Which ion, Mg2+ or Ag+, is more easily reduced in aqueous solution?\n\nWrite the half-equation for the rxn that occurs in the silver half-cell.\n\nWrite the equation for the net cell rxn.\n\nCalculate the expected voltage for this cell.
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Using the sheet from this website http://www.bced.gov.bc.ca/exams/search/grade12/english/release/additional-material/0508ch_databooklet.pdf \n1/2 rxns:\n1) 2 Ag+1 + 2 electrons --> 2 Ag(s)+.80 V\n2) Mg(s) --> 2 electrons + Mg2+ +2.37 V\nBalanced) 2  Ag+ + Mg(s) --> 2 Ag(s) + Mg2+ Exp. voltage = difference between 0.8 and 2.37 = 1.57 V
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Okay, this is trivial stuff here.\n\nThe trick is to have enough + and - to add up to zero.  so, for example:\n\na) Cu2+ and NO3-;  you have two + and one -, thus to make a neutral compound you will need twice as much -, so the formula for the salt is: Cu(NO3)2\n\nYour problem in d) is a little trickier, but it is the same process:\n\nd) Mg2+ and PO4(3-): how can you get an equal number of + and - here?  use three Mg and two PO4!  Mg3(PO4)2\n\nI will leave the rest for you to do by yourself so you understand the process.
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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
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When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
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Part A\n\nReduction: 3e^- + 4H^+ + NO3^- = NO + 2H2O    Eo = \nOxidation:                 Cu  = Cu^2+ + 2e^- Eo =\n\nMultiply the reduction half equation by 2 and the oxidation half equation by 3 (so that you have 6e^- in both) then add together to get:\n\n3Cu + 8H^+ + 2NO3^- = 2NO + 3Cu^2+ + 4H2O\n\nNow Eo cell = Eo cathode - Eo anode = 0.96 - 0.34 = +0.62V\n(Note: you do NOT multiply the Eo values by the number of electrons in the half equations because Eo is a voltage, measure of electrical pressure.\n\n\nPart B\n\nThe Nernst equation can be found at:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
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AgNO3 + NaCl => AgCl(white ppt) + NaNO3\nAgNO3 + KI => AgI( yellow ppt) + KNO3\n\nthere is a change of color because AgI is produced
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HgO ------> Hg + O2 at about 400 C\n\nMg + CuCl2 ---------->  MgCl2 + Cu(solid)\n\nMg, being higly electropositive than Cu, replaces Cu.
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So the half reaction thing is for redox equations (which I assume is given).  You basically put the like things together and start to balance.\n\nAu+ + Cl-  >  AuCl4                  NO3-  >  NO2\nBalance everything except O and H\nAu+ + 4Cl-  >  AuCl4                NO3-  >  NO2\nBalance O with H20\nAu+ + 4Cl-  >  AuCl4                NO3-  >  NO2 + H2O\nBalance H with H+\nAu+ + 4Cl-  >  AuCl4                2H+ + NO3-  >  NO2 + H2O\nBalance charges with e-\nAu+ + 4Cl-  >  AuCl4 + 3e-         e- + 2H+ + NO3-  >  NO2 + H2O\nMake the e- equal\nAu+ + 4Cl-  >  AuCl4 + 3e-/    3(e- + 2H+ + NO3-  >  NO2 + H2O)\nAu+ + 4Cl-  >  AuCl4 + 3e-/ 3e- + 6H+ + 3NO3-  >  3NO2 + 3H2O\nCancel and add\nAu+ + 4Cl- + 6H+ + 3NO3-  >  AuCl4 + 3NO2 + 3H2O\nMake sure everything balances (charges too!)\n\nOxidizing agent is the substanced being reduced (or gaining electrons) and the reducing agent is being oxidized (or losing electron)  For Au, the charge is going from a +1 to a +4, which means its losing electrongs, so the reducing agent is Au+.  The N in NO3- is going from +5 to +4, which means its gaining electrons, so NO3- is the oxidizing agent.
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Well, for a start your equation shows copper(II) sulphide, not sulphate, and your equation is wrong even then.  That makes a bit of a difference!\n\nLet us imagine that you've got the formula right, but the name wrong.\n\nThe equation is simply Cu + S ---> CuS\n\nThen 64g will give (64+32)g.\n\nThus 64g will give 96g.\n\nSo 64 x (200/96)g of Cu will give 200g of product.
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2Mn2+(aq)+14H+(aq)+5NaBiO3(aq)\n-->\n2MnO4-(aq)+5Bi3+(aq)\n+5Na+(aq)+7H2O(l)
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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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Standard electrode potential for Ag(NH3)2 = 0.37V
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If you complete the reaction:\n\nCa(OH)2 + 2HCl(aq) -----> CaCl2(s) + 2H2O\n\nThe net ionic equation is:\n\nCa(2+) + 2OH(-) + 2H(+) + 2Cl(-) ------> CaCl2(s) + 2H2O\n\nstatements a) b) are true. c)d)e) are false.\n\nHope it helps!\n\nGood luck!
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You are looking at a displacement reaction, in which the more reactive metal (Cu) is displacing the less reactive metal (Ag) from its salt solution (AgNO3).\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nHope this helps. :)
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copper metal is listed as "very unreactive".  This means it is relatively stable as a metal, and relatively unstable as an ion.\n\nMagnesium metal "reacts with acids" so it is more stable in ionic form than copper,and less stable as pure metal.\n\nThe goal of nature is to go from an net unstable form to a more stable form.  (This is called reducing the free energy, or delta G of a system).\n\nIn order to use the activity series:\n\nany metal below a given metal is less stable in ionic form and more stable as a pure metal.\n\nThis means that a pure metal can reduce the ions of any metals below it.\n\nFor example, aluminum metal can reduce manganese ions, zinc ions, and so on because they are below it.\n\nAluminum metal cannot reduce magnesium ions, calcium ions, and other metals above it.\n\nI hope you can get it now.  Ok I am sure what I typed is correct now.  Good luck!\n\nhttp://en.wikipedia.org/wiki/Activity_series
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This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)
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The answer is A\n\nCoBr2 + Pb(NO3)2 ==> PbBr2  + Co(NO3)2\nPb(C2H3O2)2 + 2KBr ===> PbBr2 + 2KC2H3O2\n\nIn both cases, the net ionic equation is:\n\nPb+2  + 2 Br-  ====> PbBr2(s)