write a balanced equation for the cell reaction, and identify which half-reactio takes place at each anode and cathode.\n\na)Cd^2+ + 2e^- = Cd\n  Ag^2+ + e^- = Ag\nb)AgBr + e^- = Ag + Br^-\n  MnO2 + 4H^+ + 2e^- = Mn^2+ + 4H2O\nc)PtCl4^2- + 2e^- = Pt + 4Cl^-\n  AgCl + e^- = Ag + Cl^-
212986
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.
484918
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
402503
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
407062
1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
374337
Cl2 + 2 KI → I2 + 2 KCl
467284
Oxidation :\n\nCu > Cu+2 + 2e-\n\nReduction :\n\n2H+ + e- + NO3- > NO2 + H2O\n\nMultiply the 2nd equation by 2, you get the complete balanced redox equation :\n\nCu + 4H+ + 2NO3- > 2NO2 + 2H2O + Cu+2
36059
balance equation:\n \nCaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2CO3(aq)\nH2CO3(aq)->  H2O(l) + CO2(g)\n\nconbining the two equations,\n\n  CaCO3(s) + 2HCl(ag) -> CaCl2 (aq) + H2O(l) + CO2(g)
377286
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
375486
HClO2 + Ba(OH)2 -> BaClO2 + H2O\n\n2(NH4)3PO4 + 3Ca(NO3)2 -> 6(NH4)(NO3) + Ca3(PO4)2\n\nFinished and balanced.
139605
reduction:  2e- + Cu2+ --> Cu      Eo = +0.34 V  cathode (+)\n\noxidation:   Zn   --> Zn2+ + 2e-   Eo = +0.76 V    anode (-)\n________________________________________\n\nZn  + Cu2+ --> Cu  +  Zn2+         Eo = +1.10V therefore cell spontaneous as written.
471696
2AgNO3 + CaCl2 --> 2AgCl + Ca(NO3)2\n\nMoles of AgNO3 given\n= (115/1000) dm3 x (0.23) mol/dm3 = 0.02645 mol\n\nMoles of CaCl2 given\n= (115/1000) dm3 x (0.14) mol/dm3 = 0.0161 mol\n\n2 moles of AgNO3 needs 1 mole of CaCl2 for complete precipitation. Hence for 0.02645 mol of AgNO3  provided, (0.02645/2 = 0.013225) mole of CaCl2 is needed.\n\n2 moles of AgNO3 produced 2 moles of AgCl. Hence 0.02645 mole of AgCl is produced.\nMass of 0.02645 mole of AgCl\n= (0.02645 x Mr of AgCl) = 0.02645 x 143.5 = 3.80g (to 3 s.f.)\n\n0.02645 mole of AgNO3 produces 0.02645 mole of Ag+ and 0.02645 mole of nitrate ions.\n0.0161 mole of CaCl2 produces 0.0161 mole of Ca2+ and (0.0161 x 2 = 0.0322) moles of Cl-.\n\nAfter precipitation, with AgNO3 as the limiting reagent,\nMoles of Ag+ left = 0 (all precipitated as AgCl)\nMoles of Cl- left = 0.0322 - 0.02645 = 0.00575\nMoles of nitrate ions left = 0.02645 (nitrate not precipitated)\nMoles of Ca2+ = 0.0161 (Ca2+ not precipitated either)\n\nTotal volume of solution is now (115 + 115 = 230 ml).\nFor concentration of each ion left, please divide the moles of each ion left by the volume (230 ml) respectively.\n\nHope this helps. :)
574455
Silver is being oxidized, nitrogen from the nitrate is being reduced.\n\nOxidation half reaction:\nAg --> Ag+  +  e\n\nReduction half reaction:\nNO3- +  4H+ + 3e  -->  NO + 2H2O\n\nNet balanced and reduced equation:\n3Ag +  NO3- +  4H+ --> 3Ag+ + NO + 2H2O
433923
First write the two half reactions:\n\n3 e(-) + Bi(OH)3 ---> Bi\nSnO2(=) ---> snO3(=) + 2 e(-)\n\nMultiply the first equation by 2 and the second by 3 and sum to get:\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi\n\n12 O in the left, 9 on the right, so add 3H2O\n\n3SnO2(=) + 2Bi(OH)3 ---> 3SnO3(=) + 2Bi + 3H2O\n\nCheck that H is OK.  Yes. Done\n\nSo the sum is 13. (d) is the answer.
492041
Na2S (ag) + 2 AgNO3 (ag) --> 2 NaNO3 (ag) + Ag2S (cr)\n\nB. Double displacement.\n\nA. Ba(NO 3)2 and H2O\n\nB. If there is a double arrow (pointing in both directions), then it is an equilibrium reaction.\n\nC. Ba(OH)2 and HNO3\n\nD. Barium\n\nE. 6 oxygen atoms. \n\nF. Ba(OH)2
78615
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
60085
:) Thanks for taking all the trouble in finding me to send me a question. I Feel flattered. I am not a chemist..but i would suggest that you try searching google/yahoo with the following word combination:\n1. "Voltaic cell"+experiment+laboratory\n2. "Voltaic cel"+"kitchen laboratory"\n\nand so on.. I am sure you will find some answers. good luck! I will try to pull out more answers and try to send them to you. by the way my email id is same as the handle..scientist_atlarge.
550738
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD