Electro chemistry practice · Easy

Option B (Reduced at cathode) is correct because reduction, which is the gain of electrons, always occurs at the cathode in an electrochemical cell.

Option A ($\mathrm{Pt}{\mathrm{H}_2}\mathop {\left| {\mathrm{HCl}} \right|}\limits_{{c_1}} \mathop {\left| {\mathrm{HCl}} \right|}\limits_{{c_2}} \mathrm{Pt}{\mathrm{H}_2}$) correctly represents a concentration cell because it has the same electrodes and electrolyte but different concentrations.

Option D (45 g) is correct because passing 5 Faradays of charge through a solution of $Al^{3+}$ ions will deposit 5/3 moles of Al, which corresponds to 45 g.

Option A ($\mathrm{C}{\mathrm{u}^{2 + }}{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{oxidises}{\kern 1pt} {\kern 1pt} \mathrm{ Fe}$) is correct because $\mathrm{C}{\mathrm{u}^{2 + }}$ has a higher reduction potential than $\mathrm{F}{\mathrm{e}^{2 + }}$, making it a stronger oxidizing agent.

Option A is correct because conductance equals conductivity when the cell constant (length/area) is equal to 1.

Option D (There is no reaction) is correct because copper is less electropositive than aluminium and cannot displace it from its salt solution.

Option C is correct because it correctly represents the galvanic cell where hydrogen gas is oxidized to H+ ions at the platinum electrode, and AgCl is reduced to Ag at the silver electrode.

Option B (From Cu to Zn outside the cell) is correct because electrons flow from the zinc electrode (anode) to the copper electrode (cathode) through the external circuit.

Option A ($E_{\mathrm{cell}}^ \circ = {\phi _R} - {\phi _L}$) is correct because the standard cell potential is calculated by subtracting the standard reduction potential of the left-hand electrode (anode) from the standard reduction potential of the right-hand electrode (cathode).

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

Option B is correct because the constant of proportionality is conductivity, and its unit is S m² mol⁻¹ when conductance is directly proportional to area and concentration and inversely proportional to length.

Option B ($\mathrm{BeC}{\mathrm{l}_2}$) is correct because beryllium has the smallest ionic size among the given options, leading to lower mobility and thus lower equivalent conductance in the fused state.

Option D (Zero) is correct because passing 2F of charge through 1 liter of 1M $CuSO_4$ will completely deposit all the $Cu^{2+}$ ions, leaving a molarity of zero.

Option A ($\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\mathrm{and}{\kern 1pt} {\kern 1pt} \mathrm{C}{\mathrm{u}^{2 + }}} \right|\mathrm{Cu}$) is correct because in a Daniel cell, the anode is zinc immersed in a zinc sulfate solution, and the cathode is copper immersed in a copper sulfate solution.

Option B is not correct because the units of electrochemical equivalent are g/Coulomb (grams per Coulomb), not g-coulomb.

Option A is correct because the limiting molar conductivity of a uni-univalent electrolyte is the sum of the limiting molar ionic conductivities of its cation and anion.

Option C is correct because, according to Kohlrausch's law, to calculate the limiting molar conductivity of CH3COOH, you need the limiting molar conductivities of CH3COONa, HCl, and NaCl.

Option D (The ${E^ \circ }$ values of the two half cells) is correct because the standard EMF of a galvanic cell is calculated from the difference in standard reduction potentials of the two half-cells.

Option A (Making it cathode) is correct because making iron the cathode in an electrochemical cell provides cathodic protection, preventing oxidation (rusting).

Option C ($\mathrm{K}\left[ {\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}} \right]$) is correct because it is the electrolyte commonly used for gold plating due to its ability to form stable gold complexes.

Option C is correct because $\mathrm{AgN}{\mathrm{O}_3} + \mathrm{NaCl} \to \mathrm{AgCl} \downarrow + \mathrm{NaN}{\mathrm{O}_2}$ is a precipitation reaction, not a redox reaction, and therefore cannot be a basis for an electrochemical cell.

Option B is correct because it uses the Nernst equation and the Ksp of Cu(OH)2 to calculate the reduction potential at pH=14.

Option C ($\mathrm{AgN}{\mathrm{O}_3}$) is correct because copper is more reactive than silver and will displace silver ions from the solution, forming copper(II) ions which give the solution a blue color.

Option C ($C{a^{2 + }} > {K^ + } > N{a^ + }$) is the correct order because ionic mobility generally increases with charge density and decreases with hydrated size.

Option D ($6.02 \times 10^{23}$) is correct because one Faraday of electricity corresponds to the charge of one mole of electrons, which is Avogadro's number.

Option B (S is wrong but E is correct) is correct because the speed of ions depends on factors other than just the nature of the ions, while the explanation correctly identifies the factors affecting ion mobility.

Option D is correct because a standard hydrogen electrode (SHE) uses a platinum electrode, which *does* adsorb hydrogen gas to facilitate the electron transfer process.

Option C is correct because at pH = 1, the Nernst equation for the hydrogen electrode gives a potential of -0.059 V.

Option C is correct because electrolytic conduction occurs due to the movement of ions in the solution, carrying the electric charge.

Option A (Ions) is correct because ions are the charge carriers within the electrolyte solution in the internal circuit of a galvanic cell.