We will use the Faraday's law
c. In a separate electroplating experiment, if a 5.00 A current was used to electroplate a...
Electroplating Postlab Questions 1. Suppose that you have the opportunity to be the first in the world to come up witha your data for the number of Coulombs of charge passed and the number of moles of value of Faraday's constant -never mind we al ready know it is 96485 C/mol e. Using copper actually lost from the copper electrode in your experiment, what would you predict for the value of Faradays constant? Show your work. What is the percent...
Draw a labelled diagram of the apparatus that includes the anode, the cathode, the electrolyte, the direction of ion flow and the direction of electron flow provided by the power source. 28. Answer (4 Marks) Cu 29. Write and identify the half-reactions and net reaction. Answer (3 Marks) Cathode Half-Reaction Anode Half-Reaction Net Reaction 30. Calculate the theoretical potential difference of this cell. Answer (2 Mark) Chemistry so Unit C Module 6 S Besides deposition of Cu(s), what other empirical...
Part A and B Please show all of the steps Analysis of Electroplating When electricity (the flow of electrons) is passed through a solution, it causes an oxidation-reduction (redox) reaction to occur. If the solution contains a metal cation such as Ag+, the flow of electrons will reduce the silver ion, causing solid silver to plate onto the electrode. The amount of metal plated depends on the number of electrons passed. The total charge of a mole of electrons is...
2. Faraday's Experiment. Because of some AV, a current of 1 amp flows through a silver solution, and after a minute, 0.06709 grams of silver are deposited on the electrodes. a. How many electrons flow in one minute? b. From this experiment, what's the mass of a Ag atom in grams? c. Determine the molar mass of Ag (i.e. g/mole)
Suppose the current flowing from a battery is used to electroplate an object with silver. Calculate the mass of silver that would be deposited by a battery that delivers 1.60 A·hr of charge. answer in years.
E.C. A constant current was passed through a solution of Cut2 for 5.00 min. The cathode increased in mass by 1.24 g. How many amperes of current was used? smin 1.245 X 13008ccords ICELAIS 300 seconds 6.5 0195 moles & 63,59 X 2.506x10 1.24 16,510 Current: 1.04A 1
Au ++ 3e → Au The picture above shows a set-up that could be used for electroplating. When the Olympic Games were held in Atlanta, the United States provided gold medals for the athletes. Calculate how many grams of gold would be plated out on a coin, if a current of 3.2 (Amp) is applied for 27 min to an electrolytic cell shown above. The following reaction is occurring: Au3+ + 3 e → Au MW for gold equals 1979...
1.) In Faraday’s electrolysis experiment, he used approximately 96,500 C of charge. A) How many moles of electrons is that? [This also explains where the unit of moles disappeared during the example in class.] B) We are going to use Faraday’s law of electrolysis to find the weight of a magnesium atom. [Magnesium has two valence electrons.] A current of 70.0 A flows through a solution of MgCl2 for 1.25 hours. Find the total charge and total number of electrons....
for question number 2, the answer cannot be negative 6. In an electrolysis experiment, a student passed 1.57 A of current through an aqueous solution of lead (II) nitrate containing a 7.63 g piece of lead foil as the cathode. What is the final mass of the lead electrode after 17 minutes? For each of the following cell notations a. Identify the cathode and anode. b. Write the net cell equation. c. Calculate the cell potential. I) Pb(s) 1 Pb2+...
Homework 3 t Gas Density and Molar Mass 6 of 14 > ReviewI Constants Periodic Table ▼ Part B Pressure and temperature affect the amount of space between gas molecules, which affects the volume and, therefore, the density of the gas since To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.5-L bulb, then filled it with the gas at 1.30 atm and 24.0 °C and weighed it again. The difference in...