albertlee Posted November 14, 2004 Share Posted November 14, 2004 I just want to ask... How do you determine which ion gives/accepts the electrons if there are more than 1 cations/anions at cathode/anode?? Please do tell me also why..... thx Albert Link to comment Share on other sites More sharing options...
YT2095 Posted November 14, 2004 Share Posted November 14, 2004 the ones least likely to be displaced are the ones that will be acted upon on the Cation side, the more reactive metals will stay put Link to comment Share on other sites More sharing options...
albertlee Posted November 14, 2004 Author Share Posted November 14, 2004 YT, what do you mean by least likely to be displaced??? Any body can help?? Albert Link to comment Share on other sites More sharing options...
albertlee Posted November 14, 2004 Author Share Posted November 14, 2004 Sorry for duplicating post.... but can any one help me?? about how to determine the ions at cathode/anode?? thx in advance Albert Link to comment Share on other sites More sharing options...
albertlee Posted November 15, 2004 Author Share Posted November 15, 2004 Can any body help according to my first post?? thx in advance Link to comment Share on other sites More sharing options...
YT2095 Posted November 15, 2004 Share Posted November 15, 2004 taken to an extreme, Sodium will displace copper. so if you had a 50/50 mix of sodium sulphate and Copper sulphate in your electrolyte, you`de get a copper deposit, and no sodium deposit. Link to comment Share on other sites More sharing options...
albertlee Posted November 15, 2004 Author Share Posted November 15, 2004 why??? Link to comment Share on other sites More sharing options...
YT2095 Posted November 15, 2004 Share Posted November 15, 2004 the more reactive a metal (or Cation) is, the harder it is to displace. Link to comment Share on other sites More sharing options...
Gilded Posted November 15, 2004 Share Posted November 15, 2004 "so if you had a 50/50 mix of sodium sulphate and Copper sulphate in your electrolyte, you`de get a copper deposit, and no sodium deposit." You mean sodium (not sulphate) and copper sulphate? Or then I'm not getting something here. Link to comment Share on other sites More sharing options...
YT2095 Posted November 15, 2004 Share Posted November 15, 2004 I meant as I posted, a 50/50 mix of each it could Magnesium sulphate instead of sodium sulphate if you wished Link to comment Share on other sites More sharing options...
Gilded Posted November 15, 2004 Share Posted November 15, 2004 Huh? Then howcome you won't have any sodium sulphate afterwards? Or the more proper question would go: Why won't you have any sodium sulphate afterwards? Because I can't see why a displacement reaction would do something like that in this case. Link to comment Share on other sites More sharing options...
albertlee Posted November 15, 2004 Author Share Posted November 15, 2004 ............ I still dont get my question answered..... Ask again!! How can you determine that one cation/anion is more likely to displace the other cation/anion at cathode/anode?? By the way, please also tell me a scientific reason for this determination... thx Albert Link to comment Share on other sites More sharing options...
jsatan Posted November 15, 2004 Share Posted November 15, 2004 I meant as I posted' date=' a 50/50 mix of each it could Magnesium sulphate instead of sodium sulphate if you wished [/quote'] why would the copper sulphate just drop the copper? So this would make a stronger sulfuric acid? Link to comment Share on other sites More sharing options...
jdurg Posted November 15, 2004 Share Posted November 15, 2004 There is a listing posted in nearly every single scientific handbook of the activity series of the metals. These listings have gold, platinum, iridium, etc. at the bottom since are not as reactive in relation to metals such as sodium, potassium, cesium, etc. (The listing usually has hydrogen in there too so you can tell if a metal will react with acid or not). I believe the metals are ranked based upon their reduction abilities. I.E. the metal that is a better reducing agent is listed at the top, and the worst reducing metals are listed at the bottom. A metal at the top of the list will replace a metal lower than it on the list in solution. (I.E. a stick of zinc metal placed into a solution of silver nitrate will replace the silver in solution, and the silver will preciptate out). This is because Zinc is more willing to give up an electron and move into solution and silver ions are more willing to accept an electron and move out of solution. This also works in determining if a metal will be dissolved by an acid. If the metal is above H+ in the listing, it will be willing to give up an electron to the H+ ion and move into solution while the H+ ion will take the electron and move out of solution. This is why magnesium will dissolve in an acid and generate hydrogen gas and why copper metal won't dissolve in an acid. (Though it will dissolve in nitric acid, but that's because of the oxidation power of nitric acid and not the H+ ion). Check out this website. Link to comment Share on other sites More sharing options...
Gilded Posted November 15, 2004 Share Posted November 15, 2004 [jamaica-accent]It's all to do with the electronegativity, mon.[/jamaica-accent] "while the H+ ion will take the electron and move out of solution." Those little buggers leave the poor metal floating around with a +1 charge and then fly off as H2 gas? I didn't realize how sad it was. :< Link to comment Share on other sites More sharing options...
chadn Posted November 15, 2004 Share Posted November 15, 2004 Since this is electolysis.... Find a table of the standard reduction potentials, if you're in a general chem class or something there should be one in the back of your book. The more positive your standard reduction is for a particular reaction, the stronger of an oxidizing agent or oxidant it is. The more negative the standard reduction the stronger it is as a reducing agent or reductant. If you have two cations in solution at the cathode the one being reduced is the strongest oxidant or the one with the more positive standard reduction. The reverse is true at the anode. So the key is FIND A TABLE OF STANDARD REDUCTION POTENTIALS Link to comment Share on other sites More sharing options...
albertlee Posted November 15, 2004 Author Share Posted November 15, 2004 Yes.... I know there is a table for that.... Does that mean the more reactive element the more willing to give up electrons?? if so, if both elements have same amount of electrons at the outer shell, how do you determine their reactivity?? can we refer this to their electronegativity?? if so, how?? thx for any respond!! Albert Link to comment Share on other sites More sharing options...
budullewraagh Posted November 15, 2004 Share Posted November 15, 2004 the word isnt "reactive" it's "active". the more active metals will be more apt to give up electrons, while the more active nonmetals will be more apt to gain electrons. activity depends on reduction potentials. reduction potentials are based on many factors including effective nuclear charge, atomic radius, etc Link to comment Share on other sites More sharing options...
chadn Posted November 15, 2004 Share Posted November 15, 2004 Does that mean the more reactive element the more willing to give up electrons?? Depends on which half-reaction your talking about. In the half-reaction for the cathode the more reactive element will be the one that gains electrons the easiest. For the anode the more reactive element is the more willing to give up electrons. if so, if both elements have same amount of electrons at the outer shell, how do you determine their reactivity?? can we refer this to their electronegativity?? if so, how?? I think its related to their ionization energies, but that may be wrong. Link to comment Share on other sites More sharing options...
budullewraagh Posted November 15, 2004 Share Posted November 15, 2004 ionization energy doesnt explain the physics of it all, however. again, effective nuclear charge is the greatest factor Link to comment Share on other sites More sharing options...
chadn Posted November 15, 2004 Share Posted November 15, 2004 ionization energy doesnt explain the physics of it all Sure it does, ionization energy combines the different factors such as atomic radii and nuclear charge into something more meaningful. again, effective nuclear charge is the greatest factor Not really, the greatest factor would be distance from the nucleus. nuclear charge would be the next major factor. Link to comment Share on other sites More sharing options...
budullewraagh Posted November 16, 2004 Share Posted November 16, 2004 Not really, the greatest factor would be distance from the nucleus. nuclear charge would be the next major factor. effective nuclear charge is calculated based on nuclear charge as well as atomic radius... Link to comment Share on other sites More sharing options...
chadn Posted November 16, 2004 Share Posted November 16, 2004 effective nuclear charge is calculated based on nuclear charge as well as atomic radius... My bad, all I saw was nuclear charge. While this may help explain the physics of electron removal I think ionization energy is still far more practical for this topic. Link to comment Share on other sites More sharing options...
budullewraagh Posted November 16, 2004 Share Posted November 16, 2004 it depends on how specific you wish to get Link to comment Share on other sites More sharing options...
albertlee Posted November 16, 2004 Author Share Posted November 16, 2004 by the way, what is the difference of "active" and "reactive"?? Albert Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now