Electro chemistry practice · Mixed difficulty

Option D ($A{g_2}S + A{g_2}O$) is correct because the tarnishing of silver is primarily due to the formation of silver sulfide ($A{g_2}S$) and silver oxide ($A{g_2}O$) on the surface.

The correct answer is 1 because the current passed can be calculated using Faraday's laws of electrolysis, relating the amount of gas produced to the charge passed.

Option B (0.32 V) is correct because it is calculated using the Nernst equation, considering the standard reduction potential of Ag+/Ag and the Ksp of AgCl.

Option A (193) is correct because it is the calculated time required to deposit the specified amount of silver using Faraday's laws of electrolysis.

Option B ($\Lambda = \frac{{2 \times 1000}}{{x \times 0.1}}$) is correct because it uses the correct formula for molar conductance considering the normality of H2SO4.

Option A (Both Assertion and Reason are true and the Reason is correct explanation of Assertion) is correct because conductivity decreases upon dilution due to the reduction in the number of ions per unit volume.

Option C (-72375 J) is correct because it is the Gibbs free energy calculated using the Nernst equation and the given standard electrode potentials and concentrations.

Option A ($4.76 \times {10^{ - 3}}\mathrm{mho}$) is correct because conductance is the reciprocal of resistance.

Option B (+1.09) is correct because it is the standard cell potential calculated using the standard free energy change and the number of electrons transferred.

Option A (Both S-I & S-II are true) is correct because powdered catalysts have a larger surface area, making them more effective.

The correct answer is B (${P_1} < {P_2}$) because for the cell reaction to be spontaneous, the cell potential must be positive, which occurs when ${P_2} > {P_1}$.

Option A ($1.08\,g$) is correct because the amount of Ag deposited is calculated using Faraday's law of electrolysis.

Option C (5) is correct because the current passed is calculated using Faraday's laws of electrolysis, relating the amount of hydrogen gas collected to the charge passed through the solution.

Option C (3) is correct because the pH is calculated using the Nernst equation and the given cell emf and standard electrode potential.

Option C (Not change) is correct because after passing the current, the ratio of concentrations remains the same, resulting in no change in EMF.

Option B (11.5 kJ) is correct because the energy expended is calculated by multiplying the current, time, and potential difference.

Option D (0.02 S cm² mol⁻¹) is correct because it is calculated using the formula for molar conductivity, considering the cell constant and resistance.

Option B ($K=\frac{L}{A} \times \frac{1}{R}$) is correct because specific conductivity (K) is defined as the product of the cell constant (L/A) and the conductance (1/R).

Option A is correct because the reaction 2CuSO4(aq) + 2Ag(s) → 2Cu(s) + Ag2SO4(aq) cannot occur since silver (Ag) is less reactive than copper (Cu) based on their standard electrode potentials.

Option D ($Mg$) is correct because Magnesium is used as a deoxidizer in metallurgy due to its high affinity for oxygen.

Option D ($\mathrm{oh}{\mathrm{m}^{ - 1}}{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} {{\left( {\mathrm{g} - \mathrm{equivalent}} \right)}^{ - 1}}$) is correct because it represents the correct units for equivalent conductivity.

Option A ($\mathrm{V}^{2+}$ and $\mathrm{Cr}^{2+}$) is correct because metals with more negative standard reduction potentials than that of $\mathrm{H}^+/\mathrm{H}_2$ (0 V) can liberate $\mathrm{H}_2$ from dilute acid.

Option A (0.2606 V) is correct because it is calculated using the Nernst equation, considering the standard electrode potentials and the concentrations of the ions.

Option B is correct. The assertion is true because different amounts of charge are required to deposit 1 mole of different metals. The reason is true, but it doesn't explain the assertion; the difference is due to the different charges on the ions.

Option A (0.42 V) is incorrect. The correct answer is D (1.1 V) because the emf of the cell is calculated by adding the standard reduction potential of the cathode and the standard oxidation potential of the anode.

Option C ($7 \mathrm{~g}$) is correct because it is the mass of nickel deposited, calculated using Faraday's laws of electrolysis.

The correct answer is A (Both assertion and reason are correct statements, and reason is the correct explanation of the assertion) because the overpotential of water oxidation favors the production of hydrogen at the cathode instead of oxygen at the anode during the electrolysis of aqueous NaCl.

Option B ($1.101 \mathrm{~V}$) is correct because the EMF of the Daniel cell is the difference between the reduction potentials of the copper and zinc electrodes.

Option A (226) is correct because the cell potential is calculated using the Nernst equation with the given standard electrode potentials and concentrations.

Option A ($2.4 \times 10^{-3} \mathrm{~ohm}^{-1}$) is correct because conductance is the reciprocal of resistance.