Given the standard electrode potentials; \(\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \quad \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}, \quad \mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}\) \(\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V}, \quad \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}\) Arrange these metals in their increasing order of reducing power.

Short Answer

Expert verified
In increasing order of reducing power: \text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K}.

Step by step solution

01

Understand the Concept of Reducing Power

Reducing power refers to the ability of an element to reduce others (to donate electrons). Elements with lower (more negative) standard electrode potentials have a greater tendency to lose electrons and are therefore stronger reducing agents.
02

Compare the Standard Electrode Potentials

Look at the given standard electrode potentials and compare their values. Lower standard electrode potentials indicate greater reducing power.
03

Arrange the Metals

List the metals in the order of their standard electrode potentials, from the most negative to the least negative. This will give their order of increasing reducing power since the more negative the electrode potential, the stronger the reducing agent.

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Key Concepts

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

Understanding Reducing Power
When it comes to understanding chemical reactions, one crucial concept is that of reducing power. Reducing power essentially represents the ability of an element or compound to donate electrons to another substance during a chemical reaction, thereby reducing that substance. In other words, an entity with high reducing power willingly gives up its own electrons to help another compound gain electrons.

In a practical sense, the reducing power of an element is often indicated by its standard electrode potential. Elements with a more negative standard electrode potential are generally better at donating electrons than those with a more positive potential. This is because a more negative potential indicates a greater tendency for the element to lose electrons and undergo oxidation. In the exercise at hand, we're asked to use this principle to arrange given metals according to their reducing power.
The Role of Reducing Agents
A reducing agent is at the heart of a reduction-oxidation (redox) reaction, serving as the electron donor in the process. This agent is itself oxidized as it reduces another substance. The strength of a reducing agent is measured by its ability to transfer electrons to other substances, which is, in essence, its reducing power.

In the provided exercise, we ranked metals by their ability to act as reducing agents, which is directly based on their standard electrode potentials. The lower the potential, the stronger the reducing agent. This is because a lower potential means that the substance more readily loses electrons, and losing electrons is what definingly characterizes a reducing agent.
Electron Donation in Chemical Reactions
Electron donation is the key mechanism by which reducing agents achieve chemical reduction. The act of donating an electron is a fundamental part of redox reactions, which are composed of two half-reactions: reduction (gain of electrons) and oxidation (loss of electrons).

During the electron donation process, the reducing agent itself loses electrons, and in doing so, oxidizes. This change allows another substance to gain those electrons and thus reduces it. The ability of a substance to donate electrons can vary greatly, and this is quantified by its electrode potential. Greater the negative value of electrode potential, the more readily the substance donates electrons and functions as a reducing agent.
Chemical Reduction Explained
Chemical reduction is a process in which a material gains electrons and thereby reduces its oxidation state. It's one half of a redox reaction, with the other half being oxidation. When a substance is reduced, it receives electrons, typically from a reducing agent, which defines the course of chemical interactions.

To place substances in order of their reducing power, as was our task in the exercise, we consider their propensity to donate electrons—and the standard electrode potentials provide a quantitative measure for this. Put simply, the better a substance is at providing electrons to others, the more powerful a reducing agent it becomes, and accordingly, a greater chemical reduction capability it has.

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Most popular questions from this chapter

$$ \text { Write equations for the electrolysis of } \mathrm{CaH}_{2} \text { in fused state. } $$

Write the nernst equation and e.m.f. of the following cells at \(298 \mathrm{~K}\) : (a) \(\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.0001 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\) (b) \(\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(0.001 M) \| \mathrm{H}^{\mathrm{i}}(1 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{s})\) (c) \(\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.050 \mathrm{M}) \| \mathrm{H}^{+}(0.020 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{s})\) (d) \(\mathrm{Pt}(\mathrm{s})\left|\mathrm{Br}_{2}(\mathrm{l})\right| \mathrm{Br}^{-}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.030 \mathrm{M}) \mid \mathrm{H}_{2}(\mathrm{~g})(\mathrm{I} \mathrm{bar}) ! \mathrm{Pt}(\mathrm{s})\) Given : \(E_{\text {OP }}^{\circ} \mathrm{Mg}=2.36 \mathrm{~V}, E_{\mathrm{OP}}^{\circ} \mathrm{Cu}=-0.34 \mathrm{~V}, E_{\mathrm{OP}}^{\mathrm{O}} \mathrm{Fe}=0.44 \mathrm{~V}\) \(E_{\mathrm{OP}}^{\circ} \mathrm{Sn}=0.14 \mathrm{~V}\) and \(E_{\mathrm{OP}}^{\circ} \mathrm{Br}_{2}=-1.09 \mathrm{~V}\) respectively.

The e.m.f. of a cell corresponding to the reaction : $$ \begin{aligned} \mathrm{Zn}(s)+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}+& \mathrm{H}_{2}(\mathrm{~g}) \\ (0.1 \mathrm{M}) &(1 \mathrm{~atm}) \end{aligned} $$ is \(0.28 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) and \(E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=0.76 \mathrm{~V}\) (i) Write half cell reactions. (ii) Calculate \(\mathrm{pH}\) of the solution at \(\mathrm{H}\) electrode.

$$ \begin{aligned} &\text { The e.m.f. of the cell } M \mid M^{n+}(0.02 M) \| \mathrm{H}^{+}(1 M) \mathrm{H}_{2(\mathrm{~g})}(1 \mathrm{~atm}) \mathrm{Pt} \text { at }\\\ &25^{\circ} \mathrm{C} \text { is } 0.81 \mathrm{~V} \text { . Calculate the valence of metal if } E^{\circ}{ }_{\ldots 1 / 2}^{n+}=0.76 \mathrm{~V} \end{aligned} $$

A metal is known to form fluoride \(M \mathrm{~F}_{2}\). When 10 ampere electricity is passed throwgh i multen salt for \(330 \mathrm{sec}, 1.95 \mathrm{~g}\) metal is deposited. Find out the atomic weight of metal. What will be the quantity of charge required to deposit the same mass of \(\mathrm{Cu}\) from \(\mathrm{CuSO}_{4}\) (aq.) (At. wt. of \(\mathrm{Cu}=63.6\) )

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