Electrochemistry Lab Handout
Electrochemistry Lab Handout Faris AspenArin Name:________________________ Background: An electrochemical cell is a device that either generates electrical energy from chemical reactions, a galvanic cell, or utilizes electrical energy to create chemical reactions, an electrolytic cell. Regardless of the type, all electrochemical cells utilize oxidation-reduction (redox) chemical reactions. In a redox reaction, both oxidation and reduction occur simultaneously. Oxidation is when an atom loses electrons while reduction is gaining electrons. Helpful acronyms: ● “LEO” the lion says “GER” ○ Lose Electrons = Oxidized… Gain Electrons = Reduced. ● OIL RIG ○ Oxidation Is Loss… Reduction Is Gain. Y IT END In many redox reactions, there is a complete transfer of electrons from the substance being oxidized to the substance being reduced. In such an instance, there are two half-reactions, an oxidation half-reaction and a reduction half-reaction. When these electrons travel through a conductor, such as a wire, electric current can be harnessed and measured in volts by a device called a voltmeter. Most often, the half-reaction takes place in different containers called half-cells. In order to complete the electrical circuit, ions (atoms or molecules with a net electrical charge) must be free to travel from one half-cell to another. This is accomplished by connecting the half-cells via a salt-bridge. A salt-bridge, which typically consists of fa neutral electrolytic solution such as potassium chloride or sodium sulfate, allows ions to migrate from one cell to another in order to maintain the cells’ charge balance. In each half-cell a conductor (a substance allowing the flow of electricity), known as an electrode, is needed to establish the electric current. In the oxidation half-cell, the electrode is called the anode. The anode will have an overall negative charge because electrons are being produced. In the reduction half-cell, the electrode is called the cathode. The cathode will have an overall positive charge because electrons are being consumed, leaving behind positive metal cations. The electrons travel from the anode (negative) to the cathode (positive) because the negative charge is attracted to positive charge. The current generated by a thermodynamically favorable, spontaneous electrochemical cell is the electromotive force, or Ecell. Current generated is measured in volts and compared relative to the standard hydrogen half-cell at standard conditions. For an electrochemical cell, standard conditions are 1.0 atm for gasses, 1.0M for solutions, and 298 K. When voltage is measured at standard conditions, it is indicated using the naught symbol (o), as follows: Eocell. The electrochemical potential can be calculated in three steps utilizing a table of standard reduction potentials and the following equation: Eocell = Eooxidation + Eoreduction. Step 1: Decide which element is being oxidized (losing electrons) and which is being reduced (gaining electrons). The more positive on the table, the easier that element is to reduce, i.e., the easier it is to gain electrons. Whichever element is more positive, this is the reducing agent. Step 2: For the reaction that you have identified as being oxidized, reverse both the sign of the Eo and the equation. Step 3: Substitute your new values into the equation. For example, look at the two half-reactions from the table of Standard Reduction Potentials. Ag+ (aq) + e- → Ag(s) Eo = 0.80V (reduced) Ni2+ (aq) + 2e- → Ni(s) Eo = -0.26V (oxidized) Ni(s) → Ni2+ (aq) + 2 Eo = 0.26V (oxidized) ← reversed the sign and equation EoCell = Eooxidation + Eoreduction = +0.26V + 0.80V = 1.06V Pre-Lab Questions: Using the supplied values from the standard reduction potential bale, answer the following questions: Au3+ (aq) + 3e- → Au(s) Eo = 1.50V Ag+ (aq) + e- → Ag(s) Eo = 0.80V Cu2+ (aq) + 2e- → Cu(s) Eo = 0.34V 2+ Pb (aq) + 2e → Pb(s) Eo = -0.13V Sn2+ (aq) + 2e- → Sn(s) Eo = -0.14V Co2+ (aq) + 2e- → Co(s) Eo = -0.28V 2+ Fe (aq) + 2e → Fe(s) Eo = -0.44V Zn2+ (aq) + 2e- → Zn(s) Eo = -0.76 IT Ag a. The silver (I) and Sh tin (II) electrodes are connected to form a battery. i. Identify which metal is the anode. Use the table of standard reduction potentials to justify your answer. Tin is the anode because it is in the negative ii. Identify which metal is the cathode. Use the table of standard reduction potentials to justify your answer. silver is the cathode positive because numbere it is b. Calculate the expected electrochemical potential of the silver (I) and tin (II) cell. 0.94 V c. Which way are the electrons going to travel? They always go negative to positive meaning it will travel from tin to silver d. Identify the oxidation half-reaction and the reduction half-reaction. Show all work. Efftite snot aq 2 Ages gnes n5 Agtfaq e x2 sureCag tze x2 9 A9 28ns Ages 2 2 aq Line notation of a cell is a shorthand version that has the following structure: Anode metal (s) | anode ion (aq) || cathode ion (aq) | cathode metal (s) a. Utilizing the following line notation, explain what is happening inside the cell. Include which metal is the cathode, which is the anode, what is being oxidized, what is being reduced, and the direction of the electron flow. i. Fe | Fe3+ || Cu2+ | Cu Fe is anodemetal Festanode ion and ca is the cathode metal Cat is thecathode ion INSPIREvalue. o cell b. Calculate the E Iron is the tone being oxidized the copper is a the one being reduced Knee copper has positive and iron has a negative voltage it moves voltage withe the electron FE Ca 0.44 0.34 while flow from 0.78 V iron to copper Investigation: How are galvanic cells constructed, and what are their properties? on Procedure: 1. Observe the three metals and metallic solutions supplied by your instructor. a. Copper (Cu) and Copper Sulfate. b. Magnesium (Mg) and Magnesium Sulfate. c. Zinc (Zn) and Zinc Sulfate. 2. Using the table of standard reduction potentials from the pre-lab section, calculate the target (expected) voltages between each of these three metals. a. Copper // Magnesium 0.34 2.37 0.34 0.76 2.71 V 1.1 v 0.76 3.13 b. Copper // Zinc c. Magnesium // Zinc 2 37 3. Check with your instructor to ensure that you correctly calculated the expected voltages of each of the reactions and record them in the data table at the end of this lab. 4. Obtain the metals. Using sandpaper, polish the surface of the metals until all of the tarnished coating has been removed (it should be shiny!). If your metals are too brittle, let your instructor know and they will supply you with a fresh metal piece. 5. Create your salt bridge. a. Obtain a piece of filter paper. b. Using the sodium sulfate solution in the dropper bottle, saturate the filter paper with the solution. 6. Create your electrochemical cell. a a. Obtain the voltmeter with appropriate connection probes and two wires with alligator clips, one red and one black. b. Connect the red wire to the middle of the three inserts and the black wire to the bottom of the three inserts in the voltmeter. This will measure the current that the electrochemical cell is producing. i. Note: the voltmeter is not a battery. It is not producing electricity, but is a means to measure the electrical current. c. Turn on the voltmeter to 20 DCV. d. Start with Copper and Zinc. e. Determine which metal will be oxidized and which will be reduced. i. Oxidized = _____________ Reduced = _____________ f. Attach the metal that is being reduced to the red alligator clip. g. Attach the metal that is being oxidized to the black alligator clip. Stop and Think: Why is the reducing metal being attached to the red alligator clip? ● Hint: look at the example electrochemical cell on the first page. h. Place the reducing metal (with attached wire) in the corresponding metallic solution. i. I.e., copper metal into copper sulfate solution, zinc metal into zinc sulfate solution. i. Place the oxidizing metal (with attached wire) into the corresponding metallic solution. j. Place one end of the salt bridge (saturated filter paper) in one of the solution jars and the other end of the bridge into the other metallic jar. Stop and Think: What is a salt bridge and why is it necessary when constructing your electrochemical cell? ● Hint: look at the example electrochemical cell on the first page. What would happen to the voltage if we removed the salt bridge? Why? Test this and record your results. k. Leave the electrodes in the solution until the voltage is steady. l. Record the voltage in your data table under the “Observed Voltage” column. m. Remove the metals from their metallic solutions, wipe them off with paper towels, and return the metals to their original plastic tubes. n. Repeat steps 6e – 6m for Copper and Magnesium, making sure to create a new salt bridge for this electrochemical cell. o. Repeat steps 6e – 6m for Magnesium and Zinc, making sure to create a new salt bridge for this electrochemical cell. 7. Clean up the workstation as directed by the laminated cards on your lab bench. Analysis Questions: 1. How did your experimental voltage compare to your theoretical, calculated voltage? 2. Calculate your percent error between your observed voltage and your theoretical voltage. Briefly describe factors that could have led to your percent error. 3. The investigation for this lab is, “How are electrochemical cells constructed and what are their properties?” Based on evidence gathered from this lab, develop an explanation that addresses this question. 4. How can you see electrochemical cells or galvanic cells be applied to a field or a career that interests you. Research potential ideas! Table 1. Electrochemical Cell Operations Electrochemical Cell Calculated Voltage Copper // Zinc Copper// Magnesium Magnesium // Zinc Percent Error = 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 − 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑂𝑏𝑠𝑒𝑟𝑣𝑒𝑑 x 100 Observed Voltage Percent Error*
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