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Vlada [557]
3 years ago
13

Study this chemical reaction: (aq)(s)(s)(aq) Then, write balanced half-reactions describing the oxidation and reduction that hap

pen in this reaction.
a. oxidation
b. reduction:
Chemistry
1 answer:
disa [49]3 years ago
6 0

The question is incomplete, complete question is:

Study this chemical reaction:

FeSO_4 (aq) + Zn (s)\rightarrow Fe (s) + ZnSO_4 (aq)

Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.

Oxidation:

Reduction:

Answer:

Oxidation taking place in given reaction :

Zn(s)\rightarrow Zn^{2+}+2e^-

Reduction taking place in given reaction;

Fe^{2+}(aq)+2e^-\rightarrow Fe(s)

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction reaction occur side by side.

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

FeSO_4 (aq) + Zn (s) \rightarrow Fe (s) + ZnSO_4 (aq)

In the given reaction, iron(II) ions are getting reduced and zinc metal is getting oxidized to zinc(II) ions.

Oxidation :

Zn(s)\rightarrow Zn^{2+}+2e^-

Reduction ;

Fe^{2+}(aq)+2e^-\rightarrow Fe(s)

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Please help me
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Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

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- log [H⁺] = pH → 6.999

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Answer:

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Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

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Best regards!

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3 years ago
Write a balanced half-reaction describing the reduction of aqueous vanadium(V) cations to aqueous vanadium(I) cations.
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Explanation:

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