Bubbles of \(\mathrm{H}_{2}\) form when metal \(\mathrm{D}\) is placed in hot \(\mathrm{H}_{2} \mathrm{O}\). No reaction occurs when \(\mathrm{D}\) is placed in a solution of a salt of metal E, but D is discolored and coated immediately when placed in a solution of a salt of metal \(\mathrm{F}\). What happens if \(\mathrm{E}\) is placed in a solution of a salt of metal F? Rank metals \(D, \mathrm{E},\) and \(\mathrm{F}\) in order of increasing reducing strength.

Short Answer

Expert verified
Metal D < Metal E < Metal F in terms of reducing strength. No reaction occurs when E is placed in a solution of a salt of F.

Step by step solution

01

- Identify the Reducing Strength Relative to Hydrogen

The metal D forms hydrogen gas bubbles when placed in hot water, indicating that it is a stronger reducing agent than hydrogen. This means metal D can reduce water to release hydrogen.
02

- Analyze the Interaction Between D and E

No reaction occurs when metal D is placed in a solution of a salt of metal E. This suggests that metal D cannot displace metal E from its salt, implying that metal E is a stronger reducing agent than metal D.
03

- Analyze the Interaction Between D and F

Metal D is discolored and coated when placed in a solution of a salt of metal F. This indicates that metal F can displace metal D from its salt, making metal F a stronger reducing agent than metal D.
04

- Analyze the Interaction between E and F

Given that metal D cannot displace metal E, but metal F can displace metal D, metal F is a stronger reducing agent than both metal D and metal E. When metal E is placed in a solution of a salt of metal F, no reaction will occur because metal F is the strongest reducing agent among the three.
05

- Rank the Metals

Based on the analyses, the order of increasing reducing strength is: Metal D < Metal E < Metal F.

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Key Concepts

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

reducing strength
In chemistry, reducing strength refers to the ability of a substance to donate electrons during a chemical reaction. The stronger a reducing agent, the more easily it loses electrons to other substances.

In our exercise, metal D forms hydrogen gas bubbles when placed in hot water. This means that metal D is a strong enough reducing agent to reduce water by giving electrons, thus releasing hydrogen gas.

Next, we saw that metal D could not react with a salt solution of metal E, indicating that metal D is a weaker reducing agent than metal E.

Finally, since metal D gets coated and discolored in a salt solution of metal F, metal F is a stronger reducing agent than metal D. By placing E in a solution of F's salt, no reaction happens, showing that metal F is still stronger.
displacement reaction
Displacement reactions happen when a more reactive metal displaces a less reactive metal from its compound. These reactions tell us a lot about the reactivity and reducing strength of metals.

When metal D is placed in hot water, it displaces the hydrogen, indicating its reactivity.
In the scenario with metal E's salt, no reaction means metal D cannot displace E, showing E is more reactive.

The discoloration of metal D in F's salt solution demonstrates a displacement reaction where metal F is clearly more reactive and reduces D ions by taking their place.
metal reactivity
The metal reactivity series helps us understand how different metals will behave in reactions. This series arranges metals in order of decreasing reactivity.

In the provided exercise:
  • Metal D reacts with hot water and releases hydrogen, showing it’s quite reactive.
  • Metal D does not cause a reaction with metal E's salt, which means E is less reactive than D.
  • Metal F is highly reactive as evidenced by displacing D’s ions resulting in discoloration.
  • When metal E is placed in the salt of F, no reaction occurs.
Thus, the overall reactivity order from least to most would be D < E < F.

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Most popular questions from this chapter

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