Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

When manganese dioxide is fused with potassium hydroxide in the presence of an oxidizing agent, it undergoes a chemical transformation to produce dark green potassium manganate, as illustrated by the reaction $\mathrm{MnO}_2 + 4\mathrm{KOH} + \mathrm{O}_2 \rightarrow 2\mathrm{K}_2\mathrm{MnO}_4 + 2\mathrm{H}_2\mathrm{O}$. Subsequently, the manganate ion can be converted into the permanganate ion through electrolytic oxidation within an alkaline medium, a process characterized at the anode by the half-reaction $\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^{-} + \mathrm{e}^{-}$. Because both of these chemical steps accurately describe the established synthesis pathways for these manganese compounds, both provided statements are verified as correct.

A calomel electrode is constructed by placing metallic mercury in contact with a paste of its insoluble salt, mercurous chloride ($Hg_2Cl_2$), which is then immersed in a potassium chloride solution containing chloride ions. Because the electrode potential is governed by the equilibrium involving the metal, the solid salt, and the common anion, it is classified as a metal-insoluble salt-anion electrode. The half-reaction occurring at this interface, $2Hg(l) + 2Cl^-(aq) \rightleftharpoons Hg_2Cl_2(s) + 2e^-$, demonstrates this structural configuration by linking the metal phase, the sparingly soluble salt phase, and the ionic solution phase.

Representing a galvanic cell requires correctly identifying the oxidation half-reaction at the anode and the reduction half-reaction at the cathode. The given reaction involves the oxidation of hydrogen gas, $\frac{1}{2}\mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{H}^+(\mathrm{aq}) + \mathrm{e}^-$, which is facilitated by an inert platinum electrode, alongside the reduction of silver chloride, $\mathrm{AgCl}(\mathrm{s}) + \mathrm{e}^- \rightarrow \mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}^-(\text{aq})$, at a silver electrode. Connecting these two half-cells requires an electrolyte containing chloride ions, such as $\mathrm{KCl}$, to maintain the electrochemical balance of the system. This leads to the cell representation $\mathrm{Pt} | \mathrm{H}_2(\mathrm{g}) | \mathrm{KCl}(\text{soln.}) | \mathrm{AgCl}(\text{s}) | \mathrm{Ag}$, where the components align perfectly with the overall redox process.

A fuel cell functions as a specific type of galvanic cell that converts chemical energy directly into electrical energy through the reaction of hydrogen and oxygen, described by the equation $2H_2 + O_2 \rightarrow 2H_2O$. Because the sole byproduct of this clean process is water, the system is classified as eco-friendly. Due to these advantages, the technology has been successfully utilized to provide reliable power in spaceships.

Electrochemical cells function based on distinct redox processes and physical designs tailored to specific operational needs. The Leclanche cell facilitates energy production through the oxidation of zinc at the anode, described by the reaction $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2 \mathrm{e}^{-}$, while the Ni-Cd cell is characterized by its ability to be recharged after depletion. Additionally, fuel cells are engineered to convert the chemical energy of combustion directly into electrical energy, whereas the mercury cell is prized for use in hearing aids because its design does not involve any ion in the electrolyte solution, ensuring a constant potential.

An electrochemical cell typically functions by converting chemical energy into electrical energy through a spontaneous redox reaction, yielding an intrinsic potential of $E_{\text{cell}}^0$. When an external source applies a potential in the opposite direction, it creates a counter-force against this natural electron flow. If the magnitude of this external potential is increased to be greater than $E_{\text{cell}}^0$, the original spontaneous process is reversed, forcing the system to drive non-spontaneous chemical changes as an electrolytic cell.

The molar conductivity ($\Lambda_m$) of an electrolyte reaches its theoretical limit, known as limiting molar conductivity, when the solution approaches infinite dilution, where dissociation is essentially complete. Since the solution provided is already at infinite dilution, the degree of dissociation and the ionic interactions have already achieved their steady state. Consequently, adding further water to an already infinitely dilute system produces no change in molar conductivity, making any alteration indistinguishable or impossible to measure accurately.

According to Faraday's laws of electrolysis, the reduction of copper ions occurs via the equation $\mathrm{Cu}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}$. This stoichiometry establishes that 2 Faradays of electricity are required to liberate 1 mole, or 1 gram atom, of copper. Passing 1 Faraday of electricity therefore results in the deposition of 0.5 gram atoms of copper. Converting this quantity into the specified expression $x \times 10^{-1}$ yields $5 \times 10^{-1}$, indicating that $x$ is 5.

Materials are classified by their electrical conductivity measured at 298.15 K, with conductors exhibiting high values, semiconductors showing intermediate values, and insulators having extremely low values. Within the provided list, the values 2.1 x 10^3, 1.2 x 10, 3.91, and 1.0 x 10^3 correspond to conductors. The values 1.5 x 10^-2 and 1 x 10^-7 are categorized as semiconductors, while 1.0 x 10^-16 represents an insulator. Counting these identified entries confirms there are 4 conductors in the set.

The Nernst equation allows us to determine the reduction potential of a half-cell under non-standard conditions by relating it to the standard hydrogen electrode potential, which is zero. For the reaction $2\text{H}^+_{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}$, the potential is given by the expression $E = E^\circ_{\text{H}^+/\text{H}_2} - \frac{0.06}{2} \log \frac{P_{\text{H}_2}}{[\text{H}^+]^2}$. Since $E^\circ_{\text{H}^+/\text{H}_2}$ is $0\text{V}$, substituting the given values of $[\text{H}^+] = 1\text{M}$, $P_{\text{H}_2} = 2\text{atm}$, and $\log 2 = 0.3$ leads to $E = 0 - \frac{0.06}{2} \log \frac{2}{1^2}$. This calculation simplifies to $E = -0.03 \times 0.3$, which results in a potential of $-0.9 \times 10^{-2}\text{V}$.

Electrolytic conductance measures the ease with which a solution carries current through the movement of dissolved ions, a property fundamentally governed by the specific nature of the electrolyte, the concentration of the solution, and the characteristics of the solvent. Because the electrodes merely act as the terminals for current entry and exit at the phase boundary, they do not participate in the bulk ionic transport process occurring within the solution. Consequently, changing the material of the electrodes has no influence on the intrinsic conductance of the electrolyte itself, as the mobility and number of ions are independent of the electrode composition.

Faraday's laws of electrolysis establish a direct relationship between the quantity of electricity passed through an electrolytic cell and the mass of the substance deposited at the electrode. Given the reduction reaction $\mathrm{Zn}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}$, the process involves the transfer of two electrons per zinc ion, meaning the n-factor is 2.

Convert the time interval from 15 minutes into 900 seconds to ensure consistent units when calculating the charge. The total mass $W$ deposited is determined using $W = \frac{M \times I \times t}{n \times F}$, where $M$ is the atomic mass 65.4, $I$ is the current 0.015 A, $t$ is 900 seconds, $n$ is 2, and $F$ represents Faraday's constant of 96500 C/mol. Substituting these values into the expression $W = \frac{65.4 \times 0.015 \times 900}{2 \times 96500}$ results in a final mass of $45.75 \times 10^{-4} \text{g}$.

The oxidizing power of a substance is inherently linked to its ability to undergo reduction, which is quantitatively measured by its standard reduction potential ($E^{\circ}$). A higher positive $E^{\circ}$ value indicates a greater spontaneity for electron gain, meaning that species with a larger reduction potential act as stronger oxidizing agents. Comparing the given potentials, where $\mathrm{BrO}_4^{-}$ has an $E^{\circ} = 1.74\ \mathrm{V}$, $\mathrm{IO}_4^{-}$ has an $E^{\circ} = 1.65\ \mathrm{V}$, and $\mathrm{ClO}_4^{-}$ has an $E^{\circ} = 1.19\ \mathrm{V}$, we observe that the oxidizing strength follows the descending order of these values, resulting in the sequence $\mathrm{BrO}_4^{-} > \mathrm{IO}_4^{-} > \mathrm{ClO}_4^{-}$.

Faraday's laws of electrolysis dictate that the same quantity of electricity will produce an equivalent number of equivalents for all substances involved in the reaction. Consequently, the equivalents of silver deposited must equal the equivalents of oxygen gas evolved. Since oxygen is formed via an oxidation process requiring four electrons per mole, the n-factor for oxygen is $4$. Using the standard molar volume of $22.4\ \text{L}$ at STP, the $5600\ \text{mL}$, or $5.6\ \text{L}$, of oxygen provided represents $5.6 / 22.4 = 0.25\ \text{moles}$. Multiplying these moles by the n-factor of $4$ gives a total of $1.0\ \text{equivalent}$ for the oxygen gas. Because the equivalent weight of silver is its atomic mass of $108\ \text{g/mol}$ divided by its valency of $1$, one equivalent of silver corresponds to a mass of exactly $108\ \text{g}$.

The relationship between molar conductivity and specific conductance is foundational for determining the extent of ionization in weak electrolytes, defined by the equation $\Lambda_m = \frac{\kappa \times 1000}{C}$. Substituting the given specific conductance of $5 \times 10^{-5} \mathrm{S cm}^{-1}$ and concentration of $0.0025 \mathrm{M}$ yields a molar conductivity of $20 \mathrm{S cm}^2 \mathrm{mol}^{-1}$. Comparing this to the limiting molar conductivity of $400 \mathrm{S cm}^2 \mathrm{mol}^{-1}$ allows us to determine the degree of dissociation $\alpha$, which is the ratio $\frac{\Lambda_m}{\Lambda_m^\circ}$, equal to $\frac{20}{400}$ or $\frac{1}{20}$. Because acetic acid is a weak electrolyte, its dissociation constant $K_a$ is determined by the equilibrium expression $K_a = \frac{C \alpha^2}{1 - \alpha}$. Inserting the values $C = 0.0025$ and $\alpha = \frac{1}{20}$ leads to the calculation $\frac{0.0025 \times (\frac{1}{20})^2}{1 - \frac{1}{20}}$, which simplifies to $\frac{0.0025}{19 \times 20}$ or $6.6 \times 10^{-6}$. Expressing this value as a multiple of $10^{-7}$ results in $66 \times 10^{-7}$.

The Nernst equation serves as the governing principle for calculating the reaction quotient when the cell potential deviates from standard conditions. In the given electrochemical setup, hydrogen undergoes oxidation at the anode while ferric ions are reduced at the cathode according to the overall reaction $\frac{1}{2}\mathrm{H}_2(g) + \mathrm{Fe}^{3+}(aq) \rightarrow \mathrm{H}^+(aq) + \mathrm{Fe}^{2+}(aq)$, which involves the transfer of a single electron. Because the standard hydrogen electrode is defined with a potential of $0\text{ V}$, the standard cell potential $E^\circ$ is equal to $0.771\text{ V}$.

Applying the relation $E = E^\circ - \frac{0.059}{n} \log Q$, where $n=1$, we substitute the known variables into the equation: 0.712 = 0.771 - \frac{0.059}{1} \log \frac{[\mathrm{Fe}^{2+}]\}{[\mathrm{Fe}^{3+}]$}$. Given that the pressure of hydrogen and the concentration of hydrogen ions are at unity, the reaction quotient simplifies solely to the ratio of the iron species concentrations. Rearranging the equation results in$0.059 \log \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} = 0.771 - 0.712$, which simplifies to$0.059 \log \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} = 0.059$. Dividing both sides by$0.059$leaves$\log \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} = 1$, which confirms that the ratio is$10^1$, or$10$.

The Nernst equation serves as the fundamental principle for relating the cell potential to the reaction quotient and standard electrode potential under non-standard conditions. The electrochemical cell reaction consists of hydrogen gas oxidation at the anode and the reduction of $\mathrm{M}^{3+}$ ions to $\mathrm{M}^{+}$ ions at the cathode, represented by the net process $\mathrm{H}_{2}(g) + \mathrm{M}^{3+}(aq) \rightarrow 2\mathrm{H}^{+}(aq) + \mathrm{M}^{+}(aq)$. This reaction involves the transfer of two electrons, meaning the value for $n$ is 2.

The standard cell potential is determined by subtracting the anode potential from the cathode potential, giving 0.2 V. Using the Nernst relation $E_{\text{cell}} = E^{\theta}_{\text{cell}} - \frac{0.059}{n} \log Q$, and substituting the standard conditions where the hydrogen ion concentration is 1 M and pressure is 1 bar, the reaction quotient simplifies to Q = \frac{[\mathrm{M}^{+}]\}{[\mathrm{M}^{3+}]$}$. Setting up the calculation with the given values results in the equation$0.1115 = 0.2 - \frac{0.059}{2} \log \frac{[\mathrm{M}^{+}]$}{[\mathrm{M}^{3+}]$}$. Given the ratio is$10^a$, the equation becomes$0.1115 = 0.2 - 0.0295 \cdot a$. Solving for$a$by subtracting 0.2 from both sides leads to$-0.0885 = -0.0295 \cdot a$, which confirms$a = 3$.

The maximum electrical work that a chemical reaction can perform at constant temperature and pressure is given by the negative of the change in Gibbs energy, represented as $w_{\text{elec}} = -\Delta_r G$. Because a spontaneous reaction is characterized by a negative $\Delta_r G$, the electrical work produced must be a positive quantity, meaning the assertion that the work is equal to the Gibbs energy itself is incorrect due to the missing negative sign. Conversely, the standard cell potential is a fixed value defined under standard conditions, and the temperature dependence of this potential is correctly described by the relationship $\frac{dE^{\circ}}{dT} = \frac{\Delta_r S^{\circ}}{nF}$. Furthermore, a cell operates reversibly when the internal potential is perfectly balanced by an opposing external source, ensuring that no net current flows through the system.

The potential of an electrochemical cell under non-standard conditions is determined by the Nernst equation, which adjusts the standard cell potential based on the reaction quotient. In this specific cell, the anode is the standard hydrogen electrode with a potential of $0\ \mathrm{V}$, while the cathode consists of the $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ redox couple. Given that the standard potential for the reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$ is $0.771\ \mathrm{V}$, the standard cell potential is $0.771\ \mathrm{V}$.

Because the hydrogen electrode operates at standard conditions of $1\ \mathrm{atm}$ for pressure and $1\ \mathrm{M}$ for ion concentration, the reaction quotient for the overall cell simplifies to the ratio of the iron species concentrations, specifically \frac{[\mathrm{Fe}^{2+}]\}{[\mathrm{Fe}^{3+}]$}$. Applying the Nernst equation with a single electron transfer ($n=1$), the expression becomes$0.712 = 0.771 - 0.06 \log \left( \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} \right)$. Rearranging this leads to$0.06 \log \left( \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} \right) = 0.771 - 0.712$, which simplifies to$0.06 \log \left( \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} \right) = 0.059$. Calculating the ratio,$\log \left( \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} \right) = \frac{0.059}{0.06} \approx 1$, which indicates the concentration ratio is$10^1$, or$10$.

The electrochemical potential of a cell under non-standard conditions is determined by applying the Nernst equation, which adjusts the standard cell potential based on the reaction quotient. From the provided cell notation $X \vert X^{2+} \Vert Y^{2+} \vert Y$, we identify the oxidation half-reaction $X \rightarrow X^{2+} + 2e^-$ and the reduction half-reaction $Y^{2+} + 2e^- \rightarrow Y$, resulting in an overall balanced equation $X + Y^{2+} \rightarrow X^{2+} + Y$ where the number of electrons transferred is $n=2$.

Determining the standard cell potential requires subtracting the standard electrode potential of the anode from that of the cathode, resulting in $E_{\mathrm{cell}}^0 = 0.36 - (-2.36) = 2.72\ \mathrm{V}$. Substituting the known concentrations into the Nernst equation $E_{\mathrm{cell}} = E_{\mathrm{cell}}^0 - \frac{0.06}{n} \log \frac{[X^{2+}]\$}{[Y^{2+}]\$}$, we obtain $E_{\mathrm{cell}} = 2.72 - \frac{0.06}{2} \log \frac{0.001}{0.01}$.

The logarithmic term simplifies to $\log(0.1) = -1$, which transforms the calculation into $2.72 - 0.03 \times (-1)$. This results \in a final value of $2.72 + 0.03 = 2.75\ \mathrm{V}$, which is equivalent to $275 \times 10^{-2}\ \mathrm{V}$.