Electro chemistry practice · Easy

Option C is correct because the number of Faradays needed is equal to the number of gram-equivalents of the substance being reduced or oxidized.

Option A ($3.2 \times {10^{ - 19}}$) is correct because the charge of a $\mathrm{C}{\mathrm{u}^{2 + }}$ ion is twice the elementary charge.

Option C ($9.65 \times {10^4}$ coulomb) is correct because the total charge on 1 mole of monovalent ions is equal to Faraday's constant, which is approximately 96500 Coulombs.

Option C is correct. The EMF of the voltaic cell is calculated by subtracting the standard reduction potential of the anode (Ni) from the standard reduction potential of the cathode (Au).

Option B is correct because Michael Faraday proposed the laws of electrolysis.

Option D (Intensive property) is correct because the standard electrode potential is independent of the amount of substance and depends only on the nature of the electrode and the temperature.

Option A is correct because zinc has a higher tendency to lose electrons than hydrogen, so it undergoes oxidation at the anode, while hydrogen ions get reduced at the cathode (SHE).

Option A (${\mathrm{H}^ + }$) is correct because hydrogen ions have exceptionally high molar conductivity due to their unique mechanism of transport in water.

Option B (This electrode potential is assumed to be zero) is correct because the standard hydrogen electrode (SHE) is defined as having a zero potential at all temperatures.

Option A is correct because 96500 Coulombs (1 Faraday) liberates 0.5 mole of O2, which occupies 5.6 liters at NTP.

Option C is correct because the electrochemical equivalent is defined as the mass of a substance deposited or liberated by the passage of 1 coulomb of charge (1 Ampere for 1 second).

Option A is correct because the transport number represents the fraction of the total current carried by a specific ion in an electrolytic solution.

Option C (Quantity of electricity passed) is correct because Faraday's laws of electrolysis state that the amount of substance produced is directly proportional to the quantity of electricity passed.

Option B (${\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8}$) is correct because at high concentrations of sulfuric acid, peroxydisulfuric acid ($H_2S_2O_8$) is formed at the anode.

Option D (Sulphuric acid) is correct because the electrolysis of cupric sulphate solution leads to the formation of sulphuric acid as the copper ions are deposited and water is oxidized.

Option A (ampere-sec) is correct because quantity of charge (Q) is defined as the product of current (I) and time (t), where current is measured in amperes and time in seconds.

Option A (Power for heat and light) is correct because hydrogen-oxygen fuel cells produce electricity, which can be used to generate heat and light in spacecraft.

Option D ($\mathrm{AuC}{\mathrm{l}_3} + \mathrm{NaCN}$) is the correct answer because the electrolytic bath for gold plating typically contains a gold salt (like AuCl3) and sodium cyanide (NaCN) to form a complex ion that facilitates smooth and even deposition of gold.

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

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 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 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.

Option B is correct because iron is more readily corroded in moist air compared to copper, silver, and nickel due to its higher reactivity and tendency to form rust.

Option A is correct because the most reactive alkali metal is the one with the most negative standard electrode potential. A more negative value indicates a greater tendency to lose electrons and undergo oxidation. Among the given options, A has the most negative value (-3.05), making it the most reactive.