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Given the standard reduction potentials \(\mathrm{Zn}^{2+} / \mathrm{Zn}=\) \(-0.74 \mathrm{~V}, \mathrm{Cl}_{2} / \mathrm{Cl}^{-}=1.36 \mathrm{~V}, \mathrm{H}^{+} / 1 / 2 \mathrm{H}_{2}=0 \mathrm{~V}\) and \(\mathrm{Fe}^{2+} / \mathrm{Fe}^{3+}\) \(=0.77 \mathrm{~V}\). The order of increasing strength as reducing agent is (a) \(\mathrm{Zn}, \mathrm{H}_{2}, \mathrm{Fe}^{2+}, \mathrm{Cl}\) (b) \(\mathrm{H}_{2}, \mathrm{Zn}, \mathrm{Fe}^{2+}, \mathrm{Cl}^{-}\) (c) \(\mathrm{Cl}^{-}, \mathrm{Fe}^{2+}, \mathrm{Zn}, \mathrm{H}_{2}\) (d) \(\mathrm{Cl}^{-}, \mathrm{Fe}^{2+}, \mathrm{H}_{2}, \mathrm{Zn}\)

Short Answer

Expert verified
Option (d) is correct: \( \mathrm{Cl}^{-}, \mathrm{Fe}^{2+}, \mathrm{H}_{2}, \mathrm{Zn} \).

Step by step solution

01

Understand Reduction Potentials

Reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. A higher (more positive) reduction potential indicates a greater ability to gain electrons.
02

List Reduction Potentials

Given potentials are: \( \mathrm{Zn}^{2+} / \mathrm{Zn} = -0.74 \, \mathrm{V} \), \( \mathrm{Cl}_{2} / \mathrm{Cl}^{-} = 1.36 \, \mathrm{V} \), \( \mathrm{H}^{+} / 1/2 \mathrm{H}_{2} = 0 \, \mathrm{V} \), and \( \mathrm{Fe}^{2+} / \mathrm{Fe}^{3+} = 0.77 \, \mathrm{V} \).
03

Identify Reverse Reactions for Reducing Agents

To find the strength as a reducing agent, consider the reverse (oxidation) reactions of the given potentials. The more negative the reduction potential, the stronger the reducing agent.
04

Compare Their Reduction Potentials

Reducing strength order based on the reverse of the potentials: - Strongest reducing agent has the most negative potential. Compare: \( \mathrm{Zn} = -0.74 \, \mathrm{V} \) as a strong reducing agent, \( \mathrm{H}_{2} = 0 \, \mathrm{V} \), \( \mathrm{Fe}^{2+} = -0.77 \, \mathrm{V} \) as medium, and \( \mathrm{Cl}^{-} = -1.36 \, \mathrm{V} \) as weakest since it prefers to be reduced as \( \mathrm{Cl}_{2} .\)
05

Arrange in Order of Increasing Reducing Strength

Increasing strength as reducing agents based on potentials from less negative to more negative: \( \mathrm{Cl}^{-} \), \( \mathrm{Fe}^{2+} \), \( \mathrm{H}_{2} \), \( \mathrm{Zn} \).
06

Verify Correct Answer from Provided Options

From the given options, the correct increasing order of reducing agent strength is option (d): \( \mathrm{Cl}^{-}, \mathrm{Fe}^{2+}, \mathrm{H}_{2}, \mathrm{Zn} \).

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Key Concepts

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

Reducing Agents
Reducing agents play a crucial role in chemistry, particularly in redox reactions. A reducing agent is a substance that donates electrons to another substance, reducing that substance in the process. This means the reducing agent itself gets oxidized. The strength of a reducing agent is determined by how easily it loses electrons.
In the context of reduction potentials, a reducing agent is stronger if it has a more negative standard reduction potential. This is because a negative reduction potential indicates a greater tendency to lose electrons and thus drive oxidation. For instance, in the exercise, zinc, with a reduction potential of -0.74 V, is considered a strong reducing agent. Zinc readily donates electrons, facilitating the reduction of other substances.
Some properties of reducing agents include:
  • They tend to have low electronegativity, which means they don't hold onto their electrons tightly.
  • They are usually metals or substances with extra electrons they can donate.
  • Their strength varies with environmental conditions, such as pH and temperature, which can affect electron transfer capabilities.
Understanding the role and strength of reducing agents helps predict and control the outcomes of chemical reactions, making it easier to harness their properties in industrial and laboratory settings.
Electrochemistry
Electrochemistry is the study of chemical reactions that involve the transfer of electrons, and it is fundamental for understanding various processes and devices, such as batteries and electroplating. At the heart of electrochemistry lies the concept of the cell potential, which is determined by the reduction potentials of the half-reactions involved in a redox reaction.
In any electrochemical cell, the cell potential, or electromotive force (EMF), is calculated from the difference in the reduction potentials of the two half-reactions. The standard cell potential is given as:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]
Here, the cathode is where reduction occurs, and the anode is where oxidation takes place. A positive cell potential indicates that the redox reaction will proceed spontaneously under standard conditions.
Electrochemistry also helps in finding the direction of electron flow in a redox reaction. Electrons flow from the reducing agent at the anode to the oxidizing agent at the cathode. This flow can be harnessed to do electrical work, as in the operation of batteries.
Moreover, standard reduction potentials can be used to predict reaction feasibility and calculate theoretical yields in electrochemical processes, playing a pivotal role in many technological applications.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, often referred to as redox reactions, are a class of reactions where the oxidation state of atoms changes due to the transfer of electrons. These reactions are foundational in understanding many chemical and biological processes.
In a redox reaction, one substance (the reducing agent) loses electrons and is oxidized, while another substance (the oxidizing agent) gains those electrons and is reduced. This transfer of electrons results in changes in the oxidation states of the substances involved. To remember this concept, the mnemonic "LEO the lion says GER" can be handy - Lose Electrons Oxidation (LEO) and Gain Electrons Reduction (GER).
Considering the steps in the original exercise, it's evident that the reversal of the process to identify reducing agents involves understanding the reverse redox reactions, where we focus on the electron-donating capacity of different species. This approach helps us to arrange the species in order of their reducing power, as was done with Cl⁻, Fe²⁺, H₂, and Zn in the exercise.
Redox reactions are pivotal not just in theoretical chemistry, but also in practical real-world applications. They drive processes in everything from biological systems, where they help convert energy, to industrial processes like metal extraction and purification.

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Most popular questions from this chapter

Identify correct statements for the following incomplete reactions. (i) \(\mathrm{KI}+\mathrm{Cl}_{2} \rightarrow\) (ii) \(\mathrm{KClO}_{3}+\mathrm{I}_{2} \rightarrow\) (iii) \(\mathrm{KNO}_{3(3)}+{ }^{3} \mathrm{~K}_{(s)} \rightarrow\) (iv) \(\mathrm{P}+\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow\) (Red) (a) All are redox reactions (b) Only I, II and III have one of the product in its element form (c) I and II are displacement reactions (d) III is disproportionation reaction

The standard electrode potential of \(\mathrm{Cu}^{2+} / \mathrm{Cu}=0.34 \mathrm{~V}\). The electrode potential will be zero, when the conc. of \(\mathrm{Cu}^{2+}\) is as \(\mathrm{x} \times 10^{-12} \mathrm{M}\). the value of \(\mathrm{x}\) is ___ .

For a spontaneous reaction the \(\Delta \mathrm{G}\), equilibrium constant \((\mathrm{K})\) and \(E_{\mathrm{ccll}}^{\circ}\) will be respectively (a) \(-\mathrm{ve},>1,+\mathrm{ve}\) (b) \(+\mathrm{ve},>1,-\mathrm{ve}\) (c) \(-\mathrm{ve},<1,-\mathrm{ve}\) (d) \(-v \mathrm{e},>1,-v e\)

\(.500 \mathrm{ml}\) of \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution is mixed with 500 \(\mathrm{ml}\) of \(1.09 \mathrm{M} \mathrm{Fe}^{2+}\) solution and the reaction is allowed to reach equilibrium at \(25^{\circ} \mathrm{C}\). \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})=\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\) For \(25 \mathrm{ml}\) of the equilibrium solution, \(30 \mathrm{ml}\) of \(0.0833\) \(\mathrm{M} \mathrm{KMnO}_{4}\) were required for oxidation. Calculate the approximate equilibrium constant for the reaction at \(25^{\circ} \mathrm{C}\).

Which of the following statements are correct? (a) \(\mathrm{KMnO}_{4}\) is a powerful oxidising agent. (b) \(\mathrm{KMnO}_{4}\) is a weaker oxidising agent than \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acid medium. (c) \(\mathrm{KMnO}_{4}\) is a stronger oxidising agent than \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acid medium. (d) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) oxidises a secondary alcohol to a ketone.

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