Identify which elements, if any, undergo oxidation/reduction in each of the following reactions?
471546
A.  Iodine is being reduced from +5 in HIO3 to 0 in I2; oxygen is being oxidized from -1 in H2O2 to 0 in O2.\n\nB.  There is no oxidation/reduction going on; this is simple metathesis, producing the barium sulfate precipitate.
470703
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
802199
Hi,, hydrogen, oxygen, and nitrogen... \n\n good luck.....
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
832296
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)
877634
Any acid, for example:\n\nHCl + H2O  --> Cl- + H3O+
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
828740
fe is the reducing agent cos fe undergo oxidation prosess changing its oxidation no from 0 to 2+
864677
carbon monoxide = CO\nIron (II) Chlorate= Fe(ClO3)2\nLead (IV) oxide = PbO4\nNitric acid = HNO3\nSulfuric Acid = H2SO4\n\nall numbers are subscripts.
457655
Iron (II) SulfIDE\n\nSodium ChlorIDE\n\nSodium HydroxIDE\n\nand Lead...umm...something. *lol*\n\nThe answer is evidently (A). =D\n\n\n\nPS okay, so yea, someone already wrote what i wrote when i was typing it. sorry!
952260
Many materials can react with ozone to form oxygen as a by-product.  Most all are reducing agents.  A more definitive answer rests on how you are using the term "element".  If you are using it the periodic table sense then your answer would include many of the metals capable of being oxidized.
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
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
329297
decomposition reaction\n\n2H2O2 -------- O2   +  2H20
880087
De = 2\nHydrogen = H\nmono = 1\nOxide = Oxygen\nOxygen = O\nH2O
589750
SO4 2-\nCO3 2-\nOH -\nPO4 3-
766197
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)