II. ELECTROCHEMISTRY
Introduction In 1812 Humphry Davy wrote: “If a piece of zinc and a piece of copper be brought in contact with each other, they will form a weak electrical combination, of which the zinc will be positive, and the copper negative; this may be learnt by the use of a delicate condensing electrometer”. One can consider what happens when electrodes (zinc and copper in Davy’s experiment) are immersed in electrolyte solutions and connected via an external metallic conductor. Such an arrangement is a typical electrochemical cell.
Electromotive force (EMF) of the cell
One can consider a cell built from a zinc electrode immersed into a solution of ZnSO 4 and a copper electrode immersed in CuSO4 (Daniel’s cell). The two solutions are separated by a porous barrier, which allows electrical contact but prevents excessive mixing of the solutions by interdiffusion. This cell can be represented by a scheme:
Zn|Zn2+|Cu2+|Cu where vertical line denotes phase boundaries. According to definition given by IUPAC the electromotive force (emf), E, is defined as follows: the emf is equal in sign and in magnitude to the electrical potential of the metallic conducting lead on the right when that of the similar lead on the left is taken as zero, the cell being open. Thus it can be written that:
E Eright Eleft 1 where Eright and Eleft would be the potentials of the right and left leads relative to some common standard. The meaning of left and right refers to the cells as written.
Measurement of emf – the potentiometer The definition of emf states that the potential difference is measured while the cell is open, i.e., while no current is being drawn from the external leads. In practice E is being measured under conditions in which the current drawn from the cell is so small as to be negligible. The
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Figure 1. Potentiometer scheme
The slide wire is calibrated with a scale so that any setting of the contact corresponds to a certain voltage. With a double through switch in the standard cell position S, slide wire is set to the voltage reading of standard cell, and rheostat is being adjusted until no current flows through the galvanometer G. At this point the potential difference between A and B, the IR along the section AB of the slide wire, just balances the emf of the standard cell. Then switch of the unknown cell is being set to the X position and slide wire is being readjusted until no current flows through galvanometer. From the new setting the emf of the cell can be read directly from the scale of the slide wire. The most widely used standard is the Weston cell written as:
Cd(Hg)|CdSO4 · 8/3 H2O|CdSO4 (sat.sol.)|Hg2SO4|Hg The cell reaction is: 8 8 Cds Hg SO s H Ol CdSO H Os 2Hgl 2 4 3 2 4 3 2 The accuracy of the compensation method for measuring an emf is limited only by the accuracy of the standard E and of the various resistances in the circuit. The precision of the method is determined mainly by the sensitivity of the galvanometer used to detect the balance between unknown and standard emf.
Reversible cells An electrode immersed in a solution is said to constitute a half cell. The typical cell is combination of two half cells. One should be primarily interested in so called reversible cells, which can be recognized by the following criterion: the cell is connected with a potentiometer arrangement for emf measurement by compensation method. The emf of the cell is measured:
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a) with a small current flowing through the cell in one direction b) then with an imperceptible flow of current c) and finally with a small flow in opposite direction If the cell is reversible, its emf changes only slightly during this sequence, and there is no discontinuity in the value of the emf at the point of balance (b). Reversibility implies that any chemical reaction occurring in the cell can proceed in either direction, depending on the flow of current, and the null point of the driving force of the reaction is just balanced by the compensating emf of the potentiometer. One of the sources of irreversibility in cells is the liquid junction, like in the Daniel’s cell presented before. Another one can be salt bridge which consists of a connecting tube filled with a concentrated solution of a salt, usually KCl. Than cell scheme can be written as:
Zn|Zn2+||Cu2+|Cu
A better way to avoid irreversible effects is to avoid liquid junctions altogether, by using single electrolyte, like in the Weston cell (CdSO4 solution saturated with sparingly soluble
Hg2SO4.
Types of half cells One of the simplest half cells consists of metal electrode in contact with solution containing ions of the metal e.g. silver and silver-nitrate solution. Gas electrodes can be constructed by placing a strip of nonreactive metal (platinum or gold) in contact with both the solution and the gas stream. The hydrogen electrode consists of a platinum strip exposed to a current of hydrogen and partly immersed in an acid solution. Overall reaction is: 1 H H e 2 2 2 In nonmetal-nongas electrodes, the inert metal passes into a liquid or solid phase e.g. - bromine-bromide half cell: Pt|Br2|Br . In an oxidation-reduction electrode an inert metal dips into a solution containing ions in two different oxidation states, e.g. ferric and ferrous ions in the half cal Pt|Fe2+|Fe3+. When electrons are supplied to the electrode the reaction is Fe3+ + e Fe2+. Since it is a function of
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ELECTROCHEMISTRY electrodes either to accept electrons from, or to donate electrons to ions in the solution they are all in the sense oxidation-reduction electrodes. Metal, insoluble salt electrodes consist of a metal in contact with one of its slightly soluble salts; in the half cell, this salt is in turn in contact with a solution containing common anion. - An example is the silver, silver chloride half cell Ag|AgCl|Cl (c1) and overall electrode reaction is AgCls e Ag s Cl . Metal, insoluble oxides electrodes are similar to the metal, insoluble salt one e.g. antimony, — antimony trioxide electrode with a scheme Sb|Sb2O3|OH and an overall reaction 1 3 Sbs 3OH Sb O H Ol 3e . An antimony rod is covered with a thin layer of 2 2 3 2 2 oxide and dips into a solution containing OH- ions.
Classification of cells When two suitable cells are connected an electrochemical cell is given. The connection is made by bringing the solutions in the half cells into contact so that ions can pass between them. If these two solutions are the same, there is no liquid junction, and one can have a cell without transference. If the solutions are different, the transport of ions across the junction will cause irreversible changes in the two electrolytes, and one can have a cell with transference. Cells in which the driving force is the change in concentration are called concentration cells. The change in concentration can occur either in the electrolyte or in electrodes. The variety of electrochemical cells is given in Figure 2.
Electrochemical cells
Chemical cells Concentrationl cells
Without With Electrolyte Electrode transference transference Concentration Concentration cells cells
Without With transference transference
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Figure 2. Electrochemical cells
Emf and standard emf of the cell For generalized cell reaction: aA bB cC dD free-energy change in terms of the activities of the reactants is:
c d 0 aC aD G G RT ln a b 3 a AaB Since G=-|z|FE, division by -|z|F gives
c d 0 RT aC aD E E ln a b 4 z F a AaB called Nernst equation. E0 is called a standard emf of the cell. Determination of this value is one of the most important procedures in electrochemistry. As an example lets consider cell consisting of a hydrogen electrode and a silver-silver chloride electrode immersed in a solution of hydrochloric acid: Pt(H2)|HCl(m)|AgCl|Ag. The overall reaction is:
1 AgCl H H Cl Ag 2 2 The emf of the cell is
RT a Ag a a E E 0 ln Cl H 5 F 1 a a 2 AgCl H 2 Setting the activities of the solid phases equal to the unity, and choosing hydrogen pressure so that aH2=1 (for ideal gas P=1atm) following reaction can be obtained:
0 RT E E ln a a 6 F Cl H
Introducing the mean activity of the ions defined by a±=±m one can obtain 2RT 2RT E E 0 ln a E 0 ln m 7 F F 2RT 2RT E ln m E 0 ln 8 F F 1/2 According to the Debye-Hückel theory, in dilute solutions ln ±=Am , where A is a constant. Hence the equation becomes:
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2RT 2RTA E ln m E 0 ln m1/ 2 9 F F If the quantity on the left is plotted against m1/2, and extrapolated to m=0, the intercept at m=0 gives value of E0.
II.A. DETERMINATION OF THE EMF OF THE DANIEL’S CELL
Goal: To know the principle of electrochemical cell functioning and the method of measuring the emf
Introduction Daniel’s cell consist of zinc and copper electrodes immersed in its sulfate solutions. The cell scheme is Zn|ZnSO4||CuSO4|Cu and the cell reaction is:
Zn + CuSO4 ZnSO4 + Cu 0 0 0 The E is E R-E L=0.337-(-0.736)=1.1V The Nernst equation for Daniel’s cell is: a 0 RT ZnSO4 E E ln 10 F a CuSO4 where is activity of zinc and copper ions. Mean activity can be expressed as a±=±m where ± is the activity coefficient and m is molality. As stated before electromotive force of the cell can be determined by means of Poggendorf method. In this technique resistance of tested and standard (Weston) cell should be chosen for which there is no current flow through galvanometer. Voltage decrease, UX, for tested cell, equal to its emf, EX, is given by equation:
EX U X I 0 RX 11 where I0 is residual current intensity and RX is resistance for which there is no current flow through galvanometer.
Value of I0 is unknown and that is why comparative measurement with standard cell (with known emf, EW=1,018V) should be carried out. Voltage decrease, UW, for standard cell, equal to its emf, EW, is given by equation:
EW UW I0 RW 12 Electromotive force of a tested cell can be calculated as:
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RX E X EW 13 RW Experimental procedure: 1. Prepare electrolyte solutions of given concentrations. 2. Connect electrochemical cell as given in the scheme
Scheme 1. Daniel’s cell scheme 3. Measure electromotive force of the tested cell. Repeat measurement 3 times. Data analysis: 1. Put emf of tested cell into the table. 2. Calculate emf of tested cell by using Nernst equation. Standard electrode potentials and mean activity coefficients are given in the tables. 3. Calculate relative error of emf measurement (%) and put it to a table.
Cell EEXPERIMENT [V] ETHEORY [V] E E EXPERIMENT THEORY 100% E THEORY
Standard electrode potentials, E0 Electrode Zn, Zn2+ Cu, Cu2+ E0 [V] -0.763 0.337
Mean activity coefficients, γ for electrolyte solutions at 25ºC Electrolyte Molality [mol/kg] 0.005 0.01 0.02 0.05 0.1 0.2 0.5 1
CuSO4 0.573 0.438 0.317 0.217 0.154 0.104 0.062 0.043
ZnSO4 0.477 0.387 0.298 0.202 0.15 0.104 0.062 0.043
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II.B. DETERMINATION OF AGNO3 CONCENTRATION IN AQUEOUS SOLUTION BY MEAN OF POTENTIOMETRIC TITRATION
Goal: To know the principle of electrochemical cell functioning and to observe the influence of the solution concentration on emf
Introduction Measurement of the potential of certain electrodes offers a convenient and accurate means for determining the end points of titrations. For example, a hydrogen electrode or a glass electrode may be used to determine the pH during the titration of an acid by the base, Figure 3 gives the emf obtained with a hydrogen electrode and calomel electrode during the titration of a solution of hydrochloric acid with a solution of sodium hydroxide.
Figure 3. Electromotive force obtained with a hydrogen electrode and calomel electrode during the titration of a solution of hydrochloric acid with a solution of sodium hydroxide.
At the beginning of the titration the pH changes slowly because a considerable amount of sodium hydroxide must be added to change the hydrogen ion concentration 10-fold. As the end point is approached the pH changes rapidly. When the solution is neutral, the activities of the hydrogen ions and hydroxyl ions are the same. The voltage of the hydrogen electrode against the calomel one is E=0.6942V. At the end point the concentration of hydrogen ions is very small in comparison to hydroxyl ions. When weak acids and bases are used the end point will not come at pH=7, because the salts produced in the neutralization are hydrolyzed and give an acid or alkaline reaction. The end points in oxidation-reduction reactions may be determined by measuring the potential difference between a platinum wire or other inert electrode and a calomel one. In Figure 4 results of a titration of ferrous sulfate with potassium dichromate are given.
Figure 4. Potentiometric titration of ferrous sulfate with potassium dichromate shoeing the use of a plot of E/ml vs. volume to obtain the end point
In order to locate the end point more accurately, it is helpful to plot the slope E/V vs. the volume of reagent added.
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For determining end points in neutralization it is possible to bubble oxygen or even air over a platinized electrode. Since the oxygen electrode is not reversible, it is not possible to set –nFE equal to F, and no theoretical significance can be attached to the absolute voltages. But when addition of reagents leads to a rapid change of OH- concentration and potential the end point can be identified. It is also possible to obtain end points in precipitation reactions. Let us consider determination of AgNO3 concentration with a concentration cell:
Ag|AgCl|AgCl (sat. sol.) (mAg+)1||AgNO3 (mAg+)2|AgCl|Ag
+ where (mAg+)1 is Ag concentration in AgCl saturated solution and (mAg+)2 is unknown concentration of AG ions in AgNO3 solution.
To the right half-cell, containing AgNO3 solution, NaCl solution of known concentration is being added:
AgNO3 + NaCl NaNO3 + AgCl(s)
Ag+ concentration in titrated solution decreases as a result of AgCl precipitation. Before titration silver ion concentration in titrated solution is higher than in saturated solution:
(mAg+)2 > (mAg+)1
In stoichiometric point concentrations of Ag ions in both solutions are equal:
(mAg+)2 = (mAg+)1
After this point Ag+ concentration in titrated solution is smaller than in saturated one, L (mAg+)1 > (mAg+)2 m Cl where L= (mAg+)1 = (mCl-)1 is AgCl solubility product. According to Nernst equation following dependency of emf in function of molality can be presented:
a m RT Ag RT Ag E ln 2 ln 2 14 F a F m Ag 1 Ag 1
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According to this equation with decrease of along with (mAg+)2 concentration cells emf decreases, whereas for (mAg+)1 > (mAg+)2 there is a change in emf sign, E<0, which decribes change in character of reaction proceeding on electrodes. Scheme of the setup is shown below.
Experimental procedure: 1. Connect electrochemical cell as given in the scheme 2. Read value of EMF on the voltage meter.
3. Add aqueous solution of NaCl (portions 0.2-0.5 mL) to AgNO3 solution and mix it precisely 4. Read value of EMF on the voltage meter after every addition of another portion of NaCl solution 5. After change of the EMF value to the negative one add another 2 mL of NaCl (portions of 0.5 mL) Data analysis:
1. Prepare a titration graph EMF=f(VNaOH) 2. For exact determination of stoichiometric point prepare derivative graph EMF f (VNaCl) VNaCl
3. From the graph read stoichiometric point and calculate number of moles of AgNO3 in titrated solution
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II.C. pH-METRIC TITRATION
Goal: Main goal of the exercise is to analyze changes in pH of the solutions during pH-metric titrations and calculation of acid concentration in its aqueous solution.
Introduction One of the most important examples of chemical equilibrium is the one that exists when acids and bases are present in solution. According to the Brønsted-Lowry classification an acid is a proton donor and a base is a proton acceptor. These definitions make no mention of the solvent. One of the properties of central interest in aqueous (one of the most important) solutions of acids and bases is the pH, which is defined as
pH log a 1 H3O + where H3O is the hydronium ion, a representation of the state of the proton in aqueous solution. At low concentrations, the activity of hydronium ions approximately equal to their molality and molar concentration, so the determination of pH is an indication of hydronium ion concentration. However, many thermodynamic observables depend on pH itself, and there is no need to make this approximation and interpretation.
Acid-base equilibrium in water An acid HA takes part in the following proton transfer equilibrium in water:
a a HAaq H Ol H O aq A aq K H3O A 2 2 3 a a HA H2O where A‾ is the conjugate base of the acid. If one confine attention to dilute solutions, the activity of water is close to 1, and the equilibrium can be expressed as: aH OaA K 3 3 a aHA where Ka is the acidity constant. To simplify the discussion if all ions are present in small concentrations one can replace activities by numerical values of the molar concentrations [J] e.g. [HA] in moles of a substance per liter: H O A K 3 4 a HA Lodz University of Technology, Faculty of Chemistry, Institute of Applied Radiation Chemistry, Laboratory of Chemistry
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It is common to report values of Ka in terms of its negative logarithm, pKa, which gives an insight into strength of the acid (high pKa => low Ka => weak acid). For a base B in water, the characteristic proton transfer equilibrium is
a a Baq H Ol HB aq OH aq K HB OH 5 2 a a B H2O where HB+ is is the conjugate acid of the base B. In dilute solutions, where activity of water is
1, one can express this equilibrium in terms of the basicity constant Kb: a HB a OH K 6 b aB Similarly to acidic constant, basicity constant can be used to assess the strength of a base. It is common to express proton transfer equilibrium involving a base in terms of its conjugate acid:
a aB H3O HB aq H 2Ol H 3O aq Baq K a 7 a HB The acidity constant of the conjugate acid HB+ is related to the basicity constant of the base
B, which may be verified by multiplying the expressions for Ka and Kb:
Ka Kb K w 8 where Kw is the autoprotolysis constant of the water:
2H 2Ol H 3O aq OH aq K w a a 9 H3O OH -14 At 25ºC, Kw=1.008 × 10 (pKw=14), showing that only a few of a water molecules are ionized. If, in analogy to pH, pOH=-log aOHˉ will be introduced, then it follows that
pK w pH pOH 10 + - and because molar concentrations of H3O and OH are equal in pure water then in 25ºC pH=½pKw 7.00.
Acid base titrations One method a chemist can use to investigate acid-base reactions is a titration. The word "titration" comes from the Latin word "titalus", meaning inscription or title. The French word, titre, also from this origin, means rank. Titration is by definition the determination of rank or concentration of a solution. The origins of volumetric analysis are in late 18th century French chemistry. Francois ANTOINE HENRI DESCROIZILLES developed the first burette (which looked more like a graduated cylinder) in 1791. JOSEPH LOUIS GAY-LUSSAC developed an improved version Lodz University of Technology, Faculty of Chemistry, Institute of Applied Radiation Chemistry, Laboratory of Chemistry
ELECTROCHEMISTRY of the burette that included a side arm, and coined the terms "pipette" and "burette" in a 1824 paper about on the standarization of indigo solutions. A major breakthrough in the methodology and popularization of volumetric analysis was due to KARL FRIEDRICH MOHR, who redesigned the burette by placing a clamp and a tip at the bottom, and wrote the first textbook on the topic, Lehrbuch der chemisch-analytischen Titrirmethode (Textbook of analytical-chemical titration methods), published in 1855. A pH titration is performed by adding small, accurate amounts of standard base to an acid of unknown concentration. The pH is recorded methodically and is plotted vs. the volume of base added to the acid solution. The result of this plot is an "S" shaped curve. The inflection point of this curve (middle of the "S") is indicative of the equivalance (stoichiometric) point. This point occurs when the acid and base in solution are stoichiometrically equivalent. An equivalance point can be very useful in determining the concentration of an acid or base. Chemists can also titrate using an indicator to determine the end point of a titration. An indicator contains a molecule that exists in at least two different forms which have different colors. The forms differ by the addition or removal of a hydrogen ion. Thus, the color of the indicator solution changes when the pH changes past a certain point.
Let one consider titration of a weak acid (such as CH3COOH) and strong base (NaOH) – the analyte (the solution being titrated). At the stoichiometric point their mixtures become an aqueous solution of the weak acid-strong base salt (sodium acetate). It also contains ions - streaming from autoprotolysis. The presence of the Brønsted base CH3COO means that pH greater then 7 can be expected (see Figure 1).
Figure 1. pH-metric titration of strong and weak acid with a strong base
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For prediction of the pH at any stage of acid-base titration one should suppose that VA volume of a solution of a weak acid with nominal molar concentration A0 is being titrated with with a solution of strong base of molar concentration B. Approximations given before are based on the fact that the acid is weak, and therefore that HA is more abundant than any A- ions in the solution. Further more, when HA is present; it provides hydronium ions that greatly outnumber any ions from the autoprotolysis of water. Similarly, when excess base is present number of OH ions is much greater than that from autoprotolysis of water. At the start of the titration, pH of the analyte solution (weak acid) can be calculated as: 1 pH pK log A 11 a 2 0 After addition of some strong base (still before stoichiometric point) concentration of A- ions stems almost entirely from the salt that is present, for a weak acid present provides only a few A- ions. Therefore, [A-]=S, the molar concentration of salt (base). The amount of the acid molecules that remains is the original amount A0V0 less the amount of HA molecules that have been converted to the salt by addition of base, so the molar concentration of acid is A’=A-S which ignores additional small loss of HA as a result of its ionization in the solution. Hence
a a a S H3O A H3O K a 12 aHA A' The derivation has made the doubtful approximation that the activity of the A- ions is close to
1. It follows that: A' pH pK a log 13 S called the Henderson-Hasselbalch equation, which in general form, after recognition that A’ is molar concentration of acid in solution and S is molar concentration of base, is: acid pH pK log 14 a base
When the molar concentrations of acid and salt are equal pH=pKa – hence pKa can be measured directly from the pH of the mixture. + At the stoichiometric point, the H3O ions in the solution stem from the influence of the OH- ions on the autoprotolysis equilibrium, and the OH- ions are produced by the proton - transfer equilibrium from H2O to A . Because only a small amount of acid is formed in this way, the concentration of A- ions is almost exactly that of the salt, and one can write that [A-
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]=S. The number of OH- ions arise from the proton transfer equilibrium greatly outnumber those produced by the water autoprotolysis, that is why one can set [HA]=[OH-]:
2 a a HA OH OH Kb 15 a S A At this point pH can be calculated as: 1 1 1 pH pK pK log S 16 2 a 2 w 2 When surplus of strong base has been added that the titration has been carried out well past + stoichiometric point, the pH is determined by the excess base present. Then, [H3O ] = - Kw/[OH ] and
pH pK w log B' 17 where B’ is the molar concentration of excess base.
Experimental procedure: 1. Calibrate a pH-meter with a phthalate buffer (pH=4) 2. Prepare an aqueous solution of strong acid (10 mL of strong acid, 20 mL of acetone and 20 mL of distilled water) 3. Fill a burette with a NaOH solution (0.2 mol/L) 4. Titrate an acid solution by adding 0.1-0.5 mL of a base 5. Prepare an aqueous solution of weak acid (10 mL of strong acid, 20 mL of acetone and 20 mL of distilled water) 6. Fill a burette with a NaOH solution (0.2 mol/L) 7. Titrate an acid solution by adding 0.1-0.5 mL of a base
Strong acid - HCl Weak acid - CH3COOH V [mL] pH ( pH ) V [mL] pH ( pH ) NaOH NaOH VNaOH VNaOH
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Data analysis:
1. Prepare a titration graph pH=f(VNaOH) for a strong and weak acid separately. Inflexion point of titration curve denotes stoichiometric point. 2. For exact determination of stoichiometric point prepare derivative graph ( pH) f (VNaOH ) VNaOH 3. Calculate number of moles and concentrations of acids in solution
VNaOH 0 VNaOH 4. For a weak acid prepare a graph log f VNaOH and determine its VNaOH
pKa. Compare it with theoretical one (hint: you can find it in chemical tables).
Problems to solve:
1. Estimate the pH of 0.1M HClO(aq). pKa=7.43 2. The stoichiometric point of a titration of 25 mL of 0.1M HClO(aq)with 0.1M NaOH(aq) occurs when the molar concentration of NaClO is 0.05M (volume of the solution increased to 50 mL). Calculate pH.
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II.D. CONDUCTOMETRIC TITRATION
Goal: Main goal of the exercise is to analyze changes in conductivity of the solutions during conductometric titrations and calculations of acid concentration in its aqueous solution.
Introduction The connection between chemistry and electricity is a very old one, going back to ALESSANDRO VOLTA'S discovery, in 1793, that electricity could be produced by placing two dissimilar metals on opposite sides of a moistened paper – Volta’s batteries consisted of a series of zinc or silver disks, arranged alternately with paper soaked in salt water between them. In 1800, Nicholson and Carlisle, using Volta's primitive battery as a source, showed that an electric current could decompose water into oxygen and hydrogen. This was surely one of the most significant experiments in the history of chemistry, for it implied that the atoms of hydrogen and oxygen were associated with positive and negative electric charges, which must be the source of the bonding forces between them. By 1812, the Swedish chemist BERZELIUS could propose that all atoms are electrified, hydrogen and the metals being positive, the nonmetals negative. In electrolysis, the applied voltage was thought to overpower the attraction between these opposite charges, pulling the electrified atoms apart in the form of ions (named by Berzelius from the Greek for “travelers”). It would be almost exactly a hundred years later before the shared electron pair theory of G.N. LEWIS could offer a significant improvement over this view of chemical bonding. Meanwhile the use of electricity as a means of bringing about chemical change continued to play a central role in the development of chemistry. HUMPHREY DAVEY prepared the first elemental sodium by electrolysis of a sodium hydroxide melt. It was left to Davey's former assistant, MICHAEL FARADAY, to show that there is a direct relation between the amount of electric charge passed through the solution and the quantity of electrolysis products. JAMES CLERK MAXWELL immediately saw this as evidence for the “molecule of electricity”, but the world would not be receptive to the concept of the electron until the end of the century
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Conductance and conductivity The electric current in an electrolyyic solution consist of a flow of ions; in a metal, it consists of a flow of electrons. The fundamental measurement used to study the motion of ions is that of the electrical resistance, R, of the solution. The standard technique is to incorporate a conductivity cell into one arm of the resistance bridge (Figure 1) and to search for the balance point.
Figure 1. Conductometer scheme
The conductance, Γ, of a solution is the inverse of its resistance R: Γ=1/R. As resistance is expressed in ohms, Ω, the conductance of a sample is expressed in Ω-1, which officially is designated as siemens, S, and 1 S = 1 Ω-1. The conductance of a sample decreases with its length l and increases with its cross-sectional area A. Therefore one can write: A 1 l where κ is the conductivity in siemens per meter, S m-1. The conductivity of a solution depends on the number of ions present, and it is normal to introduce the molar conductivity, Λm, which is defined as 2 m c where c is the molar concentration of the added electrolyte. The SI unit of molar conductivity is siemens metre-squared per mole (S m2 mol-1). The molar conductivity of an electrolyte would be independent of concentration if κ were proportional to the concentration of the electrolyte. However, in practice, the molar conductivity is found to vary with the concentration. One reason for this variation is that the number of ions in the solution might not be proportional to the concentration of the electrolyte. For instance, the concentration of ions in the solution of weak acid depends on the concentration of acid in a complicated way, and doubling the concentration of the acid added
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Strong electrolytes Strong electrolytes are substances that are virtually fully ionized in solution, and include ionic solids and strong acids. As a result of their complete ionization, the concentration of ions in solution is proportional to the concentration of strong electrolyte added. In an extensive series of measurements during the XIXth century, FRIEDRICH KOHLRAUSCH showed that at low concentrations the molar conuctivities of strong electrolytes vary linearly with the square root of the concentration:
0 m m K c 3
0 This variation is called Kohlrausch’s law. The constant m is the limiting molar conductivity, the molar conductivity in the limit of zero concentration (when the ions are effectively infinitely apart from each other and do not interact with one another). The constant K is found to depend more on the stoichiometry of the electrolyte (that is, whether it is of the form MA, or M2A, etc.) than on its specific identity. Kohlrausch was also able to show that can be expressed as the sum of contributions from individual ions. If the limiting molar conductivity of the cations is denoted and of the anions , then his law of the independent migration of ions states that
0 m 4 where and are the numbers of cations and anions per formula unit of electrolyte.
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Weak electrolytes Weak electrolytes are not fully ionized in the solution. They include weak Brønsted acids and bases, such as CH3COOH and NH3. The marked concentration dependence of their molar conductivities arises from the displacement of the equilibrium
aH OaA HAaq H Ol H O aq A aq K 3 5 2 3 a aHA towards products at low molar concentrations. The conductivity depends on the number of ions in the solution, and therefore on the degree of ionization, , of the electrolyte. The degree of ionization is defined so that, for the acid HA at molar concentration c, at equilibrium
H3O c A c HA 1c 6
If we ignore activity coefficients, the acidity constant, Ka, is approximately 2c K 7 a 1
Conductometric titration The electrical conductance of a solution serves as a means for determining the end point in chemical reactions, such as titrations of acids and bases, or precipitations. For example, when a strong acid is added to a strong base (hydrochloric acid and sodium hydroxide), the conductance decreases to a minimum, at which the base is completely neutralized, and then it increases, owing to the excess of acid, as shown in the Figure 2.
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0 0 1 2 3 4 5 6 V HCl Figure 2. Conductometric titration of strong base by strong acid
The two lines AB and CD intersect at the point E, which is the end point. The OH‾ ions of the base and the H+ ions of the acid have much greater motilities than the sodium and chloride
Lodz University of Technology, Faculty of Chemistry, Institute of Applied Radiation Chemistry, Laboratory of Chemistry
ELECTROCHEMISTRY ions, and so the conductance is least at the end point E where the acid and the base are present in exactly equivalent portions, and there is no excess of either OH‾ ions or H+ ions. In order that the lines AB and CD shall be straight and thus permit calculations from set of only two points, it is desirable to keep the volume constant throughout the titration. In order to approach this condition, the added reagent must be concentrated, whereas the solution which is being titrated must be dilute. If the same titration is carried out with a weak acid (e.g. acetic acid) instead of a strong acid, as shown in Figure 3, the excess acid beyond the end point will not cause such a sharp increase in conductance.
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0 0 1 2 3 4 5 6 V CH3COOH Figure 3. Conductometric titration of strong base by weak acid
In fact, with acetic acid a horizontal line is obtained after all the sodium hydroxide has been neutralized, because the number of ions being added in the excess of acetic acid is small, particularly in the presence of the sodium acetate. The sharp change in the slope of the line is useful, however, in determining end points. In the colored or turbid solutions, where a colored indicator cannot be used, this determination of end points by means of conductance measurements is particularly useful. Conductance measurements are used for a variety of testing and control operations, such, for example, as concentration of acids or salts by evaporation, leakage of salt solutions, hardness of water, moisture content of soil or wood, and rates of chemical reaction in which the products have a different conductance from that of the reactants.
Lodz University of Technology, Faculty of Chemistry, Institute of Applied Radiation Chemistry, Laboratory of Chemistry
ELECTROCHEMISTRY
Experimental procedure: 1. Fill a burette with a NaOH solution (0.2 mol/L) 2. Fill a beaker with 5 mL of an acid (HCl) and adequate volume of water to cover conductivity cell 3. Read and write initial value of conductivity 4. Titrate an acid solution by adding 0.1-0.5 mL of a base
5. Fill a beaker with 5 mL of an acid (CH3COOH) and adequate volume of water to cover conductivity cell 6. Read and write initial value of conductivity 7. Titrate an acid solution by adding 0.1-0.5 mL of a base
Strong acid - HCl Weak acid – CH3COOH
VNaOH [mL] Γ [S] VNaOH [mL] Γ [S]
Data analysis:
1. Prepare a titration graph Γ=f(VNaOH) for a strong and weak acid separately. Inflexion point of titration curve denotes stoichiometric point. 2. Calculate number of moles and concentrations of acids in solution
Problems to solve: A conductance cell was calibrated by filling it with a 0.02N solution of potassium chloride (κ = 0,002768 S cm-1) and measuring the resistance at 25 ºC, which was found to be 457.3 ohms.
The cell was then filled with calcium chloride solution containing 0.555 gram of CaCl2 per liter. Then measured resistance was 1050 ohms. Calculate (a) the cell constant for the cell, (b) the specific conductance of the CaCl2 at this concentration.
The following table gives the specific conductance of a solution of hydrochloric acid, to 100 ml of which have been added various amounts of an 8N solution of sodium hydroxide. If the dilution effect of the small amount of hydroxide solution added is neglected, what is the normality of the HCl solution? NaOH [ml] 0,32 0,92 1,56 2,34 Conductance [S] 0,0322 0,0186 0,0164 0,0296
Lodz University of Technology, Faculty of Chemistry, Institute of Applied Radiation Chemistry, Laboratory of Chemistry
ELECTROCHEMISTRY
Lodz University of Technology, Faculty of Chemistry, Institute of Applied Radiation Chemistry, Laboratory of Chemistry