Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Kohlrausch's law of independent migration of ions allows us to determine the limiting molar conductivity of barium sulfate by combining the molar conductivities of the given electrolytes, expressed as $\Lambda_{m}^{\circ}(\mathrm{BaSO}_{4}) = x_{1} + x_{2} - 2x_{3}$. Because molar conductivity is related to the conductivity $x$ and solubility $S$ by the equation $\Lambda_{m} = \frac{1000x}{S}$, and is further defined by the relationship $\Lambda_{m} = \alpha \Lambda_{m}^{\circ}$, the solubility $S$ can be represented as $S = \frac{1000x}{\alpha(x_{1} + x_{2} - 2x_{3})}$. Squaring this solubility value to determine the solubility product constant, $K_{sp} = S^{2}$, results in the final expression $\frac{10^{6}x^{2}}{\alpha^{2}(x_{1} + x_{2} - 2x_{3})^{2}}$.

Faraday's laws of electrolysis state that the mass of substance deposited is proportional to the electric charge passed through the solution. Since the reduction of aluminium ions follows the half-reaction $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$, three moles of electrons are required to deposit one mole of aluminium.

The total charge passed is calculated using the relation $Q = I \times t$. With a current of 2 A and a duration of 30 minutes, or 1800 seconds, the total charge passed is $Q = 2 \times 1800 = 3600$ C. Dividing this by the Faraday constant of 96500 C mol$^{-1}$ gives the moles of electrons as $n_e = 3600 / 96500 = 36/965$ mol.

Because three moles of electrons produce one mole of aluminium, the number of moles of aluminium deposited is $n_{\text{Al}} = (36/965) / 3 = 12/965$ mol. Multiplying this by the molar mass of 27 g mol$^{-1}$ gives the final mass:
$\text{Mass} = \frac{12}{965} \times 27 = 0.336\ \text{g}$

The electrolytic production of peroxodisulphuric acid, $\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8$, necessitates a high concentration of hydrogen sulphate ions at the anode. When a concentrated solution of sulphuric acid is electrolyzed, these ions are readily available to discharge and dimerize into the peroxodisulphate species, a reaction that is suppressed in dilute solutions where the electrolysis of water occurs preferentially. Therefore, the use of concentrated sulphuric acid is essential to achieve the synthesis of $\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8$.

Molar conductivity $\Lambda_m$ relates to the specific conductivity $\kappa$ and concentration $C$ through the equation $\Lambda_m = \frac{1000 \kappa}{C}$. Since the resistance $R$ of a cell is defined by $R = \frac{G^*}{\kappa}$ where $G^*$ is the constant characteristic of the cell, combining these expressions confirms that $R \cdot \Lambda_m \cdot C = 1000 G^*$ remains constant for a specific cell.

For the higher concentration where $\sqrt{C} = 0.15$, the concentration is $0.0225$, the molar conductivity is 100, and the resistance $R$ is 100. Using the relation $1000 G^* = R \cdot \Lambda_m \cdot C$, the cell constant term is $100 \cdot 100 \cdot 0.0225 = 225$. Substituting this value for the dilute solution where $\sqrt{C} = 0.1$, concentration is $0.01$, and $\Lambda_m = 150$, the resistance $x$ satisfies $x \cdot 150 \cdot 0.01 = 225$, leading to the final value $x = 150$.

Faraday's laws of electrolysis dictate that the change in concentration of an electrolyte solution depends on the amount of charge passed and the stoichiometry of the reduction reaction. For the copper cell, the passage of $1\ \mathrm{F}$ of charge reduces $0.5\ \mathrm{mol}$ of $\mathrm{Cu}^{2+}$ ions ($n=2$), resulting in a final concentration of $[\mathrm{Cu}^{2+}] = 1.5\ \mathrm{M} - 0.5\ \mathrm{M} = 1.0\ \mathrm{M}$. Similarly, for the silver cell, the passage of $0.1\ \mathrm{F}$ of charge reduces $0.1\ \mathrm{mol}$ of $\mathrm{Ag}^{+}$ ions ($n=1$), resulting in a final concentration of $[\mathrm{Ag}^{+}] = 0.2\ \mathrm{M} - 0.1\ \mathrm{M} = 0.1\ \mathrm{M}$.

When these cells are combined to form an electrochemical cell, the overall spontaneous reaction is $\mathrm{Cu}(s) + 2\mathrm{Ag}^{+}(aq) \rightarrow \mathrm{Cu}^{2+}(aq) + 2\mathrm{Ag}(s)$, with a standard cell potential of $E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag}^{+}/\mathrm{Ag}} - E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.8\ \mathrm{V} - 0.34\ \mathrm{V} = 0.46\ \mathrm{V}$. Utilizing the Nernst equation, E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{n} \log\left(\frac{[\mathrm{Cu}^{2+}]\}{[\mathrm{Ag}^{+}]^2}\right)$, where$n=2$represents the moles of electrons transferred, the final potential is calculated as$E_{\text{cell}} = 0.46 - \frac{0.06}{2} \log\left(\frac{1.0}{(0.1)^2}\right)$. This simplifies to$E_{\text{cell}} = 0.46 - 0.03 \log(100) = 0.46 - 0.06 = 0.40\ \mathrm{V}$, which corresponds to$400\ \mathrm{mV}$.

The limiting molar conductivity, represented by $\Lambda_{m}^{0}$, reflects an ion's ability to migrate through water, with the $H^{+}$ ion displaying exceptionally high values due to its unique proton-hopping mechanism that bypasses the limitations of hydrodynamic drag. For metal cations like $Ca^{2+}$, $Mg^{2+}$, $K^{+}$, and $Na^{+}$, the measured conductivity is inversely related to the size of their hydration sphere, which is strongly dictated by their charge density and the extent to which they organize solvent molecules. This interplay between solvation effects and inherent ionic properties creates a hierarchy where the divalent $Ca^{2+}$ and $Mg^{2+}$ ions exhibit superior mobility compared to the monovalent $K^{+}$ and $Na^{+}$ ions, establishing the overall trend $H^{+} > Ca^{2+} > Mg^{2+} > K^{+} > Na^{+}$.

Charging a lead storage battery is an electrolytic process that forces the chemical reactions to proceed in the reverse direction of discharge, regenerating the active electrode materials from lead sulfate. At the anode, lead sulfate $\text{PbSO}_4$ containing lead in the +2 oxidation state acts as the reactant and is reduced to form metallic lead, where the lead exists in the 0 oxidation state, resulting in a change from +2 to 0. Meanwhile, at the cathode, lead sulfate also containing lead in the +2 oxidation state undergoes oxidation to produce lead dioxide $\text{PbO}_2$. Given that oxygen typically exhibits a -2 oxidation state, the lead in $\text{PbO}_2$ must be in the +4 state to maintain molecular neutrality, resulting in a transition from +2 to +4.

Electrolysis behavior at the anode depends on whether the electrodes are inert, such as platinum, and on the oxidation potential of the ions in the solution compared to that of water molecules. When inert platinum electrodes are used, the anode does not participate in the redox reaction, forcing a competition for oxidation between the anions and water, with water undergoing oxidation to release oxygen gas if its oxidation potential is lower. This specific anodic reaction is represented by the equation $2\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^-$.

In an aqueous solution of $\mathrm{AgNO}_3$ with platinum electrodes, the nitrate ions remain stable, causing water to oxidize and release oxygen. Similarly, in a dilute $\mathrm{H}_2\mathrm{SO}_4$ solution, the sulfate ions are not easily oxidized, which again forces water to oxidize at the anode and produce oxygen gas. In contrast, using silver electrodes in a silver nitrate solution allows the silver anode to oxidize into $\mathrm{Ag}^+$ ions, which prevents the evolution of oxygen, while high concentration $\mathrm{H}_2\mathrm{SO}_4$ environments involve different chemical pathways that do not primarily yield oxygen.

Every electronic application requires a specific electrochemical cell tailored to its power requirements, size constraints, and discharge characteristics. Transistors generally utilize dry cells with an anode $-\mathrm{Zn}$ and a cathode $-\mathrm{Carbon}$, which corresponds to the match (A)-(III). Compact, portable devices like hearing aids rely on mercury cells, characterized by an anode $-\frac{\mathrm{Zn}}{\mathrm{Hg}}$ and a cathode $-\mathrm{HgO} + \mathrm{C}$, providing the link (B)-(I).

High-capacity power systems such as inverters operate using lead storage batteries, identified by an anode $-\mathrm{Pb}$ and a cathode $-\mathrm{Pb} \mid \mathrm{PbO}_2$, corresponding to match (C)-(IV). Finally, space exploration missions, such as the Apollo space ship, utilized the hydrogen fuel cell to provide a reliable energy source, completing the final pair with (D)-(II).

A strong reducing agent is characterized by its high propensity to donate electrons, a property that inversely relates to the species' standard reduction potential. Essentially, the lower or more negative the standard reduction potential of a substance, the more readily it undergoes oxidation, making it a stronger reducing agent. Given the electrochemical data, $E^{\circ}_{\mathrm{Cr}^{3+}/\mathrm{Cr}} = -0.74\,\text{V}$ is the most negative value listed, whereas the other potentials for the provided species are positive. Therefore, metallic Cr acts as the strongest reducing agent among the choices because it possesses the greatest tendency to release electrons compared to the others.

Standard electrode potentials are intensive properties and cannot be added algebraically, but the Gibbs free energy of a reaction is an extensive property that allows us to combine successive electrochemical steps. By relating the change in free energy to the electrode potential via the relation $\Delta G^\ominus = -nFE^\ominus$, we can find the potential for a composite reaction by balancing the total number of electrons involved.

In the reduction from $\mathrm{FeO}_4^{2-}$ to $\mathrm{Fe}^{3+}$, 3 electrons are transferred at a potential of $2.0\,\text{V}$, and in the step from $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$, 1 electron is transferred at a potential of $0.8\,\text{V}$. The overall reduction from $\mathrm{FeO}_4^{2-}$ to $\mathrm{Fe}^{2+}$ involves the sum of these electron transfers, resulting in a total of 4 electrons. Using the principle of additivity for Gibbs free energy, which is represented by the equation $n_{total} E^\ominus = n_1 E_1^\ominus + n_2 E_2^\ominus$, we calculate the unknown potential:
$4 \times E^\ominus = (3 \times 2.0) + (1 \times 0.8)$
This simplifies to $4 \times E^\ominus = 6.8$, and dividing by 4 yields a standard electrode potential of $1.7\,\text{V}$.

Corrosion operates as an electrochemical mechanism where the pure metal undergoes oxidation at the anode site, represented by the process $M \rightarrow M^{n+} + ne^-$, while existing impurities function as the cathode. Since an acidic environment facilitates this electron transfer, the rate of corrosion is substantially faster in acidic media than in alkaline media, rendering the second statement false.

Standard cell potential calculations for multi-step redox reactions require utilizing the additivity of Gibbs free energy, $\Delta G^{\circ}$, because electrode potentials are intensive properties that cannot simply be summed. Each half-reaction relates to the free energy via the expression $\Delta G^{\circ} = -nFE^{\circ}$, where $n$ represents the number of electrons transferred, $F$ is the Faraday constant, and $E^{\circ}$ is the standard potential.

Express the net chemical change, $\mathrm{Fe}^{2+} + \mathrm{Ag}^{+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{Ag}$, as a combination of the known half-reactions to derive the total energy change. The reduction of $\mathrm{Ag}^{+}$ to $\mathrm{Ag}$ involves one electron and provides $\Delta G^{\circ} = -xF$. The oxidation of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ is derived by subtracting the reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}$ from the reduction of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}$, resulting in a free energy change of $\Delta G^{\circ} = -2yF - (-3zF)$, which simplifies to $-2yF + 3zF$. Adding these values yields a total free energy change of $\Delta G_{total}^{\circ} = -xF - 2yF + 3zF$. Since the target reaction involves a single electron transfer, the total free energy corresponds to $-1 \cdot E^{\circ} \cdot F$, identifying the standard cell potential as $x + 2y - 3z$.

An oxidizing agent facilitates reduction by readily accepting electrons, which means a higher standard reduction potential $E^{\ominus}$ is synonymous with a greater oxidizing capacity. Among the listed potentials of 1.67 V, 1.26 V, 1.15 V, and -1.66 V, the value for $Pb^{4+}/Pb^{2+}$ stands out as the highest. Consequently, $Pb^{4+}$ exhibits the strongest tendency to accept electrons and acts as the most potent oxidizing agent of the set.

The Nernst equation allows us to determine the potential of a cell under non-standard conditions by incorporating the reaction quotient of the spontaneous cell reaction. For the cell representation $\mathrm{Mg} \mid \mathrm{Mg}^{2+}(\mathrm{aq}) \parallel \mathrm{Ag}^{+}(\mathrm{aq}) \mid \mathrm{Ag}$, the oxidation at the anode is $\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+} + 2e^{-}$ and the reduction at the cathode is $\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}$. Balancing the electrons necessitates multiplying the reduction half-reaction by two, resulting in the overall cell reaction $\mathrm{Mg} + 2\mathrm{Ag}^{+} \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{Ag}$.

With $n = 2$ electrons transferred, the reaction quotient is defined as Q = \frac{[\mathrm{Mg}^{2+}]\}{[\mathrm{Ag}^{+}]^2}$. Substituting these values into the standard Nernst equation yields:

$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{2F} \ln \frac{[\mathrm{Mg}^{2+}]\$}{[\mathrm{Ag}^{+}]^2}$

Using the property of logarithms where $-\ln(\frac{a}{b}) = \ln(\frac{b}{a})$, we can rewrite the negative sign as a positive and invert the quotient to arrive at the final form:

$E_{\text{cell}} = E_{\text{cell}}^{\circ} + \frac{RT}{2F} \ln \frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Mg}^{2+}]\$}$

The standard potential of an electrochemical cell is determined by the difference between the standard reduction potentials of the cathode and the anode, expressed by the relationship:
$E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ$
In the methanol fuel cell described, oxygen gas undergoes reduction at the cathode, while methanol is oxidized at the anode. By substituting the provided values, we establish the equation $1.21 \text{ V} = 1.229 \text{ V} - E_{\text{CO}_2/\mathrm{CH}_3\mathrm{OH}}^\circ$. Solving for the standard reduction potential of the methanol half-reaction yields:
$E_{\text{CO}_2/\mathrm{CH}_3\mathrm{OH}}^\circ = 1.229 \text{ V} - 1.21 \text{ V} = 0.019 \text{ V} = 19 \text{ mV}$

The electromotive force of an electrochemical cell is determined by the difference in reduction potentials between the cathode and anode, modified by the Nernst equation to account for the specific ionic concentrations. The standard cell potential for this arrangement is 0.1 V, calculated from 0.8 V minus 0.7 V. Applying the Nernst equation for the cell reaction, the operating potential relates to the hydrogen ion concentration at the quinhydrone electrode such that $E_{\text{cell}} = E_{\text{cell}}^{\circ} - 0.06 \log [\mathrm{H}^{+}]$. Substituting the observed potential of 0.4 V into the equation $0.4 = 0.1 - 0.06 \log [\mathrm{H}^{+}]$ simplifies to $0.3 = -0.06 \log [\mathrm{H}^{+}]$, which is equivalent to $0.3 = 0.06 \text{pH}$. This confirms that the solution pH is 5.

The pH of a salt derived from a weak base and a strong acid is described by the relation $\text{pH} = 7 - \frac{pK_b + \log C}{2}$. Inserting the calculated pH of 5 and the molar concentration 0.01 M into the formula gives $5 = 7 - \frac{pK_b + \log 0.01}{2}$. This simplifies to $-2 = -\frac{pK_b - 2}{2}$, which further rearranges to $4 = pK_b - 2$. Solving this expression results in a value of 6 for the $pK_b$.

The Nernst equation provides a relationship between the electrode potential and the concentrations of the reacting species, defined as $E = E^{\circ} - \frac{0.059}{n} \log Q$. For the reduction of dichromate to chromium(III) ions, the number of electrons involved is $n = 6$. The reaction quotient $Q$ for this process is given by \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Cr}_2\mathrm{O}_7^{2-}]\$[\mathrm{H}^{+}]^{14}}$. Given the specified ratio of$\frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Cr}_2\mathrm{O}_7^{2-}]$} = 10^{-6}$, the quotient simplifies to$Q = \frac{10^{-6}}{[\mathrm{H}^{+}]^{14}}$.

Setting the potential $E$ to zero allows us to relate the standard potential $E^{\circ}$ to the concentration of hydrogen ions. Substituting the values into the equation yields $0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\mathrm{H}^{+}]^{14}} \right)$. Using the properties of logarithms, $\log Q$ expands to $\log(10^{-6}) - 14 \log[\mathrm{H}^{+}]$, which simplifies to $-6 + 14\mathrm{pH}$ since $\mathrm{pH} = -\log[\mathrm{H}^{+}]$. The expression becomes $0 = 1.33 - \frac{0.059}{6}(-6 + 14\mathrm{pH})$. Distributing the term $\frac{0.059}{6}$ results in $0 = 1.33 + 0.059 - \frac{0.059 \times 14}{6} \mathrm{pH}$. Solving for the $\mathrm{pH}$ gives $\frac{1.389 \times 6}{0.059 \times 14}$, which equals approximately $10.1$. Rounding to the nearest integer, the required value is $10$.

During the electrolysis of aqueous metal nitrate solutions, cations with the highest standard reduction potential preferentially undergo reduction at the cathode. Given the reduction potentials of 0.80 V for silver, 0.79 V for mercury, and 0.24 V for copper, silver ions are reduced first, followed by mercury and then copper, confirming the sequence described in the first statement.

When evaluating the electrolysis of magnesium nitrate, the standard reduction potential for magnesium at $-2.37\,\mathrm{V}$ is significantly lower than that of water, which is $-0.827\,\mathrm{V}$. Consequently, water molecules are reduced at the cathode to generate hydrogen gas instead of magnesium being deposited. The assertion that oxygen gas is produced at the cathode is incorrect, as oxygen evolution is an anodic process involving the oxidation of water or anions.

Faraday's law of electrolysis establishes that the chemical change produced at an electrode is proportional to the total electric charge passed through the circuit. During the electrolysis of an aqueous solution of sodium chloride, the reduction of water at the cathode generates hydroxide ions and hydrogen gas, expressed by the reaction $2H_2O + 2e^{-} \rightarrow H_2 + 2OH^{-}$. This relationship confirms that one mole of electrons is required to generate one mole of hydroxide ions.

The pH of the solution is 12, which implies a pOH of $14 - 12 = 2$. This corresponds to a hydroxide ion concentration of $[OH^{-}] = 10^{-2}$ M. With a total volume of 600 mL, which is 0.6 L, the total number of moles of hydroxide ions formed is $0.6 \times 10^{-2} = 0.006$ moles. The required quantity of electricity is therefore 0.006 Faraday, which is equivalent to $0.006 \times 96500$ Coulombs.

The current is calculated by dividing the total charge by the time duration of 300 seconds, which corresponds to 5 minutes. The current is given by $I = \frac{0.006 \times 96500}{300} = 1.93$ A. Rounding this value to the nearest integer leads to 2 A.