Electro chemistry practice · Mixed difficulty

Option A (Violet due to formation of ${\mathrm{I}_2}$) is correct because chlorine will oxidize iodide ions to iodine, which dissolves in $\mathrm{CHC}{\mathrm{l}_3}$ giving it a violet color.

The correct answer is A (7.883 g) because it is calculated using Faraday's law of electrolysis, considering the current efficiency.

The correct answer is B (Copper is liberated at cathode, oxygen at anode) because during electrolysis of $CuSO_4$, $Cu^{2+}$ ions are reduced at the cathode and water is oxidized at the anode.

Option D is correct because the conductivity initially decreases as ions are removed by precipitation, then increases as excess Ba(OH)₂ adds more ions.

Option B is correct. Statement-1 is true because 2 Faradays of electricity will produce 11.2 L of O2 at STP. Statement-2 is also true, but it doesn't explain Statement-1.

Option C (${H}_2, {Br}_2$ and ${NaOH}$) is correct because during the electrolysis of aqueous NaBr, hydrogen gas and hydroxide ions are produced at the cathode, while bromine gas is produced at the anode.

Option C ($4.48{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{d}{\mathrm{m}^3}$) is correct because during electrolysis of water, the volume of hydrogen liberated is twice the volume of oxygen liberated.

Option C (Ions) is correct because electrolytic conduction occurs due to the migration of ions.

Option B is correct. In a hydrogen-oxygen fuel cell, the combustion of hydrogen is used to create a potential difference between the two electrodes, which drives the electric current.

Option D (${H_2}S{O_4}$ is consumed) is correct because sulfuric acid is used up in the reaction to form lead sulfate during the discharge of the battery.

Option B (Decreases by $189 \mathrm{~mV}$) is correct because decreasing the H+ concentration decreases the oxidizing power, as shown by the Nernst equation.

The correct answer is A (Increase in ionic molility ions) because the increase in equivalent conductivity of a strong electrolyte with dilution is primarily due to increased ionic mobility.

Option C (Temperature) is correct because the electrode potential is directly related to temperature through the Nernst equation.

The correct answer is 8 g of oxygen because 4F of charge liberate 32g of oxygen, therefore 1F will liberate 8g of oxygen.

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.

Option A (10 A) is correct because it is calculated using Faraday's laws of electrolysis, relating the mass of aluminum deposited to the current passed.

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 A ($6.24 \times 10^{21}$) is correct because it represents the number of electrons calculated using the formula Q/e, where Q is the total charge and e is the charge of an electron.

Option D (Al) is correct because aluminum is a strong electropositive metal and cannot be obtained by electrolysis of its aqueous salt solution.

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 A (4) is correct because the pH is calculated from the hydrogen ion concentration, which is determined using limiting molar conductivities and the degree of dissociation.

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.