Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

The stability of a metal ion in solution is directly reflected by its standard reduction potential, where a more negative value indicates greater resistance to reduction and thus higher ionic stability. Comparing the provided data, $Al^{+}/Al$ has a potential of $+0.55$ V, which signifies that $Al^{+}$ is easily reduced and therefore unstable. In contrast, $Tl^{+}/Tl$ has a potential of $-0.34$ V, indicating that $Tl^{+}$ is much more resistant to reduction. Consequently, the negative reduction potential of the thallium ion relative to the positive potential of the aluminum ion confirms that $Tl^{+}$ is more stable than $Al^{+}$.

Standard reduction potential is an intensive property and cannot be added directly; therefore, one must use the additive property of Gibbs free energy, $\Delta G^{\circ} = -nFE^{\circ}$. For the target reaction $\mathrm{Fe}^{3+} + 3e^{-} \rightarrow \mathrm{Fe}$, the total Gibbs free energy change is the sum of the free energy changes for the reactions $\mathrm{Fe}^{2+} + 2e^{-} \rightarrow \mathrm{Fe}$ and $\mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+}$, which is expressed by the relationship $n_3 E^{\circ}_3 = n_1 E^{\circ}_1 + n_2 E^{\circ}_2$. Substituting $n_1=2$, $E^{\circ}_1=-0.47\,\text{V}$, $n_2=1$, and $E^{\circ}_2=0.77\,\text{V}$ into the equation $3E^{\circ}_3 = (2)(-0.47\,\text{V}) + (1)(0.77\,\text{V})$, we calculate $3E^{\circ}_3 = -0.94\,\text{V} + 0.77\,\text{V} = -0.17\,\text{V}$. Solving for the standard reduction potential gives $E^{\circ}_3 = \frac{-0.17\,\text{V}}{3} = -0.057\,\text{V}$.

The strength of a reducing agent is determined by its tendency to undergo oxidation, which inversely correlates with the standard reduction potential of the species. Comparing the given values, the $\mathrm{Cr}^{3+}/\mathrm{Cr}$ couple has a standard reduction potential of $-0.74\,\text{V}$, while the potentials for the $\mathrm{MnO}_4^{-}/\mathrm{Mn}^{2+}$ ($1.51\,\text{V}$), $\mathrm{Cl}_2/\mathrm{Cl}^-$ ($1.36\,\text{V}$), and $\mathrm{Cr}_2\mathrm{O}_7^{2-}/\mathrm{Cr}^{3+}$ ($1.33\,\text{V}$) couples are all significantly higher and positive. Because $\mathrm{Cr}$ exhibits the most negative reduction potential, it possesses the greatest ease of oxidation and acts as the strongest reducing agent among the choices provided.

The Nernst equation for a cell reaction at 298 K is expressed as $E_{\mathrm{cell}} = E^{\ominus}_{\mathrm{cell}} - \frac{0.059}{n} \log Q$. For the cell reaction $\mathrm{M} + 3\mathrm{Ag}^{+} \rightarrow \mathrm{M}^{3+} + 3\mathrm{Ag}$, the number of transferred electrons is $n = 3$, and the reaction quotient is calculated as Q = \frac{[\mathrm{M}^{3+}]\}{[\mathrm{Ag}^{+}]^3} = \frac{10^{-3}}{(10^{-2})^3} = 10^3$.

Substituting these values into the Nernst equation gives $0.421 = E^{\ominus}_{\mathrm{cell}} - \frac{0.059}{3} \log(10^3)$, which simplifies to $0.421 = E^{\ominus}_{\mathrm{cell}} - 0.059(1)$, leading to $E^{\ominus}_{\mathrm{cell}} = 0.421 + 0.059 = 0.48\,\mathrm{V}$. Utilizing the relationship $E^{\ominus}_{\mathrm{cell}} = E^{\ominus}_{\mathrm{Ag}^{+}/\mathrm{Ag}} - E^{\ominus}_{\mathrm{M}^{3+}/\mathrm{M}}$, the equation becomes $0.48\,\mathrm{V} = 0.80\,\mathrm{V} - E^{\ominus}_{\mathrm{M}^{3+}/\mathrm{M}}$, which determines the standard electrode potential as $E^{\ominus}_{\mathrm{M}^{3+}/\mathrm{M}} = 0.80\,\mathrm{V} - 0.48\,\mathrm{V} = 0.32\,\mathrm{V}$.

A displacement reaction only proceeds spontaneously if the added metal possesses a more negative standard reduction potential than the metal cation already present in the solution. In this case, the standard reduction potential of zinc is defined as $E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}} = -0.76 \text{ V}$, while that of copper is $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34 \text{ V}$. Because the reduction potential of copper is significantly higher than that of zinc, copper is unable to displace the zinc ions from the solution, meaning the metal block remains inert and no chemical change takes place.

The electrolysis of aqueous potassium succinate involves the Kolbe reaction at the anode and the reduction of water at the cathode. At the anode, the oxidation of the succinate ion proceeds as follows:
$CH_2(COO^-)_2 \rightarrow C_2H_4 + 2CO_2 + 2e^-$
This reaction produces 3 moles of gas per 2 Faradays of charge. Simultaneously, the cathode reduction of water occurs:
$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$
This reaction yields 1 mole of hydrogen gas for every 2 Faradays of electricity passed. Consequently, the passage of 2 Faradays of electricity generates a total of $1 + 2 + 1 = 4$ moles of gas. By proportional calculation, 0.2 Faraday of electricity generates $(0.2 / 2) \times 4 = 0.4$ moles of total gas. Applying the ideal gas molar volume of 22.4 L at STP, the total volume of gas collected is $0.4 \times 22.4 = 8.96$ L.

Iron corrosion is an electrochemical process that requires the simultaneous presence of moisture and oxygen to occur at the metal surface. By creating an impermeable barrier, such as through painting or specialized surface coating, one effectively isolates the iron from these environmental reactants, which stops the oxidation process and minimizes corrosion. While methods like sacrificial protection exist, they require connecting iron to a metal with a lower reduction potential to ensure the other metal corrodes first, making the claim regarding a metal with a higher reduction potential incorrect.

Galvanization is a metallurgical process used to prevent corrosion by applying a protective coating of zinc onto iron or steel surfaces. Zinc acts as a sacrificial anode because it is more chemically reactive than iron, meaning it preferentially oxidizes and provides cathodic protection to the underlying metal even if the coating becomes scratched or damaged.

Electrolysis relies on the stoichiometry of electron transfer to determine the amount of metal deposited, as defined by the reduction reaction occurring at the cathode:

$\mathrm{Cu}^{2+} + 2e^{-} \rightarrow \mathrm{Cu}$

This reaction dictates that two moles of electrons, equivalent to 2 Faraday of electricity, are strictly required to produce one mole of solid copper. Given that the atomic mass of copper is 63.5 g/mol, passing 2 Faraday of charge results in the deposition of exactly 63.5 g of copper.

The behavior of strong electrolytes like NaCl is governed by the Debye-Hückel-Onsager equation, which quantifies how interionic attractions hinder the movement of ions in a solution. As the concentration of the electrolyte increases, these retarding electrostatic forces grow in proportion to the square root of the concentration, leading to a reduction in the equivalent conductance relative to the value at infinite dilution. This relationship is mathematically defined as $\lambda_c = \lambda_\infty - B\sqrt{C}$, illustrating that conductance drops as a function of the square root of concentration due to these ionic interactions.

The cell constant, defined as the ratio of the distance between electrodes to their cross-sectional area, is an intrinsic property of a conductivity cell and remains invariant regardless of the electrolyte solution it contains. We determine this geometric factor by utilizing the data from the first solution through the expression $R = \frac{1}{\kappa} \left(\frac{l}{A}\right)$, where the resistance of $50\,\Omega$ and specific conductance of $1.4\,\mathrm{S}\,\mathrm{m}^{-1}$ yield a cell constant of $70\,\mathrm{m}^{-1}$.

Applying this cell constant to the 0.5 M solution, which has a measured resistance of $280\,\Omega$, allows us to find the new specific conductance using $\kappa = \frac{1}{R} \times \frac{l}{A}$. Substituting these values results in $\kappa = \frac{70}{280}$, or $0.25\,\mathrm{S}\,\mathrm{m}^{-1}$. Molar conductivity is found by dividing this specific conductance by the concentration expressed in SI units, where the concentration of 0.5 M corresponds to $500\,\mathrm{mol}\,\mathrm{m}^{-3}$. Dividing $0.25\,\mathrm{S}\,\mathrm{m}^{-1}$ by $500\,\mathrm{mol}\,\mathrm{m}^{-3}$ results in $5 \times 10^{-4}\,\mathrm{S}\,\mathrm{m}^2\,\mathrm{mol}^{-1}$.

To find the standard cell potential for the disproportionation reaction $3\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn} + 2\mathrm{Mn}^{3+}$, we must decompose it into its constituent oxidation and reduction half-reactions. The reduction half-reaction involves $\mathrm{Mn}^{2+}$ gaining two electrons to form $\mathrm{Mn}$, which has an $E^{0}$ of $-1.18\,\mathrm{V}$. The oxidation half-reaction involves the conversion of two $\mathrm{Mn}^{2+}$ ions into two $\mathrm{Mn}^{3+}$ ions, releasing two electrons; this is the reverse of the provided reduction half-reaction for $\mathrm{Mn}^{3+}$, requiring us to invert the sign of its given potential from $+1.51\,\mathrm{V}$ to $-1.51\,\mathrm{V}$. Adding these two values, $-1.18\,\mathrm{V} + (-1.51\,\mathrm{V})$, yields a total $E^{0}$ of $-2.69\,\mathrm{V}$. Because the calculated cell potential is negative, the reaction is thermodynamically non-spontaneous and will not occur under standard conditions.

The strength of an oxidizing agent is directly determined by its Standard Reduction Potential (SRP), where a higher positive SRP value indicates a greater tendency for a substance to gain electrons and undergo reduction. Comparing the given values of $-0.74\;\mathrm{V}$ for $\mathrm{Cr}^{3+}/\mathrm{Cr}$, $1.33\;\mathrm{V}$ for $\mathrm{Cr}_2\mathrm{O}_7^{2-}/\mathrm{Cr}^{3+}$, $1.36\;\mathrm{V}$ for $\mathrm{Cl}_2/\mathrm{Cl}^{-}$, and $1.51\;\mathrm{V}$ for $\mathrm{MnO}_4^{-}/\mathrm{Mn}^{2+}$, it is clear that the permanganate ion possesses the highest positive potential. Consequently, because $1.51\;\mathrm{V}$ represents the strongest drive to accept electrons, $\mathrm{MnO}_4^{-}$ serves as the most powerful oxidizing agent in this set.

For a chemical reaction to proceed spontaneously, the overall Gibbs free energy change, $\Delta G = -nFE^\circ$, must be negative, requiring a positive standard cell potential $E^\circ_{\text{cell}}$. In the reaction $X+Y^{2+} \rightarrow X^{2+}+Y$, the metal $X$ serves as the anode undergoing oxidation, while $Y^{2+}$ acts as the cathode undergoing reduction, with the cell potential determined by the sum of the reduction potential of the cathode and the oxidation potential of the anode. Given the standard reduction potentials of -0.76 V for $\mathrm{Zn}$, -0.44 V for $\mathrm{Fe}$, and -0.23 V for $\mathrm{Ni}$, testing the combinations reveals which one produces a positive cell potential. Using $X = \mathrm{Zn}$ and $Y = \mathrm{Ni}$ provides an oxidation potential of +0.76 V for zinc and a reduction potential of -0.23 V for nickel, resulting in a total cell potential of +0.53 V, which confirms this combination results in a spontaneous reaction.

The reduction potential of a hydrogen electrode is governed by the Nernst equation,

$E = E^{\circ} - \frac{0.059}{1} \log \frac{P_{\mathrm{H}_2}^{1/2}}{[\mathrm{H}^{+}]\$}$

where the standard reduction potential $E^{\circ}$ is 0 V. For the electrode potential to become negative, the value subtracted from the standard potential must be positive, which occurs when the ratio of the partial pressure of hydrogen gas raised to the power of one-half divided by the hydrogen ion concentration is greater than 1. When the partial pressure of hydrogen is 2 atm and the hydrogen ion concentration is 1 M, the expression

$\frac{P_{\mathrm{H}_2}^{1/2}}{[\mathrm{H}^{+}]\$} = \frac{2^{1/2}}{1}$

yields a value of approximately 1.414. Since the logarithm of any value greater than 1 is positive, this condition results in a negative reduction potential.

The relationship between the Gibbs free energy change and the potential difference is defined by the formula $\Delta G = -nFE$, which allows us to determine the external voltage required for a non-spontaneous reaction. To find the number of electrons transferred, denoted by $n$, consider that the production of 4/3 moles of aluminum from aluminum oxide, where aluminum transitions from a +3 oxidation state to 0, means that $n = 4$. Given a Gibbs energy of 966 kJ mol$^{-1}$ and the Faraday constant of approximately 96,500 C mol$^{-1}$, we use the rearranged expression $E = \Delta G / (nF)$. Substituting the values into the formula gives $E = \frac{966 \times 10^3}{4 \times 96,500}$, leading to a required potential difference of 2.5 V to facilitate the decomposition.

The standard electrode potential $E^{\circ}$ of a metal quantifies its tendency to undergo reduction, where more negative values indicate a greater tendency for the metal to act as a reducing agent. Given the potential values $E_{\mathrm{Mn}^{2+}/\mathrm{Mn}}^{\circ} = -1.18\ \text{V}$, $E_{\mathrm{Cr}^{2+}/\mathrm{Cr}}^{\circ} = -0.9\ \text{V}$, $E_{\mathrm{Fe}^{2+}/\mathrm{Fe}}^{\circ} = -0.44\ \text{V}$, and $E_{\mathrm{Co}^{2+}/\mathrm{Co}}^{\circ} = -0.28\ \text{V}$, the magnitude of the negative electrode potential reflects the intrinsic reactivity of the metal. By ordering these values from the most negative to the least negative, we obtain the sequence $\mathrm{Mn} > \mathrm{Cr} > \mathrm{Fe} > \mathrm{Co}$ as they relate to their decreasing negative electrode potential values.

The efficiency of a fuel cell is defined as the ratio of the maximum useful electrical work, given by the change in Gibbs free energy ($\Delta G_r$), to the total energy released during combustion, represented by the change in enthalpy ($\Delta H$). To determine this efficiency, we first calculate the standard Gibbs free energy of the reaction using the provided formation values for each component:

$\Delta G_r = [\Delta G_f(\mathrm{CO}_2) + 2\Delta G_f(\mathrm{H}_2\mathrm{O})] - [\Delta G_f(\mathrm{CH}_3\mathrm{OH}) + \frac{3}{2}\Delta G_f(\mathrm{O}_2)]$

Substituting the given values into this expression results in $\Delta G_r = [-394.4 + 2(-237.2)] - [-166.2 + 0]$, which simplifies to -702.6 kJ mol⁻¹. Finally, we calculate the efficiency by comparing this energy output to the provided combustion enthalpy of -726 kJ mol⁻¹ using the ratio $\frac{702.6}{726} \times 100$, which yields 97%.

Electrode potential is an intensive property and cannot be combined directly, whereas the standard Gibbs free energy change, $\Delta G^{\circ}$, is an extensive property that is additive. To determine the standard electrode potential for the target reaction, we express the total free energy change for the transformation as a combination of the free energy changes for the known half-reactions using the relation $\Delta G^{\circ} = -nFE^{\circ}$.

The target reaction, $\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}$, is obtained by subtracting the half-reaction $\mathrm{Fe}^{2+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Fe}$ from $\mathrm{Fe}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Fe}$. Consequently, the free energy change for the target reaction is $\Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2}$, where $n$ values for the first, second, and target reactions are 3, 2, and 1, respectively. Substituting the given values into the expression $-n_{3}FE^{\circ}_{3} = -n_{1}FE^{\circ}_{1} - (-n_{2}FE^{\circ}_{2})$ gives $-1 \cdot F \cdot E^{\circ}_{3} = -(3 \cdot F \cdot -0.036) - (-2 \cdot F \cdot -0.439)$. This simplifies to $-F \cdot E^{\circ}_{3} = 0.108F - 0.878F = -0.770F$, which confirms that the standard electrode potential is $0.770\;\mathrm{V}$.

To determine the potential of an electrochemical cell under non-standard conditions, we apply the Nernst equation, which adjusts the standard cell potential based on the reaction quotient and the number of electrons transferred. Identifying the half-reactions reveals that oxidation at the chromium anode involves three electrons, while reduction at the iron cathode involves two. Balancing these necessitates a common denominator, leading to an overall transfer of $n = 6$ electrons for the reaction 2Cr + 3Fe$^{2+}$ $\rightarrow$ 2Cr$^{3+}$ + 3Fe.

The standard cell potential is determined by the difference between the cathode and anode potentials, resulting in -0.42 V - (-0.72 V) = 0.30 V. We then use the Nernst equation:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3}$
Substituting the concentrations of 0.1 M and 0.01 M into the reaction quotient leads to $Q = \frac{(0.1)^2}{(0.01)^3} = 10^4$. Subtracting the correction term from the standard potential, $0.30 - \frac{0.059}{6} \times 4$ gives $0.30 - 0.0393$, which results in a final cell potential of 0.26 V.