Electro chemistry practice · Easy

Option B (Iron) is correct because iron is known to corrode readily in moist air, forming rust, which is a hydrated iron oxide.

Option C (Zinc) is correct because galvanized iron sheets are coated with zinc to prevent rusting.

Option B (63.5 g) is correct because 2 Faradays of electricity will deposit 1 mole of Cu, which is equal to its atomic weight.

Option A (-0.236 V) is correct because it is calculated using the Nernst equation for the given hydrogen electrode conditions.

Option A (A and B) is correct because for a non-spontaneous cell reaction, the cell potential must be negative. This occurs when the electrode with the lower oxidation potential acts as the anode, which is the case with A and B.

Option C (Electrons) is correct because in metallic conductors, the current is conducted by the flow of electrons.

Option C ($\Delta {G^ \circ }$ is negative) is correct because a spontaneous cell reaction is characterized by a negative change in Gibbs free energy.

Option D (Zinc displaces tin from its solution) is correct because zinc is more reactive than tin and can displace it from its solution, as indicated by the electrochemical series.

Option C is correct because it accurately represents the standard hydrogen electrode (SHE) with platinum electrode, hydrogen gas at 1 atm, and H+ ions at unit activity.

Option A (121.6) is correct because it is calculated by applying Kohlrausch's law of independent migration of ions.

Option B (390.5) is correct because it is calculated by applying Kohlrausch's law to find the molar conductivity of acetic acid at infinite dilution.

Option B (390.5) is correct because it is calculated by applying Kohlrausch's law to find the molar conductivity of acetic acid at infinite dilution.

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

Option A ($4.76 \times {10^{ - 3}}\mathrm{mho}$) 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.

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

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 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 D ($2.6 \times 10^{-3} \mathrm{~F}$) is correct because it accurately calculates the amount of electricity passed in Faradays using the given current and time.

Option A ($2.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}$) is correct because conductivity is calculated from equivalent conductivity and concentration.

Option C (3 F) is correct because 3 moles of electrons are required to obtain one mole of aluminum from Al2O3.

Option A (${\mathrm{H}_2}\left( \mathrm{g} \right)\mathrm{and}{\kern 1pt} {\kern 1pt} {\mathrm{O}_2}$ (g)) is correct because during electrolysis of aqueous MgSO4, water is reduced at the cathode to produce H2 and oxidized at the anode to produce O2.

Option B (1 F) is correct because 1 Faraday of electricity is required to liberate 1 gram equivalent of any substance.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ molecule) is correct because at infinite dilution, weak electrolytes like acetic acid are considered to be 100% ionized, meaning there are virtually no unionized molecules left.

Option C (Electrochemical equivalent) is correct because according to Faraday's first law of electrolysis, the mass of a substance deposited is directly proportional to the quantity of electricity passed, and the electrochemical equivalent (Z) is the mass deposited by 1 coulomb (1 A for 1 s).

Option A (Voltaic cell) is correct because a voltaic cell produces electric current through spontaneous oxidation-reduction reactions.

Option D (I & III) is correct because oxidizing power is directly proportional to SRP and inversely proportional to SOP, while reducing power is directly proportional to SOP and inversely proportional to SRP.