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The standard reduction potential values of three metallic cations, \(\mathrm{X}, \mathrm{Y}\) and \(Z\) are \(+0.52,-3.03\) and \(-1.18 \mathrm{~V}\), respectively. The order of reducing power of the corresponding metals 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 is (b) Y > Z > X.

Step by step solution

01

Understand the concept of reduction potential

The standard reduction potential is a measure of the tendency of a chemical species to be reduced, and is measured in volts. A higher reduction potential means a species is more easily reduced (gains electrons), while a lower (more negative) reduction potential means the species is less easily reduced. In the context of metallic cations, a metal with a higher reduction potential acts as a better oxidizing agent than one with a lower reduction potential.
02

Compare the reduction potentials

To determine the reducing power of the corresponding metals, compare their reduction potential values. The metal that has the most negative reduction potential will be the strongest reducing agent, because it has a greater tendency to lose electrons. The reduction potentials given are: X: +0.52 V,Y: -3.03 V, Z: -1.18 V. Now, arrange these metals in order from the most negative to the least negative reduction potential.
03

Arrange the metals according to their reducing power

Following Step 2, the order from the most negative to the least negative reduction potential is: Y (-3.03 V) < Z (-1.18 V) < X (+0.52 V). Since a more negative reduction potential indicates a stronger reducing agent, the metal corresponding to cation Y will have the highest reducing power, followed by the metal corresponding to Z, and then X.

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Key Concepts

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

Standard Reduction Potential
In chemistry, the standard reduction potential (SRP) is a crucial concept that involves measuring the intrinsic tendency of a substance to gain electrons and thus be reduced. The SRP is expressed in volts and can be found in a table of standard electrode potentials, where each chemical species is listed along with its tendency to be reduced. A more positive SRP indicates a stronger tendency to become reduced; conversely, a more negative SRP suggests a lower likelihood of the species gaining electrons. However, for the metals in our exercise, these potentials reflect how willing they are to lose electrons and become oxidized (since we're considering metal cations). Understanding SRP is fundamental when predicting the outcome of electron transfer reactions, identifying oxidizing and reducing agents, and explaining the flow of electrons in both galvanic and electrolytic cells.
Chemical Species Reduction Tendency
The reduction tendency of a chemical species is intrinsic to its character and represents its willingness to gain electrons during a chemical reaction. This tendency is often captured by the SRP, with a higher (more positive) SRP indicating a higher affinity for electrons, making the species a better oxidant. Conversely, a lower (more negative) SRP suggests a lower electron affinity, indicating a pronounced propensity to lose electrons and therefore act as a reducing agent. In the given exercise, the cation with the most negative SRP is Y, which implies that its corresponding metal will most readily give up electrons, thereby showing the greatest reducing power among X, Y, and Z.
Oxidizing and Reducing Agents
Oxidizing and reducing agents are substances that can cause oxidation and reduction, respectively, in other chemicals during a chemical reaction. An oxidizing agent gains electrons and is reduced in the process, often characterized by a higher SRP. On the other hand, a reducing agent loses electrons and is oxidized, which is mirrored by a more negative SRP. The effectiveness of these agents is directly related to their reduction potentials. In electron transfer reactions, the agent with the higher SRP will typically be reduced, while the one with the lower SRP will be oxidized. Thus, in our exercise, cation Y would produce the strongest reducing agent due to its lowest (most negative) SRP, making it the most ready to part with electrons.
Electron Transfer Reactions
Electron transfer reactions are at the heart of redox processes, where electrons move from one chemical species to another. In these reactions, reducing agents donate electrons, while oxidizing agents accept them. The driving force behind these transfers is the difference in SRP between the reactants. A species with a significant difference in SRP compared to its reaction partner will tend to undergo electron transfer more readily. The importance of these differences is illustrated in our exercise with cations X, Y, and Z. As Y has the lowest SRP, it has the most substantial inclination to lose electrons and thus, it is the most effective reducing agent of the three, ready to undergo electron transfer in a redox reaction.

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Most popular questions from this chapter

When metallic copper is shaken with a solution of a copper salt, the reaction \(\mathrm{Cu}+\mathrm{Cu}^{2+} \rightleftharpoons 2 \mathrm{Cu}^{+}\) proceeds. When equilibrium is established at \(298 \mathrm{~K}\), \(\left[\mathrm{Cu}^{2+}\right] /\left[\mathrm{Cu}^{+}\right]^{2}=1.667 \times 10^{6} \mathrm{M}^{-1}\). If the standard potential of the \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\) halfcell is \(+0.3376 \mathrm{~V}\), what is the standard potential of Cu'| Cu half-cell? (Given: \(2.303 R T / F=0.06, \log 2=0.3, \log 3=0.48\) ) (a) \(-0.3732 \mathrm{~V}\) (b) \(0.6752 \mathrm{~V}\) (c) \(0.5242 \mathrm{~V}\) (d) \(0.151 \mathrm{~V}\)

We have an oxidation-reduction system: \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} \rightleftharpoons\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) \(E^{\circ}=+0.36 \mathrm{~V}\). The ratio of concentrations of oxidized and reduced from at which the potential of the system becomes \(0.24 \mathrm{~V}\), is [Given: \(2.303 R T / F=0.06\) ) (a) \(2: 1\) (b) \(1: 2\) (c) \(1: 20\) (d) \(1: 100\)

A volume of \(100 \mathrm{~m}\) of a buffer of \(1 \mathrm{M}\) \(-\mathrm{NH}_{3}\) and \(1 \mathrm{M}-\mathrm{NH}_{4}^{+}\) is placed in two half-cells connected by a salt bridge. \(A\) current of \(1.5 \mathrm{~A}\) is passed through the cell for \(20 \mathrm{~min}\). If electrolysis of water takes place only and the electrode reactions are: Right: \(2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-}\) and Left: \(2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{H}^{+}+\mathrm{O}_{2}+4 \mathrm{e}\), then, the \(\mathrm{pH}\) of the (a) right electrode will increase (b) left electrode will increase (c) both electrode will increase (d) both electrode will decrease

Which process occurs in the electrolysis of an aqueous solution of nickel chloride at nickel anode? (a) \(\mathrm{Ni} \rightarrow \mathrm{Ni}^{2+}+2 \mathrm{e}^{-}\) (b) \(\mathrm{Ni}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}\) (c) \(2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}\) (d) \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\)

What is the equilibrium constant of the reaction: \(2 \mathrm{Fe}^{3+}+\mathrm{Au}^{+} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Au}^{3+} ?\) Given \(E_{\mathrm{Au}^{+} \mid \mathrm{Au}}^{\circ}=1.68 \mathrm{~V}, E_{\mathrm{Au}^{3 \cdot} \mid \mathrm{Au}}^{\circ}=1.50 \mathrm{~V}\), \(E_{\mathrm{Fe}^{3 *} \mid \mathrm{Fe}^{\mathrm{e}}^{\circ}}=0.75 \mathrm{~V}\) and \(2.303 R T / F=0.06\) (a) \(1 \times 10^{22}\) (b) \(1 \times 10^{-22}\) (c) \(1 \times 10^{-11}\) (d) \(1 \times 10^{-72}\)

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