Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

The standard Gibbs free energy change, $\Delta G^{\circ}$, is related to the standard electrode potential $E^{\circ}$ by the equation $\Delta G^{\circ} = -nFE^{\circ}$. For the lead reactions provided, the Gibbs free energy changes are $\Delta G^{\circ}(\mathrm{Pb}^{2+} \to \mathrm{Pb}) = -2mF$ and $\Delta G^{\circ}(\mathrm{Pb}^{+} \to \mathrm{Pb}) = -nF$. The target reaction $\mathrm{Pb}^{2+} + e^{-} \rightarrow \mathrm{Pb}^{+}$ is obtained by subtracting the second reaction from the first, so the Gibbs free energy change for this reaction is $\Delta G^{\circ}(\mathrm{Pb}^{2+} \to \mathrm{Pb}^{+}) = \Delta G^{\circ}(\mathrm{Pb}^{2+} \to \mathrm{Pb}) - \Delta G^{\circ}(\mathrm{Pb}^{+} \to \mathrm{Pb}) = -2mF - (-nF) = -F(2m - n)$. Using the relation $\Delta G^{\circ} = -1 \cdot F \cdot E_{(\mathrm{Pb}^{2+} / \mathrm{Pb}^{+})}^{\circ}$, it follows that $-FE^{\circ} = -F(2m - n)$, which simplifies to $E_{(\mathrm{Pb}^{2+} / \mathrm{Pb}^{+})}^{\circ} = 2m - n$. Given the target expression is of this form, the value of $x$ is 2.

The standard electrode potential is determined by the overall free energy change of the reduction reaction $M^{+}(aq) + e^{-} \to M(s)$, which is evaluated through a thermodynamic cycle comprising the sublimation of the solid metal, the ionization of the gaseous metal atom, and the hydration of the resulting gaseous ion. Because ionization is fundamentally defined as the energy required to remove an electron from a gaseous species, the potential depends on the ionization energy of gaseous metal atoms rather than solid metal atoms. Consequently, the direct ionization of a solid metal atom is not a constituent step in the thermodynamic cycle that defines the electrode potential, as the metal must first undergo sublimation to the gaseous phase before ionization can occur.

Electrolysis of a brine solution involves the dissociation of sodium chloride into ions, where the applied electric current drives specific reduction and oxidation reactions at the electrodes. Since water has a higher reduction potential than sodium ions, the reaction at the cathode follows the equation $2\,\mathrm{H}_2\mathrm{O}_{(\ell)} + 2\,\mathrm{e}^{-} \rightarrow \mathrm{H}_2(\text{g}) + 2\,\mathrm{OH}_{\text{(aq)}}^{-}$, which directly results in the production of hydroxide ions within the solution.

A redox reaction proceeds spontaneously if the reduction potential of the oxidizing agent exceeds the reduction potential of the species being oxidized. Nitrate serves as the oxidizing agent with a standard reduction potential of $E^o = 0.97\,\text{V}$, meaning it will oxidize any metal with a standard reduction potential lower than this value.

Vanadium with $E^o = -1.19\,\text{V}$, iron with $E^o = -0.04\,\text{V}$, and silver with $E^o = 0.80\,\text{V}$ all meet this criterion by having potentials below $E^o = 0.97\,\text{V}$. Gold, which has a standard reduction potential of $E^o = 1.40\,\text{V}$, exceeds the nitrate threshold and will not be oxidized. This identifies a total of three metals that will be oxidized.

Faraday's laws of electrolysis provide the relationship between the electrical charge transferred and the quantity of metal deposited at the cathode. Determining the time required for electroplating depends on calculating the total mass of the deposited layer, relating that mass to the moles of metal, and applying the stoichiometry of the reduction reaction to find the necessary charge.

The volume of the nickel layer is found by multiplying the surface area by the thickness. Converting the thickness of $0.001\text{ mm}$ to $0.0001\text{ cm}$ leads to a volume of $100\text{ cm}^2 \times 0.0001\text{ cm} = 0.01\text{ cm}^3$. Since the density of nickel is $10\text{ g/mL}$, the mass of the layer is $0.01\text{ cm}^3 \times 10\text{ g/cm}^3 = 0.1\text{ g}$. Using the molar mass of $60\text{ g/mol}$, the number of moles of nickel to be deposited is $\frac{0.1\text{ g}}{60\text{ g/mol}} = \frac{1}{600}\text{ mol}$.

The reduction process $\mathrm{Ni}^{2+}+2 e^{-}\rightarrow\mathrm{Ni}(\text{s})$ requires $2$ moles of electrons for every mole of nickel. Consequently, the total charge in coulombs is the number of moles multiplied by $2$ and Faraday's constant, $96500\text{ C/mol}$. This calculation gives $Q = \frac{1}{600} \times 2 \times 96500 = \frac{965}{3}\text{ C}$. Given the relationship $t = \frac{Q}{I}$ where the current is $2\text{ A}$, the time is $t = \frac{965/3}{2} \approx 160.83\text{ s}$. Rounding to the nearest integer yields $161\text{ s}$.

Conductivity measures the charge-carrying capacity per unit volume, which necessarily decreases upon dilution as the density of ions falls. Molar conductivity, defined by the relationship $\Lambda_m = \frac{1000 \kappa}{C}$, increases because the proportional rise in solvent volume significantly outpaces the decrease in conductivity ($\kappa$). This behavior varies between solute types, as strong electrolytes follow the Debye-Hückel-Onsager equation while weak electrolytes exhibit a much steeper increase due to an increase in their degree of dissociation. Since dilution decreases ion concentration per unit volume, statements A, C, and D accurately describe the electrochemical principles, confirming that there are three correct statements.

Acidity in this electrochemical environment is determined by applying the Nernst equation, which relates the actual cell potential to standard conditions through the reaction quotient. For the reduction half-reaction $\mathrm{MnO}_4^{-} + 8\mathrm{H}^{+} + 5\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$, the process involves the transfer of five electrons, and the reaction quotient is dependent on the concentrations of the manganese species and the hydrogen ions.

The Nernst equation is:
$E = E^\circ - \frac{0.059}{5} \log \frac{[\mathrm{Mn}^{2+}]\$}{[\mathrm{MnO}_4^{-}]\$ \$[\mathrm{H}^{+}]^8}$
Substituting the given values into the expression yields 1.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \[\mathrm{H}^{+}]^8}$. Rearranging the equation to isolate the logarithmic term results \in$1.54 - 1.282 = \frac{0.059}{5} \log \frac{10^{-2}}{[\mathrm{H}^{+}]^8}$, which simplifies to$0.258 \cdot \frac{5}{0.059} = \log(10^{-2}) - 8 \log[\mathrm{H}^{+}]$. Because$\log(10^{-2}) = -2$and the definition of$\mathrm{pH}$is$-\log[\mathrm{H}^{+}]$, the expression becomes$21.86 = -2 + 8 \mathrm{pH}$. Solving this for$\mathrm{pH}$gives approximately$2.98$, which corresponds to a nearest integer value of$3$.

A metal ion can liberate hydrogen from a dilute acid if it acts as a strong reducing agent capable of transferring electrons to $\mathrm{H}^{+}$ ions to form $\mathrm{H}_2$ gas. This occurs when the standard reduction potential of the $\mathrm{M}^{3+}/\mathrm{M}^{2+}$ couple is negative, signifying that the oxidation half-reaction $\mathrm{M}^{2+} \rightarrow \mathrm{M}^{3+} + e^-$ is energetically favorable relative to the reduction of $\mathrm{H}^{+}$. Given that the potentials for $\mathrm{V}$ and $\mathrm{Cr}$ are $-0.26\ \mathrm{V}$ and $-0.41\ \mathrm{V}$ respectively, while those for $\mathrm{Mn}$ and $\mathrm{Co}$ are $+1.57\ \mathrm{V}$ and $+1.97\ \mathrm{V}$, only $\mathrm{V}^{2+}$ and $\mathrm{Cr}^{2+}$ provide the necessary driving force to spontaneously liberate hydrogen from an acidic medium.

A lead storage battery operates as a secondary cell where chemical reactions are reversible, and it utilizes a 38% solution of sulphuric acid as the electrolyte. In the basic construction of this battery, a lead plate serves as the anode, whereas the cathode consists of a lead grid packed with $\mathrm{PbO}_2$. When an external current is applied to charge the battery, the discharge-accumulated $\mathrm{PbSO}_4$ on the cathode is converted back into $\mathrm{PbO}_2$, and the $\mathrm{PbSO}_4$ on the anode reverts to metallic lead. Identifying the correct electrolyte composition and the specific cathode conversion process clarifies the operational characteristics of the device.

A galvanic cell operates by spatially separating oxidation and reduction processes, with the standard cell notation conventionally placing the anode on the left and the cathode on the right. In the arrangement $\mathrm{Pt} \left| \mathrm{H}_2(\mathrm{g}) \right| \mathrm{HCl}(\mathrm{sol}^{n}) \left| \mathrm{AgCl}(\mathrm{s}) \right| \mathrm{Ag}$, the hydrogen gas at the platinum anode is oxidized to $\mathrm{H}^{+}$ ions, while the solid $\mathrm{AgCl}$ at the cathode undergoes reduction to form metallic silver and chloride ions. Combining these half-reactions-specifically $\frac{1}{2} \mathrm{H}_2 \rightarrow \mathrm{H}^{+} + \mathrm{e}^{-}$ at the anode and $\mathrm{AgCl} + \mathrm{e}^{-} \rightarrow \mathrm{Ag} + \mathrm{Cl}^{-}$ at the cathode-results exactly in the balanced chemical equation provided in the question.

Electrochemical potential $E_{\text{cell}}$ is classified as an intensive property because its value is independent of the mass or size of the system, while a negative standard electrode potential $E^{\ominus}$ signifies that a redox couple acts as a stronger reducing agent than the $\mathrm{H}^{+} / \mathrm{H}_{2}$ couple due to its greater susceptibility to oxidation. The principles of electrolysis dictate that the amount of chemical reaction occurring at an electrode is directly proportional to the quantity of electricity passed through the electrolyte, and the specific amount of electricity required to drive these transformations is determined entirely by the stoichiometry defined by the electrode reaction.

Molar conductivity is the measure of the conducting power of all ions produced by one mole of an electrolyte in solution, calculated using the relationship $\Lambda_m = \frac{1000}{\rho M}$, where $\rho$ is resistivity and $M$ is molarity. Given the values $\rho = 5 \times 10^{-3} \Omega \mathrm{cm}$ and $M = 0.8 \mathrm{M}$, we can determine the conductivity directly by substitution.

$\Lambda_m = \frac{1000}{(5 \times 10^{-3}) \times 0.8} = \frac{1000}{0.004} = 250,000$

Converting this to the required form of $25 \times 10^4 \Omega^{-1} \mathrm{cm}^2 \mathrm{mol}^{-1}$, the integer value obtained is 25.

The relationship between the equilibrium constant $K_{eq}$ and the standard cell potential $E^0_{cell}$ is fundamental to understanding electrochemical processes at equilibrium. At $298\ \mathrm{K}$, these quantities are related by the equation $E^0_{cell} = \frac{0.059}{n} \log_{10} K_{eq}$, where $n$ represents the number of moles of electrons transferred in the balanced redox equation.

For the reaction $\mathrm{Zn}(\mathrm{s}) + \mathrm{Sn}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{Sn}(\mathrm{s})$, zinc undergoes oxidation and tin undergoes reduction, resulting in the transfer of $n=2$ electrons. Substituting the given equilibrium constant $K_{eq} = 1 \times 10^{20}$ into the equation, the standard cell potential is $E^0_{cell} = \frac{0.059}{2} \log_{10}(10^{20}) = 0.59\ \mathrm{V}$. Because the cell potential is determined by the difference between the reduction potentials of the cathode and the anode, we have $E^0_{cell} = E^0_{\mathrm{Sn}^{2+}/\mathrm{Sn}} - E^0_{\mathrm{Zn}^{2+}/\mathrm{Zn}}$.

Plugging in the known potential for zinc, the expression becomes $0.59\ \mathrm{V} = E^0_{\mathrm{Sn}^{2+}/\mathrm{Sn}} - (-0.76\ \mathrm{V})$, which simplifies to $E^0_{\mathrm{Sn}^{2+}/\mathrm{Sn}} = 0.59 - 0.76 = -0.17\ \mathrm{V}$. The question asks for the standard electrode potential of $\mathrm{Sn} / \mathrm{Sn}^{2+}$, which represents the oxidation potential of tin. This is the negative of the reduction potential, yielding $-(-0.17\ \mathrm{V}) = 0.17\ \mathrm{V}$. Expressed as $17 \times 10^{-2}\ \mathrm{V}$, the required magnitude is 17.

The Nernst equation provides the relationship between the reduction potential of a half-reaction and the concentrations of the species in solution. For the reduction of the dichromate ion in an acidic medium, the balanced half-reaction is:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

The electrode potential $E$ is determined by:

$E = E^\circ - \frac{0.059}{n} \log \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Cr}_2\mathrm{O}_7^{2-}]\$\$[\mathrm{H}^{+}]^{14}}$

With $n=6$, concentrations of 0.01 M for the dichromate ion, 0.1 M for the chromium(III) ion, and $10^{-3}$ M for the hydrogen ion based on the given pH, the reaction quotient $Q$ is computed as:

$Q = \frac{(0.1)^2}{(0.01)(10^{-3})^{14}} = 10^{42}$

Substituting the standard potential of 1.330 V into the Nernst equation leads to:

$E = 1.33 - \left( \frac{0.059}{6} \right) \times 42 = 1.33 - 0.413 = 0.917$

This result represents the potential in volts, which can be expressed as $917 \times 10^{-3}$ V. Therefore, the value of $x$ is 917.

Conductometric titration of a weak acid like benzoic acid with a strong base like $\mathrm{NaOH}$ relies on the changing concentration of ions in the solution as the neutralization reaction occurs, represented by $\mathrm{C_6H_5COOH} + \mathrm{NaOH} \rightarrow \mathrm{C_6H_5COONa} + \mathrm{H_2O}$. As the base is added, the undissociated benzoic acid is converted into the salt $\mathrm{C_6H_5COONa}$, which is a strong electrolyte, causing a gradual increase in ion concentration and conductance. Once the equivalence point is surpassed, the addition of further $\mathrm{NaOH}$ introduces excess $\mathrm{Na}^{+}$ and $\mathrm{OH}^{-}$ ions, and because the hydroxide ion exhibits exceptionally high molar ionic mobility, the conductance rises much more sharply after this stage.

The behavior of molar conductivity as a function of concentration depends fundamentally on the nature of the electrolyte involved. Strong electrolytes, represented here by electrolyte B, follow the Debye-Hückel-Onsager relationship defined by the equation
$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$
where plotting $\Lambda_m$ against $\sqrt{c}$ yields a straight line with an intercept on the y-axis equal to the limiting molar conductivity, $\Lambda_m^0$. Because this relationship is strictly linear for strong electrolytes, the limiting value is easily determined via extrapolation, confirming that the second statement provided is factually correct.

Conversely, weak electrolytes like electrolyte A exhibit a curve as concentration decreases, showing a rapid increase in molar conductivity that never intersects the y-axis in a linear fashion, which makes the determination of $\Lambda_m^0$ through simple extrapolation impossible. This renders the first statement incorrect. Regarding dissociation, strong electrolytes are effectively fully ionized at all concentrations, and their degree of dissociation approaches unity-not zero-at infinite dilution, making the third statement incorrect as well.

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of any electrolyte can be expressed as the sum of the individual ionic contributions, which applies universally to both strong and weak electrolytes. Since the fourth statement accurately reflects this principle, it is correct. With statements (B) and (D) being true and (A) and (C) being false, there are exactly two incorrect statements.

The reduction potential of a copper electrode in a basic solution depends on the concentration of copper ions, which is governed by the solubility product constant of $\mathrm{Cu}(\mathrm{OH})_2$. At a $\text{pH}$ of 14, the hydroxide ion concentration is $1\text{M}$, allowing us to calculate the $\mathrm{Cu}^{2+}$ concentration as $[\mathrm{Cu}^{2+}] = \frac{K_{\text{sp}}}{[\mathrm{OH}^-]^2} = \frac{1 \times 10^{-20}}{1^2} = 10^{-20}\text{M}$. Using the Nernst equation for the $\mathrm{Cu}^{2+} + 2e^{-} \rightarrow \mathrm{Cu}$ half-reaction, E = E^{\circ} - \frac{0.059}{2} \log_{10} \frac{1}{[\mathrm{Cu}^{2+}]\}$, we substitute the standard reduction potential of$0.34\text{V}$and our calculated ion concentration to find$E = 0.34 - \frac{0.059}{2} \log_{10} (10^{20})$. This expression simplifies to$0.34 - 0.59 = -0.25\text{V}$, which matches the form$-25 × 10^{-2}\text{V}$and confirms that$x$ is 25.

Potassium hydroxide ($\mathrm{KOH}$) is a highly reactive compound that readily absorbs carbon dioxide from the atmosphere to form potassium carbonate, as represented by the chemical equation $2\mathrm{KOH} + \mathrm{CO}_2 \rightarrow \mathrm{K}_2\mathrm{CO}_3 + \mathrm{H}_2\mathrm{O}$. This reaction effectively decreases the actual hydroxide ion concentration in the solution, meaning the molarity of the reagent changes over time. Because volumetric analysis relies on precise concentration values to determine unknown quantities, verifying the exact strength of the solution through standardization is necessary before it is used in any titration.

The conductivity of an electrolytic solution represents the sum of the contributions from all individual ions present, calculated as the product of each ion's molar conductivity and its molar concentration. When silver nitrate is added to a saturated silver bromide solution, the common silver ions suppress the dissociation of silver bromide, making the concentration of silver ions and nitrate ions equal to the concentration of the added silver nitrate, while the bromide ion concentration is restricted by the solubility product constant K_{\text{sp}} = \[\text{Ag}^+]$[\text{Br}^-]$. Given$[\text{Ag}^+] = 10^{-5} \text{ M}$and$[\text{NO}3^-] = 10^{-5} \text{ M}$, the concentration of$[\text{Br}^-]$is found using the relation$[\text{Br}^-] = \frac{K{\text{sp}}}{[\text{Ag}^+]$} = \frac{4.9 × 10^{-13}}{10^{-5}} = 4.9 × 10^{-8} \text{ M}$.

Converting these molarities to SI units where $1 \text{ M} = 10^3 \text{ mol m}^{-3}$, the concentration for silver and nitrate ions is $10^{-2} \text{ mol m}^{-3}$ and for bromide ions is $4.9 \times 10^{-5} \text{ mol m}^{-3}$. Applying the conductivity formula $\kappa = \lambda C$ for each ion yields the contributions: $\kappa_{\text{Ag}^+} = (6 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}) \times 10^{-2} \text{ mol m}^{-3} = 6 \times 10^{-5} \text{ S m}^{-1} = 6000 \times 10^{-8} \text{ S m}^{-1}$, $\kappa_{\text{NO}_3^-} = (7 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}) \times 10^{-2} \text{ mol m}^{-3} = 7 \times 10^{-5} \text{ S m}^{-1} = 7000 \times 10^{-8} \text{ S m}^{-1}$, and $\kappa_{\text{Br}^-} = (8 \times 10^{-3} \text{ S m}^2 \text{ mol}^{-1}) \times 4.9 \times 10^{-5} \text{ mol m}^{-3} = 39.2 \times 10^{-8} \text{ S m}^{-1}$. The total conductivity of the solution is the sum of these parts, resulting in $(6000 + 7000 + 39.2) \times 10^{-8} \text{ S m}^{-1}$, which is $13039.2 \times 10^{-8} \text{ S m}^{-1}$.

The relationship between the equilibrium constant of a redox reaction and the standard cell potential is governed by thermodynamic principles linking Gibbs free energy to both variables, specifically through the expressions $\Delta G^{\circ} = -RT \ln K$ and $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$. By combining these equations and substituting the given constant $\frac{2.303 RT}{F} = 0.06$, we derive the formula $\log K = \frac{nE^{\circ}_{\text{cell}}}{0.06}$, which provides a direct method to determine the equilibrium constant from electrochemical data.

To determine the standard cell potential for the target reaction, we treat the reduction of $\mathrm{Pd}^{2+}$ as the cathode process and the reverse of the reduction of $\mathrm{PdCl}_4^{2-}$ as the anode process. Subtracting the standard reduction potential of the anode reaction from that of the cathode reaction yields $E^{\circ}_{\text{cell}} = 0.83 - 0.65 = 0.18$ V. Since the overall stoichiometry involves the transfer of $n = 2$ electrons, we apply these values to our derived formula, resulting in $\log K = \frac{2 \times 0.18}{0.06}$, which simplifies to 6.