chemical equation for the reaction that occurs in the following cell:
Ag | Ag
|| Cu
2+
| Cu.
- Given the following E°'s, calculate the standard-cell potential for the cell in
question 2.
2 +
(aq) + 2 e
−
→ Cu(s) E° = +0.34 V
−
→ 𝐴𝑔(𝑠) E° = +0.80 V
- Calculate the voltage of the following cell:
Zn | Zn
2+
(0.10 M) || Cu
2+
(0.40 M) | Cu
- Calculate the cell potential, the equilibrium constant, and the free energy for the
following cell:
Ba(s) + Mn
2+
(aq)(l M) → Ba
2+
(aq)(l M) + Mn(s)
given the following E° values:
2 +
aq
−
→ Ba
s
E° = - 2.90 V
2 +
−
→ 𝑀𝑛(𝑠) E° = - 1.18 V
- Would you normally expect ΔH
0
to be positive or negative for a voltaic cell?
Justify your answer.
- Predict whether the following reactions are spontaneous or not.
Pd
2+
+ H
2
→ Pd + 2H
Pd
2+
−
→ Pd E° = 0.987 V
Sn
4+
+ H
2
→ Sn
2
++ 2H
Sn
4+
−
→ Sn
2+
E° = 0.154 V
Ni
2+
+ H
2
→ Ni + 2H
Ni
2+
−
→ Ni E° =-0.250 V
Cd
2+
+ H
2
→ Cd + 2H
Cd
2+
−
→ Cd E° = - 0.403 V
From your answers decide which of the above metals could be reduced by hydrogen.
- Identify the oxidizing agents and reducing agents in the reactions in question 7.
