Electro chemistry practice · Medium

The correct answer is 1/2 because at E=0, the ratio of cation concentrations is determined by the Nernst equation and the standard cell potential.

Option A is correct because galvanized iron does not rust due to the zinc coating, and zinc has a more negative electrode potential than iron, making it a sacrificial anode that corrodes instead of the iron.

Option C (18.7 s) is correct because it is calculated using Faraday's law of electrolysis, relating the mass of silver deposited to the current, time, and electrochemical equivalent.

Option C is correct because the conductivity is calculated using the cell constant and the resistance of the solution.

Option B is correct. Electrolysis of aqueous CuSO4 using graphite electrodes results in the formation of H2SO4 in the solution as Cu2+ ions are reduced to Cu at the cathode and water is oxidized to O2 and H+ at the anode.

The correct answer is D (All of the above) because the positive E° indicates a spontaneous reaction, Fe acts as the anode, and Fe would directly react with $Fe^{3+}$ if used as the cathode.

Option D ($8.6 \times 10^5$) is correct because it accurately calculates the total charge of the nitride ions based on the number of moles and Faraday's constant.

Option D (The nature of the electrode used) is correct because the electrolytic conductance of a solution does not depend on the nature of the electrode used.

Option C (2.5 moles) is correct because the number of moles of MnO4- reduced is calculated based on the number of electrons transferred in the reduction reaction.

Option A (Only gold) is correct because gold has a higher standard reduction potential than silver, so it will be deposited first.

Option D (300) is correct because charge is the product of current and time, and 5 amperes for 1 minute equals 300 coulombs.

The correct answer is that A+2 will be reduced by D because the cell potential for the reaction is negative, indicating a non-spontaneous reaction.

Option B ($-0.177 {~V}$) is correct because it's calculated using the Nernst equation with the given pH and hydrogen pressure.

The correct answer is C (1.72 × 10⁵⁴) because it is calculated using the Nernst equation and the relationship between standard cell potential and equilibrium constant.

Option B (87 $S{\kern 1pt} {\kern 1pt} c{m^2}{\kern 1pt} mo{l^{( - 1)}}$) is correct because it is calculated using Kohlrausch's Law of independent migration of ions.

Option C (AgCl is precipitated) is correct because Ag+ ions from AgNO3 react with Cl- ions from KCl to form insoluble AgCl, disrupting the salt bridge function.

The correct answer is D (${H}_2 {~S}_2 {O}_8$) because during the electrolysis of concentrated ${H}_2 {SO}_4$, peroxydisulfuric acid (${H}_2 {~S}_2 {O}_8$) is formed at the anode.

Option A is correct because molar conductance decreases with increasing concentration due to decreased ionization, and dilution increases the degree of ionization for weak electrolytes.

Option C ($X = \mathrm{Zn}, Y = \mathrm{Ni}$) is correct because the cell potential for this combination is positive, indicating a spontaneous reaction.

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.

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