Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Molar conductivity links the conductive capacity of an electrolytic solution to its concentration, expressed by the formula $\Lambda_m = \frac{\kappa \times 1000}{M}$. Here, $\kappa$ denotes the specific conductivity, which is calculated as the ratio of the cell constant, $129\ \mathrm{m}^{-1}$, to the measured resistance $R$ of the solution. Since the cell constant remains identical for both measurements, the specific conductivity of each solution is inversely proportional to its resistance.

When comparing the two KCl solutions, the ratio of their molar conductivities simplifies to $\frac{\Lambda_1}{\Lambda_2} = \frac{\kappa_1}{\kappa_2} \times \frac{M_2}{M_1}$. Because molarity $M$ is directly proportional to concentration expressed in ppm, the concentration ratio $\frac{M_2}{M_1}$ is $\frac{149}{74.5} = 2$. Using the resistance values of $100\ \Omega$ and $50\ \Omega$, the ratio $\frac{\kappa_1}{\kappa_2}$ is equivalent to $\frac{R_2}{R_1} = \frac{50}{100} = 0.5$. Multiplying these factors gives a final ratio of $0.5 \times 2 = 1$. Setting this equal to $x \times 10^{-3}$ leads to the equation $1 = x \times 10^{-3}$, which results in $x = 1000$.

Standard reduction potentials indicate the relative stability of oxidation states, and among the given pairs, only the $\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}$ couple possesses a negative reduction potential of -0.41 V. The chromium ion $\mathrm{Cr}^{3+}$ has an electronic configuration of $d^3$, which corresponds to 3 unpaired electrons. Calculating the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ represents the number of unpaired electrons, gives $\mu = \sqrt{3(3+2)} = \sqrt{15}$ B.M. Since this calculation yields approximately 3.87, the value rounded to the nearest integer is 4.

The tendency of a chemical species to undergo reduction is measured by its standard electrode potential, where more positive values indicate a higher driving force to gain electrons. By examining the values provided, the $\mathrm{Cl}_2 / \mathrm{Cl}^{-}$ couple holds the highest potential at $+1.36\,\mathrm{V}$, followed by the $\mathrm{Ag}^{+} / \mathrm{Ag}$ couple at $+0.80\,\mathrm{V}$, and the $\mathrm{I}_2 / \mathrm{I}^{-}$ couple at $+0.54\,\mathrm{V}$. In the range of negative values, the $\mathrm{Na}^{+} / \mathrm{Na}$ couple stands at $-2.71\,\mathrm{V}$, whereas the $\mathrm{Li}^{+} / \mathrm{Li}$ couple has the lowest potential at $-3.05\,\mathrm{V}$. Arranging these values in descending order yields the overall sequence $A > C > B > D > E$.

The oxidation state of iron in $\mathrm{Fe}_3\mathrm{O}_4$ is determined by balancing the negative charge of the oxygen atoms. Since four oxygen atoms contribute a total charge of $-8$, the three iron atoms must have a combined oxidation state of $+8$, resulting in an average oxidation state of $+8/3$ per iron atom. To obtain one mole of elemental iron with an oxidation state of $0$, each mole of iron must gain $8/3$ moles of electrons. Because the charge of one mole of electrons is equal to $1$ Faraday ($F$), the amount of charge required is $8/3$ Faradays. Calculating this value gives approximately $2.66 F$, which rounds to $3$ when expressed as the nearest integer.

Calculating the potential of a galvanic cell operating under non-standard conditions requires the Nernst equation, which adjusts the standard cell potential based on the reaction quotient and the concentration of species involved. The standard cell potential, $E^{\circ}_{\mathrm{cell}}$, is determined by the difference between the standard reduction potentials of the cathode and the anode. Using the provided oxidation potentials, where copper oxidation is $-0.34\,\mathrm{V}$ and zinc oxidation is $+0.76\,\mathrm{V}$, we identify the reduction potentials as $E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34\,\mathrm{V}$ and $E^{\circ}_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76\,\mathrm{V}$. This gives a standard cell potential of $E^{\circ}_{\mathrm{cell}} = 0.34 - (-0.76) = 1.10\,\mathrm{V}$.

To find the actual potential $E_{\mathrm{cell}}$, we apply the Nernst equation for a system at $298\,\mathrm{K}$ involving two electrons, which is expressed as
$E_{\mathrm{cell}} = E^{\circ}_{\mathrm{cell}} - \frac{0.059}{n} \log Q$
where $n=2$ represents the number of electrons transferred in the redox process. The reaction quotient $Q$ is calculated from the molarities of the aqueous species as Q = \frac{[\mathrm{Zn}^{2+}]\}{[\mathrm{Cu}^{2+}]$} = \frac{0.04}{0.02} = 2. Substituting these values into the expression yields $$E_{\mathrm{cell}} = 1.1 - \frac{0.059}{2} \log 2$$ Using the approximation \log 2 \approx 0.30$and calculating the term$\frac{0.059}{2} \times 0.30$, we find a subtraction value of$0.00885,\mathrm{V}$. Subtracting this from the standard potential results in$E_{\mathrm{cell}} \approx 1.1 - 0.009 = 1.09,\mathrm{V}$. Consequently, this value is equivalent to$109 \times 10^{-2},\mathrm{V}$.

The relationship between the non-standard potential of an electrochemical cell and the concentrations of its reactants is described by the Nernst equation. For the given reaction $\mathrm{Zn} + 2\mathrm{Fe}^{3+} \rightarrow \mathrm{Zn}^{2+} + 2\mathrm{Fe}^{2+}$, the standard cell potential $E^{\circ}_{\text{cell}}$ is $1.53 \text{ V}$, derived from the standard reduction potentials $0.77 \text{ V}$ and $-0.76 \text{ V}$. Applying the Nernst equation at $25^{\circ}\mathrm{C}$ with $n=2$ electrons transferred, we set up the expression $1.50 = 1.53 - \frac{0.06}{2} \log \frac{[\mathrm{Fe}^{2+}]^2}{[\mathrm{Fe}^{3+}]^2}$, where the zinc ion concentration is taken as $1 \text{ M}$.

Simplifying this equation leads to \log \frac{[\mathrm{Fe}^{2+}]\}{[\mathrm{Fe}^{3+}]$} = \frac{0.03}{0.06} = 0.5$, which yields the concentration ratio$\frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$} = 10^{0.5} = \sqrt{10}$. The fraction of iron existing as$\mathrm{Fe}^{3+}$is defined by the ratio$\frac{[\mathrm{Fe}^{3+}]$}{[\mathrm{Fe}^{3+}]$ + $[\mathrm{Fe}^{2+}]$}$, which can be rearranged to$\frac{1}{1 + \frac{[\mathrm{Fe}^{2+}]$}{[\mathrm{Fe}^{3+}]$}}$. Substituting the value of$\sqrt{10}$into this fraction results in$\frac{1}{1 + \sqrt{10}} \approx 0.2402$, or$24 \times 10^{-2}$. Consequently, the value of$x$ is 24.

The stoichiometry of the balanced reduction half-reaction determines the total amount of charge required to transform the reactant. According to the equation
$\mathrm{MnO}_4^{-} + 8 \mathrm{H}^{+} + 5 e^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_2\mathrm{O}$
each mole of $\mathrm{MnO}_4^{-}$ requires five moles of electrons to be fully reduced. Since one mole of electrons represents one Faraday of electricity, reducing a single mole of the reactant consumes five Faradays. Therefore, for five moles of $\mathrm{MnO}_4^{-}$, the total electricity required is $5 \times 5 = 25$ Faradays.

Lead belongs to group 14 and exhibits both $+2$ and $+4$ oxidation states, but the inert pair effect renders the $+2$ state significantly more stable than the $+4$ state. Because of this stability difference, compounds containing $\mathrm{Pb}^{4+}$ function as potent oxidizing agents by readily reducing to $\mathrm{Pb}^{2+}$. Among the provided options, $\mathrm{PbO}_{2}$ specifically fulfills the requirements of being an amphoteric oxide-demonstrated by its ability to react with both acids, such as $\mathrm{HCl}$, and bases, like $\mathrm{NaOH}$-while also functioning as the cathode material in lead storage batteries.

The limiting molar conductivity of an electrolyte at infinite dilution is determined by the sum of the individual contributions of its constituent ions, as defined by Kohlrausch's law. When calculating these values, the limiting molar conductivity for KCl is $73.55 + 76.3 = 149.3$ S cm²/mol, whereas for CH3COOH, it is $349.8 + 40.9 = 390.7$ S cm²/mol. Because the value for the weak electrolyte CH3COOH is significantly greater than that of the strong electrolyte KCl, the initial statement claiming that the strong electrolyte has a higher limiting value is incorrect.

Molar conductivity describes the conductive capacity of all ions generated by one mole of an electrolyte within a given volume of solution. As the concentration of the solution decreases, the dilution increases the volume available per mole, and for weak electrolytes, it specifically drives an increase in the degree of dissociation. These factors result in a higher number of conducting ions in the solution, causing molar conductivity to rise as concentration drops, which contradicts the second statement.

Faraday's laws of electrolysis provide the relationship between the mass of a substance produced and the quantity of charge passed during an electrolytic process. The stoichiometry of the reaction, $6 \mathrm{OH}^{-} + \mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-} + 3 \mathrm{H}_{2}\mathrm{O} + 6 \mathrm{e}^{-}$, reveals that 6 moles of electrons are required for every 1 mole of potassium chlorate generated.

Since the molar mass of potassium chlorate is $122.6\,\mathrm{g}\,\mathrm{mol}^{-1}$, producing 10.0 g corresponds to $\frac{10.0}{122.6}$ moles of product. Each mole requires a charge of $6 \times 96500$ C, meaning the total charge needed to produce this specific mass is $\frac{10.0 \times 6 \times 96500}{122.6}$ C. Because current represents the rate of charge flow over time, the current $x$ in amperes is determined by dividing this total charge by the duration of 10 hours, which must be converted to 36000 seconds. This calculation results in $x = \frac{10.0 \times 6 \times 96500}{122.6 \times 36000} \approx 1.31$. Rounding this result to the nearest integer gives 1.

The standard Gibbs free energy change of a chemical reaction is fundamentally linked to the standard cell potential by the equation $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$, where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant, and $E^{\circ}_{\text{cell}}$ is the standard cell potential. To find the cell potential for the reaction $2\mathrm{Fe}^{3+}(aq) + 2\mathrm{I}^{-}(aq) \rightarrow 2\mathrm{Fe}^{2+}(aq) + \mathrm{I}_2(s)$, we must first determine the standard reduction potential of the iron(III)/iron(II) couple. Because standard potentials are not additive but Gibbs energies are, we use the sequence of reductions $\mathrm{Fe}^{3+} \xrightarrow{e^-} \mathrm{Fe}^{2+} \xrightarrow{2e^-} \mathrm{Fe}$, leading to the relationship $3E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}} = 1E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} + 2E^{\circ}_{\mathrm{Fe}^{2+}/\mathrm{Fe}}$. Substituting the given values gives $3(-0.036\ \mathrm{V}) = E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} + 2(-0.440\ \mathrm{V})$, which simplifies to $E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = -0.108\ \mathrm{V} + 0.880\ \mathrm{V} = 0.772\ \mathrm{V}$.

With the iron reduction potential established, we determine the standard cell potential by subtracting the reduction potential of the iodide anode from the iron cathode potential: $E^{\circ}_{\text{cell}} = 0.772\ \mathrm{V} - 0.539\ \mathrm{V} = 0.233\ \mathrm{V}$. The balanced redox equation indicates the transfer of two electrons, so $n = 2$. Substituting these values into the Gibbs equation, $\Delta G^{\circ} = -(2 \times 96500\ \mathrm{C} \times 0.233\ \mathrm{V}) = -44969\ \mathrm{J}$. Expressing this in kilojoules and rounding to the nearest integer, the magnitude of the standard molar Gibbs free energy change is $45\ \mathrm{kJ}$.

The cell potential depends on the concentration of ions involved in the redox reaction, a relationship governed by the Nernst equation. For the spontaneous reaction occurring in this cell, $\mathrm{Cu}(\mathrm{s}) + 2\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq}) + 2\mathrm{Ag}(\mathrm{s})$, two electrons are transferred, leading to the following expression:

$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{[\mathrm{Cu}^{2+}]\$}{[\mathrm{Ag}^{+}]^2}$

Substituting the concentrations for the first cell, where $[\mathrm{Cu}^{2+}]=0.1\ \mathrm{M}$ and $[\mathrm{Ag}^{+}]=0.01\ \mathrm{M}$, gives $0.3095 = E^\circ_{\text{cell}} - 0.0295 \log \frac{0.1}{(0.01)^2}$. Since the ratio simplifies to $1000$ and $\log(1000)=3$, this equation becomes $0.3095 = E^\circ_{\text{cell}} - 0.0885$, which identifies the standard cell potential $E^\circ_{\text{cell}}$ as $0.398\ \mathrm{V}$.

Using this standard potential for the second cell with concentrations $[\mathrm{Cu}^{2+}]=0.01\ \mathrm{M}$ and $[\mathrm{Ag}^{+}]=0.001\ \mathrm{M}$, the potential is determined by $E_2 = 0.398 - 0.0295 \log \frac{0.01}{(0.001)^2}$. Because the ratio simplifies to $10000$ and $\log(10000)=4$, the expression evaluates to $E_2 = 0.398 - (0.0295 \times 4)$, resulting in $E_2 = 0.398 - 0.118 = 0.28\ \mathrm{V}$. Represented as $28 \times 10^{-2}\ \mathrm{V}$, the required integer is 28.

Every conductivity cell possesses a fixed geometry, characterized by its cell constant, $G^*$. Because this physical constant is determined solely by the dimensions of the electrodes, the product of the solution conductivity, $k$, and the observed resistance, $R$, remains identical for any electrolyte tested in that same cell, defined by the relationship $k \cdot R = G^*$. This fundamental property allows us to equate the product for the known KCl solution with the product for the unknown HCl solution.

Substituting the given values into the equality $k_{\text{KCl}} \cdot R_{\text{KCl}} = k_{\text{HCl}} \cdot R_{\text{HCl}}$, we set up the equation $0.14 \times 4.19 = k_{\text{HCl}} \times 1.03$. Solving for the conductivity of the HCl solution involves dividing the product of the KCl measurements by the HCl resistance, which is $k_{\text{HCl}} = \frac{0.14 \times 4.19}{1.03}$. Performing this calculation gives approximately 0.5695 S m^-1. To express this value in the required scientific notation, we shift the decimal to obtain 56.95 $\times 10^{-2}$ S m^-1. Rounding this result to the nearest integer provides 57.

Kohlrausch's Law of Independent Migration of Ions states that the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the conductivities of its constituent ions. The infinite dilution molar conductivity for $\mathrm{BaSO}_4$ is determined by combining the known values for $\mathrm{BaCl}_2$, $\mathrm{H}_2\mathrm{SO}_4$, and $\mathrm{HCl}$ to isolate the contributions of the barium and sulphate ions. The required relationship is $\Lambda_{\mathrm{m}}^{\infty }(\mathrm{BaSO}_4) = \Lambda_{\mathrm{m}}^{\infty }(\mathrm{BaCl}_2) + \Lambda_{\mathrm{m}}^{\infty }(\mathrm{H}_2\mathrm{SO}_4) - 2\Lambda_{\mathrm{m}}^{\infty }(\mathrm{HCl})$. Substituting the given values leads to 280 + 860 - 2(426), which simplifies to 1140 - 852 and results in a final value of 288 S cm^2 mol^-1.

The standard electrode potential reflects an element's inherent ability to undergo reduction, where negative values indicate a metal is more easily oxidized than hydrogen. While most $3d$ elements-such as $\mathrm{Fe}$ (-0.41 V), $\mathrm{Co}$ (-0.28 V), and $\mathrm{Zn}$ (-0.76 V)-possess negative values because they readily release electrons to form ions, copper stands as the exception. This occurs because copper has a lower tendency to form ions compared to hydrogen, resulting in a positive $E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}$ of +0.34 V.

The cell constant, which represents the geometric factor of an electrochemical cell, is defined as $G^* = l/A$ and is expressed in the unit of m⁻¹. Molar conductivity characterizes the total charge-carrying capacity of all ions produced by one mole of an electrolyte in a given volume of solution, measured in units of S cm² mol⁻¹. Conductivity acts as the reciprocal of resistivity to quantify how effectively a material allows the flow of electric current, with standard units of Ω⁻¹ m⁻¹. Finally, the degree of dissociation is defined as the proportion of total moles that have split into ions, and because it is a simple ratio of dissociated moles to the total amount of substance, it is inherently a dimensionless quantity.

The Nernst equation describes the relationship between a cell's potential and the concentrations of its components, adjusting the standard potential for non-standard conditions. The reaction involves the transfer of two electrons, leading to the expression E_{\text{cell}} = E^{\ominus}_{\text{cell}} - \frac{0.059}{2} \log \frac{[\mathrm{Cu}^{2+}]\}{[\mathrm{Ag}^{+}]^2}$. Substituting the specific values into the reaction quotient gives$Q = \frac{0.25}{(10^{-3})^2} = 2.5 × 10^5$.

The expression becomes $E_{\text{cell}} = 2.97 - \frac{0.059}{2} \log(2.5 \times 10^5)$. Using the provided logarithm, the term $\log(2.5 \times 10^5) = 0.3979 + 5 = 5.3979$ yields a correction factor of approximately 0.16 V. Subtracting this from the standard potential of 2.97 V results in 2.81 V, which rounds to 3.

The thermodynamic relationship connecting Gibbs free energy, enthalpy, and entropy is represented by the equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$. In electrochemical reactions, standard Gibbs free energy is linked to the standard cell potential by the equation $\Delta G^{\circ} = -nFE^{\circ}$, where $n$ represents the number of electrons transferred, $F$ is the Faraday constant, and $E^{\circ}$ is the standard cell potential. Combining these principles yields the expression $\Delta S^{\circ} = \frac{nFE^{\circ} + \Delta H^{\circ}}{T}$, which provides the means to solve for the standard entropy change.

For this reaction, the number of electrons transferred is 2, the Faraday constant is 96487 C/mol, the cell potential is 4.315 V, the enthalpy change is -825.2 kJ/mol, and the temperature is 298 K. The electrical work component is computed using the equation $nFE^{\circ} = 2 \times 96487 \times 4.315 = 832682.02$ J/mol. Converting the enthalpy change to units of J/mol gives -825200. Substituting these into the derived formula results in the equation $\Delta S^{\circ} = \frac{832682.02 - 825200}{298}$. This calculation yields 25.11 J/K/mol, which rounds to a final integer value of 25.

The electrode potential of the $\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}$ reduction couple is determined by the Nernst equation, which relates the potential to the concentration of participating ions. The reduction half-reaction is defined as $\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}$. Based on this stoichiometry, the potential can be expressed as:

$E = E^{\circ} - \frac{0.059}{5} \log \frac{[\mathrm{Mn}^{2+}]\$}{[\mathrm{MnO}_{4}^{-}]\$\$[\mathrm{H}^{+}]^{8}}$

Given that the ratio of the manganese ions remains constant, the potential change depends solely on the concentration of hydrogen ions. The magnitude of the potential change between the initial concentration of 1 M and the final concentration of $10^{-4} \mathrm{M}$ is determined by the term $\frac{0.059}{5} \log \frac{1^{8}}{(10^{-4})^{8}}$. Simplifying the logarithmic expression results in $\frac{0.059}{5} \log (10^{32})$, which evaluates to $\frac{0.059}{5} \times 32$. This calculation yields 0.3776 V, and expressing this as $3776 \times 10^{-4} \mathrm{V}$ reveals that the value of $x$ is 3776.

Conductivity relates to the physical geometry of an electrolytic cell through the product of conductance and the cell constant, defined by the formula
$\kappa = \frac{1}{R} \times \frac{l}{A}$
where $\kappa$ represents the conductivity, $R$ is the measured resistance, and $\frac{l}{A}$ is the cell constant.

Rearranging this expression to isolate the cell constant yields $\frac{l}{A} = \kappa \times R$. Substituting the given values, the calculation becomes $(1.07 \times 10^6 \, \text{S m}^{-1}) \times 0.243 \, \Omega$, which equals $260010 \, \text{m}^{-1}$. Converting this value into the form $26 \times 10^4 \, \text{m}^{-1}$ identifies that $x$ is 26.