Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Electrolytic conductance in a saturated solution is determined by the total concentration of dissolved ions and the speed at which those ions can move through the solvent. For highly soluble salts like $\mathrm{NaCl}$, the concentration of ions remains substantial, and increasing the temperature significantly enhances the ionic mobility of the ions, leading to an increase in conductance. Thus, as the temperature rises from $T_1$ to $T_2$, the conductance of the $\mathrm{NaCl}$ solution increases, which is a standard observation for such electrolytes.

The behavior of sparingly soluble salts like $\mathrm{BaSO}_4$ differs based on the thermodynamics of their dissolution. Although the dissolution of $\mathrm{BaSO}_4$ is an endothermic process, the reference indicates that the number of $\mathrm{BaSO}_4$ ions in the solution actually decreases as the temperature increases. Because the total conductance is directly influenced by the quantity of free ions present, this decrease in ion concentration leads to a corresponding decrease in the conductance of the $\mathrm{BaSO}_4$ solution at higher temperatures. Consequently, the claim that $C_{\mathrm{BaSO}_4}(T_2) > C_{\mathrm{BaSO}_4}(T_1)$ is false, as the conductance instead decreases with an increase in temperature.

The relationship between thermodynamic spontaneity and electrochemical potential is defined by the formula $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$. To calculate the cell potential, first convert the given free energy of 17.37 kJ mol⁻¹ into 17370 J mol⁻¹ to ensure consistency with the Faraday constant of 96500 C mol⁻¹. By rearranging the formula to $E^{\circ}_{\text{cell}} = -\frac{\Delta G^{\circ}}{nF}$ and substituting the transfer of 3 electrons, you obtain the potential by dividing the total energy by the product of 3 and 96500. This calculation results in -0.06 V, which corresponds to $-6 \times 10^{-2}$ V, confirming the magnitude to be 6.

The strength of an oxidizing agent is fundamentally tied to its standard reduction potential, as a higher positive value signifies a greater inclination to accept electrons and act as an oxidizing agent. By comparing the provided potential values, we find that $\mathrm{Bi}^{3+}$ has the weakest affinity for electrons at $+0.20\ \mathrm{V}$, followed by $\mathrm{Ce}^{4+}$ at $+1.61\ \mathrm{V}$, $\mathrm{Pb}^{4+}$ at $+1.67\ \mathrm{V}$, and $\mathrm{Co}^{3+}$ at the highest value of $+1.81\ \mathrm{V}$. This trend confirms that the species exhibit increasingly stronger oxidizing capabilities in the sequence $\mathrm{Bi}^{3+} < \mathrm{Ce}^{4+} < \mathrm{Pb}^{4+} < \mathrm{Co}^{3+}$.

Kohlrausch's law demonstrates that molar conductivity, $\Lambda_m$, rises as the concentration of a solution decreases, specifically when plotted against $\sqrt{C}$, because the weakening of interionic attractions allows for faster ion movement. The speed of these ions is significantly influenced by their size in the solvated state; specifically, a solvated $\mathrm{K}^{+}$ ion is smaller than a solvated $\mathrm{Na}^{+}$ ion. Because the smaller potassium ion encounters less resistance moving through the solvent, it travels faster, ensuring that $\mathrm{KCl}$ exhibits a higher molar conductivity than $\mathrm{NaCl}$ across all concentrations, including at the limit of infinite dilution.

Faraday's first law of electrolysis dictates that the mass of a substance altered at an electrode is directly proportional to the amount of electrical charge passed through the electrolyte. During the recharging of a lead-acid battery, the anodic half-cell undergoes a reduction process governed by the equation $PbSO_4 + 2e^- \rightarrow Pb + SO_4^{2-}$, which indicates that two moles of electrons, or two Faraday of charge, are required to reduce one mole of lead sulfate.

Since the molar mass of lead sulfate is $303 \text{ g mol}^{-1}$, this stoichiometric relationship implies that $2 \text{ F}$ of electricity will electrolyze $303 \text{ g}$ of the compound. Consequently, the mass electrolyzed by $0.05 \text{ F}$ is calculated by the expression $(0.05 \text{ F}) \times \frac{303 \text{ g}}{2 \text{ F}}$, which results \in $7.575 \text{ g}$, rounding to $7.6 \text{ g}$.

Conductivity ($\kappa$) measures the conducting power of ions present in a unit volume of solution; as the concentration decreases, the number of ions per unit volume inevitably falls, causing conductivity to decrease. In contrast, molar conductivity ($\Lambda_m$) represents the conductance of all ions produced by one mole of electrolyte, and as the solution is diluted, the increased volume and reduced inter-ionic attraction consistently result in higher molar conductivity.

The standard cell potential is given by $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$, where the anodic potential is $E^\circ_{\text{anode}} = -0.76 \text{ V}$. To find the maximum value of $E^\circ_{\text{cell}}$ per electron transferred, we calculate the ratio $E^\circ_{\text{cell}} / n$, where $n$ is the number of electrons involved in the balanced redox reaction between zinc and the given metal half-cell. The total number of electrons $n$ is the least common multiple of the electron stoichiometry of the cathode reduction ($x$) and the zinc oxidation (2), resulting in $n=2$ for $x=1$ or $2$, and $n=6$ for $x=3$.

For the $\text{Fe}^{3+} / \text{Fe}^{2+}$ half-cell ($x=1, n=2$), $E^\circ_{\text{cell}} = 0.77 - (-0.76) = 1.53 \text{ V}$, which results in a ratio of $1.53 / 2 = 0.765 \text{ V}$. For the $\text{Ag}^{+} / \text{Ag}$ half-cell ($x=1, n=2$), $E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56 \text{ V}$, resulting in a ratio of $1.56 / 2 = 0.78 \text{ V}$. For the $\text{Au}^{3+} / \text{Au}$ half-cell ($x=3, n=6$), $E^\circ_{\text{cell}} = 1.40 - (-0.76) = 2.16 \text{ V}$, yielding a ratio of $2.16 / 6 = 0.36 \text{ V}$. Finally, for the $\text{Fe}^{2+} / \text{Fe}$ half-cell ($x=2, n=2$), $E^\circ_{\text{cell}} = -0.44 - (-0.76) = 0.32 \text{ V}$, yielding a ratio of $0.32 / 2 = 0.16 \text{ V}$. Comparing these values, the $\text{Ag}^{+} / \text{Ag}$ half-cell yields the maximum ratio.

To determine the degree of dissociation of a weak acid like HA, we apply Kohlrausch's law, which states that the limiting molar conductivity of an electrolyte is the sum of the contributions of its individual ions. By combining the provided values for the strong electrolytes NaCl, HCl, and NaA, we find the limiting molar conductivity of HA at infinite dilution through the relation:
$\Lambda^{\circ}_{m}(HA) = \Lambda^{\circ}_{m}(HCl) + \Lambda^{\circ}_{m}(NaA) - \Lambda^{\circ}_{m}(NaCl) = 425.9 + 100.5 - 126.4 = 400 \text{ S cm}^2 \text{mol}^{-1}$
We then determine the molar conductivity at the specific concentration of 0.001 M using the given conductivity of 5 × 10$^{-5}$ S cm$^{-1}$:
$\Lambda_{m} = \frac{1000 \kappa}{M} = \frac{1000 \times 5 \times 10^{-5}}{0.001} = 50 \text{ S cm}^2 \text{mol}^{-1}$
The degree of dissociation is the ratio of the molar conductivity at this concentration to the limiting molar conductivity, calculated as:
$\alpha = \frac{\Lambda_{m}}{\Lambda^{\circ}_{m}} = \frac{50}{400} = 0.125$

The standard electrode potential of the silver-silver chloride electrode is determined by analyzing the electrochemical cell where hydrogen gas undergoes oxidation at the anode and silver chloride undergoes reduction at the cathode. The overall spontaneous process follows the stoichiometry:
$\mathrm{H}_2(g) + 2\mathrm{AgCl}(s) \rightarrow 2\mathrm{H}^{+}(aq) + 2\mathrm{Ag}(s) + 2\mathrm{Cl}^{-}(aq)$
Since the standard potential of the hydrogen electrode is $0$ V, the standard cell potential $E^{\circ}_{\text{cell}}$ is equal to the standard electrode potential of the silver-silver chloride electrode, $E^{\circ}_{\text{AgCl/Ag,Cl}^-}$.

Applying the Nernst equation for this two-electron transfer reaction gives the following relationship:
$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{2} \log_{10}([\mathrm{H}^{+}]^{2}\$[\mathrm{Cl}^{-}]^{2})$
With an HCl solution concentration of $10^{-6}$ molal, both the hydrogen ion and chloride ion concentrations are $10^{-6}$ molal. The reaction quotient $Q$ is therefore $(10^{-6})^2 (10^{-6})^2 = 10^{-24}$. Substituting the measured potential of $0.92$ V into the expression yields $0.92 = E^{\circ}_{\text{cell}} - 0.03 \log_{10}(10^{-24})$. Given that $\log_{10}(10^{-24}) = -24$, the equation simplifies to $0.92 = E^{\circ}_{\text{cell}} - 0.03 \times (-24)$, which results in $0.92 = E^{\circ}_{\text{cell}} + 0.72$. Solving for the standard potential yields a value of $0.20$ V.

The standard cell potential of a redox reaction at equilibrium is directly related to the equilibrium constant through the Nernst equation, which simplifies when the cell potential is zero to $E^{\circ}_{\mathrm{cell}} = \frac{2.303 RT}{nF} \log K_C$. For the reaction $\mathrm{Cu (s) + 2Ag^{+} (aq) \rightarrow Cu^{2+} (aq) + 2Ag (s)}$, two electrons are transferred between the copper and silver species, so the value of $n$ is $2$. The equilibrium constant $K_C$ is given as $10 \times 10^{15}$, which simplifies to $10^{16}$.

Using the standard factor $\frac{2.303 RT}{F}$ with a more precise value of $0.0592\,\text{V}$-commonly used in electrochemical calculations to achieve the accuracy required for standard results-the equation becomes $E^{\circ}_{\mathrm{cell}} = \frac{0.0592}{2} \log (10^{16})$. Substituting the logarithmic value, we get $E^{\circ}_{\mathrm{cell}} = 0.0296 \times 16$, which results in a standard cell potential of $0.4736\,\text{V}$.

The thermodynamic relationship between standard Gibbs free energy, cell potential, and the equilibrium constant allows us to calculate $K$ for any electrochemical reaction. Since $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ and $\Delta G^\circ = -RT \ln K$, equating these relations yields the primary equation:

$\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}$

For this reaction, the number of moles of electrons exchanged is $n = 2$. Substituting the values $F = 96000$ C/mol, $E^\circ_{\text{cell}} = 2$ V, $R = 8$ J/K mol, and $T = 300$ K into the expression results in $\ln K = \frac{2 \times 96000 \times 2}{8 \times 300}$. Completing the calculation leads to $\ln K = 160$, which establishes that $K = e^{160}$.

Electrolysis follows Faraday's laws, which dictate that the amount of substance deposited at an electrode is directly proportional to the total charge passed through the solution. The cathodic reduction of nickel ions is represented by the half-reaction $\mathrm{Ni^{2+} + 2e^- \rightarrow Ni}$, showing that 2 Faraday of electricity are required to deposit 1 mole of nickel. Given that 0.1 Faraday of electricity is applied, the moles of nickel deposited can be found by dividing the total charge by the charge required per mole:
$\frac{0.1}{2} = 0.05$
As a result, 0.05 moles of nickel are deposited at the cathode.

The standard reaction enthalpy is derived from the thermodynamic relationship $\Delta_r H^{\ominus} = \Delta_r G^{\ominus} + T\Delta_r S^{\ominus}$, which, by substituting the relations $\Delta_r G^{\ominus} = -nFE^{\ominus}$ and $\Delta_r S^{\ominus} = nF\left(\frac{dE^{\ominus}}{dT}\right)$, simplifies to the expression $\Delta_r H^{\ominus} = nF\left(T\frac{dE^{\ominus}}{dT} - E^{\ominus}\right)$. In the cell reaction $\text{Zn(s)} + \text{Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq)} + \text{Cu(s)}$, the number of electrons transferred is $n = 2$. By substituting the given values $n = 2$, $F = 96,000 \text{C mol}^{-1}$, $T = 300 \text{K}$, $E^{\ominus} = 2 \text{V}$, and the temperature coefficient $\frac{dE^{\ominus}}{dT} = -5 \times 10^{-4} \text{V K}^{-1}$ into the enthalpy formula, we obtain $\Delta_r H^{\ominus} = 2 \times 96,000 \times (300 \times (-5 \times 10^{-4}) - 2)$. Evaluating this calculation results in $\Delta_r H^{\ominus} = 192,000 \times (-0.15 - 2) = -412,800 \text{J mol}^{-1}$, which is equivalent to $-412.8 \text{kJ mol}^{-1}$.

The standard cell potential is determined by the principle that Gibbs free energy is an extensive state function, allowing potentials for multi-electron processes to be derived by combining intermediate reaction steps. For the reaction $\mathrm{Fe}^{2+} + \mathrm{Ag}^{+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{Ag}$, the cell potential is expressed as $E^\circ = E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} - E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}$. Because the potential for the $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ reduction is unknown, it is calculated using the Gibbs energies of the three-electron reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}$ ($z$) and the two-electron reduction of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}$ ($y$), which yields $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 3z - 2y$. Substituting this value into the initial cell equation results in $E^\circ = x - (3z - 2y)$ which simplifies to $E^\circ = x + 2y - 3z$

The oxidizing power of a chemical species is directly determined by its standard reduction potential, where a higher positive value indicates a stronger tendency to undergo reduction and accept electrons. Among the species provided, $\mathrm{S}_{2}\mathrm{O}_{8}^{2-}$ exhibits the highest reduction potential of $2.05\,\mathrm{V}$, which surpasses the potentials of $\mathrm{Au}^{3+}$ at $1.4\,\mathrm{V}$, $\mathrm{O}_{2}$ at $1.23\,\mathrm{V}$, and $\mathrm{Br}_{2}$ at $1.09\,\mathrm{V}$. Consequently, due to this significantly greater reduction potential, $\mathrm{S}_{2}\mathrm{O}_{8}^{2-}$ serves as the strongest oxidizing agent in this group.

The reducing capability of a metal is fundamentally defined by its willingness to release electrons, a process known as oxidation. Since the provided data describes standard reduction potentials, we must remember that a metal with a more negative reduction potential effectively possesses a higher oxidation potential and thus acts as a more powerful reducing agent. Essentially, the easier it is for a metal to be oxidized, the greater its reducing strength.

Applying this to the given values, the standard reduction potentials ($E^\circ = -0.25\text{ V}$ for Ni, $E^\circ = -0.76\text{ V}$ for Zn, $E^\circ = -2.36\text{ V}$ for Mg, and $E^\circ = -2.87\text{ V}$ for Ca) must be converted to oxidation potentials. Reversing the signs of these values gives oxidation potentials of +0.25 V, +0.76 V, +2.36 V, and +2.87 V respectively. Arranging these values in increasing order of their ability to oxidize, we find the sequence of increasing reducing power is $Ni < Zn < Mg < Ca$.

The standard Gibbs energy change for an electrochemical reaction is governed by the expression $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, which correlates the thermodynamic spontaneity of the cell with its electrical potential. In this redox process, zinc undergoes oxidation to $\mathrm{Zn}^{2+}$ and copper ions undergo reduction to copper, necessitating the transfer of two electrons per mole of reaction, so $n = 2$. With a standard cell potential of 2 V and a Faraday constant of $96000\text{ C mol}^{-1}$, the calculation becomes:
$\Delta G^\circ = -2 \times 96000\text{ C mol}^{-1} \times 2\text{ V} = -384000\text{ J mol}^{-1}$
Converting this result from joules to kilojoules gives $-384\text{ kJ mol}^{-1}$.

The relationship between chemical combustion and electrolysis relies on balancing the molar requirements of the reaction with Faraday's laws, which dictate that the charge passed is proportional to the chemical yield. The molar mass of diborane is approximately 27.6 g/mol, meaning the 27.66 g sample provided constitutes 1 mole of diborane. According to the stoichiometry of the combustion reaction,

$\mathrm{B_2H_6} + 3\mathrm{O_2} \rightarrow \mathrm{B_2O_3} + 3\mathrm{H_2O}$

one mole of diborane requires 3 moles of oxygen gas to undergo complete combustion.

In the electrolysis of water, the reaction $2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 2\mathrm{H_2}$ shows that the production of 1 mole of oxygen gas involves the transfer of 4 moles of electrons, which is equivalent to $4F$ of charge. Consequently, to produce the required 3 moles of oxygen, a total charge of $3 \times 4F = 12F$ must be passed through the solution. Using the constant $1F = 96500$ Coulombs, the total charge required is $12 \times 96500$ Coulombs. Applying the relationship $Q = It$, where the current is 100 Amperes, the time in seconds is found by $t = (12 \times 96500) / 100$, which equals 11580 seconds. Converting this duration into hours by dividing by 3600 results in approximately 3.2 hours.

The reduction of nitrobenzene to p-aminophenol proceeds via the reaction $C_6H_5NO_2 + 4H^+ + 4e^- \rightarrow HOC_6H_4NH_2 + H_2O$ confirming that 4 moles of electrons are required for each mole of p-aminophenol produced. The total charge passed is calculated as $Q = I × t$, which for a current of 9.65 A over 3600 seconds equals 34740 C.

The number of moles of electrons transferred is given by $n_e = \frac{Q}{F} = \frac{34740}{96500} = 0.36$. Using the stoichiometric ratio of 1:4 from the reaction, the amount of p-aminophenol produced is 0.09 mol. The mass of the product is then calculated as $0.09 \times 109 = 9.81$ g.

Faraday's laws of electrolysis establish the relationship between the quantity of electricity passed and the chemical changes occurring at the electrodes. The reduction reaction at the cathode, $2H^+ + 2e^- \rightarrow H_2$, indicates that 2 moles of electrons are required to evolve 1 mole of hydrogen gas. Since 1 mole of any gas occupies 22400 mL at N.T.P., the number of moles of hydrogen gas collected is calculated as $n = \frac{112 \text{ mL}}{22400 \text{ mL/mol}} = 0.005 \text{ mol}$. The total moles of electrons transferred is $n_e = 2 \times 0.005 \text{ mol} = 0.01 \text{ mol}$. Using Faraday's constant $F = 96500 \text{ C/mol}$, the total charge $Q$ passed is $Q = n_e \times F = 0.01 \text{ mol} \times 96500 \text{ C/mol} = 965 \text{ C}$. Given the time $t = 965 \text{ s}$, the current $I$ is calculated using the formula $Q = I \times t$, which gives $I = \frac{965 \text{ C}}{965 \text{ s}} = 1.0 \text{ A}$.