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some chemical species. Q12. Balance the following possible oxidation reactions: Using the table of reduction potentials, is one reaction much more likely than the other or are they pretty close? According to the table, water is slightly more likely to be oxidized, but the numbers are pretty close. It turns out that in a concentrated NaCl solution, chlorine is reduced in preference to water. This has to do with more complicated non-equilibrium chemistry that isn't included in the table of reduction potentials. Q13. What gas do you expect to be produced at the cathode (negative terminal)? hydrogen gas (H2) Q14. What gas should be produced at the anode (positive terminal)? chlorine gas (Cl2) Q15. What would you expect for the ratio of gases produced at each terminal? volume of gas at cathode : volume of gas at anode = ______1:1____________ It takes the same number of electrons to produce one molecule of H2 as one molecule of Cl2, so they should be produced in equal quantities. 4. After you disconnect the battery, take a look at how much gas was actually produced at each electrode. Make a mark for the current level of the water in each test-tube. Q16. Change in height at the cathode (negative terminal): _____about 1.5______ cm Change in height at the anode (positive terminal): _____almost 0______cm Do you see the ratio that you expected? No Q17. Was one gas produced in much greater quantity than the other? Make a guess as to why this might be. Much more hydrogen than chlorine is produced. That’s because most of the chlorine gas dissolves in water. 5. Chlorine gas is highly soluble in water. As the gas is produced, it will dissolve in the water until the test-tube is fully saturated. Let's calculate how much chlorine gas you produced and whether it was enough to fully saturate the tube of water. Q18. Volume of H2 gas produced at cathode (V = R2h ): around 2 mL Created by LABScI at Stanford 7PDF Image | Electrolysis splitting water
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