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  1. Accorind to the table of standard electrode potentials, it is easier to reduce a hydrogen ion than a transition metal ion, such as nickel or zinc. Yet, sources such as the Electrochemistry Encyclopedia for nickel, and here on page 9, for nickel, cobalt and zinc. So, why are the metal ions reduced as oppose to the hydrogen ion?
  2. I don't think it is a typo. After further search on google, I have found the book in which wikipedia sourcesseen here, and on page 125, it says, "Brandes finally succeded in decomposing lithia with a powerful battery...". Then from wikipedia again, you can see that lithia means lithium oxide. Still, the book has no mention on the process used... I've yet to research the sources given by this book, perhaps they will explain. In any case, the book also states, "Davy also obtained a small amount of lithium in the same manner", and since we know Davy fused his salts, it is likely that the book means Brandes also fused lithia to perform electrolysis. However, I am determined to find out what really happened.
  3. I originally looked on wikipedia for the history of lithium, and how it was first isolated. It says, "William Thomas Brande performed electrolysis on lithium oxide", but without any specifics, as in chemical equations, voltage potentials, apparatus, and if the salt was molten or not. Now, the reason I am interested, is because I *think, that William Thomas Brande isolated lithium by dissolving lithium oxide in a polar aprotic solvent; thus allowing electrolysis without fusing the lithium oxide. Anyway, back to the point, I cannot seem to find the specific's for Brande's process anywhere. On the internet, I only find the same sentence repeated on many websites, and I even checked my municipal library, but sadly, they do not have any books that seem to be relevant to my question, nor any books on William Thomas Brande himself. Does anyone know of a reliable source that does state the specifics of Brande's process?
  4. Ok, I think I see where this is going. I thought with a power supply, you could set both the voltage and current magnitudes. For example, on an AC adapter it says there's a DC output of 7.5V, 5.7A, but that's with respect to the resistance of the machine it's designed to power. In my electromagnet scenario, it would be : Stage 1. The power supply is set to 6V, and the electromagnet has a resistance of 1 ohm, therefore the current is 6A. Stage 2. The power supply is set to 30V, and thus the current is 30A. I'm sorry for my mistake. Thank you for helping me solve this.
  5. It's the same electromagnet. Would it not have the same resistance? If you need a number, let's just say it's 1 ohm.
  6. It's not homework; it's my own curiosity. What I mean by constant, is that in the example, the two different supplies provide the same current of 5A. I'll clarify my example. Stage 1. There is an electromagnet with a DC current powering it. The current is a constant 5A, and the voltage is a constant 6V. Stage 2. The electromagnet is disconnected from the power source. Stage 3. The electromagnet is connected to a second DC power supply. The supply provides a constant current of 5A, and a constant voltage of 30V. Now, will the electromagnet be stronger in Stage 3, as opposed to Stage 1. Will more heat be produced in Stage 3 than Stage 1? Is there a significant voltage drop in an electromagnet? Will this drop be significantly larger in Stage 3 than Stage 1?
  7. How does an increase in voltage with a constant DC current affect an electromagnet? Is the increased voltage just transformed into heat, or extra strength in the electromagnet? For example, at start there is an electromagnet with a current of 5A at 6V. The current is shut off, and a new supply of 5A at 30V is applied to the same electromagnet. How does the electromagnet react?
  8. I know an AC current can be rectified to a DC current with a diode bridge, but does the magnitude of the voltage and/or current change? If so, how. For example, if one had an AC current at 2 A and 12 V, and one converted it to DC, what would the current and voltage of the new DC current be? Btw, let's assume that the process is 100% efficient.
  9. dlzc, could you explain please? For the chem experts, would H2, be a strong enough reducting agent with BrO3- to produce, H2O and Br-?
  10. If I were to use aluminum or zinc, would the majority of the BrO3- be reduced, forming solid aluminum or zinc oxide, with the respective oxide remaining inert to the solution?
  11. In a solution with both BrO3- and Br-, does anyone know how to reduce the BrO3- to Br- without oxidising the pre-existing Br- to Br2? Also, only potassium ions can be introduced into the solution. My first thought is, to use hydrogen peroxide or potassium permanganate, but I believe this would oxidise the Br- to Br2. Does anyone have any ideas?
  12. Sodium iodide dissolves in acetone (see here, here and here [under "Uses"]) From there, I assume one could just electrolyze the solution, and obtain sodium metal and iodine. However, I assume iodine would coat the +ve electrode, and thus slow or even stop electrolysis. So, one would have to constantly clean the electrode.
  13. According to wikipedia the reaction of calcium oxide and water produces 63.7kJ/mol of CaO. This reaction is more exothermic than the dissolution of NaOH which produces 44.5 kJ/mol, seen here
  14. As stated in the first post, this circuit does not include a switch. In this case, how would one reverse the direction of the ac fan? A switch can be used, but what type of switch is this? Does it have a specific name?
  15. Yes, but the reaction is very exothermic (gives off heat). You can see this here. Also, I don't know if you are just inquiring or thinking of this reaction, but if you are, I HIGHLY suggest staying away from chlorine gas.
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