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The position of some metals in the electrochemical series in decreasing electropositive character is given as: \(\mathrm{Mg}>\mathrm{Al}>\mathrm{Zn}>\mathrm{Cu}>\mathrm{Ag} .\) What will happen if a copper spoon is used to stira solution of aluminium nitrate? (a) The spoon will get coated with aluminium. (b) An alloy of copper and aluminium is formed. (c) The solution becomes blue. (d) No chemical change will take place.

Short Answer

Expert verified
No chemical change will take place when a copper spoon is used to stir a solution of aluminum nitrate.

Step by step solution

01

Understanding Electrochemical Series

Analyze the given electrochemical series of metals in terms of their electropositive character. Electropositive character refers to how readily a metal atom can lose electrons to form positive ions, with a higher position in the series indicating a greater tendency to lose electrons.
02

Comparing Reactivity of Metals

Compare the positions of aluminum (Al) and copper (Cu) in the electrochemical series. Since aluminum is higher up in the series than copper, aluminum is more electropositive and thus more reactive than copper.
03

Determining the Feasibility of Reaction

Assess if a reaction is feasible when a less reactive metal (Cu) comes into contact with a solution of a more reactive metal's salt (Al(NO3)3). According to the electrochemical series, the more reactive metal (Al) can displace the less reactive metal (Cu) from its solution.
04

Predicting the Outcome of the Reaction

Since Cu is less reactive than Al, when a copper spoon is used to stir a solution of aluminum nitrate, the copper will not displace aluminum from its compound. This means there should be no chemical change observing that the electrochemically less active metal (Copper) is not able to displace the more active metal (Aluminum) from its compound.

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Key Concepts

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

Electropositive Character
Understanding the electropositive character of metals is fundamental in chemistry, especially when predicting the outcomes of reactions involving metal elements. The term 'electropositive character' refers to the ability of an element to donate electrons and form positive ions. In essence, a metal with a high electropositive character will readily lose electrons to achieve a stable electronic configuration.

In the electrochemical series, metals are arranged according to their ability to lose electrons, which is also related to their standard electrode potentials. The higher the metal is placed in this series, the more electropositive it is and the stronger its ability to be oxidized, meaning it can easily give up electrons. Magnesium (Mg), being higher in the series, is more electropositive than silver (Ag), for example.

This characteristic plays a pivotal role in many processes, including battery operation and metal extraction. For instance, metals with higher electropositive character are more likely to be oxidized during galvanic reactions, which forms the basis of how batteries generate electric current.
Reactivity of Metals
Closely tied to the concept of electropositive character is the reactivity of metals. Reactivity in this context refers to the ease and speed with which a metal can react to form compounds, especially with nonmetals such as oxygen and acids.

Reactivity is notably influenced by the placement of metals in the electrochemical series. Generally, metals that lose electrons more readily (showing higher electropositive character) are more reactive. This is because the loss of electrons signifies the tendency of a metal to react with other substances—electron donors or acceptors—to form new compounds.

As with electropositive character, a metal's reactivity is important when it comes to its uses and the ways in which it can be safely handled or stored. For example, a reactive metal such as sodium must be stored under oil to prevent it from reacting with moisture in the air. Trends in reactivity can predict if a metal can displace another in a solution, contribute to corrosion, or define how it reacts in different environmental conditions.
Chemical Displacement Reactions
Chemical displacement reactions, or single replacement reactions, are processes where one metal displaces another metal from a compound. The feasibility of such a reaction is preferentially determined by the reactivity of the metals involved; a more reactive metal can displace a less reactive metal from its compounds.

Using the electrochemical series, one can predict whether a chemical displacement reaction will occur. If a free element is more reactive than the element in a compound, the free element can replace the less reactive one. For instance, in the textbook example, when a copper spoon is used to stir a solution of aluminum nitrate, there is no chemical displacement because copper is less reactive than aluminum. Aluminum, therefore, cannot be displaced from its nitrate compound by copper.

These reactions are not only theoretically interesting but also have practical significance in areas such as metallurgy and industrial chemistry. They provide a base for extracting metals from their ores and are part of the mechanisms behind some types of corrosion and battery chemistry.

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Most popular questions from this chapter

\(\begin{array}{llll}\text { The molar conductivity of } & 0.10 & \mathrm{M}\end{array}\) solution of \(\mathrm{MgCl}_{2}\) is \(100 \mathrm{mho} \mathrm{cm}^{2} \mathrm{~mol}^{-1}\), at \(25^{\circ} \mathrm{C}\). A cell with electrodes that are \(1.50 \mathrm{~cm}^{2}\) in surface area and \(0.50 \mathrm{~cm}\) apart is filled with \(0.10 \mathrm{M}-\mathrm{MgCl}_{2}\) solution. How much current will flow when the potential difference between the electrodes is 5 volts? (a) \(0.03 \mathrm{~A}\) (b) \(3.0 \mathrm{~A}\) (c) \(0.15 \mathrm{~A}\) (d) \(15 \mathrm{~A}\)

A volume of \(100 \mathrm{~m}\) of a buffer of \(1 \mathrm{M}\) \(-\mathrm{NH}_{3}\) and \(1 \mathrm{M}-\mathrm{NH}_{4}^{+}\) is placed in two half-cells connected by a salt bridge. \(A\) current of \(1.5 \mathrm{~A}\) is passed through the cell for \(20 \mathrm{~min}\). If electrolysis of water takes place only and the electrode reactions are: Right: \(2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-}\) and Left: \(2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{H}^{+}+\mathrm{O}_{2}+4 \mathrm{e}\), then, the \(\mathrm{pH}\) of the (a) right electrode will increase (b) left electrode will increase (c) both electrode will increase (d) both electrode will decrease

What is the equilibrium constant of the reaction: \(2 \mathrm{Fe}^{3+}+\mathrm{Au}^{+} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Au}^{3+} ?\) Given \(E_{\mathrm{Au}^{+} \mid \mathrm{Au}}^{\circ}=1.68 \mathrm{~V}, E_{\mathrm{Au}^{3 \cdot} \mid \mathrm{Au}}^{\circ}=1.50 \mathrm{~V}\), \(E_{\mathrm{Fe}^{3 *} \mid \mathrm{Fe}^{\mathrm{e}}^{\circ}}=0.75 \mathrm{~V}\) and \(2.303 R T / F=0.06\) (a) \(1 \times 10^{22}\) (b) \(1 \times 10^{-22}\) (c) \(1 \times 10^{-11}\) (d) \(1 \times 10^{-72}\)

The number of Faradays required to produce 1 g-atom of \(\mathrm{Mg}\) from \(\mathrm{MgCl}_{2}\) is (a) \(\overline{1}\) (b) 2 (c) \(0.5\) (d) 4

The correct order of molar conductance at infinite dilution of \(\mathrm{LiCl}, \mathrm{NaCl}\) and \(\mathrm{KCl}\) is (a) \(\mathrm{LiCl}>\mathrm{NaCl}>\mathrm{KCl}\) (b) \(\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}\) (c) \(\mathrm{NaCl}>\mathrm{KCl}>\mathrm{LiCl}\) (d) \(\mathrm{LiCl}>\mathrm{KCl}>\mathrm{NaCl}\)

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