The standard reduction potential values of three metallic cations \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\) are \(0.52,-3.03\) and \(-1.18\) respectively. The order of reducing power of the corresponding metal is: (a) \(\mathrm{Y}>\mathrm{Z}>\mathrm{X}\) (b) \(\mathrm{X}>\mathrm{Y}>\mathrm{Z}\) (c) \(\mathrm{Z}>\mathrm{Y}>\mathrm{X}\) (d) \(Z>X>Y\)

Short Answer

Expert verified
The correct order of reducing power of the corresponding metals is Y>Z>X, which corresponds to option (c).

Step by step solution

01

Understanding Reduction Potential

The standard reduction potential indicates a substance's ability to gain electrons, i.e., be reduced. A higher reduction potential means a substance is more likely to be reduced, gaining electrons. Conversely, substances with lower reduction potentials are better at giving up electrons, i.e., they are better reducing agents.
02

Analyze the Given Reduction Potentials

Since we have been given the standard reduction potentials of the cations, these values tell us which cation is more likely to be reduced: - For cation X, the reduction potential is 0.52 V, which is positive and relatively higher.- For cation Y, the reduction potential is -3.03 V, which is the lowest among the three.- For cation Z, the reduction potential is -1.18 V, which is negative but not as low as for Y. The cation with the highest reducing power is the one that most readily gives up electrons (low or negative reduction potential), and thus will have the lowest or most negative standard reduction potential.
03

Determine the Order of Reducing Power

Since cation Y has the lowest reduction potential (-3.03 V), it has the greatest tendency to lose electrons and therefore has the highest reducing power. Cation Z, with a reduction potential of -1.18 V, would have the next highest reducing power. Cation X, with the positive reduction potential (0.52 V), will have the lowest reducing power among the three. Therefore, the order of reducing power is Y>Z>X.

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Key Concepts

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

Reducing Power
The concept of reducing power is central to understanding how various elements and compounds interact in redox reactions. When we think of reducing power, we're assessing an element's ability or tendency to donate electrons during a chemical process.

The reducing power of an element is inversely proportional to its standard reduction potential. This means that the lower or more negative the standard reduction potential, the higher the reducing power. In the context of our exercise, cation Y, with a reduction potential of -3.03 V, is most capable of acting as a reducing agent because it has the lowest reduction potential, which conveys its readiness to give up electrons.

Conversely, substances with higher or positive reduction potentials are less inclined to donate electrons and are better oxidizing agents. Cation X's positive reduction potential at 0.52 V indicates that it is the least effective reducing agent among the three metals under examination. Reducing agents are often crucial in chemical reactions, especially in industrial processes like the extraction of metals from ores and in batteries where redox reactions are fundamental.
Redox Reactions
Redox reactions are a class of chemical reactions that involve the transfer of electrons between two species. The term 'redox' is a shorthand for reduction-oxidation reaction. In such a reaction, the oxidation state of atoms is changed due to the gain or loss of electrons.

Reduction refers to the gain of electrons, leading to a decrease in the oxidation state, while oxidation involves the loss of electrons, increasing the oxidation state. For any redox reaction, there must be a substance that gets oxidized and another that gets reduced.
  • In our exercise, cation Y, with the greatest reducing power, would be the one getting oxidized as it donates electrons to another substance.
  • Cation X, with the higher reduction potential, would more likely be the one receiving electrons, hence getting reduced.
Understanding redox reactions is vital in many fields, such as biochemistry where it plays a role in cellular respiration and photosynthesis, and in environmental science where redox reactions can affect the quality of air and water.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electrical and chemical phenomena. It focuses on processes that cause electrons to move, known as redox reactions, and involves the study of both chemical changes caused by electric current and the production of electricity through chemical reactions.

In electrochemistry, the concept of standard reduction potential is a foundational component. It enables us to predict the direction of electron flow in redox reactions, and hence, the feasibility and spontaneity of these reactions. The standard reduction potentials are measured under standard conditions and are typically tabulated for reference.

Knowing the standard reduction potential of different substances allows us to build galvanic cells—devices that convert chemical energy into electrical energy—such as batteries. A substance with a high reducing power has a low standard reduction potential and will act as the anode in a galvanic cell, while the substance with a low reducing power and higher reduction potential will act as the cathode. This principle is exemplified in the exercise, where cation Y, with the highest reducing power, would be the best candidate for the anode in an electrochemical cell.

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Most popular questions from this chapter

The emf of a Daniell cell at \(298 \mathrm{~K}\) is \(\mathrm{E}_{-1}\) \(\mathrm{Zn}\left|\mathrm{ZnSO}_{4} \| \mathrm{CuSO}_{4}\right| \mathrm{Cu}\) \((0.01 \mathrm{M}) \quad(1.0 \mathrm{M})\) when the concentration of \(\mathrm{ZnSO}_{4}\) is \(1.0 \mathrm{M}\) and that of \(\mathrm{CuSO}_{4}\) is \(0.01 \mathrm{M}\), the emf changed to \(\mathrm{E}_{2}\). What is the relationship between \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2} ?\) (a) \(\mathrm{E}_{1}=\mathrm{E}_{2}\) (b) \(\mathrm{E}_{2}=\mathrm{O} \neq \mathrm{E}_{1}\) (c) \(\mathrm{E}_{1}>\mathrm{E}_{2}\) (d) \(\mathrm{E}_{1}<\mathrm{E}_{2}\)

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