Techniques: Constructing Cells Measuring Potential

Nernst Electrochemistry Faraday Cells, Half-cells, and the Nernst Equation RT E = E 0 - -------ln Q - nf Chemical reactions with electrons 1 Objectives: Examine electrical consequences of oxidation reduction reactions Use electrical measurements to determine chemically significant information Techniques: Constructing Cells Measuring Potential pparatus: ph Meter as a Voltmeter s Silver Electrodes Various Electrode Materials lligator Clips Concepts: Chemistry: Chemical Reactions, Stoichiometry, Limiting Reagents, Oxidation, Reduction, Solubility Products Electricity: Charge, rrent, Potential (Voltage), Resistance, node, Cathode, Ohm s Law, Reference Cell Electrochemistry: Half-cells, Cells, Reduction Potentials, Nernst Equation,, Standard Hydrogen Electrode Thermodynamics: Gibbs Free Energy, Chemical Equilibrium, Spontaneous Processes, Standard Free Energy of Formation, Extended Hess Law 2 3 Organization of the Lecture Basic Electrical Units and Concepts 5 Must Half Cells Involve Different 19 Substances? The Faraday 6 Summary Concentration Cells 20 The Reaction Quotient 7 Concentration Cells Which Half 21 Cell is Positive? The Nernst Equation and ΔG 8 The Exercise 22 Half Cell Potentials 9 Part 1 Nernst Equation 23 Equilibrium Constants, G, G 0 and ΔE 0 11 The Reference Electrode 26 Why Do We Need Cells & Half cells? 12 The Spreadsheet 27 The 14 Part 2 Solubility Product 29 The ph Meter 15 Part 3 n Example 33 Electrochemical Cell Representation 16 The Revisited 40 Summary Electrochemical Cells 18 Waste Disposal 41 4 Basic Electrical Units & Concepts: Units Charge, q : coulomb Potential, E : volt rrent, i : ampere = coulomb/second Resistance, R : ohm = volt/ampere Energy, W : joule = volt-coulomb = watt-second Ohm s Law i = E / R 1 ampere 1 volt of electrical potential 1 volt applied to a resistance of 1 ohm + 1 ohm will cause a current of 1 ampere to flow. 5 One more unit - The Faraday - Chemical unit We expand our concept of chemical reactions to include electrons so, we must account for their stoichiometry, just like moles of other chemical substances. To that end, we define the Faraday, F, to be the (absolute value) of the charge on vogadro s number (1 mole) of electrons: 1 F = N e = 6.02214129 X 10 23 mol -1 X 1.602176565 X 10-19 C = 96,485.3365 C/mol 96,500 C/mol is good to 0.016% 6 The Reaction Quotient The Reaction Quotient, Q, for a Chemical Reaction a + b B +.. c C + d D +.. is: [ C ] c [ D ] d When system is Q = in a state of [ ] a [ B ] b equilibrium, Q = K eq where: solids (& gases) in their standard states are assigned a concentration (activity) of 1 E.g., for (s) + 2 + (aq) 2 (s) + 2+ (aq) [ 2+ ] Q = [ + ] 2 7 The Nernst Equation and ΔG Oxidation-reduction reactions have associated potentials. Nernst showed that the potential, E, associated with oxidation-reduction reactions in solution is proportional to ΔG and depends on concentrations in the same manner as ΔG. The relation that connects G = -n F ΔE Thermochemistry and Electrochemistry G = G 0 + RT ln Q where Q is the Reaction Quotient RT E = ΔE 0 ------- ln Q nf 8 Half Reaction Potentials Oxidation/reduction reactions involve the transfer of electrons between chemical species. We can consider such a reaction occurring in separate steps in which: the electrons are lost (oxidation), and electrons are gained (reduction) with these so called Half Reactions each having an associated potential The sum of those steps is the balanced overall reaction (in which no electrons appear) and their potentials follow Hess Law just like G and H. 9 1

E 0 s are tabulated for half reactions written as reductions. M n+ (aq) + ne - M (s) E = E 0 (RT/nF) ln 1 / [ M n+ ] They are related to free energies by: G 0 = -n F E 0 E.g., 2+ + 2 e - E = 0.337 (RT/2F) ln 1 / [ 2+ ] Standard Reduction Potential for Cd 2+ + 2 e - Cd E Cd = -0.440 (RT/2F) ln 1 / [Cd 2+ ] Equilibrium Constants, G, G 0 and ΔE 0 In which direction is a reversible chemical reaction favored? The answer depends on the value of G for the reaction. For any chemical reaction at constant P and T: G = - n F E If G < 0 (E > 0 ), the forward reaction is favored If G > 0 (E < 0 ), the backward reaction is favored If G = 0 (E = 0 ), the system is at equilibrium and neither direction is favored. The sign of G is determined by the concentrations of the reactants and products and their values, through Q, compared to G 0. G = G 0 + RT ln Q t equilibrium, G = 0, E = 0, Q = K eq and, G 0 = - RT ln K eq = - n F ΔE 0 Why Do We Need Cells & Half-Cells? Consider the reaction: (s) + 2 + (aq) 2 (s) + 2+ (aq) 10 11 12 We can write the reaction in two steps: (s) + 2 + (aq) 2 (s) + 2+ (aq) 2+ (aq) + 2e - (s) E 0 = 0.337 V + (aq) + e- (s) E 0 = 0.799 V Suppose we construct two half cells as follows: e -? ΔE = E 0 - E 0 - (RT/2F) ln [ 2+ ]/[ + ] 2 NO 3 ΔE = E 0 - E 0 = 0.462 ΔG = n F ΔE = - 2 X 96500 X 0.462 ΔG = 89.2 kj The reaction should be spontaneous in the forward direction 13 Is there a way to compensate for the charge imbalance that electron transfer would create? e - The salt bridge is designed to permit anion exchange between two containers BUT, prevent mixing of their contents. The NO 3 - KNO 3 sol n NO 3 The salt bridge completes the electrical circuit, which permits the indicated reactions to occur in the two containers prevent mixing of their contents. This way we can measure the potential between the two metal electrodes by inserting a voltmeter. 14 The ph Meter in the previous slide represents a voltmeter The voltmeter we use in the exercise is the ph meter an ideal device for the task. The ph meter has an exceptionally high resistance (10 100 Mohms) so very little current flows when a potential is applied to it. Therefore, very little reaction occurs during a potential measurement the concentrations in the cells do not change much during a measurement, and the potential does not change much during a measurement 15 = X = M = 1.0 M 1.0 M M (NO X 3 ) 2 Electrochemical Cell Representation B 0.5 M B M NO BY 3 (NO X M 3 ) 2 Molar 1.0 M BY NO M B Molar 3 0.5 M B B = BY = NO 3 M B = 0.5 M 16 X M We measure the potentials of a series of cells in which the concentrations of 2+ and + are varied. (s) + 2 + (aq) NO 3 Y M 2 (s) + 2+ (aq) 2+ (X M) + (Y M) The potential should follow: [ 2+ ] E = 0.462-0.0295 log -------- [ + ] 2 X=[ 2+ ] Y = [ + ] [ 2+ ]/[ + ] 2 E 1 1 1 0.462 v 0.1 1 0.1 0.462 + 1 X 0.0295 = 0.492 0.1 0.1 10 0.462 1 X 0.0295 = 0.433 0.1 0.01 1000 0.462 3 X 0.0295 = 0.374 0.1 1.0 X 10 5 1.0 X 10 9 0.462 9 X 0.0295 = 0.197 0.1 4.67 X 10 9 4.59 X 10 15 0.462 15.66 X 0.0295 = 0 0.1 1.0 X 10 10 1.0 X 10 19 0.462 19 X 0.0295 = 0.099 In the Nernst equation, we prefer to use log 10 instead of ln. 2.303 RT/F log 10 x = RT/F ln x :t 298 K (77 o F), 2.303 RT/F = 0.0592 17 n+, C E = E B E Summary of Electrochemical Cells E e B B n+, C B + B n+ (C B ) n+ (C ) + B n+ (C ) B n+ (C B ) B node Oxidation n+ + ne - = E 0 B E 0 RT/nF ln ( C / C B ) Cathode Reduction B n+ + ne - B E = E 0 RT/nF ln (1/C ) E B = E 0 B RT/nF ln (1/C B ) G = nf E If the reaction proceeds forward as written, the potential of B will be positive relative to. * For simplicity, we choose the number of electrons to be the same in both half reactions 18 2

Must the Two Half Cells Involve Different Substances? 0.01 C M NO 3 K eq = 1 0.10 C B M NO 3 t equilibrium, + would be the same in both half-cells (s) + + (C ) (s) + + (C B ) + (0.01) + (0.1) E 0 = E 0 B so, E 0 = 0 v G 0 = n F E 0 = 0 RT C B E = 0 ------ ln ------ F C 19 +, C E = E B E Summary - Concentration Cell ( B ) E 0 B = E 0 The anode (where oxidation occurs) will be the electrode in the more dilute solution. + n+ (C B ) n+ (C ) + n+ (C ) n+ (C B ) +, C B n+ (C ) + ne - n+ (C B ) + ne- E = E 0 RT/nF ln (1/C ) E B = E 0 B RT/nF ln (1 /C B ) E = RT/nF ln (C /C B ) If C > C B, E < 0 C = C B, E = 0 C < C B, E > 0 20 Concentration Cells Which Cell is Positive? It is helpful to remember that: in a concentration cell, electrons will always flow in the direction that will tend to equalize the concentrations in the two half cells. I.e., electrons will flow in the direction that will cause a decrease in the concentration of the more concentrated cell, or what is equivalent, an increase in the less concentrated cell. e + + e NO 3 0.0 NO 3 0.10 M + + e 21 Exercise is in 3 parts: 1. Concentration dependence of the Electrochemical Potential - confirming the Nernst equation for concentration cells 2. Determining the Solubility Product of an Unknown Silver Halide (Halide = Cl -, Br -, I - ) - measuring low concentrations electrochemically 3. Determining Standard Reduction Potentials and the relative oxidation/reduction capabilities of some metals 22 Part 1: Working in groups of two, measure potentials of a series of concentration cells based on the half reaction: + (aq) + e - (s) + (0.010 M) + (XM) 0.0 NO 3 Ref The reference electrode throughout this part of the exercise will be the one with [ + ] = 0.010 M. X M NO 3 The salt bridge is filter paper moistened with NH 4 NO 3 You are asked to plot measured potential vs log 10 [ + ] Silver nitrate solutions at concentrations 10-5, 10-4, 10-3, and 10-2 M will be available. Take only what you need! 23 10 ml Beakers Filter Paper 24 The Reference Electrode! The Concentration Cell spreadsheet Reference Electrode Test Electrode Identifying the Reference Electrode in electrical measurements is important but straightforward. The Reference Electrode has no chemical significance. It refers only to the Voltmeter. ll potential measurements are relative. The Reference Electrode is the voltmeter terminal which the Voltmeter assumes is at 0 potential. The ph meter can measure both positive and negative potentials. The sign of the measured potential will always be relative to the electrode connected to the Reference Terminal. If in doubt, measure the potential of a battery with known polarity. 25 26 27 3

If you have conducted this part of the exercise correctly, the plot of E vs log [ + ] should produce a straight line which provides qualitative confirmation of the Nernst equation for concentration cells i.e., the measured potential is linearly related to the log of the concentration, and quantitative confirmation of the Nernst equation for concentration cells i.e., the slope of the straight line has the expected value. What slope do you expect? On what does the slope depend? 28 Part 2: You have seen that the logarithmic nature of the Nernst relationship means that cell potentials are a way of measuring very small concentrations e.g., In part 2, WORKING LONE, you will prepare an assigned, but unknown silver halide and measure its solubility product electrochemically. For sparingly soluble salts, a common measure of solubility* is the solubility product. It is the equilibrium constant for the solubility reaction. M p X q (s) p M m+ (aq) + q X n- (aq) K sp = [M m+ ] p [X n- ] q Ca X (s) + (aq) + X - (aq) [ + ] [ X - 3 (PO 4 ) 2 (s) 3 Ca 2+ (aq) + 2 PO 3-4 (aq) K sp = [Ca 2+ ] 3 [PO 3-4 ]] 2 * Equilibrium solubility = 2.0 X 10-29 29 Silver halides (Cl, Br and I) are examples of slightly soluble substances. Each student prepares a sample of ONE unknown silver halide by reaction of a measured volume of a solution of potassium halide, KX, of known concentration with a measured volume of 0.010 M silver nitrate. + (aq) + X - (aq) X (s) K sp = [ + ] [ X - ] Suppose KX is 0.050 M + is clearly the limiting reagent Before Reaction Final + 7.00 ml X 0.010 M = 0.070 mmol 0.070 0.070 = ~0 mol X 8.00 ml X 0.050 M = 0.40 mmol 0.400 0.070 = 0.330 mmol X (s) 0 mmol 0.070 mmol The final volume is 15.00 ml, so the final concentration of X - is 0.330 mmol/15.00 ml = 0.0220 M 30 0.010 M NO 3 How can we measure the + concentration? We measure + concentration by letting this solution be a half cell with the / + 0.010 M reference cell. From Part 1 Suppose the measured potential is -0.620 V E Z? M + E = RT/F ln (0.010 / Z?) - 0.620 = 0.0592 log (0.010 / Z?) log (0.010 / Z?) = 0.620 / 0.0592 = 10.5 (0.010 / Z?) = 10 10.5 = 2.95 X 10 10 (Z? / 0.010) = 10-10.5 = 1 /2.95 X 10 10 Z? = 0.010 / 2.95 X 10 10 = 3.4 X 10-13 31 So, the measured [ + ] is 3.4 X 10-13 M The value of the solubility product K sp = [ + ] [ X - ] is the product of the measured + concentration and the calculated X - concentration, namely 0.022 M, K sp = 3.4 X 10-13 X 0.022 = 7.5 X 10-15 Cell potentials are a way of measuring equilibrium constants, in which very small concentrations are involved. 32 In part 3, working in pairs, you will determine the standard reduction potentials of zinc and copper and establish the relative oxidation/reduction potentials of Zn, and Pb. For this part, you will assume that the reduction potential for lead is known and is - 0.13 V. Zn Zn 2+ (0.050 M) Pb 2+ (0.050 M) Pb Pb Pb 2+ (0.050 M) 2+ (0.050 M) Zn Zn 2+ (0.050 M) 2+ (0.050 M) 33 Ni +1.25 V Pd The measured potential at the Pd electrode is positive with respect to the Ni electrode. Ni Ni 2+ 0.060 M Pd 2+ 0.060 M Pd lligator Clips In this part, we replace the silver electrodes from Parts 1 & 2 with alligator clips that are able to make good contact with metal strips The logic of Part 3 of the exercise is demonstrated in the following example where the three metals are: Nickel (Ni), Palladium (Pd) and Cadmium (Cd) instead of Zinc (Zn), Copper () & Lead (Pb). Concentrations are equal 0.060 M 0.060 M Both Ni (s) half reactions + Pd 2+ (aq) involve Ni 2+ (aq) 2 electrons + Pd (s) Ni(NO 3) 2 Pd(NO 3) 2 The measured potential is positive so the Ni must, in fact, be oxidized RT [Ni 2+ ] E = E 0 (Pd/Pd 2+ ) E 0 (Ni/Ni 2+ ) ln 2F [Pd 2+ ] E = E 0 (Pd/Pd 2+ ) E 0 (Ni/Ni 2+ ) = +1.25 V [Ni 2+ ] Q = [Pd 2+ ] 34 35 36 4

Cd +1.39 V Cd Cd 2+ 0.060 M Pd 2+ 0.060 M Pd Cd (s) + Pd 2+ (aq) Cd 2+ (aq) + Pd (s) 0.060 M 0.060 M Cd(NO 3) 2 Pd(NO 3) 2 The measured potential is positive so the Cd must, in fact, be oxidized RT [Cd 2+ ] E = E 0 (Pd/Pd 2+ ) E 0 (Cd/Cd 2+ ) ln 2F [Pd 2+ ] E = E 0 (Pd/Pd 2+ ) E 0 (Cd/Cd 2+ ) = + 1.39 V Pd [Cd 2+ ] Q = [Pd 2+ ] Cd 0.060 M Cd(NO 3) 2 +0.15 V 0.060 M Ni(NO 3) 2 Cd Cd 2+ 0.060 M Ni 2+ 0.060 M Ni Cd (s) + Ni 2+ (aq) Cd 2+ (aq) + Ni (s) RT [Cd 2+ ] E = E 0 (Ni/Ni 2+ ) E 0 (Cd/Cd 2+ ) ln 2F [Ni 2+ ] E = E 0 (Ni/Ni 2+ ) E 0 (Cd/Cd 2+ ) = 0.15 V Ni The measured potential is positive so the Cd must, in fact, be oxidized Summary: E 0 (Pd/Pd 2+ ) E 0 (Ni/Ni 2+ ) = 1.25 V E 0 (Pd/Pd 2+ ) E 0 (Cd/Cd 2+ ) = 1.39 V E 0 (Ni/Ni 2+ ) E 0 (Cd/Cd 2+ ) = 0.15 V We appear to have 3 equations in 3 unknowns. Can we solve for all three unknowns? However, if we knew E 0 of Pd (relative to the standard hydrogen electrode), we could determine the actual values of the other two. The E 0 for Pd/Pd 2+ = + 0.99 V 0.99 E 0 (Ni/Ni 2+ ) = 1.25 V 0.99 - E 0 (Cd/Cd 2+ ) = 1.39 V E 0 (Ni/Ni 2+ ) = 0.26 V E 0 (Cd/Cd 2+ ) = 0.40 V 37 38 39 The The salt bridges we use throughout this exercise are a strip of filter paper wet with an ammonium nitrate solution. While it is an adequate substitute for our purposes, unlike more elaborate glass bridges, it has limitations. If left in contact with solutions for an excessive amount of time, it will permit some mixing of the solutions it is designed to keep separate. Be sure to use a fresh salt bridge each time you change either of the solutions that it connects. 40 Be sure to dispose of all metals and solutions in the waste containers provided. Used salt bridges must not be left in the sinks! RFS 4/14/2016 41 UGT POSITIONS VILBLE FOR Fall 2016 Chemistry Learning Center (CLC) for 129, 131, 132 & CHE 130 Instructors QULIFICTIONS: / in CHE 129/132 or 131/132, CHE 152 or equivalent mulative & previous semester GP 3.00 Willingness to help students succeed COMMITMENT: Spend 4 hours each week working with students ttend a weekly staff meeting for 1 1.5 hours Write a brief paper on your experience REWRDS: 3 credits for CHE 475 or 476, Undergraduate Teaching Practicum Review for GRE and MCT, recommendations TO SIGN UP: Complete a UG Teaching Practicum Permission Request Form available in the Chemistry Main Office Place completed form in faculty mailbox as below by Friday, pril 24 th Faculty contacts: CLC: Dr. Tooker, Chem. 470, Bradford.Tooker@stonybrook.edu 130: Dr. Wolfskill, Chem. 575, Troy.Wolfskill@stonybrook.edu 5