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

Short Answer

Expert verified
The metals in increasing order of reducing power are Ag < Hg < Cr < Mg < K.

Step by step solution

01

Understanding Reducing Power

Reducing power refers to the ability of a substance to donate electrons. Metals that are better reducing agents have more negative electrode potentials.
02

Listing Standard Electrode Potentials

We have the following standard electrode potentials: \( ext{K}^{+}/ ext{K} = -2.93 \, ext{V} \), \( ext{Mg}^{2+}/ ext{Mg} = -2.37 \, ext{V} \), \( ext{Cr}^{3+}/ ext{Cr} = -0.74 \, ext{V} \), \( ext{Hg}^{2+}/ ext{Hg} = 0.79 \, ext{V} \), \( ext{Ag}^{+}/ ext{Ag} = 0.80 \, ext{V} \).
03

Arranging in Increasing Order of Potential

Rearrange these in increasing order of electrode potential: \[ ext{K (-2.93 V)}, ext{Mg (-2.37 V)}, ext{Cr (-0.74 V)}, ext{Hg (0.79 V)}, ext{Ag (0.80 V)} \]
04

Determining Reducing Power Order

Metals with more negative electrode potentials are stronger reducing agents. Thus, the order of increasing reducing power is: \[ ext{Ag < Hg < Cr < Mg < K} \] and \( \text{K} \) is the strongest reducing agent.

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Key Concepts

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

Reducing Power
Reducing power is a fascinating concept in electrochemistry. It is the ability of a species to donate electrons during a chemical reaction. The greater reducing power means that a substance is better at losing electrons. This is crucial because in chemistry, electron donation is essential for reduction reactions.

In simpler terms, think of reducing power as the eagerness of an element to give away electrons. Metals like potassium ( K ) and magnesium ( Mg ) exhibit substantial reducing power. Why? Because they have very negative standard electrode potentials. When these potentials are negative, it indicates the capability to release electrons easily.

When ranking metals by reducing power, one can look at their electrode potentials. Metals with more negative electrode potential values are better at donating electrons. Hence, they are more effective reducing agents.
Electrode Potential
Electrode potential is a measure of the voltage created by a metal in contact with its ions. It is essentially an indicator of how eager a metal is to undergo reduction or oxidation. This concept is central to understanding electrochemical cells and their behavior. In these cells, the potential difference between two electrodes drives the flow of electrons, which is essentially the electric current.

Standard electrode potentials are measured under standard conditions: 1M concentration for solutions, 1 atm pressure for gases, and a temperature of 298K. By convention, the standard hydrogen electrode ( SHE ) is used as a reference, with a potential defined as 0 V. Other electrodes are then measured against this standard.

When we arrange metals based on their standard electrode potential, we can predict their behavior in redox reactions. Metals with negative electrode potentials are usually more inclined to lose electrons, acting as strong reducing agents.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the relationship between electricity and chemical reactions. It covers various processes such as electrolysis, corrosion, and galvanic cells. Electrochemistry is pivotal for understanding a wide range of scientific and industrial applications.

One core aspect of electrochemistry is redox reactions, where oxidation and reduction happen simultaneously. These reactions are responsible for driving the electron flow in battery cells and are crucial for processes like metal plating and water splitting.

In practical terms, electrochemistry is involved in designing batteries, which power countless devices from phones to electric cars. By understanding the principles of electrode potential and reducing power, we can develop more efficient and long-lasting energy storage solutions.

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Most popular questions from this chapter

Write formulas for the following compounds: (a) Mercury(II) chloride (b) Nickel(II) sulphate (c) Tin(IV) oxide (d) Thallium(I) sulphate (e) Iron(III) sulphate (f) Chromium(III) oxide

Justify that the following reactions are redox reactions: (a) \(\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})\) (c) \(4 \mathrm{BCl}_{3}(\mathrm{~g})+\) 3LiAlH \(_{4}(\mathrm{~s}) \rightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{~s})\) (d) \(2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})\) (e) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Suggest a list of the substances where carbon can exhibit oxidation states from \(-4\) to \(+4\) and nitrogen from \(-3\) to \(+5\).

Balance the following redox rcactions by ion \(-\) clcctron method : (a) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}\) \((\mathrm{s})+\mathrm{I}_{2}(\mathrm{~s})\) (in basic medium) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}\) (aq) (in acidic solution) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Fe}^{2+}\) (aq) \(\rightarrow \mathrm{Fe}^{3+}\) (aq) \(+\mathrm{H}_{2} \mathrm{O}\) (l) (in acidic solution) (d) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (in acidic solution)

The \(\mathrm{Mn}^{3+}\) ion is unstable in solution and undergoes disproportionation to give \(\mathrm{Mn}^{2+}, \mathrm{MnO}_{2}\), and \(\mathrm{H}^{+}\) ion. Write a balanced ionic equation for the reaction.

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