Electrochemistry

The molar conductivity of a weak electrolyte is plotted against the square root of its concentration:

$c$-concentration of weak electrolyte

Weak electrolytes give curved decreasing graph.

1) Molar conductivity increases sharply with increase in concentration.

Incorrect.

With increase in concentration, molar conductivity decreases and it is lean from the graph of molar conductivity vs $\sqrt{c}$.

2) Molar conductivity decreases sharply with increase in concentration.

Correct

From the graph it is clean that, molar conductivity decreases sharply with increase in concentration.

3) A small increase in molar conductivity is observed at infinite dilution.

Incorrect

There is no small increase in molas conductivity with increase in concentration for a weak electrolyte. And there is no small increase in mola conductivity. Significant increase occurs.

4) A small decrease in molar conductivity is observed at infinite dilution.

Incorrect

For a weak electrolyte, the molar conductivity significantly increases with dilution, especially at infinite dilution.

Correct answer:

Option 2) Molen conductivity decreases sharply with increase in concentration.

Oxidizing capacity is an element's tendency to donate electrons, or the atmosphere's ability to oxidize compounds.

A higher standard reduction potential indicates a stronger oxidizing capacity. A substance with a higher standard reduction potential is move likely to gain electrons and act as an oxidizing agent in a redon reaction. The more positive the reduction potential, the stronger the oxidizing power of a species.

From the given options, the higher standard reduction potential is for lead $(\mathrm{Pb}) .(+1.67)$.

so, the p-block ion with strongest oxidizing capacity is Pb (lead).

Answer: Option 3) $E_{p_b{ }^{4+} / \mathrm{Pb}^{2+}}^0=+1.67 \mathrm{~V}$

To determine the strongest reducing agent among the options, we need to consider which species most readily donates electrons (i.e., is most easily oxidized). One standard method is to compare the standard reduction potentials of the corresponding half-reactions. A lower (more negative) standard reduction potential indicates that the reduced species is less stable and more prone to oxidation.

Here’s the data provided:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7H_2O,\quad E^\circ = +1.33\,V$

$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V$

$\mathrm{MnO}_4^- + 8H^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4H_2O,\quad E^\circ = +1.51\,V$

$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V$

Now, let’s consider the species from the Options:

Option A: Cl⁻

The relevant half-reaction is:

$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-,\quad E^\circ = +1.36\,V.$

The reduced form here is Cl⁻. Its corresponding oxidation (the reverse reaction) would have an electrode potential of

$E^\circ_{\text{oxidation}} = -1.36\,V.$

Option B: MnO₄⁻

In the given half-reaction, MnO₄⁻ acts as the oxidizing agent (it is reduced to Mn²⁺). Therefore, MnO₄⁻ is not a good reducing agent.

Option C: Cr (metallic chromium)

The relevant half-reaction is:

$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr},\quad E^\circ = -0.74\,V.$

Here, Cr is the reduced species. Reversing this reaction gives:

$\mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3e^-,$

with an effective oxidation potential of

$E^\circ_{\text{oxidation}} = +0.74\,V.$

Option D: Mn²⁺

In the MnO₄⁻/Mn²⁺ couple, Mn²⁺ is the reduced form. Its reverse (oxidation) potential would be:

$E^\circ_{\text{oxidation}} = -1.51\,V.$

When comparing the oxidation potentials (remember, a higher oxidation potential means the species is more willing to lose electrons), we have:

Cl⁻: $E^\circ_{\text{ox}} = -1.36\,V$

Cr: $E^\circ_{\text{ox}} = +0.74\,V$

Mn²⁺: $E^\circ_{\text{ox}} = -1.51\,V$

Among these, metallic Cr stands out because its corresponding reduction half-reaction has the most negative potential ($-0.74\,V$). This negative value indicates that Cr (in its elemental form) is unstable relative to its oxidized form ($\mathrm{Cr}^{3+}$) and, therefore, is the most eager to lose electrons. In other words, Cr is the strongest reducing agent.

Thus, the answer is:

Option C: $\mathrm{Cr}$

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

$\begin{aligned} & \Delta \mathrm{G}_3^0=\Delta \mathrm{G}_1^0+\Delta \mathrm{G}_2^0 \\ & -3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_2^0 \\ & \Delta \mathrm{G}_2^0=3 \mathrm{Fz}-2 \mathrm{Fy} \end{aligned}$

Also $\Delta \mathrm{G}_2^0=-\mathrm{nFE}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0$

$\begin{aligned} & 3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0\right) \\ & \mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0=2 \mathrm{y}-3 \mathrm{z} \end{aligned}$

$\mathrm{E}_{\text {Cell }}^0$ for reaction will be

$\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}^0}^0+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^0 \\ & =\mathrm{x}+2 \mathrm{y}-3 \mathrm{z} \end{aligned}$

Option (2)

$\begin{gathered} \because \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ \text { Option }(1) \mathrm{E}^{\circ}=0.8+0.76 \\ =1.56 \mathrm{~V} \\ \therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 1.56 \\ =-3.12 \mathrm{~V} \\ \text { Option (2) } \mathrm{E}^{\circ}=-2.37+0.76 \\ =-1.61 \mathrm{~V} \\ \therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times(-1.61) \\ =+3.22 \mathrm{~V} \\ \text { Option (3) } \mathrm{E}^{\circ}=-2.37-0.8 \\ =-3.17 \mathrm{~V} \\ \begin{array}{c} \therefore \Delta \mathrm{G}^{\circ}=-2 \end{array} \\ =\mathrm{F} \times(-3.17) \\ =+6.34 \\ \text { Option }(4) \mathrm{E}^{\circ}=0.8-0.34 \\ =0.46 \mathrm{~V} \\ \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 0.46 \\ =-0.92 \mathrm{~V} \end{gathered}$

$\begin{aligned} & \Delta \mathrm{G}_4^{\mathrm{o}}=\Delta \mathrm{G}_1^{\mathrm{o}}+\Delta \mathrm{G}_2^{\mathrm{o}} \\ \Rightarrow & -\mathrm{n}_4 \mathrm{FE}_4^{\mathrm{o}}=-\mathrm{n}_1 \mathrm{FE}_1^0-\mathrm{n}_2 \mathrm{FE}_2^{\mathrm{o}} \\ \Rightarrow & +4 \mathrm{E}_4^{\mathrm{o}}=3 \times 2+(1 \times 0.8) \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=\frac{6.8}{4} \mathrm{~V} \\ \Rightarrow & \mathrm{E}_4^{\mathrm{o}}=1.7 \mathrm{~V} \end{aligned}$

Let’s examine both statements one by one and see whether they hold true for typical corrosion processes (e.g., iron rusting):

Statement (I)

"Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure metal as a cathode."

Corrosion is electrochemical

True. Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and reduction at another region (the cathode) on the metal’s surface.

Pure metal as anode, impurity as cathode

In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) — that is, it serves as the anode.

The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode region where reduction (e.g., oxygen reduction) occurs.

This difference in electrode potentials between the pure region and the impurity region drives the corrosion cell.

Hence, Statement (I) is generally considered true in the usual context of corrosion (like rusting of iron).

Statement (II)

"The rate of corrosion is more in alkaline medium than in acidic medium."

For most common metals (like iron), acidic media typically enhance corrosion because abundant H⁺ ions (and possibly other acidic species) can accelerate the oxidation/dissolution of the metal.

In mildly alkaline or neutral environments, metals often form protective oxide or hydroxide layers that can slow down further corrosion.

While certain strong bases can attack specific metals (e.g., Al in strong NaOH), in general for iron and many other metals, corrosion is more severe in acidic environments than in alkaline ones.

Thus, Statement (II) is false under normal corrosion scenarios (e.g., rusting of iron).

Conclusion

Statement (I): True

Statement (II): False

Therefore, the correct choice (matching these truth values) is:

(C) Statement I is true but Statement II is false.

$\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8$ is obtained by electrolysis of concentrated solution of sulphuric acid.

At anode :

$ 2 \mathrm{HSO}_4^{-} \rightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{e}^{-} $

$ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} $

The mass of aluminium deposited can be calculated using Faraday’s laws of electrolysis. The steps are as follows:

Compute the total charge passed:

$ Q = I \cdot t = 2\,\text{A} \cdot (30 \times 60\,\text{s}) = 3600\,\text{C} $

Determine the moles of aluminium deposited. Since three electrons are required to deposit one mole of aluminium, the number of moles is given by:

$ n(\text{Al}) = \frac{Q}{3F} = \frac{3600}{3 \times 96500} \approx 0.01242\,\text{mol} $

Finally, calculate the mass of aluminium using its molar mass ($M = 27\,\text{g/mol}$):

$ m(\text{Al}) = n(\text{Al}) \times M = 0.01242 \times 27 \approx 0.335\,\text{g} $

Thus, the amount of aluminium deposited at the cathode is approximately

$ 0.336\, \text{g} $

To find the correct match between List I (Cell) and List II (Use/Property/Reaction), let's first understand each item and its corresponding match:

List I (Cell):

• Leclanché cell: This is a primary cell (non-rechargeable) widely used in flashlights and other portable devices. It's not designed for recharging.


• Ni - Cd cell: Nickel-Cadmium (NiCd) batteries are rechargeable and known for being used in portable electronics, tools, and various battery-operated devices.


• Fuel cell: Converts chemical energy, from a fuel into electricity through a chemical reaction with oxygen or another oxidizing agent. It converts energy of combustion (such as hydrogen) into electrical energy directly.


• Mercury cell: A type of battery that uses mercury oxide. It's known for its stable output voltage, making it suitable for devices like hearing aids. However, the reaction described in the list II does not pertain to the mercury cell but more closely to the Leclanché cell or similar.

List II (Use/Property/Reaction):

• Converts energy of combustion into electrical energy: This directly relates to the principle behind fuel cells, which generate electricity through the chemical reaction of a fuel with oxygen.


• Does not involve any ion in solution and is used in hearing aids: This refers to the mercury cell, which is chosen for devices requiring a stable voltage, like hearing aids, although the description about ions might be misleading. The characteristic here is about stable voltage and application in hearing aids.


• Rechargeable: This trait is indicative of Ni - Cd cells, which are known for their ability to be recharged multiple times.


• Reaction at anode $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$: This anodic reaction is typical in cells like the Leclanché cell, which features a zinc anode undergoing oxidation.

Given these explanations, the correct matches would be:

• A (Leclanché cell) with IV, due to the anodic reaction involving zinc.


• B (Ni - Cd cell) with III, because it is rechargeable.


• C (Fuel cell) with I, as it converts the energy of combustion into electrical energy.


• D (Mercury cell) with II, by elimination and considering its use in devices like hearing aids and the slight misdescription regarding ions.

Therefore, the correct answer is Option C: A-IV, B-III, C-I, D-II.

The dissociation of a weak electrolyte ($\mathrm{HA}$) in solution can be represented as:

$ \mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{A}^{-}(\mathrm{aq}) $

The degree of dissociation ($\alpha$) for a weak electrolyte is given by:

$ \alpha = \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} $

Where $\Lambda_{\mathrm{m}}$ is the molar conductivity at a given concentration $C$, and $\Lambda_{\mathrm{m}}^{\circ}$ is the molar conductivity at infinite dilution.

The dissociation constant ($K_{\mathrm{a}}$) is described by:

$ K_{\mathrm{a}} = \frac{\alpha^2 C}{1 - \alpha} $

Rewriting this equation in terms of $\Lambda_{\mathrm{m}}$ and $\Lambda_{\mathrm{m}}^{\circ}$:

$ K_{\mathrm{a}} = \frac{\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C}{1 - \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}} $

By multiplying out and simplifying the above expression:

$ \left( \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right)^2 C + K_{\mathrm{a}} \left( \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right) = K_{\mathrm{a}} $

Substitute $\alpha = \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$:

$ \left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C + K_{\mathrm{a}} \left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right) - K_{\mathrm{a}} = 0 $

Rewriting by multiplying through by $\left( \Lambda_{\mathrm{m}}^{\circ} \right)^2$:

$ \Lambda_{\mathrm{m}}^2 C + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} - K_{\mathrm{a}} \left( \Lambda_{\mathrm{m}}^{\circ} \right)^2 = 0 $

Thus, the correct equation that describes the change in molar conductivity with respect to concentration for a weak electrolyte is given by:

Option D:

$ \Lambda_{\mathrm{m}}^2 C - K_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2} + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} = 0 $

To determine how the emf of the cell, $\mathrm{Tl}\left|\underset{(0.001 \mathrm{M})}{\mathrm{Tl}^{+}}\right| \underset{(0.01 \mathrm{M})}{\mathrm{Cu}^{2+}} \mid \mathrm{Cu}$, can be increased, we can use the Nernst equation. The Nernst equation for this electrochemical cell is given by:

$ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{n} \log \left( \frac{[\mathrm{Tl}^{+}]}{[\mathrm{Cu}^{2+}]} \right) $

where:

• $ E_{\text{cell}} $ is the emf of the cell.
• $ E_{\text{cell}}^\circ $ is the standard emf of the cell.
• $ n $ is the number of moles of electrons transferred in the cell reaction; here, $ n = 2 $.
• $ [\mathrm{Tl}^{+}] $ is the concentration of thallium ions.
• $ [\mathrm{Cu}^{2+}] $ is the concentration of copper ions.

Given that the emf of the cell can be expressed in terms of the concentration of the ions involved, we can see that increasing the concentration of $[\mathrm{Cu}^{2+}]$ or decreasing the concentration of $[\mathrm{Tl}^{+}]$ will affect the logarithmic term in the Nernst equation:

$ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0591}{2} \log \left( \frac{0.001}{0.01} \right) $

To increase the emf ($ E_{\text{cell}} $) of the cell, it is beneficial to have a less negative (or more positive) correction term. This can be achieved by:

• Increasing the concentration of $[\mathrm{Cu}^{2+}]$ ions: This makes the term inside the logarithm smaller (since we are dividing by a larger number), which in turn makes the logarithmic term less negative.

Therefore, the correct answer is:

Option B: increasing concentration of $\mathrm{Cu}^{2+}$ ions

To determine which galvanic cell corresponds to the given reaction:

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

we need to examine the setups of the given cells and check if they produce the same reactants and products as in the above reaction.

Analyzing the reaction, we observe:

• Hydrogen gas ($\mathrm{H}_{2(\mathrm{~g})}$) is involved at one electrode and splits into $\mathrm{H}_{(\mathrm{aq})}^{+}$ ions.
• Solid silver chloride ($\mathrm{AgCl}_{(\mathrm{s})}$) decomposes at another electrode into $\mathrm{Ag}_{(\mathrm{s})}$ and $\mathrm{Cl}_{(\mathrm{aq})}^{-}$ ions.

Let's examine each option:

Option A:

$\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text {soln.) }}\left|\mathrm{AgNO}_{3(\mathrm{aq})}\right| \mathrm{Ag}$

This cell setup does not directly involve $\mathrm{AgCl}_{(\mathrm{s})}$ solid separating into $\mathrm{Ag}_{(\mathrm{s})}$ and $\mathrm{Cl}_{(\mathrm{aq})}^{-}$. Hence, it does not match the reaction given.

Option B:

$\mathrm{Ag}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{KCl}_{(\text {(soln.) }}\left|\mathrm{AgNO}_{3 \text { (aq.) }}\right| \mathrm{Ag}$

This involves silver electrodes and does not have a setup for reaction with hydrogen gas or the splitting of $\mathrm{H}_{2(\mathrm{~g})}$ into $\mathrm{H}_{(\mathrm{aq})}^{+}$ ions. So, this does not match either.

Option C:

$\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{KCl}_{(\text {soln.) }}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}$

While this configuration includes $\mathrm{AgCl}_{(\mathrm{s})}$ and $\mathrm{H}_{2(\mathrm{~g})}$, it uses potassium chloride solution instead of hydrochloric acid solution, which alters the chemistry and does not perfectly match the required reaction mechanism.

Option D:

$\mathrm{Pt}\left|\mathrm{H}_{2(\mathrm{~g})}\right| \mathrm{HCl}_{(\text {soln. })}\left|\mathrm{AgCl}_{(\mathrm{s})}\right| \mathrm{Ag}$

This configuration fits the reaction given. Here hydrogen gas at the platinum electrode splits into $\mathrm{H}_{(\mathrm{aq})}^{+}$ ions in the hydrochloric acid solution. On the other side, the solid silver chloride separates into $\mathrm{Ag}_{(\mathrm{s})}$ and $\mathrm{Cl}_{(\mathrm{aq})}^{-}$ ions. Hence, this accurately matches the reaction.

Therefore, the correct answer is:

Option D

Statement I is indeed true. When $\mathrm{MnO}_2$ (manganese dioxide) is fused with $\mathrm{KOH}$ (potassium hydroxide) and an oxidising agent such as $\mathrm{KNO}_3$ (potassium nitrate), it forms dark green potassium manganate ($\mathrm{K}_2\mathrm{MnO}_4$). The reaction can be represented as follows:

$\mathrm{2MnO}_2 + 4KOH + \mathrm{O}_2 \rightarrow \mathrm{2K}_2\mathrm{MnO}_4 + 2H_2\mathrm{O}$

This process involves the oxidation of manganese dioxide to manganate ion ($\mathrm{MnO}_4^{2-}$) in an alkaline medium.

Statement II is also true. The manganate ion ($\mathrm{MnO}_4^{2-}$), when subjected to electrolytic oxidation in an alkaline medium, can indeed be converted into permanganate ion ($\mathrm{MnO}_4^-$), which is characterized by a deep purple color. This conversion is an example of an electron-loss (oxidation) process at the anode of an electrolytic cell. The reaction can be represented as follows:

$\mathrm{MnO}_4^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{MnO}_4^- + 2\mathrm{OH}^- + 2e^-$

Thus, both statements I and II are accurate, making the correct answer:

Option D: Both Statement I and Statement II are true.

An electrochemical cell can be converted into an electrolytic cell by applying an external potential greater than $E_{\text {cell }}^{\circ}$. The entire cell reaction will be reversed. The oxidised species will be reduced and the reduced species will be oxidised.

The molar conductivity, denoted as $\wedge_m$, is calculated using the formula:

$\wedge_m = \frac{\mathrm{K} \times 1000}{\mathrm{M}}$

When the volume of the solution is doubled by adding more water at a constant temperature, the conductivity, $\mathrm{K}$, will be reduced to half, while the molarity, $\mathrm{M}$, will also be halved.

Therefore, the molar conductivity $\lambda_{\mathrm{m}}$ will either remain the same or cannot be measured with sharp accuracy.

For an infinitely dilute solution, the molar conductivity $\wedge_{\mathrm{m}}$ is approximately equal to the molar conductivity at infinite dilution $\wedge_{\mathrm{m}}^0$, which is a constant.

To determine the quantity of silver deposited when one coulomb of charge is passed through $\mathrm{AgNO}_3$ solution, we need to refer to Faraday's laws of electrolysis, especially to the concept of the electrochemical equivalent. The electrochemical equivalent (ECE) of a substance is the amount of the substance that is deposited or dissolved at an electrode during electrolysis by one coulomb of charge.

The electrochemical equivalent of a substance can be calculated using the formula:

$E = \frac{M}{nF}$

where:

• $E$ is the electrochemical equivalent of the substance in grams per coulomb ($\mathrm{g/C}$).
• $M$ is the molar mass of the substance in grams per mole ($\mathrm{g/mol}$).
• $n$ is the number of electrons involved in the oxidation or reduction reaction (the valency).
• $F$ is the Faraday constant, approximately $96485 \, \mathrm{C/mol}$, which is the charge of one mole of electrons.

In the case of silver ($\mathrm{Ag}$) being deposited from $ \mathrm{AgNO}_3 $, when silver ion ($\mathrm{Ag}^+$) is reduced to metallic silver ($\mathrm{Ag}$), the reaction involves one electron ($n = 1$). The molar mass of silver (Ag) is approximately $107.868 \, \mathrm{g/mol}$. Using this information:

$E = \frac{107.868 \, \mathrm{g/mol}}{1 \cdot 96485 \, \mathrm{C/mol}} \approx 0.001118 \, \mathrm{g/C}$

This means that for every coulomb of charge passed through the solution, $0.001118 \, \mathrm{g}$ of silver is deposited.

Given the options:

• Option A: $0.1$ g atom of silver - This is incorrect because the calculation shows a much smaller amount is deposited per coulomb.
• Option B: 1 chemical equivalent of silver - This does not directly relate to the amount deposited per coulomb.
• Option C: $1 \mathrm{~g}$ of silver - This is incorrect as it overestimates the amount deposited by a large margin.
• Option D: 1 electrochemical equivalent of silver - This is correct, as it directly relates to the amount of a substance deposited by one coulomb of charge, based on the formula and calculation shown.

Therefore, the correct answer is Option D: 1 electrochemical equivalent of silver.

$\begin{aligned} & \mathrm{E}_{\mathrm{cell}}^{\circ}=\mathrm{E}_{\mathrm{X} \mid \mathrm{X}^{2-}}^{\circ}-\mathrm{E}_{\mathrm{M}^{2+} \mid \mathrm{M}}^{\circ} \\ & =0.34-0.46 \\ & =-0.12 \mathrm{~V} \text { (Non-spontaneous) } \end{aligned}$

So, reverse reaction will be spontaneous.

$\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}$

The compound with divalent cation $\left(\mathrm{A}^{2+}\right)$ and anion $\left(B^{2-}\right)$ will be $A B$.

Molar conductivity of its solution will be

$57+73=130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

In the cells commonly used in clocks, specifically alkaline batteries, the reaction at the cathode involves the reduction of manganese dioxide (MnO₂). The manganese in MnO₂ is initially in the +4 oxidation state, and it gets reduced to the +3 oxidation state. Thus, the correct reaction occurring at the cathode can be represented as follows:

The correct answer is:

Option B
reduction of $\mathrm{Mn}$ from +4 to +3

Fuel cells produce electricity with an efficiency of about $70 \%$.

Fuel cells are pollution free, thus, eco-friendly.

Fuel cells are type of Galvanic cells only.

The molar conductivity ($\Lambda_m$) of a strong electrolyte depends on the concentration ($c$) according to Kohlrausch's law, which can be mathematically expressed as $\Lambda_m = \Lambda_m^0 - k\sqrt{c}$, where $\Lambda_m^0$ is the molar conductivity at infinite dilution, and $k$ is a constant. The graph of molar conductivity ($\Lambda_m$) against the square root of concentration ($c^{1/2}$) is a straight line with a negative slope.

The unit of molar conductivity ($\Lambda_m$) is Siemens meter squared per mole ($\text{S cm}^2 \text{mol}^{-1}$). Concentration ($c$) has the unit moles per liter ($\text{mol L}^{-1}$), and thus its square root ($c^{1/2}$) has the unit $\text{mol}^{1/2} \text{L}^{-1/2}$.

The slope of the plot of $\Lambda_m$ against $c^{1/2}$ is derived from the expression $k$ in $k\sqrt{c}$. Therefore, to find the correct unit of the slope, we look at the division of the unit of $\Lambda_m$ by the unit of $c^{1/2}$:

$\frac{\text{S cm}^2 \text{mol}^{-1}}{\text{mol}^{1/2} \text{L}^{-1/2}} = \text{S cm}^2 \text{mol}^{-1-1/2} \text{L}^{1/2} = \text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}$

Thus, the correct unit for the slope of a plot of molar conductivity against the square root of concentration for a strong electrolyte is $\text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}$, which matches with Option D.

Calomel electrode is metal-insoluble salt – Anion electrode.

The electromotive force (emf) of a hydrogen electrode can be derived from the Nernst equation, which for the hydrogen half-cell reaction $\mathrm{H_2(g)} \rightarrow 2\mathrm{H^+(aq)} + 2\mathrm{e^-}$ is given by:

$ E = E^\circ + \frac{RT}{2F} \ln \left( \frac{[\mathrm{H^+}]^2}{P_{\mathrm{H_2}}} \right) $

where:

• E is the electrode potential.
• E^∘ is the standard electrode potential, which is 0 V for the hydrogen electrode.
• R is the gas constant, $8.314 \, \mathrm{J \cdot mol^{-1} \cdot K^{-1}}$.
• T is the temperature in Kelvin: $25^{\circ} \mathrm{C} = 298 \, \mathrm{K}$.
• F is the Faraday constant, $96485 \, \mathrm{C \cdot mol^{-1}}$.
• $[\mathrm{H^+}]$ is the concentration of hydrogen ions.
• $P_{\mathrm{H_2}}$ is the pressure of hydrogen gas.

In pure water at $25^{\circ} \mathrm{C}$, $[\mathrm{H^+}] = 10^{-7} \, \mathrm{M}$. To make the emf zero, we set E to 0 in the Nernst equation:

$ 0 = 0 + \frac{8.314 \times 298}{2 \times 96485} \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right) $

Simplifying the constants:

$ \frac{8.314 \times 298}{2 \times 96485} \approx 0.0128 $

Thus, the equation becomes:

$ 0 = 0.0128 \ln \left( \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} \right) $

Since ln term must be zero for this equation to hold true (as 0 divided by any number is still 0), the expression inside the logarithm must equal 1:

$ \frac{(10^{-7})^2}{P_{\mathrm{H_2}}} = 1 $

Simplifying this gives:

$ (10^{-7})^2 = P_{\mathrm{H_2}} $

$ 10^{-14} = P_{\mathrm{H_2}} $

Therefore, the required pressure of $\mathrm{H_2}$ to make the emf of the hydrogen electrode zero in pure water at $25^{\circ} \mathrm{C}$ is $10^{-14}$ bar.

The correct answer is Option B: $10^{-14}$.

Conductivity of electrolytic cell is affected by concentration of electrolyte, nature of electrolyte and nature of solvent.

Alkaline oxidative fusion of $\mathrm{MnO}_2$ :

$2 \mathrm{MnO}_2+4 \mathrm{OH}^{-}+\mathrm{O}_2 \rightarrow 2 \mathrm{MnO}_4^{2-}+2 \mathrm{H}_2 \mathrm{O}$

Electrolytic oxidation of $\mathrm{MnO}_4^{2-}$ in alkaline medium.

$\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^{-}+e^{-}$

Higher the value of $\oplus$ve SRP (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant.

Hence, ox. Power:- $\mathrm{BrO}_4^{-}>\mathrm{IO}_4^{-}>\mathrm{ClO}_4^{-}$

The statement that is not correct about the rusting of iron is Option C. Let's examine each option one by one:

Option A: Rusting of iron is envisaged as setting up of an electrochemical cell on the surface of iron object. This statement is correct. Rusting of iron is an electrochemical process that occurs when iron (Fe) reacts with water (H₂O) and oxygen (O₂) to form hydrated iron(III) oxide, which we commonly know as rust. Iron acts as the anode, the part where oxidation occurs as iron loses electrons. Oxygen, often dissolved in water, acts as the cathode where reduction occurs, completing the electrochemical cell.

Option B: Dissolved acidic oxides such as $\mathrm{SO}_2$ and $\mathrm{NO}_2$ in water act as catalyst in the process of rusting. This statement is correct. Acidic oxides like $\mathrm{SO}_2$ and $\mathrm{NO}_2$ can dissolve in water to form acidic solutions which can lead to the acceleration of the rate of rusting as they may lower the pH of water, thereby increasing the availability of hydrogen ions (H⁺) which may facilitate the electrochemical reactions involved in rusting.

Option C: Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off. This statement is incorrect. Tin (Sn) is used as a protective coating for iron to prevent it from rusting; however, if the tin coating starts to peel off, it exposes the iron underneath to the atmosphere, which will then be susceptible to rusting. Once the protective layer is compromised, the iron is at risk, especially if the exposed parts are more anodic than tin. Tin as a more cathodic metal can even act to accelerate the corrosion of the exposed iron—this is known as a bimetallic corrosion or galvanic corrosion.

Option D: When $\mathrm{pH}$ lies above 9 or 10, rusting of iron does not take place. This statement is generally correct. Rusting of iron is lessened in alkaline conditions, i.e., when the pH is above 9 or 10 because the increased availability of hydroxide ions (OH^-) can reduce the solubility of iron(II) and iron(III) ions, leading to their precipitation as hydroxides rather than participating in the rusting process.

Therefore, the statement from the given options that is not correct about the rusting of iron is Option C, which falsely claims that a peeling tin surface can still prevent rusting of iron.

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).