B. Determination of Avogadro\'s Number and the Faraday Constant Trial 1 I. Initi
ID: 521414 • Letter: B
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B. Determination of Avogadro's Number and the Faraday Constant Trial 1 I. Initial mass of copper anode (g) 2. Initial mass of copper cathode (g) 4, Time of electrolysis (s) 5. Current (or average cument) CA) 6. Final mass of copper anode (R) 7. Final mass of copper cathode (R) Data Analysis 1. Mass of copper oxidized at anode (R) 2. Moles of copper oxidized (mo) 3. Moles of electrons transferred (mol e 5. Electrons passed through cell (e) 6. Avogadro's number (e /mole 8. Literature value of Avogadro's number Data Analysis, 9. Percent emor 10. Faraday constant (Cmol e ii. Average Faraday constant (Chole 12. Literature value of Faraday constant 13. Percent error Laboratory Questions Circle the questions that have been assigned. the graphite electrodes, the reaction occurring at the anode may be dif- 1. Part A.2. If zinc electrodes are used instead of would remain unchanged. ferent, but the reaction occurring at the cathode 2. Part A 2. Nitrate ions, No, being anions, migrate to the anode in an e cell. Explain why you would expect water rather than nitrate ions to be oxidized at the Hint: state the nitrate 3. Part B. Repeat the calculation of Avogadro s number, the mass gain of the cathode instead of the mass loss of 4. the anode. Account for any difference in the calculated values, i would Avogadro's number be calculated as too high Part B.2. If the current is recorded as being less than it actually or too low, or would it be unaffected? Explain. 5. Part B.4. Because of an impure copper As (see Assignment the measured mass loss. greater of copper is than the actual mass anode A number be calculated as too high or too low? Explain. oxidized. result, will a 6. of molten KCI chlorine at the anode; however, in the electrolysis of an aqueous KCI What is the gas and why is it produced? Explain. solution, a gas is evolved. .7. The c refining of copper involves the oxidaton of such metals as iron and nickel to copperaD. nickel ions) at the anode and then teduction of the metal at the cathode. Explain why the and nick ions are not deposited on the ion to copper 376 Electrolytic cells, Avogadro's NumberExplanation / Answer
The oxidation half - reaction at anode for Cu is:
Cu -------> Cu2+ + 2e-
The equivalent mass of Cu is, M = molar mass of Cu/ no of moles of electron
M = 63.55/ 2 = 31.78 g Cu/ mole e-
Data Analysis 1. Mass of Cu oxidized at anode is = 8.300 g
2. moles of Cu oxidized = mass of Cu/ equivalent mass of Cu
= 8.300/ 31.78 = 0.2611 mole
3. moles of electrons transferred = 2 mole/e-
4. one ampere (A) = 1 coulomb/ seconds
coulombs passed though cell = ampere* seconds
t = 6000 second, current= 0.2 ampere
couloms passed through cell = 0.2 x 6000 = 1200 (C)
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