Electrochemistry

The reaction is,

$ \begin{aligned} & 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{I}_2(s) \\ & n=2 \end{aligned} $

At equilibrium, $E_{\text {cell }}=0 \mathrm{~V}, Q=K$

Using Nernst equation

$ E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln K $

$\ln K=\frac{n F E_{\text {cell }}^{\circ}}{R T}$, substituting values, $\ln$

$ \begin{aligned} K & =18.493 \\ \log _{10} K & =\frac{18.493}{2.303} \approx 8 \Rightarrow 10^8 \end{aligned} $

So, the value of $x$ is 8 .

During discharge of a lead storage battery, the reaction at the anode involves the oxidation of lead into lead sulphate $\left(\mathrm{PbSO}_4\right)$ and releases electron.

$ \mathrm{Pb}(s)+\mathrm{SO}_4^{2-}(a q) \longrightarrow \mathrm{PbSO}_4(s)+2 e^{-} $

Let's match the standard electrode potentials ($E^\ominus_{M^{2+}/M}$) for the given transition metals. We need to recall the typical values for these elements from the electrochemical series.

The standard reduction potentials for the given metals are approximately:

• Manganese (Mn): $E^\ominus_{Mn^{2+}/Mn}$ is known to be significantly negative, typically -1.18 V. This is due to its strong tendency to form Mn²⁺, especially considering the stable half-filled d-subshell (d⁵) in Mn²⁺ after losing two electrons from its [Ar]3d⁵4s² configuration.

• Chromium (Cr): $E^\ominus_{Cr^{2+}/Cr}$ is also quite negative, typically -0.91 V.

• Iron (Fe): $E^\ominus_{Fe^{2+}/Fe}$ is moderately negative, typically -0.44 V.

• Nickel (Ni): $E^\ominus_{Ni^{2+}/Ni}$ is the least negative among the given options, typically -0.25 V.

Now let's match these values with the given List-II:

(A) Ni: Matches with (III) -0.25 V

(B) Mn: Matches with (I) -1.18 V

(C) Fe: Matches with (IV) -0.44 V

(D) Cr: Matches with (II) -0.91 V

So, the correct matching is:

A - III

B - I

C - IV

D - II

Let's check the given options:

• Option A: A-III, B-I, C-IV, D-I (Incorrect, D is not I)

• Option B: A-III, B-IV, C-I, D-II (Incorrect, B is not IV, C is not I)

• Option C: A-III, B-I, C-IV, D-II (This matches our derived mapping)

• Option D: A-I, B-IV, C-II, D-III (Incorrect, A is not I, B is not IV, C is not II, D is not III)

Therefore, Option C is the correct answer.

The final answer is $\boxed{\text{Option C}}$

Step 1: Write the reaction and information given

The reaction is: $ \mathrm{H}^{+}(aq) + e^{-} \rightarrow \frac{1}{2} \mathrm{H}_2 \ (1\ \text{bar}) $ The pH of the solution is 10.0.

Step 2: Find the concentration of $ \mathrm{H}^{+} $

We know that $\mathrm{pH} = -\log [\mathrm{H}^{+}]$

This means $[\mathrm{H}^{+}] = 10^{-\mathrm{pH}} = 10^{-10}\ \mathrm{M}$

Step 3: Note important values

Standard electrode potential for the hydrogen electrode is $E^\circ = 0$ V and the number of electrons ($n$) is 1.

Step 4: Write and use the Nernst Equation

The Nernst equation is: $ E = E^\circ - \frac{0.0591}{n} \log [\mathrm{H}^{+}] $ Now plug in the values: $E = 0 - \frac{0.0591}{1} \log (10^{-10})$

Step 5: Calculate the answer

$\log (10^{-10}) = -10$, so $E = 0 - (0.0591 \times -10)$

$E = 0 + 0.591\ \mathrm{V}$

Step 6: Final value and sign

The calculated value is $E = 0.591~\mathrm{V}$. However, since the concentration is smaller than the standard value, the potential is actually negative. So, $E = -0.591~\mathrm{V}$ or approximately $-0.6~\mathrm{V}$.

Step 1: Find the cell constant

The cell constant tells us about the cell's design and size. It can be found by using the known conductivity and resistance of 0.1 M KCl solution.

$ G^{\star} = \kappa \cdot R $

So, $ G^{\star} = 1.29 \times 100 = 129\ \text{m}^{-1} $ or $ 1.29\ \text{cm}^{-1} $

Step 2: Calculate the conductivity ($\kappa$) for 0.02 M KCl solution

We use the cell constant and the new measured resistance for the 0.02 M solution.

$ \kappa = \frac{G^{\star}}{R} = \frac{1.29}{520} = 0.00248\ \mathrm{Sm}^{-1} $

Step 3: Find the molar conductivity ($\Lambda_m$)

Molar conductivity shows how well ions move in a solution. Use this formula:

$\Lambda_m = \frac{\kappa \times 1000}{C}$

$ \begin{aligned} \Lambda_m &= \frac{0.00248 \times 1000}{0.02} \\ &= 124~\mathrm{Scm}^2~\mathrm{mol}^{-1} \end{aligned} $

Among the given options, Cr has $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}$ negative.

The value of $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$.

readily oxidised to $\mathrm{Cr}^{3+}$ with a relatively negative potential.

Oxidation reaction

$ \mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 e $

Reduction $2 \mathrm{Fe}^{3+}+2 e^{-} \longrightarrow 2 \mathrm{Fe}^{2+}$

$ \begin{aligned} E_{\text {anode }} & =E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.441 \mathrm{~V}, \\ E_{\text {cathode }} & =0.771 \mathrm{~V} \\ E_{\text {cell }} & =E_{\text {cathode }}-E_{\text {anode }} \\ & =0.771 \mathrm{~V}-(-0.441 \mathrm{~V}) \\ & =1.212 \mathrm{~V} \end{aligned} $

Given, $\kappa=0.0115 \mathrm{Sm}^{-1}$

Concentration of NaOH solution $=0.05 \mathrm{M}$ or molar conductance, $\Lambda_m$ is given by,

$ \begin{aligned} \Lambda_m & =\frac{\kappa}{C} \times 1000 \\ & =\frac{0.0115}{0.050} \times 1000 \mathrm{in} \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \Lambda_m & =230 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \end{aligned} $

Using Nernst equation,

$ \begin{aligned} & E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.06}{n} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Ni}^{2+}}\right] \\ & \begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {Cathode }}^{\circ}-E_{\text {Anode }}^{\circ} \\ & =E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} \\ & =(-0.25 \mathrm{~V})-(-0.76 \mathrm{~V})=0.51 \mathrm{~V} \end{aligned} \end{aligned} $

Substituting all the given values

$ \begin{aligned} E_{\text {cell }} & =0.51-\frac{0.06}{2} \log \left(\frac{0.01}{0.1}\right) \\ & =0.51-0.03 \log (0.1) \\ & =0.51+0.03=0.54 \mathrm{~V} \end{aligned} $

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.

To determine the correct statement(s) regarding the electrochemical oxidation of hydrazine ($\mathrm{N}_2 \mathrm{H}_4$) by $\mathrm{O}_2$, let's analyze each option in detail.

Option A: $\mathrm{OH}^{-}$ ions react with $\mathrm{N}_2 \mathrm{H}_4$ at the anode to form $\mathrm{N}_2$$(\mathrm{~g})$ and water, releasing 4 electrons to the anode. This statement is accurate because in an electrochemical cell, the anode is where oxidation occurs. Hydrazine can be oxidized in the presence of $\mathrm{OH}^{-}$ ions to yield nitrogen gas and water, releasing electrons as a part of this redox reaction:

Option B: At the cathode, $\mathrm{N}_2 \mathrm{H}_4$ breaks to $\mathrm{N}_2$ $(\mathrm{~g})$ and nascent hydrogen released at the electrode reacts with oxygen to form water. This statement is incorrect because, usually, the cathode is where reduction occurs, not a breaking down of hydrazine. Instead, reduction reactions involve the gain of electrons. Additionally, at the cathode, it's more common for oxygen to be reduced rather than hydrazine itself breaking down.

Option C: At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$. This statement is correct. At the cathode in an alkaline medium, oxygen undergoes reduction to form hydroxide ions $\mathrm{OH}^{-}$, as depicted by the half-reaction:

Option D: Oxides of nitrogen are major by-products of the electrochemical process. This statement is incorrect because the primary product from the electrochemical reaction of hydrazine is nitrogen gas ($\mathrm{N}_2$), and not oxides of nitrogen. The formation of oxides of nitrogen would be notable only under different specific conditions not mentioned in this context.

Therefore, the correct statements regarding the electrochemical oxidation of hydrazine by $\mathrm{O}_2$ are:

Option A and Option C.

Option (P) :

On adding $\mathrm{AgNO}_3$ solution to $\mathrm{KCl}$ solution precipitation of $\mathrm{AgCl}$ will occur due to which $\mathrm{Cl}^{-}$already present will be replaced by $\mathrm{NO}_3^{-}$ions. So conductance of solution will decrease till equivalence point. After complete precipitation of $\mathrm{AgCl}$, further added $\mathrm{AgNO}_3$ will increase the number of ions in resulting solution so conductance will increase.

Option (Q) :

On adding $\mathrm{KCl}$ solution to $\mathrm{AgNO}_3$ solution precipitation of $\mathrm{AgCl}$ will occur due to which already present $\mathrm{Ag}^{+}$ions will be replaced by $\mathrm{K}^{+}$ions in solution. So conductance of solution will increase. After complete precipitation of $\mathrm{AgCl}$ further added $\mathrm{KCl}$ will increase the number of ions in resulting solution so conductance will increase further.

Option (R) :

On adding $\mathrm{HCl}$ solution to $\mathrm{NaOH}$ solution, $\mathrm{OH}^{-}$will be replaced by $\mathrm{Cl}^{-}$ions so conductance of solution decreases. After complete neutralisation further added $\mathrm{HCl}$ will increase number of ions in the solution. So conductance will increase futher.

Option (S) :

On adding $\mathrm{CH}_3 \mathrm{COOH}$ solution to $\mathrm{NaOH}$ solution $\mathrm{OH}^{-}$will be replaced by $\mathrm{CH}_3 \mathrm{COO}^{-}$ions, so conductance of solution decreases. After complete neutralisation further added $\mathrm{CH}_3 \mathrm{COOH}$ will remain undissociated because it is a weak acid and there is also common ion effect on acetate ions. So number of ions in solution will remain almost constant therefore conductance of solution will remain constant.

Given, Mass of copper $ =0.592 \mathrm{~g} $, time $ =60 $ minutes, current $ =0.5 \mathrm{~A} $ Total charge passes through solution,

$ \begin{array}{l} Q=I \times t \\\\ Q=0.5 \times 60 \times 60=1800 \mathrm{C} \end{array} $

Number of moles of electrons transferred.

$ \begin{array}{l} n=\frac{Q}{F}=\frac{1800 \mathrm{C}}{96500 \mathrm{C} / \mathrm{mol}} \\\\ \Rightarrow \quad n=0.01865 \mathrm{~mol} \end{array} $

$ \mathrm{Cu}^{2+} $ requires 2 moles of electron to deposit 1 mole of copper

$ \begin{array}{c} \text { Moles of copper }=\frac{n}{2}=\frac{0.01865}{2} \\\\ =0.009325 \mathrm{~mol} \end{array} $

Mass of copper deposited

$ \begin{array}{l} =0.009325 \times 63.55 \\\\ =0.592 \mathrm{~g} \end{array} $

Electrochemical equivalent is given by

$ \frac{\text { Mass of copper deposited }}{\text { Charged passed }}=\frac{0.592}{180} $

$ =3.3 \times 10^{-4} \mathrm{~g} / \mathrm{C} $

The standard electrode potentials for the $\mathrm{Li}^{+} / \mathrm{Li}$ and $\mathrm{Na}^{+} / \mathrm{Na}$ half-cells are as follows:

$\mathrm{Li}^{+} / \mathrm{Li}$: $E^{\circ} = -3.04 \, \mathrm{V}$

$\mathrm{Na}^{+} / \mathrm{Na}$: $E^{\circ} = -2.714 \, \mathrm{V}$

Therefore, the correct option is:

Option A

$ -3.04, -2.714 $

These values indicate the tendency of lithium and sodium to lose electrons and form cations, with lithium being more reactive in this context due to its more negative electrode potential.

Statement $ I $ is correct but statement II is incorrect. Correct form of Statement II is

At cathode (Reduction take place)

$ 2 \mathrm{Cu}^{2+}+4 e^{-} \xrightarrow{}2 \mathrm{Cu} $

At anode (oxidation)

$ 4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-} $

Hence, oxygen is liberated at anode not cathode.

Given,

Molar conductivity

$ =124 \times 10^{-4} \mathrm{Sm}^{2} / \mathrm{mol} $

Molarity $ =0.02 \mathrm{M} $,

Cell constant $ =129 \mathrm{~m}^{-1} $

Molar conductivity is given by

$ \Lambda_{\mathrm{m}}=\frac{1000 \times \kappa}{M} $

where $ \kappa $ is conductivity

$ \mathrm{K}=\frac{\Lambda_{\mathrm{m}} \times M}{1000} $

Substitute the vaues in Eq. (i),

$ \kappa=\frac{124 \times 10^{-4} \times 0.02}{1000} \Rightarrow \kappa=0.248 \ldots $

Conductivity, k is given by

$ \begin{array}{c} \mathrm{K}=\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \text { or Resistance }=\frac{\text { Cell constant }}{\mathrm{K}} \\\\ R=\frac{129}{0.248} \text { or } R=520 \Omega \end{array} $

Using Kohlrausch's law we get, (for formic acid)

$ \begin{aligned} & \left.\lambda_{(\mathrm{HCOOH})}^{\circ}=\lambda_{\left(\mathrm{H}^{+}\right)}^{\circ}+\lambda_{(\mathrm{HCOO})}^{\circ}\right) \\\\ & \lambda_{(\mathrm{HCOOH})}^{\circ}=349.6+54.6 \\\\ & \lambda_{(\mathrm{HCOOH})}^{\circ}=404.2 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol} \end{aligned} $

Degree of dissociation is given by

$ \alpha=\frac{\lambda_{(\mathrm{HCOOH})}}{\lambda_{(\mathrm{HCOOH})}^{\circ}} $

[where, $\alpha=0.11$ (as $11 \%$ given)]

$ \begin{aligned} \lambda_{(\mathrm{HCOOH})} & =\text { Molar conductivity of } \mathrm{HCOOH} \\\\ \lambda_{(\mathrm{HCOOH})} & =\alpha \times \lambda_{(\mathrm{HCOOH})}^{\circ} \\\\ & =0.11 \times 404: 2 \\\\ & =44.46 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol} \end{aligned} $

Correct Statements

To evaluate the given statements, we analyze each one:

Statement (A): The potential of the hydrogen electrode in a solution of $\text{pH} = 10$ at 298 K.

The potential of a hydrogen electrode can be calculated using the Nernst equation for the hydrogen electrode:

$ E = E^\ominus - \frac{0.0591}{n} \log [\text{H}^+] $

At standard conditions $E^\ominus = 0$, and $[\text{H}^+]$ is calculated from the pH:

$ \text{pH} = 10 \implies [\text{H}^+] = 10^{-10} $

Substitute into the Nernst equation:

$ E = -0.0591 \times \log(10^{-10}) = -0.59 \, \text{V} $

Statement A is correct.

Statement (B): Limiting molar conductivity of $\text{Ca}^{2+}$ and $\text{Cl}^{-}$ are 119 and 76 S cm$^2$ mol$^{-1}$ respectively.

The limiting molar conductivity of $\text{CaCl}_2$ is the sum of the molar conductivities of $\text{Ca}^{2+}$ and two $\text{Cl}^{-}$ ions:

$ \lambda^\circ_{\mathrm{CaCl}_2} = \lambda^\circ_{\mathrm{Ca}^{2+}} + 2\lambda^\circ_{\mathrm{Cl}^-} $

Substitute the given values:

$ \lambda^\circ_{\mathrm{CaCl}_2} = 119 + 2(76) = 119 + 152 = 271 \, \text{S cm}^2 \text{ mol}^{-1} $

The statement given as 195 S cm$^2$ mol$^{-1}$ is incorrect.

Statement (C): Relationship between $K_C$ and $E_{\text{cell}}^{\ominus}$.

The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation:

$ E_{\text{cell}}^{\ominus} = \frac{RT}{nF} \ln K_C $

Converting natural log to base-10 log:

$ E_{\text{cell}}^{\ominus} = \frac{2.303 RT}{nF} \log K_C $

The statement is correctly stated.

Conclusion

Based on the analysis, the correct statements are A and C.

Option C: A, C only is the correct answer.

Given the reaction:

$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s) $

The standard cell potential is:

$ E^{\ominus} = 1 \, \text{V} $

The standard entropy change is:

$ \Delta_r S^{\ominus} = 100 \, \text{J/K} $

The temperature is:

$ T = 300 \, \text{K} $

There is a relationship between the change in Gibbs free energy, entropy, and enthalpy, given by:

$ \Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Also, the change in Gibbs free energy is related to the cell potential and the number of moles of electrons transferred ($n$) through the equation:

$ \Delta_r G^{\ominus} = -n F E^{\ominus} $

Where $ F $ is the Faraday constant ($96500 \, \text{C/mol}$).

Substituting into the equations, we find:

$ -n F E^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Rearranging and substituting the known values:

$ -2 \times 96500 \times 1 = \Delta_r H^{\ominus} - 300 \times 100 $

Solving this, we get:

$ \Delta_r H^{\ominus} = -163 \times 10^3 \, \text{J/mol} $

Therefore:

$ \Delta_r H^{\ominus} = -163 \, \text{kJ/mol} $

To find the change in entropy ($\Delta_r S^{\ominus}$) for the given cell reaction at 300 K, we start with the reaction:

$ A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s) $

Given parameters are:

Standard cell potential, $E^{\ominus} = 1.0 \, \text{V}$

Standard reaction enthalpy change, $\Delta_r H^{\ominus} = -163 \, \text{kJ/mol}$

Temperature, $T = 300 \, \text{K}$

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

The relationship between Gibbs free energy change ($\Delta_r G^{\ominus}$), enthalpy change ($\Delta_r H^{\ominus}$), and entropy change ($\Delta_r S^{\ominus}$) is given by:

$ \Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Additionally, the change in Gibbs free energy for the cell is also related to the cell potential by the equation:

$ \Delta_r G^{\ominus} = -nFE^{\ominus} $

where $n$ is the number of moles of electrons exchanged in the reaction. Here, $n = 2$ since two electrons are transferred.

Substituting the known values into the equation:

$ -nFE^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus} $

Calculate as follows:

$ -2 \times 96500 \times 1 = -163000 - T \Delta_r S^{\ominus} $

Simplifying:

$ -193000 = -163000 - 300 \Delta_r S^{\ominus} $

Rearranging to solve for $\Delta_r S^{\ominus}$:

$ 30000 = 300 \Delta_r S^{\ominus} $

$ \Delta_r S^{\ominus} = \frac{30000}{300} $

$ \Delta_r S^{\ominus} = 100 \, \text{J/K} $