Problem 91
Question
Four metals \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) exhibit these properties: (a) Only \(\mathrm{A}\) and \(\mathrm{C}\) react with \(1.0-\mathrm{M} \mathrm{HCl}\) to give \(\mathrm{H}_{2}\) gas. (b) When \(\mathrm{C}\) is added to solutions of ions of the other metals, metallic \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{D}\) are formed. (c) Metal D reduces \(\mathrm{B}^{n+}\) ions to give metallic \(\mathrm{B}\) and \(\mathrm{D}^{n+}\) ions. On the basis of this information, arrange the four metals in order of increasing ability to act as reducing agents.
Step-by-Step Solution
Verified Answer
The order is B, D, A, C from least to most reducing ability.
1Step 1: Understand Reactivity with HCl
The metals A and C react with 1.0-M HCl to produce H2 gas, indicating they are more reactive metals capable of displacing hydrogen ions from HCl. B and D do not react with HCl.
2Step 2: Analyze Metal Reduction of Ions
When C is added to solutions containing ions of A, B, or D, the metallic form of A, B, and D are formed. This indicates that C can reduce these ions, making C a strong reducing agent compared to A, B, and D.
3Step 3: Understand Reducing Power from D
Metal D reduces B^n+ ions to produce metallic B and D^n+ ions. This shows that D is a stronger reducing agent than B, as it donates electrons to B^n+ to form metallic B.
4Step 4: Infer Order of Reducing Ability
Combining the above information, the order of reducing ability (from least to most) is B, D, A, C. B is the least effective because it is reduced by D, while A and C's reactivity with HCl and C's ability to reduce all other metals indicate they are stronger reducing agents.
Key Concepts
Reactivity with Hydrochloric AcidMetal ReductionOrder of Reducing AgentsRedox Reactions
Reactivity with Hydrochloric Acid
When metals are introduced to hydrochloric acid (
HCl
), a fascinating chemical interaction can occur. Metals like
A
and
C
, which react with
1.0-M HCl
to release
H_2
gas, exhibit a higher reactivity. This means that these metals have the ability to displace hydrogen ions (
H^+
) from the acid, resulting in the formation of hydrogen gas. This reaction
can be represented as follows:
- Metal + 2HCl -> Metal chloride + H_2 gas
Metal Reduction
The concept of metal reduction involves one metal causing another metal to return to its metallic state by donating electrons. When metal
C
is added to solutions containing ions of
A
,
B
, or
D
, the metallic forms of these metals (
A
,
B
, and
D
) are reformed. This process highlights the capability of
C
as a powerful reducing agent.
When metal
C
intervenes, it supplies electrons to the metallic ions present in the solution, allowing them to regain their neutral metallic form.
- C + Metal ions -> Cation + Metal
Order of Reducing Agents
The order of reducing agents is determined based on their ability to cause reduction (i.e., donation of electrons to other substances). In our exercise:
- Metal
C
, being able to reduce the ions of
A
,
B
, and
D
to their metallic forms, signifies that it's the strongest reducing agent.
- Metal
D
, capable of reducing
B^{n+}
ions to metallic
B
, is stronger
than
B
, though not as potent as
C.
It ranks higher than
B
due to its ability to donate electrons more effectively.
Through these observations, we craft an order for these metals based on their reducing power:
- B < D < A < C
Redox Reactions
Redox reactions are the heart of many chemical processes, involving the transfer of electrons between substances. These reactions are essential in understanding the behavior of reducing agents and oxidizing agents. In such reactions:
- The substance that loses electrons is undergoing oxidation and acts as a reducing agent.
- The substance that gains electrons is reduced and is known as the oxidizing agent.
In our scenario:
- Metal C acts as a reducing agent by losing electrons to the ions of A , B , and D , thereby helping them regain their metallic forms.
- Metal D reduces B^{n+} ions, indicating that it undergoes oxidation to donate electrons.
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