Chemistry, 05.07.2019 09:00 aj12381

Determine the oxidizing agent in the following reaction. ni(s) + 2 agclo4(aq) → ni(clo4)2(aq) + 2 ag(s)

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Answer from: hollandhogenson

The oxidizing agent in the reaction is silver perchlorate.

Given:

A chemical reaction between nickel and silver perchlorate:

$Ni(s) + 2 AgClO_4(aq) \rightarrow Ni(ClO_4)_2(aq) + 2 Ag(s)$

To find:

The oxidizing agent in the reaction.

Solution:

$Ni(s) + 2 AgClO_4(aq) \rightarrow Ni(ClO_4)_2(aq) + 2 Ag(s)$

The oxidation state of nickel in nickel metal = 0The oxidation state of nickel in nickel perchlorate = +2

The oxidation state of nickel changing from 0 to +2 which means that it had undergone an oxidation reaction.

The oxidation state of silver in silver perchlorate = +1The oxidation state of silver in silver metal = 0

The oxidation state of silver changing from +1 to 0 which means that it had undergone a reduction reaction.

The substance here undergoing reduction and oxidizing the other is a silver ion, hence, silver perchlorate is an oxidizing agent.The substance here undergoing oxidation and reducing the other is nickel, hence, nickel metal is a reducing agent.

The oxidizing agent in the reaction is silver perchlorate.

Learn more about an oxidizing agent and a reducing agent here:

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Answer from: dej33

Silver perchlorate $\text{AgClO}_4$ serves as the oxidizing agent in this reaction.

Explanation

Both silver perchlorate $\text{AgClO}_4$ in the reactants and nickel (II) perchlorate $\text{Ni}(\text{ClO}_4)_2$ in the products exists as ionic salt dissolved in an aqueous solution. Rewriting the chemical equation as an ionic one might help eliminate spectator ions and make changes in oxidation states more apparent.

Ionic equation:

$\text{Ni} \; (s) + 2 \; \text{Ag}^{+} \; (aq) + 2\; \text{ClO}_4 \; (aq) \to \text{Ni}^{2+} \; (aq) + 2\; \text{ClO}_4^{-} \; (aq) + 2 \; \text{Ag} \; (aq)$

Two perchlorate ions are present on both sides of the equation. These ions act as spectator ions that take no part in the overall reaction. Thus it is possible to determine the oxidizing and reducing agent without knowing the oxidation state of atoms in the perchlorate ion. Marking oxidation states on top of each element gives the following equation:

$\stackrel{0}{\text{Ni}} \; (s) + 2 \; \stackrel{+1}{\text{Ag}^{+}} \; (aq)\to \stackrel{+2}{\text{Ni}^{2+}} \; (aq) + 2 \; \stackrel{0}{\text{Ag}} \; (aq)$

There is a decrease in the oxidation state of silver, from +1 to 0. Silver atoms, therefore, gain electrons and are reduced. The reactant containing silver ions, silver perchlorate, thus acts as the oxidation agent of this reaction.

Answer from: Quest

answer: cellular respiration takes in food and uses it to create atp, a chemical which the cell uses for energy. usually, this process uses oxygen, and is called aerobic respiration. it has four stages known as glycolysis, link reaction, the krebs cycle, and the electron transport chain.

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.