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The following reactions occur spontaneously. $$ \begin{aligned} 2 \mathrm{Y}+3 \mathrm{Ni}^{2+} & \longrightarrow 2 \mathrm{Y}^{3+}+3 \mathrm{Ni} \\ 2 \mathrm{Mo}+3 \mathrm{Ni}^{2+} & \longrightarrow 2 \mathrm{Mo}^{3+}+3 \mathrm{Ni} \\ \mathrm{Y}^{3+}+\mathrm{Mo} \longrightarrow & \mathrm{Y}+\mathrm{Mo}^{3+} \end{aligned} $$ List the metals Y, \(\mathrm{Ni}\), and Mo in order of increasing ease of oxidation.

Short Answer

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The metals in order of increasing ease of oxidation are \text{Ni}, \text{Y}, \text{Mo}.

Step by step solution

01

Identify the Oxidation and Reduction Reactions

First, we identify the oxidation and reduction in the given reactions. Oxidation is the loss of electrons (increase in oxidation state), and reduction is the gain of electrons (decrease in oxidation state). In the first reaction, Y is oxidized as it goes from Y to Y^{3+}, while Ni^{2+} is reduced to Ni. In the second reaction, Mo is oxidized as it goes from Mo to Mo^{3+}, while Ni^{2+} is again reduced to Ni. In the third reaction, Y^{3+} is reduced to Y, while Mo is oxidized to Mo^{3+}.
02

Analyze the Spontaneity of Reactions

Since all reactions are given to be spontaneous, we know that the species that go from a neutral state to a charged state (oxidation) do so spontaneously, indicating they are easier to oxidize than the species being reduced in that reaction.
03

Order the Metals by Ease of Oxidation

From the first equation, Y is easier to oxidize than Ni^{2+}. From the second equation, Mo is easier to oxidize than Ni^{2+}. From the third equation, since Y^{3+} is being reduced and Mo is being oxidized, Mo is harder to oxidize than Y. Therefore, the ease of oxidation is Ni < Y < Mo.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
Understanding oxidation states is crucial in identifying which atoms are participating in redox reactions and determining if the reaction can occur spontaneously. An oxidation state, often known as oxidation number, is a measure of the degree of oxidation of an atom in a chemical compound. It is an indicator of the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic.

An atom's oxidation state can change during a reaction as electrons are transferred between atoms. For instance, when an atom loses an electron during a reaction, its oxidation state increases, indicating oxidation. Conversely, when an atom gains an electron, its oxidation state decreases, which is a sign of reduction. In the given exercise, you can see this evidenced in the reactions where Y, Mo, and Ni change their oxidation states as they undergo oxidation or reduction.

Identifying Oxidation States

First, look at the reactants and products to determine the initial and final oxidation states. For the metals in the exercise (Y, Ni, and Mo), the oxidation states change as follows: Y goes from 0 to +3, Ni goes from +2 to 0, and Mo goes from 0 to +3. Recognizing these changes allows us to follow electron transfer in the reactions and provides a foundation for predicting the reactivity of these metals.
Chemical Spontaneity
Chemical spontaneity is a concept denoting whether a reaction can proceed without needing to be driven by external energy. Spontaneous reactions occur due to the system's natural tendency to achieve a state of lower energy and increased disorder or entropy. In the context of redox reactions, spontaneous processes are often characterized by an exchange of electrons that naturally progresses towards equilibrium.

A reaction's spontaneity is influenced by the Gibb's free energy change, \( \Delta G \) - a thermodynamic quantity. If \( \Delta G \) is negative, the process is spontaneous, and if \( \Delta G \) is positive, the reaction is non-spontaneous and requires outside energy to proceed. For the reactions presented, it is indicated they occur spontaneously, implying that without intervention, the atoms of Y, Ni, and Mo will naturally undergo oxidation and reduction as described.

Factors Affecting Spontaneity

Several factors can affect the spontaneous nature of a redox reaction, including the relative ease of oxidation or reduction of the participating elements, their concentrations, and the environmental conditions such as temperature and pressure. By evaluating these factors, one can predict the likelihood of a reaction, which is especially helpful in applications like battery technology and corrosion prevention.
Oxidation and Reduction
Oxidation and reduction are two halves of a whole in redox (reduction-oxidation) chemistry. These processes involve the transfer of electrons between chemical species, altering their oxidation states and driving the overall reaction.

Oxidation refers to the loss of electrons by a molecule, atom, or ion, resulting in an increase in its oxidation state. It's easy to remember with the phrase 'LEO' from 'Loss of Electrons is Oxidation'. Conversely, reduction refers to the gain of electrons, which you can recall with 'GER' from 'Gain of Electrons is Reduction'. The two processes always occur together because when one species loses electrons, another must necessarily gain them.

Tracking Electron Movement

In the exercise, we observe that Y and Mo are oxidized as they lose electrons and their oxidation states increase, while Ni is reduced as it gains electrons and its oxidation state decreases. Redox reactions play a significant role in many biological processes, industrial applications, and everyday chemical occurrences (like rusting). By studying instances of oxidation and reduction in various reactions, we gain a deeper understanding of the chemical changes taking place and can utilize this knowledge in technologies like batteries, where control of redox reactions is essential for storing and releasing energy.

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Most popular questions from this chapter

A copper bar with a mass of \(12.340 \mathrm{~g}\) is dipped into \(255 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{AgNO}_{3}\) solution. When the reaction that occurs has finally ceased, what will be the mass of unreacted copper in the bar? If all the silver that forms adheres to the copper bar, what will be the total mass of the bar after the reaction?

Both calcium chloride and sodium chloride are used to melt ice and snow on roads in the winter. A certain company was marketing a mixture of these two compounds for this purpose. A chemist, wanting to analyze the mixture, dissolved \(2.463 \mathrm{~g}\) of it in water and precipitated calcium oxalate by adding sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) The calcium oxalate was carefully filtered from the solution, dissolved in sulfuric acid, and titrated with 0.1000 \(M \mathrm{KMnO}_{4}\) solution. The reaction that occurred was \(6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{MnO}_{4}^{-} \longrightarrow\) $$ 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O} $$ The titration required \(21.62 \mathrm{~mL}\) of the \(\mathrm{KMnO}_{4}\) solution. (a) How many moles of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) were present in the calcium oxalate precipitate? (b) How many grams of calcium chloride were in the original \(2.463 \mathrm{~g}\) sample? (c) What was the percentage by mass of calcium chloride in the sample?

Write balanced molecular, ionic, and net ionic equations for the reactions of the following metals with hydrochloric acid to give hydrogen plus the metal ion in solution. (a) Cobalt (gives \(\mathrm{Co}^{2+}\) ) (b) Cesium (gives \(\left.\mathrm{Cs}^{+}\right)\) (c) Zinc (gives \(\mathrm{Zn}^{2+}\) )

Sulfites are used worldwide in the wine industry as antioxidant and antimicrobial agents. However, sulfites have also been identified as causing certain allergic reactions suffered by asthmatics, and the FDA mandates that sulfites be identified on the label if they are present at levels of 10 ppm (parts per million) or higher. The analysis of sulfites in wine uses the "Ripper method" in which a standard iodine solution, prepared by the reaction of iodate and iodide ions, is used to titrate a sample of the wine. The iodine is formed in the reaction $$ \mathrm{IO}_{3}^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O} $$ The iodine is held in solution by adding an excess of \(\mathrm{I}^{-}\), which combines with \(\mathrm{I}_{2}\) to give \(\mathrm{I}_{3}^{-}\). In the titration, the \(\mathrm{SO}_{3}^{2-}\) is converted to \(\mathrm{SO}_{2}\) by acidification, and the reaction during the titration is $$ \mathrm{SO}_{2}+\mathrm{I}_{3}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{SO}_{4}^{2-}+3 \mathrm{I}^{-}+4 \mathrm{H}^{+} $$ Starch is added to the wine sample to detect the end point, which is signaled by the formation of a dark blue color when excess iodine binds to the starch molecules. In a certain analysis, \(0.0421 \mathrm{~g}\) of \(\mathrm{NaIO}_{3}\) was dissolved in dilute acid and excess NaI was added to the solution, which was then diluted to a total volume of \(100.0 \mathrm{~mL}\) A \(50.0 \mathrm{~mL}\) sample of wine was then acidified and titrated with the iodine- containing solution. The volume of iodine solution required was \(2.47 \mathrm{~mL}\). (a) What was the molarity of the iodine (actually, \(\left.\mathrm{I}_{3}^{-}\right)\) in the standard solution? (b) How many grams of \(\mathrm{SO}_{2}\) were in the wine sample? (c) If the density of the wine was \(0.96 \mathrm{~g} / \mathrm{mL}\), what was the percentage of \(\mathrm{SO}_{2}\) in the wine? (d) Parts per million (ppm) is calculated in a manner similar to percent (which is equivalent to parts per hundred). $$ \mathrm{ppm}=\frac{\text { grams of component }}{\text { grams of sample }} \times 10^{6} \mathrm{ppm} $$ What was the concentration of sulfite in the wine, expressed as parts per million \(\mathrm{SO}_{2} ?\)

In an acidic solution, permanganate ion reacts with tin(II) ion to give manganese(II) ion and tin(IV) ion. (a) Write a balanced net ionic equation for the reaction. (b) How many milliliters of \(0.230 M\) potassium permanganate solution are needed to react completely with \(40.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) tin(II) chloride solution?

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