what is the electric current used in Titanium Reduction using FFC method???
99748
about 3.2VDC using CaCl3 for the electrolytic salt.\nThe voltage needed must be enough to reduce the oxide, but not so much as to break down the CaCl3; the details are described in the link below:
258321
Your question is a little hazy.  Are you looking for titanium?  Many different "elements", as in the periodic table, are in an engine.
360112
Why dont you just try going to google search and just type "NaCl Electrodialization". You should get more info. than you think you'ld get.
713269
Faraday's Law!\n\nMass Plated = t * I * MW / (n * F)\n\nt = time\nI = current (Amps)\nMW = mol weight of metal\nn = valence charge of metal\nF = Faraday's Constant (96,500 Coulombs/mol e-)\n\nSo....\n\nt = 2 days * 24 hours/day * 3600s/hour = 172800s\nI = 7.60 A\nMW = 51.94 g/mol\nn = 3 mol e- / mol M\n\nPlug it all in and you should get 235.62g of Cr.
141606
There is an application to convert an AC current to DC.\nAssume that f is a sigmoid function, then integration of\nf^2 over one period gives a constant number which is considered as flattened sigmoid and is analogous with converting AC to DC.
528106
since reduction occurs at the cathode, it would be the Ag+, but it is relative to what your anode is.
420687
Without applying a potential, nothing.  The net voltage of the cell is less than zero (silver's oxidation is -0.80V, while aluminum's reduction is -1.68V).  Total voltage would be -2.48V, which is non-spontaneous.
892848
One of the most frequently used is E=IR    ;E =voltage    I=current   R= resistance. U can change this every way depending on the unknown.\n    Good luck
304984
I would need more information to help you answer that question.\n\nWhat is the mass of the metal being used? What was the original temperature of the metal?\n\nThe formula you need is:\n\nq = mc(Tf - Ti)\n\nq = energy change\nm = mass\nc = specific heat\nTf - Ti = change in temperature
437234
More than likely DC.  By using direct current you are always keeping one end the anode and the other the cathode, one side always is "giving" and the other end always "recieving."  DC systems have a tendancy to corrode one end and slowly eat away at it.
976128
I believe the answer is conductive.  This is the process in which things are silver plated.  \n\nHere's how it goes.You fill a tray that has been lined with aluminum foil with salt water ( or is it plain water ? ).  You run a current through the aluminum foil.  The electrons in the the foil get excited due to the electrical current and start releasing aluminum ions ( Al-, I think )  Now, take your rusty, dull ' silver ' spoon.  Place it in the bath for about 2 hours.  Be careful,  electricity, water and living things don't mix well.  After 2 hours, remove the electrical current source and remove the spoon.  The free ions in the water would have coated the spoon, making it shining again.\n\nActually, toothpaste to clean the spoon is easier.
1184412
Yes, in fact that has been done throught the concept of "Thermo-coupling" but again this method is economically infeasible, because the current generated in the process is in nano amperes.\nor in very small qty.
144775
The melting point of titanium is approx. 1660 Celcius
136930
Try the Journal of Electrochemical Society. You may have to go to a local university library to get access to it. Or, try a Google search.
343278
Half reactions are:\nOxidation: Cu ----> Cu2+ + 2e    Eo = -0.34V\nReduction: NO3- + 4H+ + 3e --->  NO + 2H2O   Eo = +0.96V\n\nAdd the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.\n\nThe Nernst equation will take this form:\nE = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}\n\nn will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).\n\nThe concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278.  Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.
364875
Not sure what you mean, are you talking about using 2 different metals and generating electricity from a potato?
964894
IF U ARRANGE THE IONS IN ORDER OF increasing oxidation potentials then it's first Mg+2 , Zn+2, Cu+2\nso in a soln of these 3 ions Cu ions have highest reduction potential i.e.\nhigh tendency to accept e~ (to get reduced)\nso Cu will get first discharged & not Mg\nif Cu ions get exhausted then Zn will get deposited then Mg