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For the electrochemical cell \(\mathrm{M}\left|\mathrm{M}^{+} \| \mathrm{X}^{-}\right| \mathrm{X}\) \(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{0}=0.44 \mathrm{~V}, \mathrm{E}_{\mathrm{X} / \mathrm{X}^{-}}^{0}=0.33 \mathrm{~V}\) from this data one can deduce that (a) \(\mathrm{M}+\mathrm{X} \longrightarrow \mathrm{M}^{+}+\mathrm{X}^{-}\)is spontaneous reaction (b) \(\mathrm{M}^{+}+\mathrm{X}^{-} \longrightarrow \mathrm{M}+\mathrm{X}\) is spontaneous reaction (c) \(\mathrm{E}_{\text {cell }}=0.77 \mathrm{~V}\) (d) \(\mathrm{E}_{\text {cell }}=-0.77 \mathrm{~V}\)

Short Answer

Expert verified
The correct answer is (b) \( \text{M}^+ + \text{X}^- \rightarrow \text{M} + \text{X} \) is a spontaneous reaction.

Step by step solution

01

Determine the Overall Cell Reaction

The electrochemical cell is written as \( \text{M} | \text{M}^+ || \text{X}^- | \text{X} \). This means that at the anode, \( \text{M} \) is oxidized to \( \text{M}^+ \), and at the cathode, \( \text{X}^- \) is reduced to \( \text{X} \). Therefore, the overall cell reaction is \( \text{M} + \text{X}^- \rightarrow \text{M}^+ + \text{X} \).
02

Calculate the Standard Cell Potential

The standard cell potential \( E_{\text{cell}}^0 \) can be calculated using the standard electrode potentials given: \( E_{\text{M}^+ / \text{M}}^0 = 0.44 \, \text{V} \) and \( E_{\text{X} / \text{X}^-}^0 = 0.33 \, \text{V} \). \( E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0 = 0.33 \, \text{V} - 0.44 \, \text{V} = -0.11 \, \text{V} \).
03

Determine Spontaneity of the Reaction

Since the standard cell potential \( E_{\text{cell}}^0 = -0.11 \, \text{V} \) is negative, the reaction \( \text{M} + \text{X}^- \rightarrow \text{M}^+ + \text{X} \) in the cell is not spontaneous. Instead, the reverse reaction \( \text{M}^+ + \text{X} \rightarrow \text{M} + \text{X}^- \) would be spontaneous, as it would produce a positive cell potential.
04

Compare with Options (c) and (d)

The calculated \( E_{\text{cell}}^0 = -0.11 \, \text{V} \) does not match either \( 0.77 \, \text{V} \) or \( -0.77 \, \text{V} \). Therefore, options (c) and (d) regarding \( E_{\text{cell}} \) are not correct.

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Key Concepts

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

Standard Cell Potential
The standard cell potential, denoted as \(E_{\text{cell}}^0\), is a key factor in determining the ability of an electrochemical cell to produce an electric current. It is derived from the standard electrode potentials of the half-reactions occurring at the anode and cathode.
The standard cell potential is calculated using the formula:
  • \(E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0\)
In the given electrochemical cell, the potentials were \(E_{\text{M}^+ / \text{M}}^0 = 0.44 \, \text{V}\) for the oxidation at the anode and \(E_{\text{X} / \text{X}^-}^0 = 0.33 \, \text{V}\) for the reduction at the cathode. Plugging in these values, we find:
  • \(E_{\text{cell}}^0 = 0.33 \, \text{V} - 0.44 \, \text{V} = -0.11 \, \text{V}\)

A negative standard cell potential indicates that the reaction, as written, does not proceed spontaneously under standard conditions.
Spontaneous Reaction
A spontaneous reaction in an electrochemical cell is one where the reaction naturally progresses to produce electrical energy. This occurs when the standard cell potential, \(E_{\text{cell}}^0\), is positive.
In essence, the spontaneity of a reaction is guided by:
  • Positive \(E_{\text{cell}}^0\): Reaction is spontaneous as written.
  • Negative \(E_{\text{cell}}^0\): Reaction is non-spontaneous as written.
For the reaction \(\text{M} + \text{X}^- \rightarrow \text{M}^+ + \text{X}\), the standard cell potential was found to be \(-0.11 \, \text{V}\). This means the reaction is not spontaneous. However, the reverse reaction, \(\text{M}^+ + \text{X} \rightarrow \text{M} + \text{X}^-\), would indeed be spontaneous as it would yield a positive cell potential.
Understanding the direction of spontaneity is crucial in deciding the feasibility and energy efficiency of chemical processes.
Electrode Potential
Electrode potential refers to the potential difference between an electrode and its surrounding solution, and it plays a vital role in electrochemical cells. Each half-cell has a standard electrode potential, \(E^0\), which is measured under standard conditions and compared against the standard hydrogen electrode.
The electrode potentials guide the flow of electrons in the cell:
  • Higher electrode potential at the cathode indicates a greater tendency for reduction.
  • Lower electrode potential at the anode indicates a greater tendency for oxidation.
In the provided electrochemical cell, the electrode potential for the M/M+ half-cell is \(0.44 \, \text{V}\), and for the X/X- half-cell, it is \(0.33 \, \text{V}\). When designing electrochemical cells, selecting half-reactions with appropriate electrode potentials enables the desired flow of electrons and determines the overall cell performance.
Electrode potentials are thus central to predicting the course of a reaction and its ability to generate electrical power.

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Most popular questions from this chapter

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below: $$ \begin{aligned} &\mathrm{MnO}_{4}^{-}(\mathrm{aq} .)+8 \mathrm{H}^{+}(\mathrm{aq})+5 \mathrm{e}^{-} \rightarrow \\ &\mathrm{Mn}^{2+}(\mathrm{aq} .)+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ &\mathrm{E}^{0}=1.51 \mathrm{~V} \\ &\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\text {aq. })+6 \mathrm{e}^{-} \rightarrow \\ &2 \mathrm{Cr}^{3+}(\text { aq. })+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ &\mathrm{E}^{0}=1.38 \mathrm{~V} \\ &\mathrm{~F} \mathrm{e}^{3+}(\mathrm{aq} .)+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} \text { (aq.) } \quad \mathrm{E}^{0}=0.77 \mathrm{~V} \\ &\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \text {(aq.) } \quad \mathrm{E}^{0}=1.40 \mathrm{~V} \end{aligned} $$ Identify the only incorrect statement regarding the quantitative estimation of aqueous \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\). (a) \(\mathrm{MnO}_{4}^{-}\)can be used in aqueous \(\mathrm{HCl}\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) can be used in aqueous \(\mathrm{HCl}\) (c) \(\mathrm{MnO}_{4}^{-}\)can be used in aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) can be used in aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Calculate the boiling point of a one molar aqueous solution (density \(=1.04 \mathrm{~g} \mathrm{~mL}^{-1}\) ) of potassium chloride, \(\mathrm{K}_{\mathrm{b}}\) for water \(=0.52 \mathrm{~kg} \mathrm{~mol}^{-1}\). Atomic masses of \(\mathrm{K}=39, \mathrm{Cl}=35.5\) (a) \(107.28^{\circ} \mathrm{C}\) (b) \(103.68^{\circ} \mathrm{C}\) (c) \(101.078^{\circ} \mathrm{C}\) (d) None of these

Given: \(2 \mathrm{Br} \rightarrow \mathrm{Br}_{2}+2 \mathrm{e}^{-} \mathrm{E}^{\circ}=-1.09 \mathrm{~V}\) \(\mathrm{I}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-} \quad \mathrm{E}^{\circ}=0.54 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e} \rightarrow \mathrm{Fe} \quad \mathrm{E}^{\circ}=-0.44\) Which of the following reactions will not be spontaneous? (a) \(\mathrm{Fe}+\mathrm{Br}_{2} \rightarrow \mathrm{FeBr}_{2}\) (c) \(\mathrm{Fe}+\mathrm{I}_{2} \rightarrow \mathrm{FeI}_{2}\) (c) \(\mathrm{I}_{2}+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{Br}_{2}\) (d) \(\mathrm{Br}_{2}+2 \mathrm{I}^{-} \rightarrow 2 \mathrm{Br}+\mathrm{I}_{2}\)

Using the following Latimer diagram for Bromine, \(\mathrm{pH}=0 ; \mathrm{BrO}_{4}^{-} \stackrel{182 \mathrm{~V}}{\longrightarrow} \mathrm{BrO}_{3}^{-} \stackrel{1.50 \mathrm{~V}}{\longrightarrow} \mathrm{HBrO}\) \(\stackrel{1.595 \mathrm{~V}}{\longrightarrow} \mathrm{Br}_{2} \stackrel{1.0632 \mathrm{~V}}{\longrightarrow} \mathrm{Br}\) the species undergoing disproportionation is (a) \(\mathrm{BrO}_{4}^{-}\) (b) \(\mathrm{BrO}_{3}^{-}\) (c) \(\mathrm{HBrO}\) (d) \(\mathrm{Br}_{2}\)

Molar conductances of \(\mathrm{BaCl}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{HCl}\) are \(\mathrm{x}_{1}, \mathrm{x}_{2}\) and \(\mathrm{x}_{3} \mathrm{~S} \mathrm{~cm}^{2}\) equiv \(^{-1}\) at infinite dilution. If specific conductance of saturated \(\mathrm{BaSO}_{4}\) solution is of \(\mathrm{y}\) \(\mathrm{S} \mathrm{cm}^{-1}\), then \(\mathrm{k}_{\text {ep }}\) of \(\mathrm{BaSO}_{4}\) is: (a) \(\frac{10^{3} y}{2\left(x_{1}+x_{2}-2 x_{3}\right)}\) (b) \(\frac{10^{6} y^{2}}{\left(x_{1}+x_{2}-2 x_{3}\right)^{2}}\) (c) \(\frac{10^{6} y^{2}}{4\left(x_{1}+x_{2}-2 x_{3}\right)^{2}}\) (d) \(\frac{x_{1}+x_{2}-2 x_{3}}{10^{6} y^{2}}\)

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