Electro chemistry practice · Mixed difficulty

Option D (${\mathrm{I}_2} + 2\mathrm{B}{\mathrm{r}^ - } \to 2{\mathrm{I}^ - } + \mathrm{B}{\mathrm{r}_2}$) is correct because the overall cell potential for this reaction is negative, indicating a non-spontaneous process.

Option D (All of the above) is correct because, based on the reduction potentials, electrons flow from B to A, the EMF is 0.56V, and A is the positive electrode.

Option A ($\frac{l}{a}$) is correct because the cell constant is defined as the ratio of the distance between the electrodes (l) to the area of cross-section of the electrode (a).

Option D (-0.177 V) is correct because it is the calculated potential of the hydrogen electrode using the Nernst equation with the given pH.

Option A (Violet due to formation of $I_2$) is correct because chlorine will oxidize iodide ions to iodine, which dissolves in chloroform to give a violet color.

Option C (1.5) is correct because the deposition of 1 mole of Al requires 3 Faradays, and 13.5g of Al is 0.5 moles, thus requiring 1.5 Faradays.

Option C ($Cu + F{e^{2 + }} \to C{u^{2 + }} + Fe \downarrow$) is correct because copper cannot displace iron from its solution due to copper having a lower reduction potential than iron.

Option B is the correct answer because in a dry cell, manganese is reduced from +4 to +3 state, not +5 to +4.

Option C ($Z{n^{2 + }}$) is correct because hydrogen cannot reduce ions of metals that are more electropositive than itself (i.e., have a more negative standard reduction potential).

Option A is correct because the cell emf is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode: E°cell = E°cathode - E°anode.

Option A (Anions move towards anode, cations towards cathode) is correct because anions are negatively charged and move towards the positive electrode (anode), while cations are positively charged and move towards the negative electrode (cathode).

Option A ($F{e_2}{O_3} \cdot x{H_2}O$) is correct because rust is hydrated iron(III) oxide, where 'x' represents a variable number of water molecules.

Option C (Temperature) is correct because according to the Nernst equation, electrode potential is directly dependent on temperature.

Option D is correct because the standard electromotive force (e.m.f.) of a dry cell is approximately 1.5 V.

Option C (${Au}, {Ag}, {Cu}$) is correct because the metal with the highest reduction potential will be deposited first on the cathode.

Option A (Paste of $\mathrm{HgO}$ and carbon) is correct because this mixture is used as the cathode in mercury cells.

Option B ($H_2$) is correct because during the electrolysis of aqueous $Na_2SO_4$, water is reduced at the cathode to produce hydrogen gas.

Option B (0.317 g) is correct because the mass of copper deposited can be calculated using the ratio of equivalent weights and the mass of silver deposited.

Option C ($E = {E_R} - {E_L}$) is correct because the EMF of a cell is the difference between the reduction potential of the right-hand electrode (cathode) and the left-hand electrode (anode).

The correct answer is A because Zinc has a positive oxidation potential and Silver has a negative oxidation potential, making Zinc oxidation and Silver reduction favorable.

Option C ($388.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{g}{\kern 1pt} {\kern 1pt} \mathrm{equi}{\mathrm{v}^{ - 1}}$) is correct because it applies Kohlrausch's law of independent migration of ions.

Option D (+0.4 V) is correct because it represents the standard EMF of the cell calculated from the difference in standard reduction potentials.

Option A ($1M{\kern 1pt} {\kern 1pt} \mathrm{NaCl}$) is correct because NaCl dissociates into one cation and one anion, so its molar mass and equivalent mass are the same.

Option C is correct because the given reduction potentials indicate that MnO2 can be reduced to Mn2+ and oxidized to MnO4- in acidic solution, thus disproportionating.

Option D (1.0 ×${10^{10}}$) is correct because it is calculated using the relationship between standard EMF and the equilibrium constant, $K = 10^{\frac{{nE^0}}{{0.0591}}}$.

Option A (2.71 V) is correct because the EMF of the cell is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.

Option D ($0.1M{\kern 1pt} {\kern 1pt} \mathrm{HCl}$) is correct because HCl is a strong acid and dissociates completely, leading to a higher concentration of ions and thus higher conductance.

Option C is correct. The volume of ${O_2}$ liberated at STP is calculated using Faraday's laws of electrolysis, considering the number of moles of electrons and the stoichiometry of the oxygen evolution reaction.

Option B is correct because Manganese dioxide acts as a depolarizer by oxidizing hydrogen gas produced at the cathode, preventing polarization and maintaining the cell's voltage.

Option C ($Zn$ can reduce $A{g^ + }$ and $C{u^{2 + }}$) is correct because Zn has a more negative standard electrode potential than both Ag and Cu, making it a stronger reducing agent.