![]() ![]() Another technique is to 'seed' the solution by adding a small crystal of the desired pure solid to the cooled solution. The scratch increases the glass surface area, providing a roughened surface on which the solid can crystallize. It's possible to aid crystallization by scratching the flask with a glass rod at the air-solvent junction (assuming you are willing to purposely scratch your glassware). #"Pb"_ ((s)) + "Sn"_ ((aq))^(2+) -> "N.R.Slower cooling may lead to a higher purity product, so it's common practice to allow the solution to cool to room temperature before setting the flask in an ice bath or refrigerator.Ĭrystals usually begin forming on the bottom of the flask. #"Sn"_ ((aq))^(2+)-># a stronger oxidizing agent is being converted to #"Sn" _((s))#, a weaker reducing agent #"Zn"_ ((s)) -># a stronger reducing agent is being converted to #"Zn"_ ((aq))^(2+)#, a weaker oxidizing agent Looking at this reaction, you can say that you have You can write out the net ionic equation for the reaction between zinc and aqueous tin(II) chloride This implies that the lead(II) cations are actually stronger oxidizing agents than the zinc cations. Option (E) is also #color(red)("incorrect")# because zinc is oxidized to zinc cations in solution, whereas lead is not. Option (D) is #color(red)("incorrect")# because zinc does not act as an oxidizing agent in the reaction, so saying that tin is a stronger oxidizing agent than a reducing agent doesn't really make sense to me. ![]() Option (C) is #color(green)("correct")# because zinc manages to reduce the tin(II) cations to tin metal but lead does not, and so zinc is indeed a stronger reducing agent than lead. Option (B) is also #color(red)("incorrect")# because tin does not act as a reducing agent when paired with zinc, it acts as an oxidizing agent. Option (A) is #color(red)("incorrect")# because lead metal is actually a weaker reducing agent than zinc metal. In other words, lead metal is unable to reduce tin(II) cations to tin metal, or tin(II) cations are unable to oxidize lead metal to lead(II) cations, #"Pb"^(2+)#. Moreover, judging from the options given to you, I'd say that when this happens, no reaction takes place. Now, I think that the second part of the question features a lead rod being placed in a tin(II) chloride aqueous solution. In other words, the tin(II) cations act as an oxidizing agent because they oxidize zinc metal to zinc cations. So zinc acts as a reducing agent because it reduces tin(II) cations to tin metal. This means that it's gaining electrons, which must mean that it's being reduced. Likewise, tin is going from a #2+# oxidation state in tin(II) chloride to being precipitated as tin metal. You know that when a zinc rod is placed in a tin(II) chloride, #"SnCl"_2#, aqueous solution, tin will precipitate on the zinc rod and zinc will go into solution as zinc cations, #"Zn"^(2+)#.Įven without writing a balanced chemical equation, you can look at the info given and say that because zinc is losing electrons to form zinc cations, that must mean that it's being oxidized. Now, you can answer the question intuitively. No mention of lead's possible reaction was made in the question, yet lead is mentioned in the answers, so I can only assume that you're missing some information. Based on the answer options given to you, it appears as though the question is incomplete.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |