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For the electrochemical cell, \(M\left|M^{+} \| X^{-}\right| X, E^{\circ}\left(M^{+} / M\right)=0.44 \mathrm{~V}\) and \(E^{\circ}\left(X / X^{-}\right)=0.33 \mathrm{~V}\) From this data one can deduce that [2000S] (a) \(M+X \rightarrow M^{+}+X^{-}\)is the spontaneous reaction (b) \(M^{+}+X^{-} \rightarrow M+X\) is the spontaneous reaction (c) \(\mathrm{E}_{\text {cell }}=0.77 \mathrm{~V}\) (d) \(\mathrm{E}_{\mathrm{ccll}}=-0.77 \mathrm{~V}\)

Short Answer

Expert verified
(b) \(M^{+}+X^{-} \rightarrow M+X\) is the spontaneous reaction.

Step by step solution

01

Understand the Cell Diagram

The cell diagram given is \( M | M^+ || X^- | X \). This indicates that \( M \) is the anode and \( X \) is the cathode. Therefore, the anode reaction is \( M \rightarrow M^+ + e^- \) and the cathode reaction is \( X + e^- \rightarrow X^- \).
02

Recall the Standard Electrode Potential Convention

For a cell represented as anode | anode ion || cathode ion | cathode, the cell potential \( E_{cell} \) is calculated as \( E_{cell} = E^{ ext{cathode}} - E^{ ext{anode}} \). The more positive the electrode potential, the stronger the oxidizing agent presents at the cathode.
03

Calculate the Standard Cell Potential

Using the given standard reduction potentials, \( E^{ ext{cathode}} = E^ ext{o}(X/X^-) = 0.33 \text{ V} \) and \( E^{ ext{anode}} = E^ ext{o}(M^+/M) = 0.44 \text{ V} \), calculate the cell potential: \[ E_{cell} = E^{ ext{cathode}} - E^{ ext{anode}} = 0.33 \text{ V} - 0.44 \text{ V} = -0.11 \text{ V} \]
04

Interpret the Cell Potential

A negative \( E_{cell} \) means the cell reaction, as written (\( M \rightarrow M^+ \) and \( X \rightarrow X^- \)), is non-spontaneous. To make it spontaneous, the reverse direction (\( M^+ + X^- \rightarrow M + X \)) would need to be considered as it would make \( E_{cell} \) positive.
05

Select the Correct Option

With \( E_{cell} = -0.11 \text{ V} \), the original reaction is not spontaneous \((M+X\rightarrow M^++X^-)\). Thus, option (b), the reverse reaction \((M^+ + X^- \rightarrow M + X)\), is the spontaneous one. Also, options (c) and (d) are incorrect because they incorrectly calculate \( E_{cell} \) as \( 0.77 \text{ V} \) or \( -0.77 \text{ V} \).

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Key Concepts

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

Standard Electrode Potential
The concept of Standard Electrode Potential is central to understanding electrochemical cells. This refers to the voltage, or potential difference, measured under standard conditions for a specific electrochemical reaction at the electrode. Essentially, it's a measure of how readily a chemical species will gain or lose electrons.

Each substance has a standard reduction potential value, often represented as \( E^{\circ} \). This value indicates how likely an element is to undergo reduction. More positive values suggest a greater tendency to gain electrons and act as an oxidizing agent. Conversely, more negative values indicate a preference to lose electrons and act as a reducing agent.

In the given electrochemical cell, the standard electrode potential for the cathode \( E^{\circ}(X/X^{-}) \) is 0.33 V, and for the anode \( E^{\circ}(M^{+}/M) \) is 0.44 V. This allows us to predict the direction of electron flow and determine the cell's reaction potential, which is crucial in assessing whether a reaction will happen spontaneously.
Cell Potential
In an electrochemical cell, the cell potential, often referred to as the cell voltage, is the measure of the electrochemical reaction’s driving force. This is the difference in potential energy between the two electrodes connected in a cell, calculated by subtracting the anodic potential from the cathodic potential.

For the electrochemical cell, the cell potential \( E_{cell} \) is calculated using the formula: \[E_{cell} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
Using the values provided, we have:
  • \( E^{\circ}_{\text{cathode}} = 0.33 \text{ V} \)
  • \( E^{\circ}_{\text{anode}} = 0.44 \text{ V} \)
Thus, the cell potential is: \[E_{cell} = 0.33 \text{ V} - 0.44 \text{ V} = -0.11 \text{ V} \]
A negative cell potential means that, as-written, the cell reaction will not proceed spontaneously. For a reaction to be spontaneous, the cell potential should be positive, indicating a thermodynamically favorable reaction.
Spontaneous Reaction
Spontaneity in electrochemistry refers to a reaction's natural tendency to occur without external intervention. For a reaction in an electrochemical cell to be spontaneous, the cell potential \( E_{cell} \) must be positive.

The given exercise involves determining the spontaneous reaction between substances \( M \) and \( X \) in a cell with potentials calculated to be \( -0.11 \text{ V} \). This indicates that the forward reaction, as described \( (M + X \rightarrow M^+ + X^-) \), is not spontaneous on its own.

Instead, the spontaneous process occurs when the reaction direction is reversed: \( (M^+ + X^- \rightarrow M + X) \). This reversal changes the outlook of the cell potential, making the reaction positive and indicating spontaneous behavior. Understanding these factors is vital, as it influences which reactions can occur naturally and which require additional energy.

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Most popular questions from this chapter

Potassium chlorate is prepared by the electrolysis of \(\mathrm{KCl}\) in basic solution \(6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}\) If only \(60 \%\) of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce \(10 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\) using a current of \(2 \mathrm{~A}\) is (Given : \(\mathrm{F}=\overline{96,500 \mathrm{C}} \mathrm{mol}^{-1} ;\) molar mass of \(\mathrm{KClO}_{3}=122 \mathrm{~g} \mathrm{~mol}^{-1}\) )

During the discharge of a lead storage battery, the density of sulphuric acid fell from \(1.294\) to \(1.139 \mathrm{~g} / \mathrm{mL}\). Sulphuric acid of density \(1.294\) \(\mathrm{g} / \mathrm{mL}\) is \(39 \%\) by weight and that of \(1.139 \mathrm{~g} / \mathrm{mL}\) is \(20 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) by weight. The battery holds \(3.5\) litres of the acid and the volume remained practically constant during the discharge. Calculate the number of ampere- hours for which the battery must have been used. The charging and discharging reactions are : [1986-5 Marks] Anode : \(\mathrm{Pb}+\mathrm{SO}_{4}^{2-}=\mathrm{PbSO}_{4}+2 \mathrm{e}^{-}\)(discharging) Cathode : \(\mathrm{PbO}_{2}+4 \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}+2 \mathrm{e}^{-}=\mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) (discharging) Note : Both the reactions take place at the anode and cathode respectively during discharge. Both reaction get reverse during charging.

How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing silver in \(+1\) oxidation state for a period of \(8.0\) hours at a current of \(8.46\) amperes? What is the area of the tray if the thickness of the silver plating is \(0.00254 \mathrm{~cm}\) ? Density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\).

An oxidation-reduction reaction in which 3 electrons are transferred has a \(\Delta \mathrm{G}^{0}\) of \(17.37 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). The value of \(\mathrm{E}_{\text {cell }}^{0}\) (in \(\mathrm{V}\) ) is \(\times 10^{-2}\) \(\left(1 \mathrm{~F}=96,500 \mathrm{C} \mathrm{mol}^{-1}\right)\)

A solution of copper sulphate \(\left(\mathrm{CuSO}_{4}\right)\) is electrolysed for 10 minutes with a current of \(1.5\) amperes. The mass of copper deposited at the cathode (at. mass of \(\mathrm{Cu}=63 \mathrm{u}\) ) is: [Main Online April 25, 2013] (a) \(0.3892 \mathrm{~g}\) (b) \(0.2938 \mathrm{~g}\) (c) \(0.2398 \mathrm{~g}\) (d) \(0.3928 \mathrm{~g}\)

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