How many coulombs of electrical charge must pass through an electrolytic cell to reduce (check my answer)?
931517
The answer is b however I don't quite follow your reasoning\n\nFaraday's constant is approximately 96500 coulomb/mole of electrons\n\nEach Ca2+ needs 2 electrons so 0.82 mole Ca2+ need 2*0.82 mole of elecrons, so a total charge of 2*0.82*F=2*0.82*96500= 1.6*10^5
95444
Some tips...\n--Use a strong electrolyte (substances that will break up into ions [almost] completely in solution), the higher the concentration, the better.  Also you need to be careful on which electrolyte you choose.\n--Try not to use an over volttage (a voltage over the minimum necesary to electrolisize the water.  Insread, try to use a very high current, higher current = more production of gasses.  To increase current without increasing voltage, reduce resistance.\n--Place the electrodes close together to reduce internal resistance in the cell.\n\nEven by doing these things, you cannot hope to get much better than about 50% efficency doing water electrolysis, maybe 65% if you are good.\n\nThere are yahoo groups out there devoted to just such a tast as Hydrogen production through water electrolysis (I have one infact), they can also give advice based on experience.
1162596
Electricity through the wire because it is electron movement which is way faster than the ion in a solution-just consider the difference in the sizes of the two and friction.
808587
so that it gets ionised during electrolysis and the ions help in the conduction of electricity....
495744
Well, if you run a current through saltwater (NaCl in water solution), two half-reactions are taking place in an electrolytic cell: \n2Cl- ---> Cl2 + 2e-  anode, oxidation (+)\n2H20 + 2e- ---> H2 + 2OH-  reduction, cathode (-)\nThe negative end is producing H2 gas, which bubbles out of solution, and hydroxide ions, which are basic.  So the stuff coming off of the negative end is basic, not acidic.
200995
by electrolisis\nH2O------>H2+O2\ni.e by passing electric current
942137
I have two problems with this: First, this is not an avenue for high school kids to get their homework done for them and second, the thought that someone would take yahoo answers 'points' as payment for doing a highschool kids homwork makes me feel ill. \n\n(i)All that aside, the reducing agent is the compound/element that looses electrons. \n\n(ii) First balance the equation, then find then number of moles of potassium permagenate used in the titration, then convert to the equivalent number of mols of the Be complex. \n\n(iii) multiply answer for (ii) by 50\n\n(iv) use the difference between your asnwer for (iii) and the given sample mass to find mass percent.
127709
Coulombs law.\nWhen the distance is made 3 times longer (for instance),\nthe force between the charges becomes 3^2=9 times smaller.
536437
In order to overcome the strength of the  oxygen-hydrogen bond of water, a force must be used to electrochemically decompose a sample of water.  This is called a decomposition voltage and it is roughly about 1.2 volts at 25 degrees C.  In the lab a little more voltage is necessary to have the reaction occur at an acceptable rate.\nThis requires a cathode and an anode to produce hydrogen and oxygen gas, and of course a voltage.  \nElectrolysis of water:\n4H+ (aq) +4e- ------ 2H2(g) reduction at cathode\n2H2(l) ------- O2(g) + 4H+(aq) + 4e- oxidation at anode\n\n2H2O(l) ------- 2H2(g) + O2(g) net reaction\n\nEnergy: DeltaGrxn = +474.4 KJ\n\nGood night and Good Luck.
1057959
You are confused.  Electrons don't conduct electricity.  They, collectively, make electricity.  They do work by going from high potential energy to low potential energy.  The potential energy is usually caused by concentration of charge, because like charges repel each other.  The movement of electrons is what we call electricity.  Charge is a measure of the difference between number of protons and number of electrons, if you multiply the difference by the charge for one electron you get the charge.\nA coulomb is a unit designed to be more usefull than the charge on an electron because it is large enough to work easily with..
980496
This is true The electrolyte in the cell is inert unless driven by external voltage into a redox reaction with the anode and cathode.
537229
I dont know that there is a limitation for this. Probably the main reason is the Columbic repulsion factor. Since repulsions will compete with the stabilizion energy of the system, the charge transfer may not takesplace beyond  1A (angstrom)\nI am just guessing anyway!\n\nhope this reason will give some idea!\n\ngood luck!
870424
in electrochemical cells we have chemical reaction . since conversion of this chemical energy to electric energy is mainly becoz of combination of ions ..we hav electric on account of this chemical energy till we have free ions to combine once all ions of solution combines  no more energy conversion is there
825689
CuCO3 + H2SO4 --> CuSO4 + CO2 + H2O\n\nCu(OH)2 + H2SO4 --> CuSO4 + 2H2O\n\nFor the electrolysis:\n\nCuSO4 = Cu+2 (aq) + (SO4)-2 (aq)\n\nCu+2 (aq) + 2 e-1 --> Cu (metal)
631579
2 electrodes and a battery
689565
Wow, brilliant question.  I don't think they do.  A battery moves electrons through a circuit in one direction (hence the name Direct Current) but for every electron, which has mass (9.1091 x 10 to the minus 28 grams), that goes out end of the battery, another electron with the same mass comes in.
270148
To restate my previous answer given in your other question,\nhttp://answers.yahoo.com/question/;_ylt=AlzxSKBO35xmSJCjWkIsqtfsy6IX?qid=1006032314244\n\nElectric current is given as the charge flowing through a given section per unit time.\n\nThe SI unit of charge is the Coulomb. The SI unit of time is the second. Therefore, the SI unit for current is the Coulomb per second (AKA, the amp).\n\n1 Amp = 1 Coulomb / second\nSo in 1 second, 1 Coulomb of charge has flowed past a given section of wire at 1 amp.\n\nOne electron has a charge of 1.6021765314 ×10-19 Coulombs.\nIf in 1 second, 1 Coulomb of charge is flowing through a section of wire, then there must be about 6.241 E18 electrons flowing past that section.