# 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?

## If a lead rod is put into the same tin(II)chloride solution, no precipitation is observed. ${\text{Zn"_ ((s)) + "SnCl"_(2(aq)) -> "Sn"_ ((s)) + "ZnCl}}_{2 \left(a q\right)}$ 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

Jun 13, 2016

Here's what I got.

#### Explanation:

Based on the answer options given to you, it appears as though the question is incomplete.

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.

Now, you can answer the question intuitively. You know that when a zinc rod is placed in a tin(II) chloride, ${\text{SnCl}}_{2}$, aqueous solution, tin will precipitate on the zinc rod and zinc will go into solution as zinc cations, ${\text{Zn}}^{2 +}$.

Even 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.

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.

So zinc acts as a reducing agent because it reduces tin(II) cations to tin metal. In other words, the tin(II) cations act as an oxidizing agent because they oxidize zinc metal to zinc cations.

Now, I think that the second part of the question features a lead rod being placed in a tin(II) chloride aqueous solution. Moreover, judging from the options given to you, I'd say that when this happens, no reaction takes place.

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, ${\text{Pb}}^{2 +}$.

If this is the case, you can say that

• Option (A) is $\textcolor{red}{\text{incorrect}}$ because lead metal is actually a weaker reducing agent than zinc metal.

• Option (B) is also $\textcolor{red}{\text{incorrect}}$ because tin does not act as a reducing agent when paired with zinc, it acts as an oxidizing agent.

• Option (C) is $\textcolor{g r e e n}{\text{correct}}$ 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 $\textcolor{red}{\text{incorrect}}$ 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 $\textcolor{red}{\text{incorrect}}$ 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.

You can write out the net ionic equation for the reaction between zinc and aqueous tin(II) chloride

${\text{Zn"_ ((s)) + "Sn"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Sn}}_{\left(s\right)}$

Looking at this reaction, you can say that you have

${\text{Zn}}_{\left(s\right)} \to$ a stronger reducing agent is being converted to ${\text{Zn}}_{\left(a q\right)}^{2 +}$, a weaker oxidizing agent

${\text{Sn}}_{\left(a q\right)}^{2 +} \to$ a stronger oxidizing agent is being converted to ${\text{Sn}}_{\left(s\right)}$, a weaker reducing agent

$\text{Pb"_ ((s)) + "Sn"_ ((aq))^(2+) -> "N.R.}$
${\text{Pb}}_{\left(s\right)} \to$ a weaker reducing agent is not being converted to a stronger oxidizing agent
${\text{Sn}}_{\left(a q\right)}^{2 +} \to$ a weaker oxidizing agent is not being converted to a stronger reducing agent