Electrochemistry – NEET UG ChemistryPractice Questions & PYQs

Standard reduction potentials dictate that species with high positive reduction potentials have a strong tendency to undergo reduction, thereby acting as powerful oxidizing agents rather than reducing agents. Thallium in the plus three oxidation state possesses a high positive reduction potential, making it an exceptional oxidizing agent that readily accepts electrons to form a lower oxidation state. Conversely, lower oxidation states or metals with negative potentials exhibit reducing characteristics. Therefore, claiming that the trivalent thallium ion functions as a reducing agent contradicts its fundamental electrochemical behavior.

The standard electromotive force of a galvanic cell is calculated by taking the difference between the reduction potentials of the cathode and the anode according to $E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ$. Assigning the half-cells to ensure a positive overall potential for a feasible reaction involves subtracting the lower potential from the higher potential. This sets up the calculation as $1.81 - (-1.66)$, which directly evaluates to 3.47 V.

Spontaneity in an electrochemical context requires a positive standard cell potential, determined by subtracting the anode potential from the cathode potential according to $E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ$. Evaluating the silver and copper reaction yields a potential of 0.34 V minus 0.80 V, resulting in a negative value of -0.46 V. Because this negative potential indicates a non-spontaneous process, this specific reaction cannot take place under standard conditions.

The relationship between standard Gibbs energy change and standard cell potential is expressed through the equation $\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$, where $n$ denotes the total number of moles of electrons transferred during the redox reaction. For this particular cell, two electrons are exchanged, establishing $n = 2$. Inserting the respective values for $n$, the Faraday constant, and the cell potential into the formula produces $-212270 \text{ J mol}^{-1}$, which simplifies to $-212.27 \text{ kJ mol}^{-1}$.

The actual cell potential under non-standard conditions is calculated using the Nernst equation, expressed as $E = E^\circ - \frac{0.059}{n} \log Q$. For this redox process, the number of electrons transferred is 2 and the reaction quotient evaluates to 1000, making the logarithmic term equal to 3. Substituting these values along with the standard cell potential into the equation gives $1.05 - \frac{0.059}{2} × 3$, which simplifies directly to 0.9615 V.

Kohlrausch's law of independent migration of ions dictates that the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar conductivities of its constituent ions. For acetic acid, the limiting molar conductivity is calculated using the expression $\Lambda_m^\circ (\mathrm{CH_3COOH}) = \Lambda_m^\circ (\mathrm{CH_3COONa}) + \Lambda_m^\circ (\mathrm{HCl}) - \Lambda_m^\circ (\mathrm{NaCl})$. Substituting the given values of $91.0$, $426.16$, and $126.45$ into this relationship correctly balances the ionic contributions. Performing the addition of the sodium acetate and hydrochloric acid terms followed by the subtraction of dilatometric chloride terms produces $390.71 \text{ S cm}^{2} \text{ mol}^{-1}$.

The degree of dissociation of a weak electrolyte is determined by the ratio of its molar conductivity at a given concentration to its molar conductivity at infinite dilution. First, the molar conductivity at infinite dilution for acetic acid is found by summing the limiting ionic conductivities of the hydrogen ion and the acetate ion, which totals $400 \text{ S cm}^{2} \text{ mol}^{-1}$. Dividing the given molar conductivity of $20 \text{ S cm}^{2} \text{ mol}^{-1}$ by this infinite dilution value gives a dissociation degree of one-twentieth. Substituting this degree of dissociation and the concentration of $0.007 \text{ M}$ into the dissociation constant expression $K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$ yields the final value of $1.75 × 10^{-5} \text{ mol L}^{-1}$.

During the electrolysis of dilute sulphuric acid using inert platinum electrodes, water molecules undergo oxidation at the anode preferentially over sulphate ions due to their lower standard oxidation potential. This oxidation process involves the breakdown of water molecules to release oxygen gas alongside hydrogen ions and electrons. Consequently, oxygen gas is continuously evolved at the positive electrode while hydrogen gas is liberated at the negative cathode.

The quantity of charge required in Faraday to deposit a given mass of a substance is equal to its number of chemical equivalents, which is given by the relation $\text{Number of equivalents} = \frac{\text{Mass}}{\text{Equivalent mass}}$. The equivalent mass of calcium is determined by dividing its atomic mass of $40$ by its oxidation state of $2$, resulting \in an equivalent mass of $20$. Dividing the specified mass of $20$ by this equivalent mass yields a value of $1$. Thus, exactly one Faraday of charge is required to produce this amount of calcium.

The relationship between standard Gibbs energy and standard cell potential is given by the formula $\Delta_{r} G^{\ominus} = -nFE^{\ominus}_{\text{cell}}$. From the balanced redox reaction, the number of electrons transferred is 2, while the standard cell potential is 0.24 V and the Faraday constant is 96500 C mol^-1. Substituting these numerical values into the equation yields $-2 \times 96500 \times 0.24$ J mol^-1, which calculates to $-46320$ J mol^-1. Converting this energy value from joules per mole to kilojoules per mole results in $-46.32$ kJ mol^-1.

Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte equals the sum of the ionic conductivities of its constituent ions. To obtain the conductivity of acetic acid, we combine the conductivities of potassium acetate, sulfuric acid, and potassium sulfate in a specific stoichiometric proportion. Specifically, adding the conductivity $z$ of potassium acetate to half the difference between the conductivity $x$ of sulfuric acid and the conductivity $y$ of potassium sulfate eliminates the spectator ions and leaves the exact molar conductivity expression $\frac{(x-y)}{2}+z$.

Reducing power is inversely proportional to standard reduction potential, meaning that chemical species with more negative reduction potentials act as stronger reducing agents. Potassium exhibits the most negative reduction potential of -2.93 V, followed by aluminum at -1.66 V, chromium at -0.74 V, and silver with a positive potential of 0.80 V. Arranging these metals from the most negative to the most positive standard potential establishes the correct decreasing order of reducing power as K > Al > Cr > Ag.

The Nernst equation relating cell potential to the equilibrium constant is expressed as $\log K_{\text{eq}} = \frac{nE^{\circ}_{\text{cell}}}{0.059}$. At electrochemical equilibrium, substituting the given values of 1 for the number of electrons and 0.59 V for the standard cell potential gives $\log K_{\text{eq}} = \frac{1 \times 0.59}{0.059}$, which simplifies to 10. Taking the antilogarithm of this logarithmic result yields the equilibrium constant value of $1.0 \times 10^{10}$.

Molar conductivity is determined by the relation $\Lambda_m = \frac{\kappa \times 1000}{M}$. Multiplying the given electrolytic conductivity by 1000 and dividing by the molarity of $0.5$ mol/dm$^3$ calculates the required value. Evaluating $\frac{5.76 \times 10^{-3} \times 1000}{0.5}$ yields $11.52$ S cm$^2$ mol$^{-1}$.

Standard reduction potentials dictate that a metal with a more negative electrode potential acts as the sacrificial anode and oxidizes more readily than one with a less negative potential. Zinc possesses a significantly more negative standard reduction potential compared to iron, allowing it to preferentially corrode and effectively protect the underlying iron surface. Consequently, zinc readily coats and protects iron, whereas the reverse plating process is thermodynamically unfavorable.

The Nernst equation for the hydrogen half-cell is given by $E = E^{\circ} - \frac{0.059}{n} \log \frac{P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}$. In pure water at 298 K, the hydrogen ion concentration is 10^-7 M, and the electrode potential is required to be zero. Substituting the standard potential and ion concentration into the equation gives $0 = 0 - \frac{0.059}{2} \log \frac{P_{\mathrm{H}_2}}{(10^{-7})^2}$. Equating the logarithm to zero results in a hydrogen gas pressure of 10^-14 atm.

The oxidation of chloride ions at the anode follows the reaction $2\mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 + 2e^-$, establishing that producing one mole of chlorine gas requires two moles of electrons. Applying Faraday's laws through the relation $w/E = \frac{It}{96500}$, the required charge is determined for $0.10$ moles of chlorine. Substituting the current of $3$ amperes and solving for time gives approximately $6433$ seconds, which converts to $107.2$ minutes or roughly $110$ minutes.

The total electrical charge passing through a circuit is calculated using the product of current and time via $Q = It$. Dividing this total charge by the elementary charge of a single electron yields the absolute number of electrons transferred. Multiplying the current of $1$ ampere by the time of $60$ seconds gives a charge of $60$ coulombs, which when divided by $1.60 \times 10^{-19}$ coulombs produces $3.75 \times 10^{20}$ electrons.

Galvanic cells that continuously convert the chemical energy of combustion of conventional fuels like hydrogen and methane directly into electrical energy without intermediate thermal steps are known as fuel cells. These electrochemical devices maintain a constant supply of reactants to the electrodes, allowing significantly higher energy conversion efficiencies compared to traditional thermal power plants.

According to Faraday's law of chemical equivalence, the same quantity of charge displaces equivalent amounts of different substances. At STP, one gram-equivalent of oxygen gas corresponds to 5600 mL, which equals 8 grams and represents one chemical equivalent. Since an equivalent of oxygen is equivalent to one equivalent of any other element, the same quantity of electricity will displace one gram-equivalent of silver. Given that the atomic weight of silver is 108, the displaced mass is 108.0 grams.