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Which one of the following metals will not reduce \(\mathrm{H}_{2} \mathrm{O} ?\) (a) Li (b) \(\mathrm{Fe}\) (c) \(\mathrm{Cu}\) (d) \(\mathrm{Ca}\)

Short Answer

Expert verified
Copper (Cu) will not reduce water.

Step by step solution

01

Determine Reactivity Series

Identify the reactivity of the metals listed in terms of their ability to react with water. Metals that are higher in the reactivity series can reduce water to produce hydrogen gas.
02

Reactivity Analysis: Li and Ca

Lithium (Li) and calcium (Ca) are highly reactive metals. They are capable of reacting with water to produce hydrogen gas, hence they can reduce water.
03

Reactivity Analysis: Fe

Iron (Fe) is less reactive compared to lithium and calcium, but it can still react with steam (not cold water) to produce hydrogen gas. Therefore, it can reduce water under certain conditions.
04

Reactivity Analysis: Cu

Copper (Cu) is a low reactivity metal and is below hydrogen in the reactivity series. It does not react with water or steam to produce hydrogen gas.
05

Identify the Metal that Does Not Reduce Water

Since copper (Cu) cannot react with water to produce hydrogen, it will not reduce H_2O. The other metals (Li, Fe, and Ca) can react with water or steam to some degree to produce hydrogen gas.

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Key Concepts

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

Metal Reactivity with Water
Understanding how different metals react with water is essential in chemistry. Metals vary in their reactivity, and this determines whether they will interact with water to form hydrogen gas. The reactivity series ranks metals based on their ability to displace hydrogen from water.

Here's how it works: metals at the top of the reactivity series, like lithium (Li) and calcium (Ca), react vigorously with water, even at room temperature. These reactions produce hydrogen gas and metal hydroxides.
  • Lithium + Water \( ightarrow\) Lithium Hydroxide + Hydrogen gas
  • Calcium + Water \( ightarrow\) Calcium Hydroxide + Hydrogen gas
Less reactive metals, such as iron (Fe), don't react with water at room temperature. However, they react with steam. This is because the higher temperature provides the energy needed for the reaction.

Copper (Cu), a low reactivity metal, does not react with water or even steam. It means it is more stable and remains unaffected in water conditions. Remembering these patterns helps predict metal behavior in chemical reactions.
Hydrogen Production
Hydrogen gas is produced during the reaction of highly reactive metals with water. This type of reaction is characterized by a visible fizzing or bubbling as hydrogen gas is released.

The basic reaction looks like this:
  • Metal + Water \( ightarrow\) Metal Hydroxide + Hydrogen gas
Hydrogen production during these reactions is important for various applications. For instance:
  • Energy Storage: Hydrogen is a clean fuel source, often used in fuel cells to power vehicles.
  • Chemical Synthesis: It's a key reactant in making ammonia and methanol, substances used industrially.
Keep in mind that not all metals can produce hydrogen this way. Metals lower in the reactivity series, such as copper, do not engage in these reactions under normal conditions.
This connection between reactivity and hydrogen production is crucial for many scientific and industrial processes.
Reduction by Metals
Reduction is a chemical process where an element gains electrons, often from a metal. In terms of water ( \(H_2O \)), it involves the transformation into hydrogen gas. When a metal reduces water, it donates electrons to hydrogen ions (from water), turning them into hydrogen gas.

Key points to remember:
  • Only metals higher than hydrogen in the reactivity series can reduce water to hydrogen gas.
  • Metals like copper, which are below hydrogen, cannot perform this reduction as they are less willing to give up electrons.
Reduction by metals is foundational for understanding many electrochemical reactions, including how batteries and corrosion work. By knowing which metals can perform reductions, we can determine their potential uses in various chemical processes.

Reactivity and reduction capacity help us predict how different metals will behave in reactions, providing insights into their practical applications.

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Most popular questions from this chapter

The correct order of ionic radii of \(\mathrm{Y}^{3+}, \mathrm{La}^{3+}, \mathrm{Eu}^{3+}\) and \(\mathrm{Lu}^{3+}\) is (a) \(\mathrm{Lu}^{3+}<\mathrm{Eu}^{3+}<\mathrm{La}^{3+}<\mathrm{Y}^{3+}\) (b) \(\mathrm{La}^{3+}<\mathrm{Eu}^{3+}<\mathrm{Lu}^{3+}<\mathrm{Y}^{3+}\) (c) \(\mathrm{Y}^{3+}<\mathrm{La}^{3+}<\mathrm{Eu}^{3+}<\mathrm{Lu}^{3+}\) (d) \(\mathrm{Y}^{3+}<\mathrm{Lu}^{3+}<\mathrm{Eu}^{3+}<\mathrm{La}^{3+}\)

The outer electron configuration of Gd (atomic number \(=64\) ) is: (a) \(4 f^{7} 5 \mathrm{~d}^{1} 6 \mathrm{~s}^{2}\) (b) \(4 \mathrm{f}^{3} 5 \mathrm{~d}^{2} 6 \mathrm{~s}^{2}\) (c) \(4 \mathrm{f}^{5} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{2}\) (d) \(4 \mathrm{f}^{4} 5 \mathrm{~d}^{4} 6 \mathrm{~s}^{2}\)

Excess of \(\mathrm{KI}\) react with \(\mathrm{CuSO}_{4}\) solution and then \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution is added to it. Which of the following statements is incorrect for this reaction? (a) \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) is formed (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is oxidized (c) \(\mathrm{CuI}_{2}\) is formed (d) evolved \(\mathrm{I}_{2}\) is reduced

Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being (a) 4 f-orbitals more diffused than the 5 f-orbitals (b) Lesser energy difference between \(5 \mathrm{f}\) and 6 d than between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\) orbitals (c) More energy difference between \(5 \mathrm{f}\) and \(6 \mathrm{~d}\) than between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\) orbitals (d) More reactive nature of the actinoids than the lanthanoids.

Among the properties (A) reducing (B) oxidizing (C) complexing, the set of properties shown by CN ion towards metal species is (a) A, B (b) \(\mathrm{B}, \mathrm{C}\) (c) \(\mathrm{C}, \mathrm{A}\) (d) A, B, C

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