Electro chemistry practice · Easy

Option D is correct because Faraday's laws are fundamental and hold true under all normal conditions, including varying pressures and solvents.

Option D is correct because, according to Kohlrausch's law, the molar conductivity of CH3COOH at infinite dilution is the sum of the molar conductivities of CH3COONa and HCl, minus the molar conductivity of NaCl, resulting in 390.5 S cm² mol⁻¹.

Option A (Anode) is correct because in the electrolysis of fused CaH2, hydride ions (H-) migrate to the anode and are oxidized to form hydrogen gas.

Option D ($Cu$ will reduce $B{r_2}$) is correct because copper has a lower reduction potential than bromine, meaning it can be oxidized (act as a reducing agent) to reduce bromine.

Option D (Either of these) is correct because both water voltameter, silver voltameter and coulometer convert electrical energy into chemical energy.

Option C (pH < 7.0) is correct because the electrolysis of CuSO4(aq) produces H2SO4, which is acidic.

Option B is correct. The EMF of the cell is equal to the standard reduction potential of the cathode minus the standard reduction potential of the anode. In this case, the standard hydrogen electrode (SHE) is the cathode, and the zinc electrode is the anode.

Option B (Voltmeter) is correct because a voltmeter is used to measure the potential difference between two electrodes, which is how standard electrode potential is determined.

Option A is correct. During the electrolysis of aqueous $\mathrm{MgSO}_4$, water is reduced at the cathode to produce ${H_2(g)}$, and water is oxidized at the anode to produce ${O_2(g)}$ because ${Mg^{2+}}$ and ${SO_4^{2-}}$ ions are relatively stable and harder to reduce/oxidize than water.

Option A (Right electrode will increase) is correct because the electrolysis of water at the right electrode produces hydroxide ions, increasing the pH.

Option B ($- B$) is correct because in the Debye-Hückel-Onsager equation, the slope of the plot of molar conductivity vs. square root of concentration is represented by -B.

Option C ($\mathrm{pH} > 7$) is correct because the electrolysis of aqueous LiCl produces $H_2$ at the cathode and $Cl_2$ at the anode, leaving behind $OH^-$ ions, which increases the pH.

Option A is correct because the amount of an ion discharged during electrolysis depends on the current, time, and electrochemical equivalent, but not directly on the resistance of the solution.

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

Option B (Pb) is correct because Pb has a standard reduction potential between that of Ag+ and Ni2+, allowing it to reduce Ag+ but not Ni2+.

Option A ($\Delta {G^ \circ } = - nFE_{\mathrm{cell}}^ \circ$) is the correct expression relating the standard Gibbs free energy change to the standard cell potential.

Option B (+ 0.76 V) is correct because the cell potential is calculated by reversing the sign of the zinc half-cell potential since it's being oxidized.

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

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 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 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 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 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 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 B ($1:\frac{1}{2}:\frac{1}{3}$) is correct because the ratio of moles deposited is inversely proportional to the charge on the ion.