Electro chemistry practice · Medium

Option C ($\Delta G = - nFE$) is correct because the relationship between Gibbs free energy change and EMF of a reversible electrochemical cell is given by the equation $\Delta G = -nFE$.

Option A (125) is correct because the time required is calculated using Faraday's laws of electrolysis, considering the volume, density, and current.

Option D (${Al}$) is the correct answer because Aluminum has the most negative Standard Reduction Potential (SRP) value among the given options, making it the strongest reducing agent.

Option A is correct because pure water is a weak electrolyte, and adding acid or alkali increases its ionization, making it a better conductor of electricity.

Option B (${KNO}_3$) is correct because in the electrolysis of aqueous ${KNO}_3$, neither ${K}^+$ nor ${NO_3}^-$ participate in the electrode reaction, so their concentrations remain constant.

The correct answer is $\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{c} \pi_{\mathrm{c}}^2}{\pi^{\infty}(\pi^{\infty}-\pi_{\mathrm{c}})}$. This formula relates the equilibrium constant to the concentration and molar conductivity at a given concentration and at infinite dilution.

Option B (Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion) is correct because while both statements are true, Kohlrausch's law doesn't directly explain *why* the molar conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally.

Option C ($400 \mathrm{~ohm}^{-1} \mathrm{~cm}^2 \mathrm{~eq}^{-1}$) is correct because it is calculated using the specific conductance and the normality of the solution.

Option C ($\mathrm{C}_1=0.15 \mathrm{M}, \mathrm{C}_2=0.25 \mathrm{M}$) is correct because for a concentration cell, $\Delta G < 0$ when $\mathrm{C}_2 > \mathrm{C}_1$.

Option C (Mercury) is correct because mercury is below hydrogen in the electrochemical series and does not react with water or steam to produce hydrogen gas.

Option B ($\mathrm{H}_2$ and $\mathrm{O}_2$) is correct because during the electrolysis of dilute $\mathrm{H}_2\mathrm{SO}_4$, water is electrolyzed to produce hydrogen gas at the cathode and oxygen gas at the anode.

Option B (Nernst's equation) is correct because the Nernst equation relates the electrode potential (E) to the standard electrode potential (E°) and the concentrations of the ions in solution.

Option C ($5.6 \mathrm{~mL}$) is correct because the volume of oxygen evolved is calculated using Faraday's laws of electrolysis, relating the mass of silver deposited to the volume of oxygen produced.

Option A (Decreases EMF of the cell) is correct because according to the Nernst equation, increasing the concentration of product ($A^{2+}$) decreases the cell potential.

Option B (Voltmeter) is correct because a voltmeter is used to measure the standard electrode potential.

Option B ($200.8$) is correct because it is calculated by summing the limiting molar conductivities of the constituent ions.

Option C is correct because, according to the Nernst equation, the cell potential ($E_{\text{cell}}$) increases when the concentration of $\mathrm{Zn}^{2+}$ ions decreases.

Option B ($0.75 \mathrm{~M}$) is correct because 0.25 moles of $Cu^{2+}$ are deposited, leaving 0.75 moles in the 1L solution.

Option D is correct because the relationship between equivalent conductivity and molar conductivity is given by $\Lambda_{eq} = \Lambda_m / n$, where n is the valence factor. For $Fe_2(SO_4)_3$, n = 6.

Option C ($20 \mathrm{~F}$) is correct because the total charge required is calculated based on the change in oxidation state of Mn during the reaction.

Option D (Zinc plating) is correct because zinc has a higher oxidation potential than iron, providing sacrificial protection against corrosion.

Option C (6:3:2) is correct because, according to Faraday's second law, the gram equivalents are equal, and the molar ratio is found by dividing by the n-factor.

Option A (0.00133 M) is correct because it is the calculated hydrogen ion concentration based on the given data and the assumption of a weak acid.

Option A (350) is correct because it is the sum of the ionic conductances of H+ and CH3COO- at infinite dilution.

Option A (0.112 L) is correct because it correctly calculates the volume of oxygen formed at NTP using Faraday's laws of electrolysis.

Option A (1930) is correct because it is calculated using Faraday's law of electrolysis, relating the mass of copper deposited to the amount of electricity consumed.

Option D (Assertion is wrong, but reason is correct statement) is correct because a more negative electrode potential indicates a greater tendency to undergo oxidation, thus a stronger reducing agent, not oxidizing agent.

Option C is the correct representation of the Nernst equation for single electrode potential, relating it to the standard electrode potential and the activity of the metal ion.

Option B (1.29) is correct because the cell constant is calculated by multiplying the specific conductivity by the resistance.

Option C (Ag) is correct because silver (Ag) has the highest electrical conductivity at room temperature compared to the other metals listed.