i am desperate for an explaination in the world of redox?
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!
825248
Basically it is\n\n2 H2O2 -> O2 +2 H2O,\n\nwith catalase written above the reaction arrow.\n\nA more detailed formula would show intermediary reactions of H2O2 with catalase, and then how catalase returns to the original state giving up O2 and H2O.
17774
It doesn't give me room to list them all and explain them on here so let me direct you to a website I found that does a pretty good job:\nhttp://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
176528
http://www.chem1.com/CQ/ionbunk.html\n\nskim this article.
738458
1. What is it? the 2nd stage of celluar respiration that prepares high energy electrons for the electron tansport chian\n2. Where does it take place? in the membrane of the mitochondrion\n3. What are the chemicals used to make it happen? Acetyl-CoA + 3 NAD+ + FAD + GDP + Pi + 2 H2O + 1 CoA-SH\n4. What are the products?2 CoA-SH + 3 NADH + 3 H+ + FADH2 + GTP + 2 CO2 + 1 H2O
806654
F2 + HCl ---> HF + Cl2\n\nThe oxidation numbers are:\nF2 :  0\nHCl: H: +1, Cl: -1\n\nHF: H: +1, F: -1\nCl2 : 0\n\nOxidation is loss of electrons.\nReduction is gain of electrons.\nF2 ---> F(-) gains electrons so it is reduced.  If it is reduced then it is the oxidation agent.\nCl(-1) ---> Cl2 loses electrons so it is oxidized.  If it is oxidized then it is the reducing agent.\n\nTo balance write the two half reactions and balance the atoms undergoing redox:\n\nF2 ---> 2F(-)\n2Cl(-1) ---> Cl2\n\nAdd the electrons to balance the charge:\n\nF2 + 2e ---> 2F(-)\n2Cl(-1) ---> Cl2 + 2e-\n\nSum the two equations:\n\nF2 + 2Cl(-) ---> 2F(-) + Cl2\n\nAdd 2 hydrogens to both sides\n\nF2 + 2HCl ---> 2HF + Cl2\n\nand that's it.
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
836198
OK....You need to understand how this is all working, and why.  So, the formula equation for this reaction would be something like:\n\nKOH(aq) + HNO3(aq) --> H2O(l) + KNO3(aq)\n\nSince KOH, HNO3 and KNO3 are all strong electrolytes, they are totally ionized when they're dissolved in water.  So, written in complete ionic form, that equation would be:\n\nK+(aq) + OH-(aq) + H+(aq) + NO3-(aq) --> H2O(l) + K+(aq) + NO3-(aq)\n\nTo get to the net ionic equation, you just cross out the things that are the same on both sides of the arrow.  K+(aq) and NO3-(aq) are on both sides, so the net ionic equation is:\n\nH+(aq) + OH-(aq) --> H2O(l)
696705
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.
535845
A solute cannot undergo reversible oxidation/reduction if it diffuses away from the electrode. The diffusion coefficient simply measures how quickly this diffusion happens. \n\nWhen deciding how reversible a particular redox reaction is by CV, you have to consider both the intrinsic reactive properties of the species in solution as well as how much material is lost to analysis as it diffuses away from your electrode.
339643
You need a reference table and some instruction. Go to this site and scroll down to electrochemistry. Good luck!\nhttp://spiderpong.sitesled.com/chemistry_12_handouts.htm
989986
Fe2O3+3CO--->2Fe+3CO2:  The Fe2O3 (oxidizing agent) is reduced to 2Fe, and the 3CO (reducing agent) is oxidized to form 3CO2.\n\nSnO2+C--->Sn+CO2:  The SnO2 (oxidizing agent) is reduced to form Sn, and the C (reducing agent) is oxidized to form CO2.\n\nRemember, oxidation means gaining oxygen (and losing electrons), and reduction means losing oxygen (and gaining electrons).
109521
I remember learning about this very complex process, its called Gluconeogenesis and if you do a simple search you can find the step by step processes!
383377
Oh Biology, how do I miss thee!!!!!  Ah that was some good times.  Cytochrome c oxidase is part of the respiration cycle and it accepts electrons.  That about all I can remember but thanks for the points and a trip down memory lane.
839183
http://en.wikipedia.org/wiki/Oxidative_phosphorylation\n \nthis site should be useful for that
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
836016
Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html