If a small zinc rod is put into a solution of tin(II) chloride, #"SnCl"_2#, tin will precipitate on the zinc rod and zinc will go into solution as #"Zn"^(2+)# ions.
(There's more) What is your conclusion?

Question

If a lead rod is put into the same tin(II)chloride solution, no precipitation is observed.

#"Zn"_ ((s)) + "SnCl"_(2(aq)) -> "Sn"_ ((s)) + "ZnCl"_ (2(aq))#

What is your conclusion?

(A) Lead is a stronger reductor than zinc
(B) Tin is a stronger reductor than zinc
(C) Zinc is a stronger reductor than lead
(D) Tin is a stronger oxidator than zinc
(E) Zn2+ ions are a stronger oxidator than Pb2+ ions

Jeremiah Adams · Accepted Answer

Here's what I got.

It seems that the question is not fully answered based on the options that you were provided. I can only assume that you're missing some information because lead is mentioned in the answers but was not mentioned in the question regarding lead's potential reaction. Now, you can answer the question intuitively. 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+)#. You can infer from the information provided that zinc is being oxidized because it is losing electrons to form zinc cations, even in the absence of a balanced chemical equation. Likewise, tin is going from a #2+# oxidation state in tin(II) chloride to being precipitated as tin metal. This means that it's gaining electrons, which must mean that it's being reduced. In other words, the tin(II) cations act as an oxidizing agent because they oxidize zinc metal to zinc cations, and zinc acts as a reducing agent because it reduces tin(II) cations to tin metal. Now, based on the options provided, I believe that the second part of the question involves the placement of a lead rod in an aqueous solution containing tin(II) chloride. Furthermore, I believe that no reaction occurs in this scenario. 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+)#. In such a scenario, you could state that Option (A) is #color(red)("incorrect")# because lead metal is actually a weaker reducing agent than zinc metal. 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 (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 (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 (E) is also #color(red)("incorrect")# because zinc is oxidized to zinc cations in solution, whereas lead is not. This implies that the lead(II) cations are actually stronger oxidizing agents than the zinc cations. The net ionic equation for the reaction of zinc with aqueous tin(II) chloride can be written down. #"Zn"_ ((s)) + "Sn"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Sn"_ ((s))# Considering this response, you could say that you have #"Zn"_ ((s)) -># a stronger reducing agent is being converted to #"Zn"_ ((aq))^(2+)#, a weaker oxidizing agent #"Sn"_ ((aq))^(2+)-># a stronger oxidizing agent is being converted to #"Sn" _((s))#, a weaker reducing agent You would have to have lead #"Pb"_ ((s)) + "Sn"_ ((aq))^(2+) -> "N.R."# This indicates that you've #"Pb"_ ((s)) -># a weaker reducing agent is not being converted to a stronger oxidizing agent #"Sn"_ ((aq))^(2+) -># a weaker oxidizing agent is not being converted to a stronger reducing agent

Henry Antrim · Answer

undefined The conclusion is that when a small zinc rod is placed into a solution of tin(II) chloride, tin will precipitate onto the zinc rod while zinc ions will be released into the solution. This reaction demonstrates the principle of displacement, where a more reactive metal displaces a less reactive metal from its compound in solution.

If a small zinc rod is put into a solution of tin(II) chloride, #"SnCl"_2#, tin will precipitate on the zinc rod and zinc will go into solution as #"Zn"^(2+)# ions. (There's more) What is your conclusion?