Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Weak electrolytes do not fully dissociate into ions at high concentrations, meaning their conductivity is heavily dependent on the degree of dissociation. As the concentration of the solution increases, the equilibrium shifts to favor the undissociated molecular form, leading to a significant reduction in the number of free ions available to carry charge. This causes the molar conductivity $\Lambda_m$ to decrease sharply as the concentration rises, creating a steep curve when plotted against the square root of the concentration.

Molar conductivity defines the conducting power of all ions produced by one mole of an electrolyte in a solution and is calculated as the ratio of electrolytic conductivity to molar concentration. To find the correct value, it is essential to ensure that the units for concentration and conductivity are converted consistently to align with the final requested units.

A 0.2% (w/v) NaOH solution contains 0.2 g of solute in 100 mL of solution. Given the molar mass of NaOH is 40 g/mol, the molarity $M$ is calculated as $M = \frac{0.2}{40 \times 0.1} = 0.05\ \text{mol dm}^{-3}$. Since conductivity $\kappa$ is the reciprocal of resistivity, using a resistivity of $870.0\ \text{m}\Omega\ \text{m}$ (which equals $0.870\ \Omega\ \text{m}$) yields $\kappa = \frac{1}{0.870} = 1.15\ \text{S m}^{-1}$. Applying the relationship $\Lambda_m = \frac{\kappa}{M}$ and performing the necessary unit conversion from $\text{S m}^2 \text{mol}^{-1}$ to $\text{mS dm}^2 \text{mol}^{-1}$ results in $\Lambda_m = 2300\ \text{mS dm}^2 \text{mol}^{-1}$, which is equivalent to $23 \times 10^{2}\ \text{mS dm}^2 \text{mol}^{-1}$.

The relationship between the standard Gibbs free energy change and the standard cell potential is given by the equation $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$, where $n$ represents the number of moles of electrons transferred, $F$ is Faraday's constant, and $E^{\circ}_{\text{cell}}$ is the standard cell potential. Because $n$ and $F$ are positive constants, $\Delta G^{\circ}$ becomes most negative when $E^{\circ}_{\text{cell}}$ is the largest positive value. The standard cell potential is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode.

For the first cell configuration, $\mathrm{Zn} | \mathrm{Zn}^{2+} \| \mathrm{Ag}^{2+} | \mathrm{Ag}$, the cell potential is $E^{\circ}_{\text{cell}} = 0.80\,\text{V} - (-0.76\,\text{V}) = 1.56\,\text{V}$, which leads to a highly negative $\Delta G^{\circ}$ value. In the cases involving magnesium in the second and third options, the calculated $E^{\circ}_{\text{cell}}$ values are negative, resulting in positive $\Delta G^{\circ}$ values and non-spontaneous reactions. For the final cell configuration, $\mathrm{Cu} | \mathrm{Cu}^{2+} \| \mathrm{Ag}^{2+} | \mathrm{Ag}$, the potential is $E^{\circ}_{\text{cell}} = 0.80\,\text{V} - 0.34\,\text{V} = 0.46\,\text{V}$. Since the cell potential of $1.56\,\text{V}$ is greater than $0.46\,\text{V}$, the first option yields the most negative $\Delta G^{\circ}$ value.

The potential of a hydrogen electrode is governed by the Nernst equation, which relates the emf to the concentration of hydrogen ions and the partial pressure of hydrogen gas. For a hydrogen electrode operating at $25^{\circ} \mathrm{C}$, the standard electrode potential is zero, which allows the expression to be simplified into the following form:

$E = 0 - \frac{0.059}{2} \log \frac{P_{\mathrm{H}_2}}{[\mathrm{H}^{+}]^2}$

To make the emf zero, the logarithmic component must equal zero, which mathematically necessitates that the ratio $\frac{P_{\mathrm{H}_2}}{[\mathrm{H}^{+}]^2}$ equals one. Given that the concentration of hydrogen ions in pure water is $10^{-7} \mathrm{M}$, squaring this value results in $10^{-14}$. Therefore, to satisfy the condition where the emf is zero, the partial pressure $P_{\mathrm{H}_2}$ must be exactly $10^{-14} \mathrm{bar}$.

Faraday's first law of electrolysis states that the mass of a substance liberated at an electrode is directly proportional to the total electric charge passed, expressed by the formula $W = ZQ$. Here, $W$ is the mass of the substance deposited, $Z$ represents its electrochemical equivalent, and $Q$ denotes the charge in coulombs. By passing a charge of 1 coulomb into this relationship, the equation simplifies to $W = Z$, meaning the total amount of silver deposited is precisely its electrochemical equivalent.

Kohlrausch's law of independent migration of ions states that the molar conductivity of an electrolyte solution is the sum of the individual molar ionic conductivities of the constituent ions, adjusted by their respective stoichiometric coefficients. In this specific case, the electrolyte is composed of a divalent cation and a divalent anion in a 1:1 stoichiometric ratio, meaning the stoichiometric numbers for both are one. Therefore, the total molar conductivity is simply the arithmetic sum of the provided ionic values, $57\,\mathrm{S}\,\mathrm{cm}^2\,\mathrm{mol}^{-1} + 73\,\mathrm{S}\,\mathrm{cm}^2\,\mathrm{mol}^{-1}$, which yields a final result of $130\,\mathrm{S}\,\mathrm{cm}^2\,\mathrm{mol}^{-1}$.

The variation of molar conductivity $\Lambda_m$ with the square root of concentration $\sqrt{C}$ depends on how effectively an electrolyte dissociates as it is diluted. Strong electrolytes, represented by curve B, exhibit a linear relationship described by the equation $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$, which allows for a straightforward extrapolation to the vertical axis to determine the limiting molar conductivity. Conversely, weak electrolytes like A display a sharp, non-linear increase in molar conductivity as the concentration approaches zero because their degree of dissociation significantly increases with dilution, preventing the determination of the limiting value through simple linear graphical extension.

Rusting is fundamentally an electrochemical process where the surface of the metal acts as a series of galvanic cells. While a tin coating provides protection by isolating the metal, this shielding is only effective while the layer remains perfectly intact. Once the tin coating is peeled off, the underlying iron is exposed to the atmosphere and, because iron is more reactive than tin, the defect actually accelerates the corrosion process instead of preventing it.

Spontaneity in an electrochemical cell is dictated by the standard cell potential $E_{\text{cell}}^{\circ}$, where a positive value indicates a spontaneous process. The total cell potential is determined by combining the standard oxidation potential at the anode with the standard reduction potential at the cathode. For the given cell $\mathrm{M} | \mathrm{M}^{2+} \parallel \mathrm{X} | \mathrm{X}^{2-}$, the metal $\mathrm{M}$ undergoes oxidation, providing an oxidation potential of $-0.46\ \mathrm{V}$, which combines with the reduction potential of $0.34\ \mathrm{V}$ for $\mathrm{X}$ to yield $E_{\text{cell}}^{\circ} = -0.46\ \mathrm{V} + 0.34\ \mathrm{V} = -0.12\ \mathrm{V}$. Because this resulting potential is negative, the forward reaction $\mathrm{M} + \mathrm{X} \rightarrow \mathrm{M}^{2+} + \mathrm{X}^{2-}$ is non-spontaneous, confirming that the reverse reaction $\mathrm{M}^{2+} + \mathrm{X}^{2-} \rightarrow \mathrm{M} + \mathrm{X}$ is the spontaneous process.

A spontaneous redox reaction occurs in an electrochemical cell when the overall cell potential is positive, which requires the reduction potential of the oxidizing agent to be greater than the reduction potential of the metal being oxidized. The dichromate ion $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ serves as the oxidizing agent in this scenario with a standard reduction potential of 1.33 V, meaning it will spontaneously oxidize any metal with a standard reduction potential lower than this value. Comparing the provided values, the metals $\mathrm{Fe}$ (-0.04 V), $\mathrm{Ni}$ (-0.25 V), and $\mathrm{Ag}$ (0.80 V) all possess reduction potentials below 1.33 V, while the reduction potential of $\mathrm{Au}$ (1.40 V) is higher. Consequently, exactly three metals, namely $\mathrm{Fe}$, $\mathrm{Ni}$, and $\mathrm{Ag}$, will undergo oxidation in this solution.

The connection between the standard cell potential of a redox reaction and its equilibrium constant relies on the thermodynamic relationship $\Delta G^{\circ} = -RT \ln K = -nFE_{\text{cell}}^{\circ}$. At standard conditions, this relates to the Nernst equation at equilibrium by the expression $\log K = \frac{n E_{\text{cell}}^{\circ}}{0.059}$ where $n$ represents the total number of electrons transferred \in the balanced reaction. First, determine the standard cell potential by combining the half-reactions: the reduction of permanganate ions ($+1.51 \text{ V}$) and the oxidation of oxalic acid. Given the reduction potential of oxalic acid is $-0.49 \text{ V}$, its oxidation potential is $+0.49 \text{ V}$, making the total cell potential $E_{\text{cell}}^{\circ} = 1.51 + 0.49 = 2.00 \text{ V}$.

Next, identify the electron transfer balance by examining the change in oxidation states. The reduction of $\mathrm{MnO}_4^{-}$ to $\mathrm{Mn}^{2+}$ involves a gain of $5$ electrons, whereas the oxidation of $\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4$ to $\mathrm{CO}_2$ involves a loss of $2$ electrons per molecule. To balance these half-reactions, the entire system requires an exchange of $n = 10$ electrons. Substituting these values into the equilibrium equation, the calculation becomes $\log K = \frac{10 \times 2.00}{0.059} \approx 338.98$ Since the equilibrium constant is expressed as $10^x$, the value of $x$ is $339$ when rounded to the nearest integer.

Faraday's first law of electrolysis states that the mass of a substance deposited at an electrode is proportional to the charge passed, defined by the equation $\frac{w}{E} = \frac{Q}{1\,\text{F}}$. Since the gold in the $\mathrm{AuCl}_4^-$ ion is in the $+3$ oxidation state, its equivalent mass is calculated as $E = \frac{197}{3}$ grams per equivalent. By substituting the mass of 1.314 grams into the relation $\frac{1.314}{197/3} = Q$, we can determine the total charge passed through the solution. This calculation yields $Q = 2 \times 10^{-2}$ Faradays.

The oxidation of water can be understood by examining the balanced half-reaction $2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2 + 4 \mathrm{H}^{+} + 4 \mathrm{e}^{-}$. This equation shows that 4 moles of electrons are released for every 2 moles of water, meaning that the oxidation of 1 mole of $\mathrm{H}_2 \mathrm{O}$ requires the removal of 2 moles of electrons. By applying Faraday's laws, the total charge $Q$ is calculated by multiplying the number of moles of electrons by Faraday's constant, which is approximately $96500 \text{ C/mol}$. Consequently, $Q = 2 \times 96500 \text{ C}$, which equals $193000 \text{ C}$, or $1.93 \times 10^5 \text{ C}$.

The electrochemical potential of a hydrogen electrode is determined by applying the Nernst equation to the reduction reaction of hydrogen ions. Because the standard electrode potential for a hydrogen half-cell is zero, the potential simplifies to a function of the solution pH. At 25°C, with the provided Nernst constant, this relationship is given by $E = -0.059 \times \mathrm{pH}$.

By substituting the given pH value of 3 into the formula $E = -0.059 \times 3$, we find the potential to be -0.177 Volts. Expressing this in the required notation results in -17.7 × 10⁻² Volts. The magnitude of this value, when rounded to the nearest integer as requested, corresponds to 18.

The EMF of a cell is governed by the Nernst equation, which for this reaction is expressed as E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{2} \log \frac{[\mathrm{Tl}^{+}]^2}{[\mathrm{Cu}^{2+}]\}$. Because the term involving the concentrations is subtracted from the standard cell potential, increasing the concentration of$\mathrm{Cu}^{2+}$ ions in the denominator decreases the overall value of the reaction quotient within the logarithm. This reduction in the subtracted term directly results in an increase in the measured EMF of the cell.

The Kohlrausch law describes the variation of molar conductivity $\Lambda_{m}$ with concentration through the equation $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{C}$. Because the terms in a physical expression must have consistent dimensions, the product $A\sqrt{C}$ must match the units of molar conductivity, which are S cm^2 mol^-1. Given that concentration is measured in mol L^-1, its square root has units of mol^0.5 L^-0.5, and dividing the units of molar conductivity by these concentration units determines the dimension of constant A as S cm^2 mol^-1.5 L^0.5.

Kolbe's electrolysis of a mixture of sodium carboxylate salts involves the generation of alkyl free radicals through decarboxylation, which then undergo random recombination to produce alkanes. With a mixture of $\mathrm{CH_3COONa}$ and $\mathrm{C_2H_5COONa}$, the system generates both methyl radicals and ethyl radicals, leading to three possible coupling pathways:

$2 \mathrm{CH_3^{\bullet}} \rightarrow \mathrm{CH_3-CH_3}$
$2 \mathrm{C_2H_5^{\bullet}} \rightarrow \mathrm{CH_3-CH_2-CH_2-CH_3}$
$\mathrm{CH_3^{\bullet}} + \mathrm{C_2H_5^{\bullet}} \rightarrow \mathrm{CH_3-CH_2-CH_3}$

Since these three processes occur concurrently, the electrolysis yields a total of 3 distinct alkanes.

The dry cells commonly found in clocks function through an electrochemical reaction where the cathode is the site of reduction. During this process, manganese in the $MnO_2$ acts as the active material, gaining an electron to shift its oxidation state from +4 to +3. This specific change in the oxidation state is the defining characteristic of the cathode reaction in these power sources.

When $\mathrm{MnO}_2$ is subjected to alkaline oxidative fusion, it reacts with hydroxide ions and oxygen to generate the manganate ion, $\mathrm{MnO}_4^{2-}$, which serves as A. Subjecting this intermediate to electrolytic oxidation in an alkaline solution triggers a redox process that converts the manganate ion into the purple permanganate ion, $\mathrm{MnO}_4^{-}$, which is identified as B.

For weak electrolytes, the relationship between the dissociation constant $K_a$, concentration $C$, and degree of dissociation $\alpha$ is governed by the expression $K_a = \frac{\alpha^2 C}{1-\alpha}$. Because the degree of dissociation is equivalent to the ratio of molar conductivity at a given concentration $\Lambda_m$ to the limiting molar conductivity $\Lambda_m^{\circ}$, we substitute $\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$ into the equilibrium formula. Clearing the fraction and rearranging the resulting terms gives the equation:

$\Lambda^2_m C - K_{\mathrm{a}} \Lambda_m^{\circ 2} + K_{\mathrm{a}} \Lambda_m \Lambda_m^{\circ} = 0$