Lab 9 CHM 116 Electroplating of Metals PDF

Title Lab 9 CHM 116 Electroplating of Metals
Author Jordan Dehart
Course General Chemistry II
Institution Arizona State University
Pages 15
File Size 461 KB
File Type PDF
Total Downloads 44
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Lab 9 chem 116 Electroplating of Metals lab 9...


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Name: Jaziah Gomez

Date:4/9/2021

Lab Partners: Analy Granados Electrochemistry Laboratory – Electroplating of Metals By Thomas Cahill Background: Electroplating is a common means to deposit a thin layer of some expensive metal like gold or silver onto an item made from a cheaper metal such as steel. The advantage is to impart the desired characteristics of the surface metal, such as aesthetics and corrosion resistance, without making the entire item from the expensive metal. Electroplating is commonly conducted on silverware where a stainless steel base is coated with a thin layer of silver to give it a desired appearance. Electroplating is also commonly used for electrical connections where gold is plated onto the connectors to prevent corrosion and to improve electrical conduction. Electrochemistry is abundant in metallurgy, but that use of electrochemistry is not often appreciated. Many of the common elements are produced from electrolysis of salts and solutions. Aluminum metal is probably the greatest use of electrolysis, and the production of aluminum metal used to account for about 6 to 8% of the total electricity used in the United States. Electroplating is also commonly used as a purification mechanism for some of the more exotic metals such as gold. In this case, the impure gold mixture is dissolved in a cyanide solution that is selective for dissolving gold. The solution is then transferred to an electrochemical cell to electroplate the pure metal out of solution. Your laboratory exercise is basically identical to this last application of electrochemistry. You will recover pure metals from salt solutions. Most of your salts will be metal nitrate or metal acetate salts such as silver nitrate & copper acetate. You will not get the ideal thin, uniform layers of metals that are the typical image of electroplating, but you will get metal crystals and aggregates that you will collect, dry and quantify. Processes involved: Electrochemistry involves the use of direct current (DC) electricity, such as that from a battery, to precipitate a pure metal. Normal power outlets in your house are alternating current (AC) and cannot perform this type of process. The basic process is that two electrodes are placed in a salt solution containing metal cations. The electrodes are attached to a DC power supply that makes one of the electrodes negatively charged and the other electrode positively charged. The metal cations are attracted to the negatively-charged electrode (called the cathode). When the ions reach the cathode, they acquire enough electrons to make them a neutral atom. Once the metal has a neutral charge, it will precipitate on the cathode, which gives the pure metal that we want. Reactions: The electroplating that you are conducting in lab is the result of two redox reactions. The metal ion is being reduced at the cathode to form metal, so the metal ion is gaining electrons. However, to have a complete circuit, we need to oxidize something at the anode to give up electrons. Since we have a pure water + metal salt system, water is oxidized at the anode to form oxygen. Therefore, the two half-reactions are as follows (using silver nitrate example): Cathode: Anode:

Ag+(aq) + e− 2H2O(l)

 Ag(s)  O2(g) + 4H+(aq) + 4 e−

Ered = +0.799 V Ered = +1.23 V Ered = −0.43 V 1

Since the net energy is negative, this reduction is not spontaneous, thus a voltage must be applied to make this reaction proceed. This voltage is supplied by the battery. The end result of the reactions is Ag(s) and H+(aq), which means that you end up with a rather acidic solution at the end of the reaction, so handle it with care. Battery capacity: The easiest means to determine the number of electrons that pass through the electroplating cell is to use the rated charge capacity of the battery and then allow the battery to completely discharge before collecting the metal. The charge capacity of a battery is generally given as milli-ampere hours (mAh). The units of milli-ampere hours tells you that the battery has a charge equivalent to the stated current (in milli-amperes) for one hour. For example, AA batteries typically have a capacity of 1800 mAh, which means they have a charge equivalent to 1800 mA for a single hour. Since an ampere is a coulomb per second (A= C/s), the coulombs in the battery can be determined (C = A·s) by converting the milli-amperes into amperes by dividing by 1000 and by multiplying by 3600 sec/hr. To calculate the number of electrons that passed through the cell, we divide the coulombs by Faraday’s constant (1 F = 96,485 C/mol e−) Mole e− = (mAh)(1A/1000mA)(3600s/1h)(1/F) For example, the charge capacity of an 1800 mAh battery is: (1800 mAh)(1A/1000mA)(3600s/1h)(1/96485C/mol e−) = 0.0671 mole electrons You will use this calculation in your experiment since your batteries will be completely discharged at the end of the experiment. The rated battery capacity is the total charge the battery can give up. Unfortunately, only about 85% of this charge is actually useable in our experiments. Therefore, you will need to multiply the rated battery capacity by 0.85 to find the useable charge in this experiment. As the battery gets weaker, the voltage drops to the point where the battery cannot drive the reaction, so the reaction stops. This is a combination of both the battery getting weaker and the “back voltage” of the electroplating cell increasing as you reduce the metal salts in solution. You can measure this back voltage at the electrodes when the battery has been disconnected. Multiple Batteries in Series: You will use multiple batteries in series (two or more batteries end to end) in some of your experiments. When batteries of the same type are put in series, the total voltage will double but the total electron capacity will remain the same. You will have the same number of electrons, but each electron will have twice the energy. Therefore, you will use the capacity (mAh) from a single battery to calculate the moles of electrons. Moles of Metal Condensed: Once the moles of electrons that passed through the cell are known, then it is a simple matter to determine the moles of the metal condensed. The moles of metal condensed: Moles metal = (moles electrons) (charge of metal ion)

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Therefore, it takes more electrons to condense a +2 ion element (e.g. Cu 2+) compared to a +1 element (e.g. Ag+). The mass of metal condensed (in grams) is the moles of metal condensed multiplied by the molar mass of the element. For example, the amount of Cu2+ condensed by an 1800 mAh AA battery is: Coulombs: 1800mAh × (A/1000mA) × (3600 sec/hr) = 6480 C Mole e− = Coulombs/F = (6480C)/(96,485C/mol e−) = 0.0672 mol e− Grams Cu = 0.0672 mole e− × 63.5 g/mol Cu = 2.13 g copper metal 2 Part A: Laboratory Experiment Quantitative recovery of pure metals from metal solutions. You will be working in groups of 4 or 5 students for this part of the lab. Materials: (per single setup) AAA rechargeable batteries (number varies depending on solution) Two stainless steel spatulas as electrodes Short (20 to 30 cm) length of flexible wire with bare ends Wooden electrode holder Battery holder (you can use tape, but it is not preferred because it leaves a sticky residue) Volt meter (with alligator clip connections if possible) One 250 mL beaker 200 mL of a metal salt solution (silver or copper) Graduated cylinder (to measure the volume of the solution) Three small glass vial for the final product (two for copper, one for silver) Sticky labels or tape to identify your electroplating setups.

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Diagram of Experimental Apparatus (using the silver nitrate solution as an example)

You will be testing three solutions: #1 - 200 mL of 0.080 M copper(II) acetate with two AA batteries in series. #2 - 200 mL of 0.175 M copper acetate with three AAA batteries in series. #3 - 200 mL of 0.05 M silver nitrate with one AAA battery. PROCEDURE: WEEK ONE There will be three experimental setups for each group. The first two are for copper and the third is for silver. The procedure described below describes a single experimental setup. 1) Add 200 mL of the metal solution (AgNO 3, or Cu(C2H3O2)2) in the 250 mL beaker. Use a graduated cylinder to measure the volume of the solution. Do not rely on the gradations on the beaker; they are often highly inaccurate! Record the volume of the solution and the molarity of the solution on your data sheet. Silver nitrate stains skin and other materials very easily, so wear gloves when handling the silver nitrate and treat it carefully. Clean up any spills (even a few drops) quickly. 2) Label each beaker with the correct solution concentration, place the beakers back in the plastic prep bin and assemble the rest of the equipment. The reactions will be stored in these containers until week two, so label the container with your group member names and lab section day/time. 3) Put the stainless steel spatulas through each of the holes in the wooden electrode holder to act as the electroplating electrodes. 4) Put the wooden electrode holder on top of the beaker. Keep the electrodes as upright as possible. The electrodes can rest on the bottom of the beaker since glass is a good insulator. 5) Insert the batteries into the respective battery holder. The copper experiment #1 (0.08M) requires two AA batteries in series (end-to-end, see example in lab); the copper experiment #2 (0.175M) requires three AAA batteries in series and the silver experiment will use one AAA battery. Some of the batteries have different capacities, so make sure that all batteries used in series have the same capacity (mAh). 4

6) Record the battery charge capacity on your data sheets. This value is often expressed as milliampere hours (mAh). This value will be either written on the battery or your lab instructor will tell you this value. The batteries used in this experiment have a usable charge of about 85% before the voltage drops below the voltage necessary to drive the electroplating. Therefore the “useable battery charge” is just the reported charge multiplied by 0.85. 7) Make sure all the contacts are secure to get the full current. Put a rubber band around the holder lengthwise to make sure the metal screen is in contact with the + end of the battery. 8) Attach the two wires from the battery holder to the two electrodes. Try to keep the electrodes vertical so the separation distance is constant. 9) Measure the voltage at the two electrodes. Your laboratory instructor will demonstrate how to use the voltmeter. The voltage is measured by connecting the alligator clips from the meter to each of the electrodes going into the solution at the same time. Be sure to measure the voltage on the electrodes and not the battery or wires. The meter needs to be set on “DCV” (direct current voltage). Check to make sure you are getting the correct voltage reading for each experiment based on the following ranges: 1.3 to 1.5 V for one battery 2.5 to 2.9 V for two batteries in series 3.8 to 4.0 V for three batteries in series 10) If you are not getting a value in the correct range, then check your electrical connections. Record the voltage on the data sheets. The voltage measurements are not used in any calculations, but they simply test that all the connections are good and the proper voltage is reaching the electrodes. 11) The cations (positively charged ions) will be attracted to the negatively-charged cathode electrode (which is the electrode attached to negative end of the battery). When the cations reach the negatively charged cathode, they will gain an electron and condense on the electrode as the metal. If the formation of silver metal occurs slowly, as is done in this experiment, shiny silver crystals will form over the course of 36 to 48 hours. 12) If everything is working properly, you should see tiny ( 0.031715 moles of Ag capable by the battery. b) What was your product yield (%) using the limiting resource? 0.03715 moles Ag  107.87 g Ag = 4.00737 g of Ag 3.24 / 4.00737 = 0.80851  100 = 80.851% ______ of 5 pts Total ______ of 25 pts

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