write the reduction and oxidation half-equations for this cell?
411418
Ox reaction:  Fe ---> Fe(2+) + 2e\nRed reaction:  Ni(2+) + 2e ---> Ni\n\nGiven that Fe/Fe(2+) is +0.44V and Ni(2+)/Ni is -0.26V, the net voltage is +0.18V.
727458
2. 2 Na + 2 H2O --> H2 + 2 NaOH\n3. 2 HNO3 + Ca(OH)2 --> Ca(NO3)2 + H2O\n4. C5H12 +  8 O2 --> 6 H2O + 5 CO2
747348
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?
407297
I2O5 + 5CO --> 5CO2 + I2
868500
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
422969
AgNO3(aq)+2Na2S2O3(aq)--> [Ag(S2O3)2]3-(aq) + 4Na+(aq)+ NO3-(aq)
733447
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
974937
please complete the equation
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
942551
Well Im not sure what youre exactly looking for but the basics of it is oxidation passes the electron and reduction accepts the electron.  There are some basic rules for assigning oxidation number and its best to memorize them for a test or something
929785
You must figure out the oxidation number of each element in an equation.  The substance oxidized goes up in oxidation number (becomes more positive) and the substance reduced goes down in oxidation number (becomes more negative).  The substance that is oxidized is the reducing agent (causes reducing in the other element) and the substance that is reduced is the oxidizing agent (causes oxidation in the other element).\nSO . . .\nin (a) carbon on the left side is -4 and on the right side is +4 so it is oxidized and CH4 is the reducing agent\n     oxygen on the left side is zero (all elements are zero) and on the right side is -2 so it is reduced and O2 is the oxidizing agent\nin (b) zinc on the left is zero and is +2 on the right, so it's oxidized and Zn is the reducing agent\n     hydrogen on the left is +1 and is 0 on the right, so it's reduced and HCl is the oxidizing agent\nI'll leave the rest for you!
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
427800
In each half-cell is a chemical undergoing either oxidation or reduction. In a full electrochemical cell, one side must be losing electrons (oxidation) in to its electrode while the other half-cell gains electrons (reduction). If the atoms/ions involved in the reaction are metal, the same metal is used for each electrode. If the atoms/ions involved in the reaction at each half-cell are not metal, obviously no electrode can be constructed out of it. Nonreactive metals such as platinum are used as a substitute. Finally, a salt bridge is necessary to provide electrical contact between the cells—but without the solutions mixing. This can simply be a strip of filter paper soaked in saturated potassium nitrate (V) solution.\n\nDifferent choices of substances for each half-cell results in varying potential differences. Each reaction is undergoing an equilibrium reaction between different oxidation states of the ions—when equilibrium is reached the cell cannot provide further voltage. In the half-cell which is undergoing oxidation, the closer the equilibrium lies to the ion/atom with the more positive oxidation state the more potential this reaction will provide. Similarly, in the reduction reaction, the further the equilibrium lies to the ion/atom with the more negative oxidation state the higher the potential.
795154
2CO(g) + O2(g) --> 2CO2(g)
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
15061
K2Cr2O7 + 2Pb(C2H3O2)2 + H2O --> 2PbCrO4 + 2HC2H3O2 + 2KC2H3O2