Chemistry, 06.04.2020 23:31 stevenabdullatif16

For the given reactions, classify the reactants as the reducing agent, oxidizing agent, or neither.
F2 + H2 > 2HF
2Mg + O2 > 2MgO

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Answer from: duhitzmay4693

For $F_{2} + H_{2} \rightarrow 2HF$ , $F_{2}$ is the oxidizing agent and $H_{2}$ is the reducing agent.

For $2Mg + O_{2} \rightarrow 2MgO$ , Mg is the reducing agent and $O_{2}$ is the oxidizing agent.

Explanation:

An oxidizing agent is defined as the substance which tends to oxidize other substances by itself gaining electrons. As a result, there will occur a decrease in its oxidation state.

On the other hand, a substance which tends to reduce other substances by itself losing electrons is known as an reducing agent. There will occur an increase in oxidation state of a reducing agent.

For example, $F_{2} + H_{2} \rightarrow 2HF$

Oxidation: $H_{2} \rightarrow 2H^{+} + 2e^{-}$

Reduction: $F + 1e^{-} \rightarrow F^{-}$

Hence, here $F_{2}$ is the oxidizing agent and $H_{2}$ is the reducing agent.

Also,

$2Mg + O_{2} \rightarrow 2MgO$

Oxidation: $Mg \rightarrow Mg^{2+} + 2e^{-}$

Reduction: $O_{2} + 2e^{-} \rightarrow O$

Hence, here Mg is the reducing agent and $O_{2}$ is the oxidizing agent.

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.

Answer from: Quest

Hello.the answer is hydrosphere, geosphere, and biosphere have a nice day

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