Electrochemistry – NEET UG ChemistryPractice Questions & PYQs

Electrode potentials are intensive properties and cannot be added or subtracted directly, requiring instead that calculations be performed through their corresponding Gibbs free energy changes via the relation $\Delta G^\circ = -nFE^\circ$. The desired third half-reaction can be obtained by subtracting the second given reaction from the first, which translates to finding the free energy change difference where $\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$. Substituting the number of electrons and standard potentials into this energy balance gives $-1 \times F \times E^\circ_3 = (-2 \times F \times 0.337) - (-1 \times F \times 0.153)$. Solving this equation yields the standard reduction potential of 0.52 V for the cuprous ion reduction step.

The mass of a substance deposited during electrolysis is calculated using Faraday's laws through the formula $W = \frac{M \times I \times t}{n \times F}$. For the reduction of aluminium, each ion requires three electrons, meaning $n = 3$ and the Faraday constant is 96500. Substituting the molar mass of 27, a current of 40000 amperes, and a time of 21600 seconds into the expression yields a total mass of 81000 grams, which corresponds to $8.1 \times 10^4\text{ g}$.

The strength of an oxidizing agent depends directly on its standard reduction potential ($E^\circ$), where a higher positive value indicates a greater tendency to accept electrons. Evaluating the relevant half-cell processes shows that the reduction potential for $[\mathrm{Fe}]^{3+}$ to $[\mathrm{Fe}]^{2+}$ is $+0.77\text{ V}$, whereas the reduction potential for $[\mathrm{Fe}(\mathrm{CN})_{6}]^{3-}$ to $[\mathrm{Fe}(\mathrm{CN})_{6}]^{4-}$ is $+0.35\text{ V}$. Because $+0.77\text{ V}$ represents the higher positive potential, $[\mathrm{Fe}]^{3+}$ readily gains electrons and emerges as the strongest oxidizing agent.

The standard cell potential of a fuel cell is determined by relating the standard Gibbs free energy change of the overall reaction to electrical work using the equation $\Delta G^\circ = -nFE^\circ$. First, the standard free energy change for the combustion of pentane is calculated by subtracting the free energies of formation of the reactants from those of the products, taking stoichiometry into account for carbon dioxide and water. For the complete oxidation of one mole of pentane, the balanced half-reactions involve a total transfer of 32 electrons, establishing $n = 32$. Substituting the calculated free energy change of -3387 kJ along with $n$ and the Faraday constant 96500 into the rearranged thermodynamic equation yields a standard cell potential of 1.0968 V.

Kohlrausch's law establishes that at infinite dilution, where complete dissociation of an electrolyte occurs, every ion contributes a fixed and characteristic value to the total conductance of the solution. This individual ionic contribution remains entirely independent of the nature of the accompanying counterion present in the electrolyte. Consequently, the overall conductance of an electrolyte can be precisely determined by summing the individual contributions of its constituent cations and anions.

The relationship between the standard cell potential and the equilibrium constant at 298 K is given by the formula
$E^{\circ}_{\text{cell}} = \frac{0.0591}{n} \log K_c$

Examining the overall redox reaction reveals that 2 electrons are transferred, which establishes the value of $n$ as 2. Substituting the standard cell potential of 0.46 V into the equation yields $\log K_c = \frac{2 \times 0.46}{0.0591}$, resulting in 15.57. Calculating the antilogarithm of this value provides the final equilibrium constant of $4.0 \times 10^{15}$.

The thermodynamic efficiency of a fuel cell represents the proportion of total reaction energy that can be successfully converted into useful electrical work. In an ideal electrochemical conversion, the maximum non-expansion work obtainable corresponds to the change in $\Delta G$, while the total energy released by the reaction is represented by $\Delta H$. Consequently, the operational efficiency is expressed directly as the ratio of $\Delta G$ to $\Delta H$.

In the given overall reaction, metallic iron undergoes oxidation to form ferrous ions, serving as the anode with a standard reduction potential of -0.441 V. Simultaneously, ferric ions are reduced to ferrous ions, acting as the cathode with a standard reduction potential of 0.771 V. Subtracting the anode potential from the cathode potential yields 1.212 V for the standard electromotive force of the cell.

In any standard electrochemical cell, oxidation takes place at the anode with the release of electrons, whereas reduction occurs at the cathode where cations accept those electrons. The given half-reaction shows species $A$ losing an electron to form $A^+$, identifying it as the anodic oxidation process. To maintain a positive cell potential, the cathodic process must involve the reduction of $B^+$ ions into solid $B$. Adding these balanced half-reactions together directly produces the overall spontaneous cell equation of $A + B^+ \rightarrow A^+ + B$.

Stoichiometric relationships in the Hall-Héroult process dictate that the moles of carbon anode consumed are directly proportional to the moles of aluminium produced based on the balanced reduction reaction. Production of 270 kg of aluminium corresponds to 10,000 moles of $\mathrm{Al}$ based on its atomic mass of 27. According to the reaction stoichiometry where carbon is oxidized by oxygen derived from alumina, the production of 2 moles of aluminium consumes 3 moles of carbon. Specifically, balancing the reaction such that 10,000 moles of $\mathrm{Al}$ are formed requires 15,000 moles of carbon. Multiplying this by the molar mass of carbon of 12 g/mol gives 180,000 g, which equals 180 kg.

According to Faraday's law of electrolysis, a specific quantity of electric charge liberates an equivalent number of gram-equivalents for any substance. First, find the equivalent weight of aluminium by dividing its atomic mass of 27 amu by its valence of 3, resulting in an equivalent weight of 9. Dividing the deposited mass of 4.5 g by this equivalent weight shows that 0.5 equivalents of charge were utilized. Applying this same charge to hydrogen ions yields 0.5 equivalents of hydrogen gas. Multiplying this by the molar volume equivalent factor of 11.2 liters per equivalent at standard temperature and pressure gives a final volume of 5.6 liters.

The equilibrium constant of an electrochemical cell reaction is logarithmically related to its standard cell potential through the Nernst equation at equilibrium. At chemical equilibrium, the net cell potential becomes zero, allowing the Nernst equation to simplify into a direct relationship between the standard electromotive force, the number of electrons transferred, and the equilibrium constant. Substituting the given values where $n = 2$ and the standard emf is 0.295 V into the derived logarithmic expression gives an exponent term of $\frac{2 \times 0.295}{0.0591}$. Evaluating this fraction yields a value of 10, and taking the antilogarithm results in an equilibrium constant of 1.0 x 10^10.

The relationship between Gibbs free energy change and cell potential is given by the equation $\Delta G = -nFE^\circ$. Substituting $\Delta G = -827000$, $n = 2$, and $F = 96500$ into the rearranged formula $E^\circ = \frac{-\Delta G}{nF}$ gives an intermediate potential of 4.28 V. Dividing this value by the stoichiometric coefficient of two for the aluminum-oxygen reaction yields the minimum required emf of 2.14 V.

The dependence of cell potential on ionic concentrations is described by the Nernst equation expression E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\mathrm{Zn}^{2+}]}}{[\mathrm{Cu}^{2+}]${} \right)$. For the initial cell configuration, the concentration ratio inside the logarithm is 0.01 divided by 1.0, which produces a negative logarithmic value that adds a positive contribution to the standard potential. When the concentrations are inverted in the second cell setup, the concentration ratio becomes 1.0 divided by 0.01, producing a positive logarithmic value that subtracts from the standard potential instead. Evaluating these opposing concentration effects reveals that the first emf is greater than the second emf.

The selective discharge of ions during electrolysis is governed by their respective discharge potentials and electrode overvoltages. Employing a mercury cathode introduces a high hydrogen overvoltage that significantly increases the potential required to discharge hydrogen ions, allowing sodium ions to reduce instead and immediately form an amalgam with the mercury.

The spontaneity of a redox interaction between multiple species is determined by comparing their standard reduction potentials, where the species with the higher potential acts as the oxidizing agent. Iron metal spontaneously reduces ferric ions to ferrous ions while being oxidized itself into ferrous ions. Consequently, keeping iron metal alongside both iron(II) and iron(III) ions drives a reaction that consumes the ferric species, leading to a direct decrease in its concentration.

The equivalent conductance of an electrolyte at infinite dilution is determined by Kohlrausch's law, expressed as $\lambda_{\infty } = \frac{1}{n_+} \lambda_+^\infty + \frac{1}{n_-} \lambda_-^\infty$. Applying this principle to barium chloride involves weighting the barium ionic conductance by its valence factor of one-half and adding the chloride ionic conductance. Substituting the respective values of 127 and 76 into this expression yields $63.5 + 76$, which equals 139.5.

The standard Gibbs free energy change is related to standard cell potential through the thermodynamic expression $\Delta G^\circ = -nFE^\circ$, allowing free energy values to be combined for coupled half-reactions. By calculating the free energy change for the reduction of cuprous ions to solid copper, a potential of 0.53 V is derived for that half-cell. Subtracting the reduction potential of the cupric-to-cuprous half-reaction from this newly derived value gives the net standard cell potential of 0.38 V for the disproportionation process.