Electro chemistry practice · Mixed difficulty

Option A is correct because when an opposite potential greater than 1.1V is applied, the electron flow reverses, and Zn is deposited at the anode and Cu at the cathode, which is the reverse of the spontaneous reaction.

Option C ($\mathrm{I}_2$ will be reduced to $\mathrm{I}^{-}$) is correct because the E° for Fe3+/Fe2+ is greater than E° for I2/2I-, so Fe3+ will oxidize I- to I2, and iodine will be reduced to iodide.

Option B is correct. Electrolyte Y is a strong electrolyte because its molar conductivity increases less upon dilution compared to X, indicating it's already highly dissociated.

Option C (D) is correct because the element with the most negative reduction potential will displace all others from their solutions.

Option A is correct. In very dilute NaCl solution, electrolysis produces H+ and OH- ions, changing the pH near the electrodes.

Option B (Is oxidised to $\mathrm{C}{\mathrm{u}^{2 + }}$) is correct because copper loses electrons and forms copper(II) ions, which give the solution a blue color.

Option A (96500) is correct because the deposition of one mole of a monovalent metal like silver requires one Faraday of charge, which is approximately 96500 Coulombs.

Option D is correct because the cell constant is the ratio of the distance between the electrodes to the area of the electrode, which is 2.5 cm / 5 cm² = 0.5 cm⁻¹.

Option D (1 g equivalent of Cu) is correct because one Faraday of electricity deposits one gram equivalent of a substance.

Option B is correct because in a Daniel cell, zinc is oxidized at the anode and copper ions are reduced at the cathode.

Option D (All of these) is correct because the speed of ions depends on the nature of the ion, the potential gradient, and the dilution of the solution.

Option B (Oxygen gas) is correct because during the electrolysis of dilute sulfuric acid, water is oxidized at the anode to produce oxygen gas.

Option A (Metal - Insoluble Salt - Anion electrodes) is correct because a calomel electrode consists of mercury (metal), mercurous chloride (insoluble salt), and chloride ions (anion).

Option K is correct. The question is incomplete and lacks context. Assuming the question is related to conductivity, 'K' likely refers to conductivity which decreases on dilution.

Option A ($6.24 \times 10^{21}$) is correct because it accurately calculates the number of electrons that flow when a current of 5A is passed through a solution for 200s.

Option D (${AgNO}_3$) is correct because copper will displace silver from the $AgNO_3$ solution, forming $Cu^{2+}$ ions, which give the solution a blue color.

Option C (${10^{10}}$) is correct because it is calculated using the relationship between standard EMF and the equilibrium constant, $K = 10^{\frac{{nE^0}}{{0.0591}}}$.

Option D (Al) is correct because aluminum has a high reduction potential and water is reduced in preference to aluminum ions during electrolysis of aqueous solutions.

Option B (1, 3 and 2) is correct because the valencies are determined by comparing the ratios of deposited masses to atomic weights, based on Faraday's laws.

Option B is correct because Siemens per meter (S/m) or ohm^-1 cm^-1 is the standard unit for conductivity.

Option C (Remains same) is correct because in an electrolytic cell with silver electrodes and silver nitrate electrolyte, silver ions are deposited at the cathode and silver from the anode dissolves into the solution, maintaining a constant concentration.

Option C ($Mn\mathrm{ } > Cr\mathrm{ } > \mathrm{ }V\mathrm{ }:\,{E^0}\left( {{{M^{3 + }}/{M^{2 + }}}} \right)$) is the incorrect order because the actual order is Cr > Mn > V for the given reduction potential.

Option A ($\mathrm{KCl}$) is correct because strong electrolytes like KCl show a linear relationship between molar conductivity and the square root of concentration.

Option D is correct. The assertion is incorrect because copper sulphate cannot be stored in a zinc pot due to a redox reaction. The reason is correct because zinc is more reactive than copper.

Option B ($P^H$ of the solution decreases) is correct because during the electrolysis of $AgNO_3$, $H^+$ ions are produced at the anode, leading to a decrease in pH.

Option D (Galvanization) is correct because it is the process of coating iron with zinc to prevent corrosion.

Option A (A and B) is correct because a non-spontaneous cell reaction occurs when the cell potential is negative, which happens when a lower oxidation potential is paired with a higher oxidation potential as the anode and cathode, respectively. In this case, A (-0.03V) and B (+0.108V) result in a negative cell potential if A is oxidized and B is reduced.

Option C is correct. Statement I is incorrect because the oxidation of 1 mole of H2O to O2 requires 4F of electricity. Statement II is correct because to get 40g Ca from molten CaCl2 requires 2F.

Option B is correct because a reaction is possible if the standard oxidation potential of the element being oxidized (Al) is greater than that of the element being reduced (Fe).

Option D ($\mathrm{C}{\mathrm{l}^ - },\mathrm{F}{\mathrm{e}^{2 + }},{\mathrm{H}_2},\mathrm{Zn}$) is correct because reducing strength increases as the standard reduction potential decreases.