Electro chemistry practice · Mixed difficulty

Option C (Temperature) is correct because the electrode potential is directly affected by temperature, as described by the Nernst equation.

Option A (0.3175 g) is correct because the mass of copper deposited is calculated using Faraday's laws, relating current, time, and the electrochemical equivalent of copper.

Option C (x-0.01M,y-1M) is correct because a lower concentration of $Zn^{2+}$ and a higher concentration of $Cl^-$ will maximize the EMF.

Option B ($\wedge _m^\mathrm{o}\left( {HN{O_3}} \right) + \wedge _m^\mathrm{o}\left( {NaN{O_3}} \right) - \wedge _m^\mathrm{o}\left( {NaOH} \right)$) is correct because it does not correctly represent the molar conductivity of water based on Kohlrausch's Law.

Option C (Cu) is correct because Cu is less reactive than H and cannot liberate H2 from HCl, but it is more reactive than Ag and can displace it from AgNO3 solution.

Option A is correct because the strongest reducing agent is the one with the most negative standard reduction potential. Among the given options, Li has the most negative value (-3.05 V), indicating it has the highest tendency to be oxidized and thus is the strongest reducing agent.

Option A (1) is correct because the current can be calculated using Faraday's laws of electrolysis, relating the volume of hydrogen gas produced to the charge passed.

Option A is correct because it represents the overall cell reaction based on the given cell notation and the standard reduction potentials.

Option D (Al) is correct because aluminum is more reactive than zinc and can displace it from its salt solution.

Option C ($\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + \mathrm{Fe}{\left( {\mathrm{OH}} \right)_2}$) is correct because rust is a mixture of iron oxide and iron hydroxide.

Option C (Hydrochloric acid) is correct because it is a strong electrolyte that dissociates completely into ions in aqueous solution, allowing for good electrical conductivity.

Option B (2 F) is correct because the reduction of one mole of $CuCl_2$ to copper metal requires 2 moles of electrons, which is equivalent to 2 Faradays of electricity.

Option B (9.65 A) is correct because it correctly calculates the current required to produce the given volume of hydrogen using Faraday's laws of electrolysis.

Option D ((i) and (iv)) is correct because Leclanche cell and Mercury cell are primary cells, meaning they are non-rechargeable.

Option A (+0.46 V) is correct because it is the calculated EMF of the cell using the given standard reduction potentials.

Option B is correct because the electrode potential of a glass electrode is primarily dependent on the concentration of hydrogen ions (pH) in the solution it is immersed in. This is due to the ion-exchange process between the glass membrane and the hydrogen ions in the solution.

Option A is correct. The standard EMF of the cell is calculated by adding the standard oxidation potential of the anode (Zn) and the standard reduction potential of the cathode (Ag).

Option A (54 g) is correct because the mass of silver displaced is calculated using the equivalent weights and volumes of hydrogen.

Option D (Rate constant and standard cell potential) is correct because there is no direct mathematical relationship between the rate constant of a reaction and the standard cell potential.

Option A ($E_{\mathrm{oxid}\mathrm{.}}^ \circ$ will increase by a factor of 0.65 from $E_{\mathrm{oxid}\mathrm{.}}^ \circ$) is correct because the oxidation potential of glucose increases as the pH increases due to the complexation of $Ag^+$ with ammonia.

Option D (In none of the above cases) is correct because Faraday's laws of electrolysis are fundamental and do not fail under the conditions listed (temperature increase, inert electrodes, or a mixture of electrolytes).

Option B (15.9 g) is correct because the mass of copper liberated is calculated using the ratio of equivalent weights and the mass of hydrogen liberated.

Option D ($\mathrm{Pt}\left| {{\mathrm{H}_2}\left( \mathrm{g} \right)|{\rm{HCl {\kern 1pt} {\kern 1pt}solution}}} \right|\left| {\mathrm{AgCl}\left( \mathrm{s} \right)} \right|\mathrm{Ag}$) is correct because it correctly represents the cell notation for the given redox reaction.

Option D (${E^ \circ } = - \frac{{RT}}{F}\mathrm{ln}{K_w}$) is incorrect. The correct answer is A because it relates the standard reduction potential to the equilibrium constant $K_w$ for water dissociation.

Option B is correct because the pH is calculated using the degree of dissociation (α) and the concentration of the acid, resulting in a pH of approximately 1.3.

Option A (0.0836 litre) is correct because it represents the volume of hydrogen gas produced at NTP, calculated using Faraday's laws of electrolysis.

Option C (${\Omega ^{ - 1}}{\kern 1pt}{\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt}\mathrm{equi}{\mathrm{v}^{ - 1}}$) is correct because it represents the units of conductance (inverse of resistance) multiplied by area and divided by equivalents.

Option A is correct. After consumption of 4.825 coulombs of charge, the pH of the solution is 9.

Option B ($5.6{\kern 1pt} {\kern 1pt} \mathrm{d}{\kern 1pt} {\mathrm{m}^3}$) is correct because 1 F of electricity passed through acidulated water evolves 0.25 moles of $O_2$, which occupies 5.6 L (or $5.6{\kern 1pt} {\kern 1pt} \mathrm{d}{\kern 1pt} {\mathrm{m}^3}$) at STP.

Option A (Only gold) is correct because gold has a higher reduction potential than silver, meaning silver will deposit first, and the applied voltage is sufficient to deposit gold.