Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Conductivity ($\kappa$) is determined by the ratio of the cell constant ($G^*$) to the resistance ($R$) of the solution. Given a cell constant of 1.14 cm$^{-1}$ and a resistance of 1500 $\Omega$, the conductivity is calculated as $\kappa = \frac{1.14}{1500}$, which equals 0.00076 S cm$^{-1}$. Molar conductivity ($\Lambda_m$) is derived by dividing this conductivity by the molar concentration ($C$) of 0.001 M. Because molar conductivity is expressed in units of S cm$^{2}$ mol$^{-1}$ while concentration is given in mol L$^{-1}$, a conversion factor of 1000 is applied to the calculation.

$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.00076 \times 1000}{0.001} = 760$

The molar conductivity of the potassium chloride solution is 760 S cm$^{2}$ mol$^{-1}$.

To find the nitric acid concentration at which the two nitrate reduction pathways are thermodynamically equivalent, we equate the electrochemical potentials of the two corresponding half-cells. The first reaction, involving the reduction of nitrate to nitric oxide, has a standard potential $E_1^{\circ} = 0.96 - (-0.34) = 1.3\,\mathrm{V}$ and a potential governed by the Nernst equation $E_1 = 1.3 - \frac{0.059}{6} \log Q_1$, where Q_1 = \frac{[\mathrm{Cu}^{2+}]^{3} p_{\mathrm{NO}}^{2}}{[\mathrm{NO}_3^{-}]^{2} \[\mathrm{H}^{+}]^{8}}$. The second reaction, reducing nitrate to nitrogen dioxide, has a standard potential$E_2^{\circ} = 0.79 - (-0.34) = 1.13,\mathrm{V}$and potential$E_2 = 1.13 - \frac{0.059}{2} \log Q_2$, where$Q_2 = \frac{[\mathrm{Cu}^{2+}]$ p_{\mathrm{NO}_2}^{2}}{[\mathrm{NO}_3^{-}]^{2} $[\mathrm{H}^{+}]^{4}}$.

Setting $E_1 = E_2$ leads to the equation $0.17 = \frac{0.059}{6} (\log Q_1 - 3 \log Q_2)$, which simplifies to $0.17 = \frac{0.059}{6} \log \left( \frac{Q_1}{Q_2^3} \right)$. Upon substituting the reaction quotients and applying the condition $p_{\mathrm{NO}} = p_{\mathrm{NO}_2}$, the copper and partial pressure terms cancel out, reducing the expression to 0.17 = \frac{0.059}{6} \log ([\mathrm{H}^{+}]^{4} \[\mathrm{NO}_3^{-}]^{4})$. With the approximation$[\mathrm{H}^{+}] = $[\mathrm{NO}_3^{-}]$ = [\mathrm{HNO}_3]$, this becomes$0.17 = \frac{0.059}{6} \log ([\mathrm{HNO}_3]^8)$. Solving this yields$\log [\mathrm{HNO}_3] = 2.16$, establishing$x = 2.16$. Doubling this value results \in$4.32$, which rounds to 4.

The electrode potential of a metal is fundamentally governed by the energy change associated with the conversion of a solid metal atom, $M(s)$, into its hydrated ion, $M^{n+}(aq)$, in solution, a process typically evaluated through a thermodynamic cycle. This transformation involves the energy required for sublimation to create gaseous atoms, the energy required for ionisation to remove electrons, and the energy released through hydration to stabilize the ions in an aqueous environment. Because these three thermodynamic quantities-sublimation enthalpy, ionisation enthalpy, and hydration enthalpy-directly determine the Gibbs free energy change for the overall process, they are the factors that influence the reduction potential, bringing the total count to 3.

The ionization constant for a weak acid relates directly to its degree of dissociation, which is determined by the ratio of molar conductivity at a specific concentration to the molar conductivity at infinite dilution. Calculating the molar conductivity $\Lambda_m$ at the concentration of 0.001 mol L^-1 involves the expression $\Lambda_m = \frac{1000 \kappa}{C}$, which results in 20 S cm^2 mol^-1 using the given conductivity $\kappa = 2.0 \times 10^{-5}$ S cm^-1. Dividing this result by the given $\Lambda_{m}^{o}$ of 190 S cm^2 mol^-1 gives the degree of dissociation $\alpha = \frac{20}{190} = \frac{2}{19}$. Applying the equilibrium expression $K_a = \frac{C \alpha^2}{1-\alpha}$ allows for the final calculation by substituting the concentration and the value of $\alpha$, yielding $K_a = \frac{0.001 \times (\frac{2}{19})^2}{1 - \frac{2}{19}}$. This calculation produces $12.3 \times 10^{-6}$, and rounding to the nearest integer provides 12.

Electrochemical cells generate electrical potential through spontaneous redox reactions. In this specific cell, the oxidation of zinc occurs at the anode while the reduction of silver ions takes place at the cathode, represented by the net reaction $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$. Because two electrons are transferred in this process, we identify $n = 2$. The standard cell potential $E^{\circ}_{\text{cell}}$ is determined by the difference between the reduction potentials of the cathode and anode, calculated as $E^{\circ}_{\text{cell}} = 0.80 - (-0.76) = 1.56$ V.

To determine the cell potential under non-standard conditions, we first evaluate the reaction quotient Q = \frac{[\text{Zn}^{2+}]\}{[\text{Ag}^+]^2}$. Substituting the given molar concentrations, we get$Q = \frac{0.1}{(0.01)^2} = 10^3$. We then apply the Nernst equation,$E_{\text{cell}} = E^{\circ}{\text{cell}} - \frac{0.059}{n} \log Q$, which allows us to find the actual potential of the cell at 298 K. Inserting our known values yields$E{\text{cell}} = 1.56 - \frac{0.059}{2} \log 10^3$.

Solving the logarithmic term results in $E_{\text{cell}} = 1.56 - 0.0295 \times 3$, which simplifies to $E_{\text{cell}} = 1.56 - 0.0885 = 1.4715$ V. Since the cell emf is given as $x \times 10^{-2}$ V, we express our result as $147.15 \times 10^{-2}$ V. By rounding this value to the nearest integer as required, we obtain $x = 147$.

Molar conductivity quantifies the conducting power of all ions produced by one mole of an electrolyte in a solution, defined by the relationship $\Lambda_{m} = \frac{\kappa}{C}$. To determine the conductivity $\kappa$, we multiply the measured conductance of $0.55\ \mathrm{mS}$ by the cell constant of $1.3\ \mathrm{cm}^{-1}$. Using the molar concentration of $5.0\ \mathrm{mmol\ dm^{-3}}$ and applying the necessary unit conversion factor of $1000$, the calculation is set up as $\Lambda_{m} = \frac{0.55 \times 1.3 \times 100}{1000 \times 5 \times 10^{-3}}$. This operation yields $14.3$, which rounds to $14$ when expressed as the nearest integer.

Faraday's laws of electrolysis state that the same quantity of electricity passed through an electrolytic cell will produce an equal number of chemical equivalents of substances at different electrodes. Given that 108 g of silver is deposited and its molar mass is 108 g mol$^{-1}$, the number of gram-equivalents of silver is calculated as $\frac{108}{108} = 1$. This implies that exactly 1 gram-equivalent of oxygen gas must be evolved at the anode.

The electrochemical oxidation of water proceeds according to the equation $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$, where 4 moles of electrons are required to generate 1 mole of oxygen gas. This means the n-factor for oxygen is 4, so 1 gram-equivalent corresponds to $\frac{1}{4}$ or 0.25 moles. Since the molar volume of an ideal gas at 273 K and 1 bar is 22.7 L, the total volume of oxygen produced is determined by the calculation $0.25 \times 22.7 = 5.675$ L.

Gibbs free energy change, denoted as $\Delta G^\circ$, is an extensive property, meaning it depends on the number of electrons involved in a half-reaction, unlike electrode potentials which remain constant regardless of scaling. To determine the standard potential for the reduction of $\text{Cu}^{2+}$ to $\text{Cu}^{+}$, we analyze the reaction $\text{Cu}^{2+} + e^{-} \rightarrow \text{Cu}^{+}$ as the difference between the reduction of $\text{Cu}^{2+}$ to $\text{Cu}$ and the reduction of $\text{Cu}^{+}$ to $\text{Cu}$.

The standard Gibbs free energy change for the reduction of $\text{Cu}^{2+}$ to $\text{Cu}$ is calculated as $\Delta G_1^\circ = -nFE^\circ$, which equals $-2 \times F \times 0.34$. Similarly, the change for the reduction of $\text{Cu}^{+}$ to $\text{Cu}$ is $\Delta G_2^\circ = -1 \times F \times 0.522$. Subtracting the second equation from the first provides the Gibbs energy for the target reaction, $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ$. Substituting the values results in the following expression:

$\Delta G_3^\circ = -0.68F - (-0.522F) = -0.158F$

Since $\Delta G_3^\circ = -nFE^\circ$ for a one-electron reduction, the standard potential $E^\circ$ is calculated as $-0.158F / (-F)$, resulting in 0.158 V.

The applied potential of 2V exceeds the standard reduction potentials of both silver ions ($E^{\circ} = 0.80\ \mathrm{V}$) and gold ions ($E^{\circ} = 1.69\ \mathrm{V}$), ensuring that both metal cations reduce simultaneously at the cathode. According to Faraday's second law of electrolysis, the passage of the same quantity of electricity through the solution deposits metal masses proportional to their respective equivalent weights. Because both $\mathrm{Ag}^{+}$ and $\mathrm{Au}^{+}$ undergo a single-electron reduction, their equivalent weights are numerically equal to their atomic weights, which results in the deposited masses being in direct proportion to those atomic weights.

The energy of an incident photon must account for both the metal's work function and the kinetic energy of the ejected electron, which is neutralized by the cell's stopping potential. The provided galvanic cell is governed by the reaction $\frac{1}{2}\mathrm{H}_2(\mathrm{g}) + \mathrm{AgCl}(\mathrm{s}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}^{-}(\mathrm{aq})$. The potential of this cell follows the Nernst equation E = E^\circ - 0.06 \log([\mathrm{H}^{+}]\$[\mathrm{Cl}^{-}]$ / P_{\mathrm{H}_2}^{1/2} )$. For the sodium electrode, with a pH of 1, the concentrations are$[\mathrm{H}^{+}] = 10^{-1}$M and$[\mathrm{Cl}^{-}] = 10^{-1}$M. Substituting these, the potential is$E = 0.22 - 0.06 \log( (10^{-1} \cdot 10^{-1}) / 1 ) = 0.34$ V.

This 0.34 V potential acts as the stopping potential for the sodium surface. Because the incident photon energy is conserved, the photon energy equals the sum of the work function and the stopping potential, totaling 2.3 eV + 0.34 eV = 2.64 eV. When shifting to the potassium surface with a work function of 2.25 eV, the new stopping potential required is the difference between the photon energy and the new work function, yielding 0.39 V. Equating the cell potential to this required 0.39 V leads to the Nernst relationship $0.39 = 0.22 - 0.06 \log([\mathrm{H}^{+}]^2 / 1 )$. This expression simplifies to $0.39 = 0.22 + 0.12 \mathrm{pH}$, where $0.12 \mathrm{pH} = 0.17$. Dividing these values gives a pH of approximately 1.42, which corresponds to 142 $\times 10^{-2}$.

The equilibrium constant for a redox process is fundamentally connected to the standard cell potential, as the electrochemical work performed at chemical equilibrium reflects the thermodynamic driving force of the reaction. Viewing this disproportionation reaction as the superposition of two distinct half-reactions involving copper species allows for the determination of the total cell potential by subtracting the reduction potential of the copper(I) to copper(II) step from the reduction potential of the copper(I) to copper(0) step. This yields:

$E^{\circ}_{\text{cell}} = 0.52\ \mathrm{V} - 0.16\ \mathrm{V} = 0.36\ \mathrm{V}$

Because the balanced disproportionation reaction involves the exchange of one electron, the number of transferred electrons is one. Utilizing the relationship between standard cell potential and the equilibrium constant, the natural logarithm of the constant is defined by:

$\ln K = \frac{n E^{\circ}_{\text{cell}}}{\frac{RT}{F}}$

Substituting the provided values into this equation results in:

$\ln K = \frac{1 \times 0.36}{0.025} = 14.4$

Expressing this result in the required format of $144 \times 10^{-1}$ confirms that the value is 144.00.

The standard cell potential for the Zn-Cu system is determined by the difference between the reduction potentials, $E^{0}_{\text{cell}} = 0.34 - (-0.76) = 1.1\,\text{V}$. When an external opposing potential $E_{\text{ext}}$ exceeds $1.1\,\text{V}$, the galvanic cell is forced to operate as an electrolytic cell, which reverses the direction of electron flow and the chemical reactions occurring at the electrodes. Under these conditions, the spontaneous process of zinc dissolution and copper deposition is inverted, making the claim that zinc dissolves and copper deposits while $E_{\text{ext}} > 1.1\,\text{V}$ factually incorrect.

The electrode potential of a half-cell under non-standard conditions is governed by the Nernst equation, which accounts for the influence of ion concentrations on the cell potential. Since the provided reaction describes the oxidation of water, we must use the standard oxidation potential, which is the negative of the given standard reduction potential, equal to $-1.23\,\mathrm{V}$.

Using the Nernst equation at $298\,\mathrm{K}$, expressed as $E = E^{\circ} - \frac{0.059}{n} \log Q$, we account for the $4$ electrons transferred and the reaction quotient Q = \[\mathrm{H}^{+}]^4$, as the oxygen gas is at$1,\mathrm{bar}$and water acts as a pure liquid. Given that$\mathrm{pH} = 5$corresponds to an$[\mathrm{H}^{+}]$concentration of$10^{-5},\mathrm{M}$, the expression becomes:

$E = -1.23 - \frac{0.059}{4} \log(10^{-5})^4$

Simplifying the logarithmic term to $4 \times (-5)$ allows the expression to resolve as $E = -1.23 - \frac{0.059}{4} \times (-20)$, which further simplifies to $-1.23 + 0.059 \times 5$. Completing this calculation yields a final electrode potential of $-0.935\,\mathrm{V}$.

The efficiency of an electrolytic process is defined as the ratio of the actual mass of the product obtained to the theoretical mass calculated via Faraday's laws, reflecting the degree to which the electric charge successfully drives the desired chemical transformation. First, determine the total charge passed through the solution by multiplying the current of 2 A by the time of 8 minutes, converted to 480 seconds, which results in a total charge of 960 C. Dividing this total charge by the Faraday constant of 96000 C indicates that the number of moles of electrons participating in the reaction is $\frac{960}{96000} = 0.01$ mol.

The reduction half-reaction at the electrode is described by the following equation:
$\mathrm{Cr}_2\mathrm{O}_7^{2-}+14\mathrm{H}^{+}+6\mathrm{e}^{-}\rightarrow2\mathrm{Cr}^{3+} +7\mathrm{H}_2\mathrm{O}$
According to this stoichiometry, 6 moles of electrons are required to produce 2 moles of $\mathrm{Cr}^{3+}$. Therefore, the theoretical moles of $\mathrm{Cr}^{3+}$ expected is $\frac{2}{6} \times 0.01$ mol. Multiplying this value by the atomic mass of chromium, 52 g/mol, provides a theoretical yield of $\frac{2}{6} \times 0.01 \times 52 = 0.1733$ g. Dividing the actual yield of 0.104 g by this theoretical mass and multiplying by 100 confirms that the process efficiency is 60%.

An electrochemical cell reaches equilibrium when the net potential difference drops to zero, meaning the chemical driving force for the reaction is balanced. For the spontaneous reaction occurring in this cell, $\mathrm{Sn}(s) + \mathrm{Pb}^{2+}(aq) \rightarrow \mathrm{Sn}^{2+}(aq) + \mathrm{Pb}(s)$, the standard cell potential is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode, resulting in $E^\circ_{\text{cell}} = -0.13\,\mathrm{V} - (-0.14\,\mathrm{V}) = 0.01\,\mathrm{V}$.

The Nernst equation relates this standard cell potential to the reaction quotient at equilibrium. Given that $n = 2$ electrons are transferred in this process, the equilibrium condition is defined by 0 = E^\circ_{\text{cell}} - \frac{0.06}{2} \log \left( \frac{[\mathrm{Sn}^{2+}]\}{[\mathrm{Pb}^{2+}]$} \right)$. Substituting the previously calculated standard potential yields$0 = 0.01,\mathrm{V} - 0.03 \log \left( \frac{[\mathrm{Sn}^{2+}]$}{[\mathrm{Pb}^{2+}]$} \right)$. Rearranging this equation leads to$\log \left( \frac{[\mathrm{Sn}^{2+}]$}{[\mathrm{Pb}^{2+}]$} \right) = \frac{0.01}{0.03} = \frac{1}{3}$. Calculating the antilog, we find the final ratio is$\frac{[\mathrm{Sn}^{2+}]$}{[\mathrm{Pb}^{2+}]$} = 2.15$.

Kohlrausch's law establishes that the limiting molar conductivity of an electrolyte is the sum of the independent contributions of its constituent ions, so subtracting one electrolyte from another that shares a common ion isolates the difference between the remaining ions. Because the left side of the first equation results in the difference between bromide and iodide ions while the right side yields the difference between potassium and sodium ions, the two sides are not equivalent. The valid relationship requires balancing the same ionic differences on both sides, which is expressed as $(\Lambda^0_m)_{\mathrm{NaBr}} - (\Lambda^0_m)_{\mathrm{NaI}} = (\Lambda^0_m)_{\mathrm{KBr}} - (\Lambda^0_m)_{\mathrm{KI}}$.

The Gibbs energy change of a reaction under specific conditions is determined by the relationship $\Delta G = \Delta G^{\circ} + RT \ln Q$. Because the concentrations of both the $\mathrm{Sn}^{2+}$ and $\mathrm{Cu}^{2+}$ ions are 1 M, the reaction quotient $Q$ equals 1, causing the logarithmic term to become zero and making $\Delta G$ equivalent to the standard Gibbs energy change, $\Delta G^{\circ}$. The standard Gibbs energy change is calculated using the formula $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$, where $n$ is the number of electrons transferred, $F$ is Faraday's constant, and $E^{\circ}_{\text{cell}}$ is the standard cell potential. The cell potential is the difference between the reduction potential of the cathode, where tin is reduced, and the oxidation potential of the anode, where copper is oxidized. With the cathode potential at -0.16 V and the anode potential at 0.34 V, the standard cell potential is -0.50 V. Using $n = 2$ and $F = 96500$ C/mol, the calculation is:

$\Delta G = -2 \times 96500 \times (-0.50) = 96500$

The total Gibbs energy change for this reaction is 96500 J.

Faraday's laws of electrolysis dictate that the total charge $Q$ required is $Q = n_{\text{electrons}} \times F$, where $n_{\text{electrons}}$ is the moles of electrons transferred and $F$ is Faraday's constant. The number of moles of $\mathrm{KClO}_{3}$ to be produced is $\frac{10}{122} \text{ mol}$. According to the balanced equation $6\mathrm{OH}^{-} + \mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-} + 3\mathrm{H}_{2}\mathrm{O} + 6\mathrm{e}^{-}$, the production of one mole of $\mathrm{KClO}_{3}$ requires 6 moles of electrons, meaning the theoretical charge required is $\frac{6 \times 10 \times 96500}{122} \text{ C}$.

With a current efficiency of $60\%$, the actual charge delivered by a current $I$ over time $t$ is expressed as $Q = I \times t \times 0.60$. Equating the actual charge delivered to the theoretical charge required gives $2 \times t \times 0.60 = \frac{60 \times 96500}{122}$. Solving for time in seconds results in $t = \frac{60 \times 96500}{122 \times 2 \times 0.60} \approx 39549.18 \text{ s}$. Converting this duration into hours by dividing by 3600, $t \approx 10.986 \text{ hours}$, which rounds to 11.00 hours.

The relationship between molar conductivity and concentration behaves differently depending on whether an electrolyte is strong or weak. For strong electrolytes, molar conductivity increases only slightly with dilution following a predictable linear path, whereas weak electrolytes show a sharp, dramatic increase in molar conductivity at low concentrations because their degree of dissociation rises significantly. Since $CH_3COOH$ is a weak electrolyte, it follows this characteristic non-linear curve, while strong electrolytes like $KNO_3$, $HCl$, and $NaCl$ do not.

The change in Gibbs energy for an electrochemical cell is given by $\Delta G = -nFE_{\text{cell}}$, which indicates that the process is spontaneous and $\Delta G$ is negative only when the cell potential $E_{\text{cell}}$ is positive. For the concentration cell $\mathrm{Cu(s)}|\mathrm{Cu}^{2+}(C_1) || \mathrm{Cu}^{2+}(C_2)|\mathrm{Cu(s)}$, the overall cell reaction is $\mathrm{Cu}^{2+}(C_2) \rightarrow \mathrm{Cu}^{2+}(C_1)$, and the potential is expressed by the Nernst equation as:
$E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{C_1}{C_2} \right)$
Since $E^0_{\text{cell}} = 0$ for any concentration cell, the expression simplifies to $E_{\text{cell}} = \frac{0.0591}{2} \log \left( \frac{C_2}{C_1} \right)$. A positive cell potential requires $\log \left( \frac{C_2}{C_1} \right) > 0$, which mathematically implies that $\frac{C_2}{C_1} > 1$, or simply $C_2 > C_1$. Among the given options, $C_2 = \sqrt{2}C_1$ is the only condition that satisfies $C_2 > C_1$, thereby ensuring $E_{\text{cell}} > 0$ and consequently $\Delta G < 0$.