If an aqueous solution of KI is electrolyzed, what is reduced (@ the cathode) and what is oxidized?
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reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
989986
Fe2O3+3CO--->2Fe+3CO2:  The Fe2O3 (oxidizing agent) is reduced to 2Fe, and the 3CO (reducing agent) is oxidized to form 3CO2.\n\nSnO2+C--->Sn+CO2:  The SnO2 (oxidizing agent) is reduced to form Sn, and the C (reducing agent) is oxidized to form CO2.\n\nRemember, oxidation means gaining oxygen (and losing electrons), and reduction means losing oxygen (and gaining electrons).
779077
Please recognize first that this is an oxidation-reduction ("redox") reaction.  Therefore, something gets oxidized, and something gets reduced.  We must identify those first and write their half reactions...\n\nH2O  + 2e- ---->  H2  \n+1                         0 \n\nThe hydrogen in water is reduced to elemental hydrogen gas, resulting in a gain of 2 electrons and a reduction of +2 in oxidation state (one per hydrogen).\n\nAl ---->  AlCl3  \n0           +3\n\nMetallic (ox. st. 0) aluminum gets oxidized to +3 in AlCl3, resulting in a loss of 3 electrons.   \n\nWe must make the electron transfer equal so electrons are conserved (no excess gain or loss).  In order to balance them out, we must multiply the H2's by 3 and the Al's by 2.  Let's see where that leaves the chemical equation at the moment:\n\nCuCl2  + 3H2O and 2Al ---> 3H2 + 2AlCl3 + CuO\n\nThis should result in 6 electrons gained and lost; therefore conserved.  The reaction however is not yet balanced, as the\nCl atoms are not conserved.  Let's multiply the CuCl2 term by 3 (as well as the CuO term to keep the Cu's in balance).\n\n3CuCl2  + 3H2O and 2Al ---> 3H2 + 2AlCl3 + 3CuO\n\nA quick inspection leads us to think that our reaction is now balanced.   However, in order to complete our redox balance we must check TWO  THINGS:\n1.  Mass balance\n2.  Charge balance\n\nMass balance is what we have been doing; making sure there are the same number of atoms of each side of the reaction.\nLet's be sure that is the case:\n3 Cu on both sides\n6 Cl on both sides (3 Cl2 vs. 2Cl3; OK)\n6 H on both sides \n3 O on both sides\n2 Al on both sides\nMass balance is correct.\n\nWe must also see that the net CHARGE on both sides is the same.  (Does not have to be zero!)  We are lucky here that we have no ions, no charged species.  So the charge on both sides is zero (OK).\n\nWe're finished.
357536
First you will need to work out what the two half equations are (Hint, have a look at a table of standard reduction potentials - these not only give Eo for each half - equation, but also the half - equation of course). URL below if you need it.\n\nSecond, record the values of each half equation, then use the formula:\n\n Eo cell = Eo cathode - Eo anode \n\nto get the cell voltage. (If you are not sure how to do this, try the link below or re post)\n\nThirdly, use the Nernst equation (should be either in your lecture notes or your text, if not try Googling "Nernst Equation") to calculate Eo under the conditions stated in part b of your question.\n\nI hope this helps.
699795
.....and the CORRECT answer is ELECTROLYSIS
198671
Electrolysis. By applying electricity, you can break H2O into H2 and O2 gas.\n\n2H2O --> 2H2 + O2
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F2 + HCl ---> HF + Cl2\n\nThe oxidation numbers are:\nF2 :  0\nHCl: H: +1, Cl: -1\n\nHF: H: +1, F: -1\nCl2 : 0\n\nOxidation is loss of electrons.\nReduction is gain of electrons.\nF2 ---> F(-) gains electrons so it is reduced.  If it is reduced then it is the oxidation agent.\nCl(-1) ---> Cl2 loses electrons so it is oxidized.  If it is oxidized then it is the reducing agent.\n\nTo balance write the two half reactions and balance the atoms undergoing redox:\n\nF2 ---> 2F(-)\n2Cl(-1) ---> Cl2\n\nAdd the electrons to balance the charge:\n\nF2 + 2e ---> 2F(-)\n2Cl(-1) ---> Cl2 + 2e-\n\nSum the two equations:\n\nF2 + 2Cl(-) ---> 2F(-) + Cl2\n\nAdd 2 hydrogens to both sides\n\nF2 + 2HCl ---> 2HF + Cl2\n\nand that's it.
780730
K2(C2O4)*H2O is Potassium oxalate monohydrate.  The oxalate is oxidized to CO2 (Carbon dioxide).\n\nKMNO4 is Potassium permanganate.  The permanganate is reduced to Mn+2 (Manganese II cation).  This requires an acid (H+).\n\n(C2O4)-2 (aq) + (MnO4)-1 (aq) + 8 H+ (aq) --> 2 CO2 (g) + Mn+2 (aq) + 4 H2O (l)
179426
well i think \nCH3COOH -kmno4--> 2COOH\n         \nAS KMNO4 IS AN OXIDISING AGENT\ni hope so ask ur teach and plz mail me with a correct answer
499916
An ionic compound is dissolved with an appropriate solvent, or otherwise melted by heat, so that its ions are available in the liquid. An electrical current is applied between a pair of inert electrodes immersed in the liquid. The negatively charged electrode is called the cathode, and the positively charged one the anode. Each electrode attracts ions which are of the opposite charge. Therefore, positively charged ions (called cations) move towards the cathode, while negatively charged ions (termed anions) move toward the anode. The energy required to separate the ions, and cause them to gather at the respective electrodes, is provided by an electrical power supply. At the probes, electrons are absorbed or released by the ions, forming a collection of the desired element or compound.
882235
In an electrolytic cell you can plate most any metal that doesn't quickly react with the solvent and its solutes.  The force of the copper ions gaining electrons at a negative charged zinc electrode is certainly possible.\n\nSince the copper would receive two electrons and go from a +2 to a zero oxidation state, you would say that it is being reduced.
797121
The answers are KCl, CaF2, and Na2O. Potassium has one electron on the outer ring while chlorine has seven. To make chlorine's ring "full" potassium gives the one electron to chlorine. They are "happy" now. :) That's what my science teacher told me. He's weird.
772155
this is the electrolytic dissociation of acetic acid in water\nCH3COOH --> CH3COO- + H+
818167
Just answered it in your other post, but perhaps this: Oxidation-reduction is a chemical reaction in which an atom or ion loses electrons to another atom or ion.
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Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
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OK....You need to understand how this is all working, and why.  So, the formula equation for this reaction would be something like:\n\nKOH(aq) + HNO3(aq) --> H2O(l) + KNO3(aq)\n\nSince KOH, HNO3 and KNO3 are all strong electrolytes, they are totally ionized when they're dissolved in water.  So, written in complete ionic form, that equation would be:\n\nK+(aq) + OH-(aq) + H+(aq) + NO3-(aq) --> H2O(l) + K+(aq) + NO3-(aq)\n\nTo get to the net ionic equation, you just cross out the things that are the same on both sides of the arrow.  K+(aq) and NO3-(aq) are on both sides, so the net ionic equation is:\n\nH+(aq) + OH-(aq) --> H2O(l)
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Electrochemical Cells are of two types:\n\nVoltaic Cells also called Galvanic Cells\nElectrolytic Cells\n\nAnions will move towards anode and cations will move towards cathode.\n\nAt anode oxidation will occur and at cathode redution will occur\nThe reactions in Volyaic cell are spontaneous and in Electrolytic cell are non-spontaneous\n\nFurther more info: plz check this foolwing web. They gave explanatons with diagrams\nhttp://chemistry.about.com/library/weekly/aa082003a.htm