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Chapter 21: Problem 50

When metal \(\mathrm{A}\) is placed in a solution of a salt of metal \(\mathrm{B}\), the surface of metal A changes color. When metal \(\mathrm{B}\) is placed in acid solution, gas bubbles form on the surface of the metal. When metal \(\mathrm{A}\) is placed in a solution of a salt of metal \(\mathrm{C},\) no change is observed in the solution or on the surface of metal A. (a) Will metal C cause formation of \(\mathrm{H}_{2}\) when placed in acid solution? (b) Rank metals \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) in order of decreasing reducing strength.

Short Answer

Expert verified
Yes, metal C will cause formation of \text{H}_2 in acid solution. Ranking: C > A > B.

Step by step solution

01

Identify the reactivity of metals A and B

When metal A is placed in a solution of a salt of metal B, it changes color, indicating a reaction. Therefore, metal A is more reactive than metal B. Additionally, when metal B is placed in an acid solution, gas bubbles form, indicating hydrogen gas formation and thus, metal B is more reactive than hydrogen.
02

Analyze the reactivity of metal A with metal C

When metal A is placed in a solution of a salt of metal C, no change is observed. This indicates that metal A is not more reactive than metal C, meaning metal C is more reactive than metal A.
03

Determine if metal C reacts with acid

Since metal C is more reactive than metal B, and metal B reacts with acid to form hydrogen gas, metal C will also react with acid, causing the formation of hydrogen gas.
04

Rank the metals in order of reducing strength

Based on the analysis, the decreasing order of reducing strength (reactivity) of the metals is: C > A > B.

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Key Concepts

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

reactivity series
The reactivity series is a way to rank metals based on their reactivity with other substances. This series lists metals from the most reactive to the least reactive. For example, potassium and sodium are at the top, meaning they react very easily, while metals like gold and silver are at the bottom because they are very unreactive.

In our exercise, we deduce the relative reactivity of metals A, B, and C. By observing how they react with each other and with acids, we can place them in a mini-reactivity series. This is crucial because it allows us to predict their behavior in chemical reactions.
metal-acid reactions
When metals react with acids, they usually produce a salt and hydrogen gas. The rate and occurrence of this reaction depend on the metal's position in the reactivity series.

For example:
  • If a metal is highly reactive, like Zinc or Magnesium, it will react quickly and produce hydrogen gas.
  • Metals such as Copper or Silver do not react with dilute acids because they are less reactive.
In our case, metal B reacts with acid to form hydrogen gas, meaning it is more reactive than hydrogen. Given that metal C is more reactive than B, we can confidently say that it will also react with acid to produce hydrogen.
displacement reactions
Displacement reactions involve a more reactive metal displacing a less reactive metal from its compound. This is often seen in reactions between metals and salt solutions.

In our problem, when metal A is placed in a solution of a salt of metal B, a reaction happens, suggesting metal A is more reactive than metal B. Conversely, no reaction occurs when metal A is placed in a salt solution of metal C, indicating metal C is more reactive than metal A.
  • Reactive metal + less reactive metal salt → reactive metal salt + less reactive metal
Understanding displacement reactions helps us establish the reactivity order: C > A > B.
reducing strength
The term reducing strength refers to a metal's ability to donate electrons and reduce other substances. A metal with high reducing strength is also highly reactive.

By understanding which metals can displace others and react with acids, we can determine their reducing strength.
  • More reactive metals like metal C have a stronger reducing strength than less reactive metals like B.
  • Thus, the order of metals based on reducing strength from our problem is: C > A > B, with C having the highest reducing strength.
Knowing reducing strength can help predict the outcomes of various redox reactions and refining processes in chemistry.

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Most popular questions from this chapter

A voltaic cell consists of \(\mathrm{A} / \mathrm{A}^{+}\) and \(\mathrm{B} / \mathrm{B}^{+}\) half-cells, where A and \(B\) are metals and the A electrode is negative. The initial \(\left[\mathrm{A}^{+}\right] /\left[\mathrm{B}^{+}\right]\) is such that \(E_{\text {cell }}>E_{\text {cell }}^{\circ}\) (a) How do \(\left[\mathrm{A}^{+}\right]\) and \(\left[\mathrm{B}^{+}\right]\) change as the cell operates? (b) How does \(E_{\text {cell }}\) change as the cell operates? (c) What is \(\left[\mathrm{A}^{+}\right] /\left[\mathrm{B}^{+}\right]\) when \(E_{\text {cell }}=E_{\text {cell }}^{\circ} ?\) Explain. (d) Is it possible for \(E_{\text {cell to be less than } E_{\text {cell }}^{\circ} \text { ? Explain. }}\)

In basic solution, \(\mathrm{Se}^{2-}\) and \(\mathrm{SO}_{3}^{2-}\) ions react spontaneously: \(2 \mathrm{Se}^{2-}(a q)+2 \mathrm{SO}_{3}^{2-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow\) $$2 \mathrm{Se}(s)+6 \mathrm{OH}^{-}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \quad E_{\mathrm{cell}}^{\circ}=0.35 \mathrm{~V}$$ (a) Write balanced half-reactions for the process. (b) If \(E_{\text {sulfite }}^{\circ}\) is \(-0.57 \mathrm{~V},\) calculate \(E_{\text {selenium }}^{\circ}\)

A voltaic cell is constructed with an \(\mathrm{Sn} / \mathrm{Sn}^{2+}\) half- cell and a \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half-cell. The zinc electrode is negative. (a) Write balanced half-reactions and the overall cell reaction. (b) Diagram the cell, labeling electrodes with their charges and showing the directions of electron flow in the circuit and of cation and anion flow in the salt bridge.

What does a negative \(E_{\mathrm{cell}}^{\circ}\) indicate about a redox reaction? What does it indicate about the reverse reaction?

In Appendix \(D,\) standard electrode potentials range from about \(+3 \mathrm{~V}\) to \(-3 \mathrm{~V}\). Thus, it might seem possible to use a half- cell from each end of this range to construct a cell with a voltage of approximately 6 V. However, most commercial aqueous voltaic cells have \(E^{\circ}\) values of \(1.5-2 \mathrm{~V}\). Why are there no aqueous cells with significantly higher potentials?
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