Amperometry, Polarography, Electro-Deposition, and Conductometry PDF

Amperometry Amperometry involves the measurement of current at a fixed potential The magnitude of the current is linearly related to the concentration of the electroactive species. Since the potential is not scanned, amperometry does not lead to a voltammogram  Amperometric Electrodes...

Amperometry Amperometry involves the measurement of current at a fixed potential The magnitude of the current is linearly related to the concentration of the electroactive species. Since the potential is not scanned, amperometry does not lead to a voltammogram  Amperometric Electrodes  Example 1: (Clark amperometric sensor = O2 sensor)  A gas-permeable membrane at the end of the sensor.  The working electrode is a Pt disk cathode.  The counter electrode is Ag ring anode.  Although several gases can diffuse across the membrane (O2, N2, CO2), only O2 is reduced at the cathode. Clark amperometric Sensor for the Determination of Dissolved O2 Example 2: Amperometric enzyme electrode (glucose sensor) Enzymes are often utilized in a chemically modified electrode layer to impart the selectivity needed. An example of an amperometric enzyme electrode is the glucose sensor, illustrated in Figure below:  The enzyme glucose oxidase is immobilized in a gel (e.g., acrylamide) and coated on the surface of a platinum wire cathode. Glucose and oxygen from the test solution diffuse into the gel where their reaction is catalyzed to produce H2O2; part of this diffuses to the platinum cathode where it is oxidized to give a current that is proportional to the glucose concentration. The enzyme glucose oxidase catalyzes the aerobic oxidation of glucose as follows: Quantitative Analysis using polarography  The principal use of polarography is in quantitative analysis.  Since the magnitude of the diffusion current is proportional to the concentration of analyte, the height of a polarographic wave tells how much analyte is present. 1. One Standard Method  It is assumed that a linear relationship holds for the concentration and the wave height.  Assuming that the wave heightes for the standard and the analyte were h1 and h2 and the concentrations were Cstandard and Canalyte then,  (Id) standard / (Id)analyte = Cstandard / Canalyte Example: 50 mL of a liquid diet supplement treated and riboflavin (C17H20N4O6) in the sample was determined polarographically and gave 0.28 µA of diffusion current. Standard 40 ppm riboflavin gave 0.45 µA of diffusion current in the same polarographic cell. Calculate riboflavin content in the diet sample 2. Standard calibration curve method  The most reliable, but tedious, method of quantitative analysis is to prepare a series of known concentrations of analyte in otherwise identical solutions. Procedures  A polarogram of each solution is recorded, and a graph of the diffusion current versus analyte concentration is prepared.  Finally, a polarogram of the unknown is recorded, using the same conditions.  From the measured diffusion current and the standard curve, the concentration of analyte can be determined.  The figure below shows an example of the linear relationship between diffusion current and concentration. Standard curve for polarographic analysis of Al(III) in 0.2 M sodium acetate, pH 4.7. 3. Standard addition method  The standard addition method is most useful when the sample matrix is unknown or difficult to duplicate in synthetic standard solutions.  This method is faster but usually not as reliable as the method employing a standard curve. Procedures  First, a polarogram of the unknown is recorded.  Then, a small volume of concentrated solution containing a known quantity of the analyte is added to the sample.  The increase in diffusion current of this new solution can be used to estimate the amount of unknown in the original solution. The diffusion current of the unknown will be proportional to the concentration of unknown, Cx: ld(unknown) = kCx  where k is a constant of proportionality.  Let the concentration of standard solution be CS.  When VS mL of standard solution is added to Vx mL of unknown,  The diffusion current is the sum of diffusion currents due to the unknown and the standard. rearrange and solve for Cx R= (id)x+s / (id)x Example: 50 mL of sample of plant tissue extract is analyzed polarographically with a diffusion current for Zn of 0.6 μA. Then 5 mL of standard Zn (1.5×10-2 M) is added to sample solution to give diffusion current of 0.8 μA for Zn. Calculate the Zn concentration in the plant tissue sample  Electro-deposition (Electro-gravimetry) 1. Electrolytic precipitation or electrodeposition has been widely used for the determination of metals. 2. The metal is deposited on a weighed platinum cathode and the increase in mass is determined. 3. There are two types of electro-gravimetric methods:  Constant current electrolysis  Constant potential electrolysis Constant Current Electrolysis Electrodeposition is carried out by keeping the current constant and gradual increases in the applied potential occurs. As shown in Fig, the apparatus for constant current electrolysis consists of: 1. platinum- anode 2. platinum gauze- cathode 3. a direct current (dc)source (6-12 V battery) 4. ammeter to indicate the current 5. voltmeter to indicate the applied voltage 6. a resistor for controlling the voltage Apparatus for constant current electrolysis Typical applications of constant current electrolysis Constant Potential Electrolysis By controlled potential electrolysis, it is possible to separate two elements whose deposition potentials differ by a few tenths of a volt. The controlled potential electrolysis apparatus is shown in Fig. It consists of two independent electrode circuits as following: A. The electrolysis circuit consists of: 1. a dc power source ( 12V battery) 2. a potential divider that permits continuous variation in the potential applied across the working electrode 3. a counter Pt- electrode 4. ammeter B. The reference circuit consists of: 1. reference electrode (SCE) 2. voltmeter 3. working electrode The 2 circuits share a common electrode which is the working electrode The electrical resistance of the reference circuit is very large, so the current in the reference electrode circuit is zero at all times The electrolysis circuit supplies the current for the deposition. Apparatus for controlled-potential electrolysis Electrodes for Electro-deposition Platinum gauze electrodes are used since they have the following advantages: 1. non-reactive 2. Can be ignited to remove any organic matter or gases that would have a harmful effect on the physical properties of the deposit Some metals like Bi and Zn cannot be deposited directly on to platinum, because they cause damage to the electrode. So, a protective coating of copper is always deposited on a platinum electrode.  Electrodeposition is governed by Faraday's laws First law: The mass (m) of any substance formed at an electrode is directly proportional to the quantity of electricity (Q) passed through the solution. 𝒎∝𝑸 or 𝒎 = 𝑬𝒒. 𝒘𝒕 × 𝑸 Where Eq.wt is the constant of proportionality known as equivalent weight of the substance. If a current of I (Amperes, A) is passed through an electrolyte solution for t (Seconds, s), we have: 𝑸=𝑰×𝒕 Therefore: 𝒎 = 𝑬𝒒. 𝒘𝒕 × 𝑰 × 𝒕 The equivalent weight of a substance is its mass that is deposited when a current of 1 A is passed for 1 second. So, 𝒎 = 𝑬𝒒. 𝒘𝒕 To produce w grams or (w/Ew) equivalents of an element by electrolysis, we need (w/Ew) × F Coulombs of charge to pass through the solution. Mathematically: 𝒘 𝑸=𝑰×𝒕=( )×𝑭 𝑬𝒒. 𝒘𝒕 𝑴𝒘𝒕 We know that: 𝑬𝒒. 𝒘𝒕 = 𝒏 Finally we can write: 𝒘 𝑰×𝒕=( )×𝑭 𝑴𝒘𝒕/𝒏 or 𝑰 × 𝒕 × 𝑴𝒘𝒕 𝒘(𝒈) = 𝑭×𝒏 Second law: When the same quantity of electricity is passed for the same time through different electrolytes connected in series, the ratio of masses of different substances deposited at the electrode will be in the proportional of their equivalent weights. 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑯𝟐 𝒍𝒊𝒃𝒆𝒓𝒂𝒕𝒆𝒅 𝑬𝒘 𝒐𝒇 𝒉𝒚𝒅𝒓𝒐𝒈𝒆𝒏 = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝒖 𝒅𝒆𝒑𝒐𝒔𝒊𝒕𝒆𝒅 𝑬𝒘 𝒐𝒇 𝒄𝒐𝒑𝒑𝒆𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑪𝒖 𝒅𝒆𝒑𝒐𝒔𝒊𝒕𝒆𝒅 𝑬𝒘 𝒐𝒇 𝒄𝒐𝒑𝒑𝒆𝒓 = 𝒎𝒂𝒔𝒔 𝒐𝒇 𝑨𝒈 𝒅𝒆𝒑𝒐𝒔𝒊𝒕𝒆𝒅 𝑬𝒘 𝒐𝒇 𝒔𝒊𝒍𝒗𝒆𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇𝑨𝒈 𝒅𝒆𝒑𝒐𝒔𝒊𝒕𝒆𝒅 𝑬𝒘 𝒐𝒇 𝒔𝒊𝒍𝒗𝒆𝒓 = 𝒎𝒂𝒔𝒔 𝒐𝒇𝑯𝟐 𝒍𝒊𝒃𝒆𝒓𝒂𝒕𝒆𝒅 𝑬𝒘 𝒐𝒇 𝒉𝒚𝒅𝒓𝒐𝒈𝒆𝒏 Problems 1. How many grams of copper are deposited by a current of 2 A in 10 minutes during the electrolysis of aqueous copper sulfate? (molar mass of Cu=63). 2. 0.4 g of analyte is deposited by a current of 0.2 A in 100 min. calculate the eq.wt of analyte? 3. How long will it take to produce each of the following by electrolysis with a current of 80 A? 50 g of Ni(s) from Ni2+ (molar mass of Ni=58) 5 moles of Ag(s) from Ag+ (molar mass of Ag=107) 4. In an electrolysis experiment, a current was passed for 30 min through two cells in series. The first cell contains a solution of gold salt and the second cell contains copper salt. If 9.5 g of Au (molar mass of Au=197) was deposited in the first cell, find: 1. amount of copper (molar mass of Cu=63) deposited on the cathode in second cell 2. calculate current in ampere  Conductometry This is a method of analysis based on measuring electrical conductance What is the Conductance (G)? It is the ability of the solution to conduct the electric current (electricity). 𝟏 −𝟏 𝑮=  𝒐𝒓(𝑺) 𝑹 What is the Resistance (R)? It refers to the opposition to the flow of current. Electrodes Solution 𝒍 𝑹=𝝆 𝒂  : Specific resistance (resistivity) l What is the Specific conductance (Conductivity) (k)? It is the inverse of Specific resistance. 𝟏 𝒍 𝜿= =𝑮 (𝑺. 𝒄𝒎−𝟏 ) 𝝆 𝒂 Factors affecting conductivity 1. Nature of electrolyte (number of ions) strong electrolytes have high conductance. weak electrolytes have low conductance. 2. Nature of ions The velocity by which ions move towards the electrodes carrying the electric current depends on: 1. size 2. molecular weight 3. number of charges present on the ion 3. Temperature With increasing temperature, the conductivity of an electrolyte increases. What is the Molar Conductivity? The conductivity of one mole of the electrolyte in each 1 cm3 of the solution (S cm2 mol-1) 𝟏𝟎𝟎𝟎 𝜿 𝒎 = 𝑪 Where (C) is the molar concentration (mol/L). (1000) is a factor where: 1 L= 1000 mL. (𝐤) is the conductivity (𝑺. 𝒄𝒎−𝟏 ) Problem A solution of CuSO4 with concentration (C) of 0.02 mole/L and has a specific conductance (k) of 2.89×10-3 S.cm-1. Calculate the molar conductivity (Ʌm) in cm2 /Ω.mole unit  Conductivity of strong electrolytes Kohlaursch showed that at low concentrations, the m of a strong electrolyte varied in a linear manner with the square root of the electrolyte concentration. °𝒎 : Limiting molar conductivity B : Constant Problem The concentration of AgNO3 (C) is 0.05 mol/L, and has molar conductivity (m) of 13.5 mS.m2.mol-1 at 25°C. Calculated the limiting molar ° conductivity (𝐦 ) (molar conductivity at infinite dilution) of AgNO3 at this temperature if (B =0.5)? Conductivity of Weak Electrolytes molar conductivity of weak electrolyte is normal at diluted solutions, but falls sharply as the concentration increases. Degree of ionization (α) for Weak electrolyte: The ratio of the amount of ions being formed in the solution and the amount of electrolyte added to the solution m = 𝒎 ° Problem Molar conductivity at infinite ° dilution at (𝐦 ) 25°C of 0.01 M 2 NH4OH equal 108.9 S.cm mol. -1 If Molar conductivity (m) equal 9.5 S.cm2mol-1 calculate degree of ionization (α) of NH4OH at this temperature. Applications of Conductivity 1. Direct OR absolute measurements 2. Conductometric titration  The principle of conductometric titration is based on the fact that: 1. During the titration, one of the ions is replaced by the other and invariably these two ions differ in the ionic conductivity with the result that conductivity of the solution varies during the course of titration. 2. The equivalence point may be located graphically by plotting the change in conductance as a function of the volume of titrant added. Some Typical Conductometric Titration Curves are: 1. Strong Acid with a Strong Base, e.g. HCl with NaOH:  Before NaOH is added, the conductance is high due to the presence of highly mobile hydrogen ions.  When the base is added, the conductance falls due to H+ ions react with OH − ions to form water.  This decrease in the conductance continues till the equivalence point.  At the equivalence point, the solution contains only NaCl.  After the equivalence point, the conductance increases due to the large conductivity of OH- ions Conductometric titration of a strong acid (HCl) vs. a strong base (NaOH) 2. Weak Acid with a Strong Base, e.g. acetic acid with NaOH: Initially the conductance is low due to the weak ionization of acetic acid. On the addition of base, there is decrease in conductance due to the replacement of H+ by Na+ and suppresses the dissociation of acetic acid due to common ion acetate. Then, the conductance increases on adding NaOH as NaOH react with CH3COOH to form CH3COONa. The increase in conductance continues raise up to the equivalence point. After the equivalence point, conductance increases more rapidly with the addition of NaOH due to the highly conducting OH− ions  Introduction to Chemical Separation (Solvent Extraction)  Introduction Separations are extremely important in: synthesis, industrial chemistry, biomedical sciences, chemical analyses. Separation methods Separation Principles 1. Complete separation A mixture of four components is completely separated so that each component occupies a different spatial region. 2. Partial separation In the partial separation, species A is isolated from the remaining mixture of B, C, and D.  Separation by Precipitation 1. Separations Based on Control of Acidity (pH) These separations can be grouped in three categories:  those made in relatively concentrated solutions of strong acids,  those made in buffered solutions at intermediate pH values,  those made in concentrated solutions of sodium or potassium hydroxide. 2. Sulfide Separations With the exception of the alkali metals and alkaline-earth metals, most cations form sparingly soluble sulfides whose solubilities differ greatly from one another. 3. Separations by Inorganic Precipitants  Hydroxide (OH-) and sulfide ions (S2-) are generally useful for separations.  Phosphate, carbonate, and oxalate ions are often used as precipitants for cations, but they are not selective.  Chloride and sulfate are useful because of their highly selective behavior. 4. Separations by Organic Precipitants Some example of organic precipitants: Dimethylglyoxime are useful because of their remarkable selectivity in forming precipitates with only a few ions. 8-hydroxyquinoline, yield slightly soluble compounds with many different cations. The selectivity of this type of precipitant is due to: 1. the wide range of solubility among its reaction products 2. the precipitating reagent is usually an anion that is the conjugate base of a weak acid. 5. Separation of Species Present in Trace Amounts by Precipitation Several difficulties or problems can accompany the separation of a trace element by precipitation: 1. supersaturation which delays formation of the precipitate, 2. coagulation of small amounts of a colloidally dispersed substance 3. losing an appreciable fraction of the solid during transfer and filtration To minimize these difficulties, a quantity of some other ion that also forms a precipitate with the reagent is often added to the solution. The precipitate from the added ion is called a collector. A collector is used to remove trace constituants from solution. 6. Separation by Electrolytic Precipitation a. In the electrolytic precipitation, the more easily reduced species, either the wanted or the unwanted component of the sample, is isolated as a separate phase. b. The method becomes particularly effective when the potential of the working electrode is controlled at a predetermined level. c. The mercury cathode has found wide application in the removal of many metal ions prior to the analysis of the residual solution. 7. Salt-Induced Precipitation of Proteins (Salting-out)  A common way to separate proteins is by adding a high concentration of salt.  This procedure is termed salting out the protein. The solubility of protein molecules depends on: pH, temperature, ionic strength, the nature of the protein, the concentration of the added salt.  Separation by Distillation Distillation is based on differences in the boiling points of the materials in a mixture. Distillation is widely used to: separate volatile analytes from nonvolatile interferents, separate components in mixtures for purification purposes in organic chemistry. There are many types of distillation: Vacuum distillation is used for compounds that have very high boiling points. Molecular distillation occurs at very low pressure such that the lowest possible temperature is used with the least damage to the distillate. Pervaporation is a method for separating mixtures by partial volatilization through a non-porous membrane. Flash evaporation is a process in which a liquid is heated and then sent through a reduced pressure chamber. The reduction in pressure causes partial vaporization of the liquid.  Separation by solvent extraction 1. Solvent extraction is a powerful separation technique. 2. Liquid-liquid extraction, is a technique in which a solution is brought into contact with a second solvent, essentially immiscible with the first, in order to bring the transfer of one or more solutes into the second solvent. 3. Advantages of this method are: a. Simple. b. Convenient. c. Rapid to perform. d. Applicable to both, trace and macro levels In analytical applications solvent extraction may serve the following three purposes:  Pre-concentration of trace elements.  Elimination of matrix interference.  Differentiation of chemical species. Principles The partition of a solute between two immiscible phases is an equilibrium process that is governed by the distribution law. If the solute species A is allowed to distribute itself between water and an organic phase, the resulting equilibrium may be written as: The ratio of activities for A in the two phases will be constant and independent of the total quantity of A: The equilibrium constant K is known as the distribution constant. Separating Metal Ions as Chelates Many organic chelating agents are weak acids that react with metal ions to give uncharged complexes that are highly soluble in organic solvents such as (ethers, hydrocarbons, ketones), but of limited solubility in water. Most uncharged metal chelates are insoluble in water. Example Equilibria in the extraction of an aqueous cation M2+ into an immiscible organic solvent containing 8-hydroxyquinoline (HQ) Four equilibria are shown: (1) distribution of HQ between the organic and aqueous layers. (2) acid dissociation of HQ to give H and Q ions in the aqueous layer. (3) complex-formation reaction giving MQ2. (4) distribution of the chelate (MQ2) between the two phases. The overall equilibrium is the sum of these four reactions The equilibrium 17 constant for this reaction is  Important laws of solvent extraction 1. Calculation of remaining concentration of analyte The concentration of A remaining in an aqueous solution is given by the equation: Where :  [A]i is the concentration of A remaining in the aqueous solution after extracting  Vaq (mL) is the volume of the solution  [A]0 is the original concentration of A  i are the portions of the organic solvent,  Vorg is the volume of the organic solvent.  K is the equilibrium constant or the distribution constant Example Solution  Note the increased extraction efficiencies that result from dividing the original 50 mL of solvent into two 25-mL or five 10-mL portions. It is always better to use several small portions of solvent to extract a sample than to extract with one large portion.  2. Calculation of percent extraction (%E) of analyte Distribution Ratio (D) The distribution of a solute between two immiscible solvents can be described by the distribution ratio “D”. Where [A] represents the stoichiometric or formal concentration of a substance A and the subscripts 1 and 2 refer to the two phases. Since in most cases, two-phase system is of analytical interest, an organic solvent and aqueous are involved, D will be understood to be; The subscripts Org and Aq refer to the organic and aqueous phases respectively Percent Extraction (%E) The more commonly used term for expressing the extraction efficiency by analytical chemist is the percent extraction “E”, which is related to “D” as: Where V represent solvent volume and the other quantities remain as previously defined. The percent extraction may be seen to vary with: the volume ratio of the two phases D For example: for value of D below 0.001, the solute may quantitatively retained in the aqueous phase for value of D (500 to 1000), the value of “E” changes only from 99.5 to 99.9%. Example If 10 cm3 of water (Vaq) containing an analyte shacked with 30 cm3 of CCl4 (Vorg), and the distribution (D) ratio was found to be 0.9, calculate percent extraction (%E) of analyte Solution: try by your self  Separation Factor (γ) The separation factor γ is related to the individual distribution ratios as follows: where A and B represent the respective solutes.  If the distribution ratios one is very small and the other is large, complete separations can be quickly and easily achieved.  If the separation factor is large but the smaller distribution ratio is sufficiently large then less separation of both components occurs. Methods of Extraction Batch extraction 1. The simplest technique and most employed in the laboratory for analytical separations. 2. Rapid, Simple and Clean separation method. 3. More beneficial when the distribution ratio (D) of the solute of interest is large. Continuous extraction 1. Continuous extractions are used when the distribution ratio (D) is relatively small. 2. Devices consists of distilling the extracting solvent from a boiler flask and condensing it and passing it continuously through the solution being extracted. Countercurrent extraction 1.The separation depends on the density difference between the fluids in contact. 2.In vertical columns, the more dense phase enters at the top and flows downwards while the less dense phase enters at the bottom and flows upwards. 3.The method has the advantage for separating materials for purification purposes and is extensively used in engineering problems Important Factors Influencing the Extraction Efficiency 1. Choice of solvent 2. Acidity of the analyte solution 3. Presence of salting-out agents 4. Oxidation state 5. Presence of masking agents