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goldfiish [28.3K]
3 years ago
11

Rusting is a chemical process that changed the strength and integrity of objects made of iron or iron alloys. Which of the follo

wing statements correctly describes the rusting process below?
4 Fe(s) + 3O2 (g) 2 Fe2O3 (s)

Iron is reduced to iron ions in this reaction.

The oxidizing agent in this reaction is iron metal.

Iron oxide is the reducing agent in this reaction.

Oxygen was reduced over the course of this reaction.
Chemistry
2 answers:
pishuonlain [190]3 years ago
6 0
Among the choices provided, the statement that correctly describes the rusting process below is that "Oxygen was reduced over the course of this reaction" as <span> iron can't be the oxidizing agent, because it is the one being oxidized.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
s2008m [1.1K]3 years ago
4 0

Answer : The correct option is, Oxygen was reduced over the course of this reaction.

Explanation :

Iron rusting : A chemical process in which the an iron nail react with water and oxygen to give iron oxide as a product. Rusting of iron is an oxidation-reduction reaction in which iron losses electrons to oxygen atom.

Oxidation reaction : A substance looses its electrons. In this oxidation state increases.

Reduction reaction : A substance gains electrons. In this oxidation state decreases.

The balanced chemical reaction for rusting of irons is,

4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)

Half reactions of oxidation and reduction are :

Oxidation : Fe(s)\rightarrow Fe^{3+}+3e^-

Reduction : \frac{1}{2}O_2+2e^-\rightarrow O^{2-}

From this reaction we conclude that the electrons are getting transferred from iron to oxygen.  That means in this reaction, oxygen is an oxidizing agent and iron is a reducing agent.

Hence, the correct option is, Oxygen was reduced over the course of this reaction.

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Answer:

3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-

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Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

BiO_3^-\rightarrow Bi^{3+}

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

(Bi^{5+}O_3)^-\rightarrow Bi^{3+}

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6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

6OH^-+6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-

Then, we accommodate the waters to obtain:

3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-

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