Electrochemistry

$ \text { The cell reaction is, } $

$ \begin{aligned} & \therefore \quad Q=\frac{\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2}{\mathrm{p}_{\mathrm{Cl}_2}} \\ & \text { and } E_{\text {cell }}=E^{\circ}-\frac{0.0591}{2} \log Q \\ & \text { So, } E_{\text {cell }} \propto \frac{1}{Q} \end{aligned} $

(a) $Q=\frac{1 \times 1^2}{1}=1$;

(b) $Q=\frac{0.01 \times 1^2}{1}=10^{-2}$

(c) $Q=\frac{1 \times(0.01)^2}{1}=10^{-4}$;

(d) $Q=\frac{0.01 \times(0.01)^2}{1}=10^{-6}$

$\Rightarrow$ The order of the values of

$ Q: d

$\Rightarrow$ The order of emf of galvanic cells:

$ d>c>b>a $

The emf is maximum in

$ \begin{aligned} & \mathrm{Mg}\left|\mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| 2 \mathrm{Cl}^{-}(0.01 \mathrm{M})\right| \\ & \mathrm{Cl}_2(1 \mathrm{~atm}) \mid \mathrm{Pt} \end{aligned} $

Let's analyze the data step by step.

We are given the following standard electrode reactions and their potentials:

For aluminum (Al):

$\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}, \quad E^\circ = -1.66 \text{ V}$

$\mathrm{Al}^{+} + e^- \to \mathrm{Al}, \quad E^\circ = +0.55 \text{ V}$

For thallium (Tl):

$\mathrm{Tl}^{3+} + 3e^- \to \mathrm{Tl}, \quad E^\circ = +1.26 \text{ V}$

$\mathrm{Tl}^{+} + e^- \to \mathrm{Tl}, \quad E^\circ = -0.34 \text{ V}$

A key idea in interpreting these potentials is that a more positive reduction potential means the ion is a stronger oxidizing agent and is more eager to gain electrons (i.e., it is less “stable” in its oxidized form). In contrast, a more negative reduction potential indicates that the species is less likely to gain electrons and is more stable in that oxidation state.

Let’s apply this reasoning:

• For aluminum:

The reaction $\mathrm{Al}^{3+} \to \mathrm{Al}$ has a very negative potential (–1.66 V), meaning that once aluminum is in the +3 state, it does not easily pick up electrons to revert to the metal. This makes $\mathrm{Al}^{3+}$ a very stable oxidation state.

The reaction involving $\mathrm{Al}^{+}$ has a positive potential (+0.55 V), indicating that $\mathrm{Al}^{+}$ is readily reduced to aluminum, and thus is less stable in the +1 state.

• For thallium:

The reaction $\mathrm{Tl}^{3+} \to \mathrm{Tl}$ has a highly positive potential (+1.26 V), so $\mathrm{Tl}^{3+}$ is not very stable; it tends strongly to gain electrons.

In contrast, the reaction involving $\mathrm{Tl}^{+}$ has a negative potential (–0.34 V), which indicates that $\mathrm{Tl}^{+}$ does not easily accept an electron, making it a relatively stable oxidation state.

Now, let’s look at the options:

Option A: “$\mathrm{Tl}^{3+}$ is more stable than $\mathrm{Al}^{3+}$.”

• Because $\mathrm{Tl}^{3+}$ (with +1.26 V) is eager to gain electrons (a strong oxidizing agent), it is less stable compared to $\mathrm{Al}^{3+}$ (with –1.66 V). Thus, Option A is incorrect.

Option B and Option C: “$\mathrm{Al}^{+}$ is more stable than $\mathrm{Al}^{3+}$.”

• Our data indicates that $\mathrm{Al}^{+}$ (with +0.55 V) is less stable (more readily reduced) than $\mathrm{Al}^{3+}$ (with –1.66 V). So, these statements are not correct.

Option D: “$\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.”

• Comparing the two, the reduction reaction for $\mathrm{Tl}^{+}$ has $E^\circ = -0.34\text{ V}$ while that for $\mathrm{Al}^{+}$ has $E^\circ = +0.55\text{ V}.$ The more negative potential for $\mathrm{Tl}^{+}$ shows it is less likely to be reduced to elemental Tl, meaning $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.

Thus, Option D is the correct statement.

In summary, the correct answer is:

Option D: $\mathrm{Tl}^{+}$ is more stable than $\mathrm{Al}^{+}$.

Let's analyze each statement using the given standard reduction potentials:

$2H^+ + 2e^- \rightarrow H_2,\quad E^\circ = 0.0\text{ V}$

$Cu^{2+} + 2e^- \rightarrow Cu,\quad E^\circ = +0.34\text{ V}$

$Zn^{2+} + 2e^- \rightarrow Zn,\quad E^\circ = -0.76\text{ V}$

For nitrate reducing to nitric oxide (in acidic solution):

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O,\quad E^\circ = +0.97\text{ V}$

Now, let’s consider each statement one by one.

Statement I: “$H^+$ does not oxidise Cu to $Cu^{2+}$”

The possible reaction would be:

$Cu(s) + 2H^+ \rightarrow Cu^{2+} + H_2(g)$

To check spontaneity, write the half-reactions:

Oxidation (copper):

$Cu(s) \rightarrow Cu^{2+} + 2e^-,$

(reverse of the given reduction reaction with $E^\circ = -0.34\text{ V}$)

Reduction (hydrogen):

$2H^+ + 2e^- \rightarrow H_2,\quad E^\circ = 0.0\text{ V}$

The cell potential is:

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.0\text{ V} - 0.34\text{ V} = -0.34\text{ V}$

A negative cell potential means the reaction is non-spontaneous.

Thus, Statement I is correct.

Statement II: “Zn reduces $Cu^{2+}$ to Cu”

The reaction is:

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Write the half-reactions:

Oxidation (zinc):

$Zn(s) \rightarrow Zn^{2+} + 2e^-,$

The oxidation potential for Zn is the reverse of its reduction: $+0.76\text{ V}$.

Reduction (copper):

$Cu^{2+} + 2e^- \rightarrow Cu(s),\quad E^\circ = +0.34\text{ V}$

Combine the potentials:

$E^\circ_{\text{cell}} = 0.34\text{ V} + 0.76\text{ V} = +1.10\text{ V}$

A positive cell potential indicates the reaction is spontaneous.

Thus, Statement II is correct.

Statement III: “$NO_3^-$ oxidises Cu to $Cu^{2+}$”

In reactions with nitric acid (which provides $NO_3^-$ in acidic medium), copper is oxidized. The half-reactions are:

Oxidation (copper):

$Cu(s) \rightarrow Cu^{2+} + 2e^-,\quad E^\circ_{\text{ox}} = -0.34\text{ V}$

Reduction (nitrate in acid):

$NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O,\quad E^\circ = +0.97\text{ V}$

Even though the electrons do not match directly (2 vs. 3 electrons), the overall potential difference can be determined by the difference in the standard electrode potentials:

$E^\circ_{\text{net}} = 0.97\text{ V} - 0.34\text{ V} = +0.63\text{ V}$

A positive net potential shows that the oxidation of Cu by $NO_3^-$ in acidic conditions is thermodynamically favored.

Thus, Statement III is correct.

Since all three statements (I, II, and III) are correct, the correct option is:

Option B: I, II, III

A reducing agent is a substance that donates electrons during a chemical reaction. The more negative the standard reduction potential (given by the half-reaction), the stronger the tendency of the substance to lose electrons—making it a stronger reducing agent.

Let’s compare the given elements by their half-reactions and approximate standard reduction potentials:

For sodium (Na):

$\mathrm{Na^+ + e^- \to Na} \quad E^\circ \approx -2.71\,V$

For rubidium (Rb):

$\mathrm{Rb^+ + e^- \to Rb} \quad E^\circ \approx -2.98\,V$

For magnesium (Mg):

$\mathrm{Mg^{2+} + 2e^- \to Mg} \quad E^\circ \approx -2.36\,V$

For strontium (Sr):

$\mathrm{Sr^{2+} + 2e^- \to Sr} \quad E^\circ \approx -2.89\,V$

Since the strength of a reducing agent corresponds to having a very negative standard reduction potential, we compare the values:

Rb: approximately -2.98 V

Sr: approximately -2.89 V

Na: approximately -2.71 V

Mg: approximately -2.36 V

Rubidium (Rb) has the most negative reduction potential. Therefore, it is the strongest reducing agent among the given elements.

The correct answer is: Option A (Rb).

The standard reduction potentials for the following half-reactions are:

$2 \mathrm{H}^{+} / \mathrm{H}_2$: 0.0 V

$\mathrm{Cu}^{2+} / \mathrm{Cu}$: +0.34 V

$\mathrm{Zn}^{2+} / \mathrm{Zn}$: -0.76 V

$\mathrm{NO}_3^{-}, \mathrm{H}^{-} / \mathrm{NO}$: 0.97 V

Let's consider the reactions:

$\mathrm{Zn} + \mathrm{HCl} \rightarrow$

$\mathrm{Cu} + \mathrm{HCl} \rightarrow$

$\mathrm{Cu} + \mathrm{HNO}_3 \rightarrow$

We are tasked with identifying which of these reactions do not liberate $\mathrm{H}_2(g)$.

Reactions II and III do not produce $\mathrm{H}_2$ because they are not spontaneous. This conclusion is based on their cell potentials ($E_{\text{cell}}^{\circ}$) being negative ($E_{\text{cell}}^{\circ} < 0$).

Calculation of Cell Potentials:

Reaction II: $\mathrm{Cu} + \mathrm{HCl}$

$ \begin{align*} E_{\text{cell}}^{\circ} &= E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \\ &= 0 - 0.34 \\ &= -0.34 \, \text{V} \end{align*} $

Reaction III: $\mathrm{Cu} + \mathrm{HNO}_3$

$ \begin{align*} E_{\text{cell}}^{\circ} &= E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \\ &= 0 - 0.97 \\ &= -0.97 \, \text{V} \end{align*} $

Since both reactions have negative cell potentials, they are non-spontaneous and thus do not release $\mathrm{H}_2(g)$.

To determine the mass of copper deposited at the cathode during electrolysis, we first need to gather the given data and use the formula for electrolysis. The relevant parameters are:

Current, $ I = 2 \, \text{A} $

Time, $ t = 10 \, \text{minutes} $

Convert the time from minutes to seconds for consistency with the formula:

$ t = 10 \times 60 = 600 \, \text{seconds} $

Next, calculate the total charge $ Q $ passed during electrolysis using the formula:

$ Q = I \times t = 2 \times 600 = 1200 \, \text{C} $

The electrochemical reaction for copper deposition is:

$ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) $

Using the formula for the mass of substance deposited during electrolysis:

$ \text{Mass of copper deposited} = \frac{\text{Molar mass} \times \text{Charge}}{\text{Number of electrons transferred} \times F} $

Given:

Molar mass of Cu = 63 g/mol

Faraday constant, $ F = 96500 \, \text{C/mol} $

Number of electrons transferred for copper, = 2

Substitute these values into the formula:

$ \text{Mass of copper deposited} = \frac{63 \times 1200}{2 \times 96500} = 0.39 \, \text{g} $

Thus, the weight of copper deposited at the cathode is 0.39 grams.

To find the atomic mass of the metal $ M $, given that 2.644 g of metal was deposited with a total charge of 8040 Coulombs, we follow these steps:

Calculate the moles of electrons used:

$ \text{Moles of electrons} = \frac{\text{Total charge}}{\text{Faraday's constant}} = \frac{8040 \, \text{C}}{96500 \, \text{C/mol}} = 0.0833 \, \text{mol} $

Determine the moles of metal deposited:

The reaction $ M\mathrm{F}_2 \rightarrow M^{2+} + 2\mathrm{F}^- $ implies that 2 moles of electrons are needed to deposit 1 mole of metal.

$ \text{Moles of metal} (M) = \frac{\text{Moles of electrons}}{2} = \frac{0.0833}{2} = 0.04165 \, \text{mol} $

Calculate the atomic mass of metal $ M $:

$ \text{Atomic mass of } M = \frac{\text{Mass of metal}}{\text{Number of moles of metal}} = \frac{2.644 \, \text{g}}{0.04165 \, \text{mol}} \approx 63.47 $

Thus, the atomic mass of metal $ M $ is approximately 63.47 u.

In the electrolytic refining process of copper:

The anode is made up of impure copper. During the process, oxidation occurs at the anode, meaning the impure copper dissolves into the electrolyte.

The cathode is made of pure copper. Here, reduction takes place; copper ions gain electrons and deposit as pure copper on the cathode.

Mathematically, the reactions are as follows:

At the anode (oxidation):

$\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$

At the cathode (reduction):

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

Therefore, the correct option is:

Option B: impure copper, pure copper.

The given standard electrode potential is:

$ E^{\circ}_\text{{cell}} = 0.3 \, \text{V} $

According to the Nernst equation:

$ E_\text{{cell}} = E^{\circ}_\text{{cell}} - \frac{2.303RT}{nF} \log \left[\frac{\text{[Cu}^{2+}]}{\text{[M}^{2+}]}\right] \tag{1} $

In this context:

$ n = 2 $ (since it involves the transfer of 2 electrons)

$ E_\text{{cell}} = 0 $ (when the cell voltage is zero)

$\text{[M}^{2+}] = 0.1 \, \text{M} $

$\frac{2.303RT}{F} = 0.06 $

Substituting these values into equation (1) gives:

$ 0.3 = -\frac{0.06}{2} \log \left[\frac{\text{[Cu}^{2+}]}{0.1}\right] $

Solving for the log term:

$ -\frac{0.6}{0.06} = \log \left[\frac{\text{[Cu}^{2+}]}{0.1}\right] $

Taking the antilog:

$ 10^{-10} = \frac{\text{[Cu}^{2+}]}{0.1} $

This implies:

$ \text{[Cu}^{2+}] = 10^{-10} \times 10^{-1} = 10^{-11} \, \text{mol/L} $

Therefore, the concentration of $\text{Cu}^{2+}$ required to make the cell potential zero is $\boxed{10^{-11} \, \text{mol/L}}$.

Let's review each statement:

A. During charging of battery, PbSO₄ on anode is converted into PbO₂.

• This statement is incorrect. During charging, PbSO₄ on the anode (lead plate) is converted to Pb, not PbO₂.

B. During charging of battery, PbSO₄ on cathode is converted into PbO₂.

• This statement is correct. During charging, PbSO₄ on the cathode (lead dioxide plate) is converted back to PbO₂.

C. Lead storage battery consists of grid of lead packed with PbO₂ as anode.

• This statement is incorrect. The anode in a lead-acid battery is made of lead (Pb), not lead dioxide (PbO₂). The cathode is made of lead dioxide (PbO₂).

D. Lead storage battery has ∼38% solution of sulphuric acid as an electrolyte.

• This statement is correct. The electrolyte in a lead-acid battery is a solution of about 35-38% sulfuric acid (H₂SO₄) in water.

Therefore, the correct answer is:

B, D only.

The provided reaction is:

$\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

This reaction involves the following half-reactions:

1. Oxidation of hydrogen gas to H+ ions: $\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + e^-$


2. Reduction of AgCl to Ag: $\mathrm{AgCl}(\mathrm{s}) + e^- \rightleftharpoons \mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}^{-}(\mathrm{aq})$

Looking at the options provided:

Option A: Doesn't involve H₂ gas, so it can't be correct.

Option B: This includes the necessary elements - H₂, AgCl, and Ag.

Option C: Doesn't involve AgCl, so it can't be correct.

Option D: Also includes the necessary elements - H₂, AgCl, and Ag.

However, looking closely, we can see that Option B represents the galvanic cell for this reaction. The reaction requires the oxidation of H₂ to H⁺, which occurs at the anode. The reaction also requires the reduction of AgCl to Ag and Cl-, which occurs at the cathode.

In Option B, the anode (on the left) is where H₂ is being oxidized to H⁺. The cathode (on the right) is where AgCl is reduced to Ag and Cl^-. The salt bridge or ion exchange component is HCl, which allows for the flow of ions to balance charge in the cell.

Therefore, the reaction occurs in the galvanic cell represented by Option B.

The standard electrode potential (E°) of a metal ion M⁺/M in aqueous solution is calculated under standard conditions and relates to the tendency of the M⁺ ion to gain an electron to form the neutral metal atom, M.

This is indeed a redox (reduction-oxidation) process, and the standard electrode potential is defined for the reduction half-reaction. E° values are based on the energies involved in the processes of ionization and hydration.

However, the standard electrode potential does not depend on the state (solid, liquid, gas) of the atom being ionized. Ionization of a solid metal atom (Option C) would not directly impact the electrode potential as it is not inherently part of the process of reduction at the electrode that the E° values are measuring.

The correct answer should be Option C : Ionisation of a solid metal atom.

In volumetric analysis, the concentration of a solution is a crucial factor in determining the accuracy of the results. In the case of an aqueous solution of KOH, the concentration can change over time due to the absorption of atmospheric CO₂. This occurs because KOH is a basic (alkaline) solution and reacts with carbon dioxide (CO₂) from the air to form potassium carbonate (K₂CO₃).

The reaction between KOH and CO₂ is given by :

KOH + CO₂ $ \to $ K₂CO₃

This reaction will change the concentration of KOH in the solution, which in turn will impact the accuracy of the results obtained in volumetric analysis. For example, if the concentration of KOH is lower than it was when the solution was prepared, then more solution will be needed to reach the endpoint in a reaction, and the results will be inaccurate.

That's why it is important to check the concentration of the KOH solution before using it for volumetric analysis, as stated in the assertion (A). And the reason (R) provides the explanation for why the concentration should be checked. Hence, both the assertion and the reason are correct and the correct answer is (D). Both (A) and (R) are correct and (R) is the correct explanation of (A).

The half-cell reaction will be

$ \begin{aligned} \mathrm{Fe}^{3+}(0.02 \mathrm{M})+e^{-} \longrightarrow & \mathrm{Fe}^{2+}(2.0 \mathrm{M}) ; \\ & E^{\circ}{ }_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=0.771 \mathrm{~V} \end{aligned} $

According to Nernst's equation,

$ E=E_{\mathrm{Cell}}^{\circ}-\frac{2303 R T}{n F} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} $

[Number of electrons involved in reduction, $n=1$ ]

$ \begin{gathered} E=0.771 \mathrm{~V}-\frac{2303 R T}{F} \log \frac{2}{0.02} \\ =0.771 \mathrm{~V}-0.059 \log 100 \\ =0.771-0.118=0.653 \mathrm{~V} \end{gathered} $

Hence, reduction potential of half-cell is 0.653 V .

Current $=15 \mathrm{~A}$

$ \begin{aligned} & \text { Time }=45 \times 60=2700 \mathrm{~s} \\ & \text { Since, } Q=I \times t \\ & \\ & \\ & Q=15 \times 2700=40,500 \mathrm{C} \end{aligned} $

Now, for reduction of chlorine gas

$ \mathrm{Cl}_2+2 e^{\ominus} \longrightarrow 2 \mathrm{Cl}^{\ominus} $

or, 2 mole of $e^{\ominus}$ one required for 1 mole of $\mathrm{Cl}_2$

or, $2 \times 96500 \mathrm{C}$ of is required for 22.4 L of $\mathrm{Cl}_2$ electricity.

$\therefore 40,500 \mathrm{C}$ will require ' $x$ '.

$ \Rightarrow \quad x=\frac{40500 \times 22.4}{2 \times 96500}=4.7 \mathrm{~L} $

$ \begin{aligned}Given,\,\, & E_{A^{+} / A}^{\circ}=-0.5 \mathrm{~V} \\ & E_{B^{+} / B}^{\circ}=0.5 \mathrm{~V} \end{aligned} $

For reaction,

$ \begin{aligned} A(s)+B^{+}(a q) & \rightleftharpoons A^{+}(a q)+B(s) \\ E_{\text {cell }}^{\circ} & =+0.5-(-0.5)=+1.0 \\ E_{\text {cell }}^{\circ} & =\frac{2.303 R T}{n F} \log K_C \quad[\because n=1] \\ 1 & =\frac{0.06}{1} \log K_C \\ & \quad\left[\text { Given, } \frac{2.303 R T}{F}=0.06 \mathrm{~V}\right] \\ \frac{1}{0.06} & =\log K_C \text { or } \frac{100}{6}=\log K_C \end{aligned} $

Degree of dissociation $(\alpha)=$

Molar conductivity at $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$

Molar conductivity at infinity dilution

$ \begin{aligned} 0.02 & =\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{\Lambda_{\mathrm{m}}}{400 \mathrm{Scm}^2 \mathrm{~mol}^{-1}} \\ \Lambda_{\mathrm{m}} & =8 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \end{aligned} $

Also,

$ \begin{aligned} \Lambda_{\mathrm{m}} & =\frac{\text { Conductivity }(\kappa) \times 1000}{\text { Concentration of solution }(C)} \\ 8 & =\frac{\kappa \times 1000}{0.1} \end{aligned} $

Conductivity, $\kappa=8 \times 10^{-4} \mathrm{Scm}^{-1}$

To determine the degree of dissociation of acetic acid, follow these steps:

Concentration of Acetic Acid:

$ C = 0.01 \, \text{mol dm}^{-3} = 10 \, \text{mol m}^{-3} $

Conductivity ($\kappa$):

$ \kappa = 19.5 \times 10^{-5} \, \text{S cm}^{-1} = 19.5 \times 10^{-3} \, \text{S m}^{-1} $

Limiting Molar Conductivity ($\lambda_m^\circ$):

$ \lambda_m^\circ = 390 \, \text{S cm}^2 \text{mol}^{-1} = 390 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} $

Calculate Molar Conductivity ($\lambda_m$):

$ \lambda_m = \frac{\kappa}{C} = \frac{19.5 \times 10^{-3}}{10} = 19.5 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} $

Degree of Dissociation ($\alpha$):

$ \alpha = \frac{\lambda_m}{\lambda_m^\circ} = \frac{19.5 \times 10^{-4}}{390 \times 10^{-4}} = 0.05 = 5 \times 10^{-2} $

Therefore, the degree of dissociation of acetic acid is $5 \times 10^{-2}$.

Let's analyze the problem step by step.

The chlorine electrode involves the half‐reaction

$Cl_2(g) + 2e^- \rightarrow 2Cl^-.$

Its potential at 25°C is given by the Nernst equation

$E = E^\circ - \frac{0.059}{2}\log \frac{[Cl^-]^2}{P_{Cl_2}},$

where $P_{Cl_2}$ is the partial pressure of chlorine gas (in atm) and $[Cl^-]$ is the activity (or concentration) of the chloride ion.

Note the following:

For fixed $P_{Cl_2}$ (usually taken as 1 atm), the electrode potential $E$ is a decreasing function of the logarithm of $[Cl^-]^2$.

Therefore, to get the maximum electrode potential, the ratio $\frac{[Cl^-]^2}{P_{Cl_2}}$ must be as small as possible—that is, $[Cl^-]$ must be minimized.

In practice the “chlorine electrode” is prepared in a solution with no added chloride. However, when chlorine gas is bubbled into pure water, it undergoes the following hydrolysis:

$Cl_2 + H_2O \leftrightarrow HCl + HOCl.$

Since HCl is a strong acid, it dissociates completely:

$HCl \rightarrow H^+ + Cl^-.$

In the absence of any added chloride, the only source of $[Cl^-]$ in the solution is from this hydrolysis. Experiments have shown that under these conditions (i.e. when the solution is free of extrinsic chloride) the concentration of chloride ion attains a minimum value.

It turns out that this minimum (and hence the maximum electrode potential) corresponds to a chloride concentration of approximately

$2.5 \times 10^{-3}\; \text{mol L}^{-1}.$

Thus, among the options given, the maximum potential is achieved when

$X = 2.5 \times 10^{-3}\; \text{mol L}^{-1}.$

So, the correct answer is Option A.