Voltaic cells based on the following pairs of half-reactions are constructed. For each pair, ...?
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First you need to compare the standard reduction potentials of each half-reaction. (See URL below if you don't have values.) The more positive value will be the reduction half-equation and must therefore be the cathode (and the positive terminal for a voltaic cell). The other half equation will be the oxidation half-equation, and must therefore be the anode (and the negative terminal). You need to reverse this equation. Before you can add the two half-equations together, you must ensure that each half-equation has the same number of electrons.\n\na) has a problem; it is not balanced. Either the silver atom on the right hand side should have a single positive charge OR there should be 2 electrons on the left hand side.\n\nb) The MnO2 reduction half-equation has a more positive value than AgBr reduction half-equation, therefore is the cathode etc ie\n\nReduction: MnO2 + 4H^+ + 2e^- = Mn^2+ + 4H2O (Cathode, +ve)\nOxidation: Ag + Br^- =  AgBr + e^-           (Anode, -ve)\n\nNow multiply the oxidation half-equation by 2 so that the number of electrons in each half equation is the same: \n\n2Ag + 2Br^- =  2AgBr + 2e^-\nMnO2 + 4H^+ + 2e^- = Mn^2+ + 4H2O \n---------------------------------\n2Ag + 2Br^- + MnO2 + 4H^+ =  2AgBr + Mn^2+ + 4H2O \n\nNote: the electrons cancel out.\n\nc)  i hope that the above will allow you to complete this one on you own.
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CoCl2(aq) + FeSO4(aq) = CoSO4 (aq) + FeCl2(aq)
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solar cells produce electricity through a process known as photoelectricity.Photoelectricity means a process in which light is shone (fall on) photocells which bring about the release of electrons.i.e.production of electricity .
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on a rad new cell that work by positive energy
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they fill the s orbitals with 2 electrons
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A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
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If you are into electrochemistry you could make your own solar cell or a flat panel solar cell. A solar cell is a device for converting energy from the sun into electricity. This project is fairly easy and there are plenty of pics to show you exactly how its done. \n\nAlso read the first source listed first because it tells you things you need to know for the second experiment.
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by electrolisis\nH2O------>H2+O2\ni.e by passing electric current
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Just the usual stuff.  Use an electrolyte (H2SO4/HNO3), a couple of electrodes, a reduction table of half cell potentials, and some electricity.
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voltametrty is measurement of voltage (which control type of reaction)\nwhile coulommitery is measuerement of electricity quantity (which control the time of  reaction)
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H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
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dobereiner's like that mendeleev guy right? if my memory is right, all  of the elements above have similar chemical properties, that's why he placed them in his "triads". This means that all of them act similarly. Remember that it is the VALENCE electrons that causes the specific reaction of an element. So, as all three elemnts have the same number of valence electrons, they act similarly.
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CONCLUSION\n\nIn the course of these experiments, you have become familiar with various concepts such as the conductivity of solutions by way of dissolved ions. you have seen redox reactions at work, and that they can be reversed by running current through the circuit. You have learned to measure potentials of reduction of various materials and to predict the voltage generated by a battery when you know it's material components. You can build a working battery from various materials, including lemons, tomatoes, potatoes, almost any fruit or vegetable. Up until now, batteries may have seemed mysterious, but now we are among the few that have a working knowledge of how and why they work. You can amaze your friends by building a battery out of almost anything, but more important than this, now the study and mastery of this subject will be easier and more meaningful\n\nsee the full answer here:
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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.
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That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
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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.
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In order to do this you will most likley have to also apply a voltage to the wire in order to drive the reaction.  Also, the bridge between the 2 beakers must be conductive, so there has to be some sort of electrolyte soaked into the bridge.  i would suggest using a simple copper sulfide solution in the plating side of the cell, and using an iron electrode in the other beaker.  electroplating is sometimes quite complex so you may want to consult a chemist regarding your final setup.