![]() (a) Physical Appearance: Colour and Smell Some preliminary tests needs to be done before doing the analysis of cations. ![]() Qualitative Analysis of Cations Preliminary Tests Let us discuss about the Qualitative Analysis of Cations. There are separate procedures for detecting cations and anions, called the Cation Analysis and Anion Analysis. Testing with various reagents gives characteristic reaction of certain ions, which may be a colour change, a solid formation or any other visible changes. The common procedure for testing any unknown sample is to make its solution and test this solution with various reagents for the ions present in it. It is mainly concerned with the detection of ions in an aqueous solution of the salt. Qualitative analysis is a method of Analytical chemistry that deals with the determination of elemental composition of inorganic salts. #"Pb"_ ((s)) + "Sn"_ ((aq))^(2+) -> "N.R.Our objective is to determine the cation present in a given salt. #"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.
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