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Experiment 7: Electrochemistry Experiment 7: Electrochemistry 89 Electrochemistry Introduction Objectives • To work with and understand voltaic half-cells. • To determine the Nernst equation experimentally. Background Electrochemistry is the study of the movement of electrons by chemical processes. This branch of chemistry is found in everyday life from your jewelry to your laptop. However, the most common application is the voltaic cell. Voltaic cells produce electrical energy (voltage) via a spontaneous oxidation-reduction reaction. The primary measurement in electrochemistry is the voltage (V) of an electrochemical cell. The voltage describes the relative energies of electrons on different atoms and/or ions. The energy difference, or potential difference, between two electrons is measured in volts (joules/coulomb). These cells are made up of two half-cells. Half-cells are produced with a metal electrode placed into a solution containing a cation of the metal and attached to a voltameter. A porous barrier (a salt bridge) normally separates the two half-reactions. Here, the salt bridge will be several drops of aqueous NaNO3 placed on the filter paper between the two half cells. This experiment implements a micro-scale version of voltaic cells. The (+) lead of the voltmeter connects with one metal and the (–) lead with another. If a positive voltage is recorded on the screen, you have connected the cell correctly. The metal attached to the (+) lead is the cathode (reduction) and has a higher, more positive, reduction potential. 90 The metal attached to the (–) lead is the anode (oxidation) and has the lower, more negative, reduction potential. If you get a negative voltage reading, then you must reverse the leads. By comparing the voltage values obtained for several pairs of half-cells and recording which metal made contact with the (+) and (–) leads, you can establish the reduction potential sequence for the five metals in this Lab. Thermodynamics can predict if electrons would prefer to be transferred from one species to another based on the free energy change of the system. The cell voltage is proportional to ΔG, the change in Gibbs free energy. The ΔG of the reaction could be harnessed to do useful electrical work. The value of ΔG is given by this thermodynamic equation. ∆𝐺 = ∆𝐺° + 𝑅𝑇 ln 𝑄 ΔG° is the change in Gibbs free energy for the process when all the reactants and products are in their standard states. Q is the reaction quotient for the reaction. Q has the same form as the equilibrium constant, K, but the concentrations and/or gas pressures in Q are the instantaneous values in the non-equilibrium reaction vessel, rather than a set of values that characterize equilibrium. The relationship between cell voltage, E, and ΔG for the cell reaction is given by the following equation. ∆𝐺° = −𝑛𝐹𝐸 The nF term is related to the coulombs of negative charge transferred in the balanced redox reaction. The number of moles of electrons transferred is given by n. F is Faraday’s constant, which is the total charge (in coulombs) of a mole of electrons. 𝐹 = 96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 𝑚𝑜𝑙𝑒 𝑒 − A process will spontaneously occur if there is a decrease in free energy, G, or ΔG < 0. This implies spontaneity is associated with a positive E. −𝑛𝐹𝐸 = −𝑛𝐹𝐸° + 𝑅𝑇 ln 𝑄 𝐸 = 𝐸° − 𝑅𝑇 ln 𝑄 𝑛𝐹 Rearranging some of the equations above gives us the Nernst equation. The Nernst equation includes a term that reflects the potential difference between products and reactants under standard conditions. 2.303𝑅𝑇 log10 𝑄 𝑛𝐹 This simplifies for the special case of a cell at 25°C. 𝐸 = 𝐸° − (0.0591 𝑣𝑜𝑙𝑡𝑠) log 𝑄 𝑛 E is the voltage measured for the cell. E° is the voltage measured when Q = 1. The value 0.0591 is the same for all cells at 25°C. E° is also related to the electron energy differences of the oxidation and reduction reaction. 𝐸 = 𝐸° − 𝐸°(𝑐𝑒𝑙𝑙) = 𝐸°1/2(𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛) − 𝐸°1/2(𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛) = 𝐸°1/2(𝑐𝑎𝑡ℎ𝑜𝑑𝑒) − 𝐸°1/2(𝑎𝑛𝑜𝑑𝑒) 91 E°1/2(reduction) is the standard reduction half potential of the reduction reaction, and E°1/2(oxidation) is the reduction half potential of the oxidation reaction. For a spontaneous reaction, the species with the most positive value of E°1/2 will be reduced and the other species will be oxidized. If the concentrations of reactants and products are those characteristics of chemical equilibrium, then it is not possible to get any useful electrical work out of the cell. E equals zero. Setting E = 0 in the Nernst equation results in the equation below. 𝐸° = (0.0591 𝑣𝑜𝑙𝑡𝑠) log 𝐾𝑒𝑞 𝑛 Here, Keq is the equilibrium constant for the reaction as written. Keq can be determined directly from a measured E° under standard conditions. Although this experiment does not deal with the above equation. 92 Electrochemistry Safety Always wear proper personal protective equipment (i.e., your goggles and lab coat). Always wear appropriate attire: long pants/skirt with no holes, shoes that completely cover your feet, and keep long hair tied back. Gloves must be worn while handling chemicals. However, gloves must not be worn in the hallways or when interacting with non-chemical entities (e.g., door handles, eyes, or laptops/cell phones). Remove gloves before exiting the lab unless transporting chemicals with the aid of an ungloved partner to open doors. Metal nitrates are oxidizers. Do not have sparks or open flames in the lab. Alert your GSA if there are any spills. Waste Disposal Dispose of all solutions into the oxidizer waste container (OX). Dispose of the used filter paper in the solid waste container. Disposable face masks can be thrown in the regular trash. DO NOT THROW ELECTRODES AWAY. Please return them CLEAN. 93 Electrochemistry Experimental Procedures Standard Reduction Potentials 𝐴𝑔+ + 𝑒 − → 𝐴𝑔 𝐶𝑢2+ + 2𝑒 − → 𝐶𝑢 𝑃𝑏 2+ + 2𝑒 − → 𝑃𝑏 𝑍𝑛2+ + 2𝑒 − → 𝑍𝑛 𝑁𝑖 2+ + 2𝑒 − → 𝑁𝑖 𝐸° = +0.80 𝐸° = +0.34 𝐸° = −0.13 𝐸° = −0.76 𝐸° = −0.23 To create the salt bridge 1. Fill a small beaker with 20-30 mL of 1.0 M NaNO3. 2. Take a strip of filter paper and fold it in half. 3. Carefully soak the filter paper in the NaNO3. 4. Gently blot the strip on a Chem-Wipe. You don’t want the concentrations of your solutions to be contaminated by solution from the salt bridge. Part 1: Determinations of E⁰ for voltaic cells Half Cells 𝐶𝑢(𝑠) |𝐶𝑢2+ (𝑎𝑞) 𝑃𝑏(𝑠) |𝑃𝑏 2+ (𝑎𝑞) 𝑍𝑛(𝑠) |𝑍𝑛2+ (𝑎𝑞) 𝑁𝑖(𝑠) |𝑁𝑖 2+ (𝑎𝑞) 𝐴𝑔(𝑠) |𝐴𝑔+ (𝑎𝑞) Cell Diagram Example Anode Salt Bridge Cathode 𝐶𝑢(𝑠) |𝐶𝑢2+ (𝑎𝑞) ‖𝐴𝑔+ (𝑎𝑞) |𝐴𝑔(𝑠) 1. Add 1-2 mL of each metal nitrate [Cu(NO3)2, Zn(NO3)2, Pb(NO3)2, Ag(NO3)2, and Ni(NO3)2] solutions to a different wells in a well plate. 2. Turn on the multimeter by turning the know to the left until it points at the 20 V minimum measurement. 3. Choose two half cells. Connect the alligator clip probes to the metal wires for each half cell. 4. Place each end of a created salt bridge in the well of the metal nitrate solutions for each of the chosen half cells. 5. Place the metal wires in their respective solutions. 6. With a positive voltage displayed, wait about five seconds to take a voltage reading, and record the value. Be sure that you correctly label your data as you collect it. Also record which metal is the positive/anode (+) terminal and which is negative/cathode (–). 94 7. Repeat steps 2-5 measuring the potential of the remaining half cell combinations. You should have a total of 10 measurements when you are done. 8. When you have finished, rinse each piece of metal with tap water. Dry it and return it to the correct container. Remove the filter paper from the well plate and discard it in the solid waste container provided. 9. Be sure to clean your well plate extremely well to avoid contamination. Part 2: Dependence of E on ion concentrations In a concentration cell, both electrodes are made of the same metal and the solutions in the cells are of the same ion as the electrodes. The difference between the cells lies in the metal ion concentration. In this case, the chemical reaction is given below. 𝑃𝑏(𝑠) + 𝑃𝑏 2+ (𝑐𝑜𝑛𝑐) ⇌ 𝑃𝑏(𝑠) + 𝑃𝑏 2+ (𝑑𝑖𝑙) E⁰ is zero for a concentration cell because the half reactions are identical in both directions. The Nernst equation becomes the following. 𝑃𝑏 2+ (𝑑𝑖𝑙) (0.0591 𝑣𝑜𝑙𝑡𝑠) log 2+ 𝑛 𝑃𝑏 (𝑐𝑜𝑛𝑐) The spontaneous direction for the reaction is from left to right when 𝑃𝑏 2+ (𝑐𝑜𝑛𝑐) (C1) is greater than 𝑃𝑏 2+ (𝑑𝑖𝑙) (C2). In cell 1, ions accept electrons from the metal, plate out on the electrode, and lower the concentration of the ions. In cell 2, atoms in the metal leave electrons behind and enter the solution, thus raising the ion concentration. The reaction occurs in such a direction as to equalize the concentrations of the two solutions. 𝐸=− Serial dilution Serial dilution involves the process of taking a sample and diluting it through a series of standard volumes. The concentration factor is the initial volume divided by the final solution volume. The dilution factor in a serial dilution can be determined either for an individual solution or can be calculated as a total dilution factor in the entire series. 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑠𝑎𝑚𝑝𝑙𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟(𝐷𝐹)𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 1 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟(𝐷𝐹) 𝑇𝑜𝑡𝑎𝑙 = 𝐷𝐹1 × 𝐷𝐹2 × 𝐷𝐹3 × … … × 𝐷𝐹𝑛 95 0.10 M Pb(NO3)2 1. Make four serial dilutions from 0.10 M Pb(NO3)2. You will only need to make 10 mL of each solution. Use a sample volume of 1 mL. 2. Label a paper towel to place your well plate on. You will need two wells of 0.10 M and one well for each of the four dilutions. 3. Transfer 1-2 mL of the solutions to your well plate. Be careful not to contaminate the adjacent wells. 4. Connect two of the wells with a salt bridge. 5. Attach two Pb wires to the alligator clips on the multimeter. 6. Place one of the wires in a well with 0.10 M Pb(NO3)2. Place the other wire in another well. 7. Wait about five seconds to take a voltage reading and record the value. Be sure that you correctly label your data as you collect it. 10. Repeat steps 4-7 measuring the potential of the remaining half cell combinations. You should have a total of 5 measurements when you are done. 11. Measure the temperature of the room. 12. Empty all the waste into the waste container labeled OX. 13. Be sure to clean your well plate extremely well to avoid contamination. 96 Answer Questions: Experiment: Electrochemistry 5 Question Need to be Answer 1. A voltaic cell is constructed with a 𝐶𝑢(𝑠) |𝐶𝑢2+ (𝑎𝑞)half-cell and a 𝐴𝑔(𝑠) |𝐴𝑔+ (𝑎𝑞) halfcell. If the observed potential of the cell, Ecell, is 0.40 V, calculate the ratio of Cu2+ to Ag+ present using the Nernst Equation. (10 pts) 2. Does the slope found from the graph in part 2 agree with the slope predicted by the Nernst equation? (5 pts) 3- Write Data & Results for each table (Tables are appropriately labeled (each correctly identified, appropriate title, and units are included. Required figures are also included.)( you have to write under each table, provide a onesentence description of what the graph is showing the reader) 4- write Analysis (Includes a separately labeled section with an explanation of results, trends, and what they mean.) 5 paragraphs 5- Complete the 2 tables, also need Sample calculations (You must show calculations for the theoretical potentials, percent difference, dilution concentrations for part 2, and for [log (C1/C2)] part 2.) Table1 Voltaic Cell Cu/Ag Cu/Ni Cu/Pb Cu/Zn Ag/Ni Ag/Pb Ag/Zn Ni/Pb Ni/Zn Pb/Zn Metal at Anode Cu Cu Pb Zn Ni Pb Zn Pb Zn Zn Metal at Cathode Ag Ni Cu Cu Ag Ag Ag Ni Ni Pb Experimental cell potential 0.36 0.22 0.52 0.75 0.10 0.87 1.14 0.76 1.4 0.37 Teoretical cell potential % difference Table2 C1 (M) 0.10 0.10 0.10 0.10 0.10 C2 (M) 0.10 Experimental cell potential 0.00 0.02 0.04 0.12 0.16 log (C1/C2) Theoretical cell potential % difference Electrochemistry Calculations Part 1: Determinations of E⁰ for voltaic cells Step 1: Calculate theoretical cell potential. 𝐸°(𝑐𝑒𝑙𝑙) = 𝐸°1/2(𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛) − 𝐸°1/2(𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛) = 𝐸°1/2(𝑐𝑎𝑡ℎ𝑜𝑑𝑒) − 𝐸°1/2(𝑎𝑛𝑜𝑑𝑒) Use the standard reduction potentials listed above and the equation to calculate all ten theoretical cell potentials. Step 2: Calculate the percent difference for each voltaic cell. |𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑒𝑙𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 − 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑐𝑒𝑙𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙| × 100 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑐𝑒𝑙𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 Part 2: Dependence of E on ion concentrations Step 1: Calculate the concentrations for the dilutions. (C1V1 = C2V2) For each dilution, take the concentration of the sample as C1, the volume taken from the sample (1 mL) as V1, and the volume of DI water added (9 mL) as V2. 𝑪 Step 2: Calculate 𝐥𝐨𝐠 𝑪𝟏. 𝟐 Step 3: Calculate theoretical cell potential. 𝑃𝑏 2+ (𝑑𝑖𝑙) (0.0591 𝑣𝑜𝑙𝑡𝑠) 𝐸=− log 2+ 𝑛 𝑃𝑏 (𝑐𝑜𝑛𝑐) 𝐶 Use the log 𝐶1 calculated above and the equation to calculate all five theoretical cell 2 potentials. Step 4: Calculate the percent difference for each voltaic cell. |𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑒𝑙𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 − 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑐𝑒𝑙𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙| × 100 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑐𝑒𝑙𝑙 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 102
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