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Electrolysis. Drill. What is the color of the following ion in solution? Nickel Ans: green Copper Ans: blue Cobalt Ans: pink Iron (II) Ans: light blue. Drill. Iron (III) Ans: yellow/brown Permanganate Ans: Deep purple Dichromate Ans: orange Chromate
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Drill • What is the color of the following ion in solution? • Nickel • Ans: green • Copper • Ans: blue • Cobalt • Ans: pink • Iron (II) • Ans: light blue
Drill • Iron (III) • Ans: yellow/brown • Permanganate • Ans: Deep purple • Dichromate • Ans: orange • Chromate • Ans: colorless (sometimes yellow)
Objectives • SWBAT • Determine the mass of metal plated out during an electrolysis process.
Electrolytic Cell • This cell uses electrical energy to produce chemical change. • Electrolysis involves forcing a current through a cell to produce a chemical change for which the cell potential is negative. • (Electrical work causes a non-spontaneous chemical reaction to occur)
How to Create Hydrogen • Electrolysis is the process of running a direct current through an electrolyte. • The process involves two electrodes placed into an electrolyte, either an acidic or a basic solution. • The electrodes are connected to a power source which tries to send a current through the water. • Because of the electrolyte current can pass through the solution and go to the other electrode. • As the current passes through bubbles of hydrogen and oxygen form separately on the two electrodes. • This gas is collected and stored. http://www.facstaff.bucknell.edu/mvigeant/univ_270_03/John/sources.htm
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Electrolysis is used for… • Charging a battery • Chrome plating
In an electrolytic cell an external power source forces electrons through the cell in the opposite direction of a galvanic cell. • When the electron flow is changed, the anode and cathode are also reversed. • Ion flow through the salt bridge is opposite too.
Electroplating • Plating is depositing the neutral metal on the electrode by reducing the metal ions in solution. • (chrome plating of car parts, or gold plating of jewelry)
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Electroplating is a process of coating deposition on a part, immersed into an electrolyte solution and used as a cathode, when the anode is made of the depositing material, which is dissolved into the solution in form of the metal ions, traveling through the solution and depositing on the cathode surface. • A scheme of the electroplating of copper in the aqueous solution of copper sulfate (CuSO4) is shown in the picture. • Positively charged copper ion moves towards the negative cathode and when it reaches the cathode surface it accepts two electrons, converts to the copper atom and deposits on the cathode surface.
Vocab Ampere (amp) is A Coulomb is C 1A = 1 C/sec
Stoichiometry of Electrolytic Processes • We are determining how much chemical change occurs with the flow of a given current for a specified time.
Example • Determine the mass of copper that is plated out when a current of 10.0 amps is passed for 30.0 min. through a solution containing Cu+2. Cu+2(aq) + 2e- Cu (s)
Determine the mass of copper that is plated out when a current of 10.0 amps is passed for 30.0 min. through a solution containing Cu+2. Cu+2 (aq) + 2e- Cu (s) This process occurs at the cathode. Step 1: Use amps and time to determine the total Coulombs of charged passed into the Cu+2 solution. Coulombs of charge = 10 C/s x 60 s/min x 30 min = 1.8 x 104 C Step 2: Calculate the # of moles of e- required to carry 1.8 x 104 C of charge. 1.8 x 104 C x (1 mol e-) / 96,485 C = 1.87 x 10-1 mol e-
Determine the mass of copper that is plated out when a current of 10.0 amps is passed for 30.0 min. through a solution containing Cu+2. Cu+2 (aq) + 2e- Cu (s) Step 3:Determine the # of moles of Cu produced (plated onto the cathode). 1.87 x 10-1 mol e- x (1 mol Cu)/(2 mol e-) = 9.35 x 10-2 mol Cu Step 4: Determine the mass of Cu produced. 9.35 x 10-2 mol Cu x (63.55g/1 mole) = 5.94 g Cu
Try this… How long must a current of 5.00A be applied to a solution of Ag+1 to produce 10.5 g of silver metal?
Solution: 10.5 g Ag x 1 mol Ag/107.87 g Ag = 9.73 x 10-2 mol Ag Ag+1 + e- Ag 9.73 x 10-2 mol Ag x (96,485 C/mol e-) = 9.39 x 103 C 5 C/s x (time) = 9.39 x 103 mol Ag Time = 1.88 x 103 s = 31.3 min
Electrolysis of Water • Combining hydrogen and oxygen to create water is a spontaneous process. • The reverse process is non-spontaneous and can be forced by electrolysis.
AP Problem to Try • Try 2007 #3
Textbook Problems • Old Book • 6, 7, 9, 14, 24, 25 a,b, 27, 29, 31, 36, 43, 44, 46, 49, 51, 52, 56, 59
Wrap up • Tonight, read through the Electrolysis section of Ch 20