Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Kohlrausch's law of independent migration of ions posits that at infinite dilution, an electrolyte's conductivity is simply the sum of its independent ionic conductances. Finding the limiting equivalent conductance for a weak electrolyte like acetic acid, $\Lambda^{\circ}_{\mathrm{CH}_3\mathrm{COOH}}$, requires isolating the acetate ion, $\mathrm{CH}_3\mathrm{COO}^-$, and the hydrogen ion, $\mathrm{H}^+$. With the given values of $\Lambda^{\circ}_{\mathrm{CH}_3\mathrm{COONa}} = 91.0\ \mathrm{S}\ \mathrm{cm}^2/\text{equiv}$ and $\Lambda^{\circ}_{\mathrm{HCl}} = 426.2\ \mathrm{S}\ \mathrm{cm}^2/\text{equiv}$, adding these two parameters provides the sum of acetate, sodium, hydrogen, and chloride ions, effectively resulting in the combined contribution of all four ions.

Because the sodium ion, $\mathrm{Na}^+$, and the chloride ion, $\mathrm{Cl}^-$, are extraneous to the target molecule, they must be removed from the calculation. This is accomplished mathematically by subtracting the limiting equivalent conductance of sodium chloride, $\Lambda^{\circ}_{\mathrm{NaCl}}$, which represents the combined contribution of $\mathrm{Na}^+$ and $\mathrm{Cl}^-$. Consequently, knowing the value of $\Lambda^{\circ}_{\mathrm{NaCl}}$ is essential to cancel these unwanted ions, as expressed by the relationship $\Lambda^{\circ}_{\mathrm{CH}_3\mathrm{COOH}} = (\Lambda^{\circ}_{\mathrm{CH}_3\mathrm{COONa}} + \Lambda^{\circ}_{\mathrm{HCl}}) - \Lambda^{\circ}_{\mathrm{NaCl}}$.

When a galvanic cell reaches a state of total discharge, the chemical system has attained equilibrium, causing the cell potential to drop to zero. According to the Nernst equation, this equilibrium state is represented by the relationship $E_{\text{cell}}^\circ = \frac{0.059}{n} \log \left( \frac{[\mathrm{Zn}^{2+}]\$}{[\mathrm{Cu}^{2+}]\$} \right)$ where $n = 2$ represents the number of electrons transferred during the redox process. Substituting the standard potential of 1.10 V and $n = 2$ into this formula leads to 1.10 = \frac{0.059}{2} \log \left( \frac{[\mathrm{Zn}^{2+}]\}{[\mathrm{Cu}^{2+}]$} \right)$. By isolating the logarithmic term, we find$\log \left( \frac{[\mathrm{Zn}^{2+}]$}{[\mathrm{Cu}^{2+}]$} \right) = \frac{2 × 1.10}{0.059} = 37.3$. Removing the logarithm by taking the base-10 exponent provides the final concentration ratio as$10^{37.3}$.

The geometric properties of a conductivity cell are defined by the cell constant, which remains fixed regardless of the specific electrolyte solution it contains. This constant represents the relationship between conductivity and resistance, determined by the formula $G^* = \kappa R$. Using the initial values of $1.29 \, \mathrm{S \, m}^{-1}$ and $100 \, \Omega$, the cell constant is calculated to be $129 \, \mathrm{m}^{-1}$.

With the cell constant established, the conductivity of the second solution is found by rearranging the same relationship to $\kappa = \frac{G^*}{R}$. Substituting the measured resistance of $520 \, \Omega$, the conductivity becomes $\frac{129}{520} \, \mathrm{S \, m}^{-1}$. Molar conductivity is then derived by dividing this conductivity by the concentration, ensuring that the concentration is expressed in SI units where $1 \, \mathrm{M}$ is equivalent to $1000 \, \mathrm{mol \, m}^{-3}$. For the concentration of $0.2 \, \mathrm{M}$, which equals $200 \, \mathrm{mol \, m}^{-3}$, the molar conductivity calculation is $\Lambda_m = \frac{129}{520 \times 200} \, \mathrm{S \, m}^2 \, \mathrm{mol}^{-1}$. This simplifies to $0.00124 \, \mathrm{S \, m}^2 \, \mathrm{mol}^{-1}$, which is equivalent to $12.4 \times 10^{-4} \, \mathrm{S \, m}^2 \, \mathrm{mol}^{-1}$.

According to Kohlrausch's law, the limiting molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions. By adding the known values of $\Lambda^{\circ}_{\mathrm{NaOAc}}$ (91.0 S cm^2/mol) and $\Lambda^{\circ}_{\mathrm{HCl}}$ (426.2 S cm^2/mol), we sum the contributions of the constituent ions $\mathrm{Na}^+$, $\mathrm{OAc}^-$, $\mathrm{H}^+$, and $\mathrm{Cl}^-$. Because we need to determine the value for $\Lambda^{\circ}_{\mathrm{HOAc}}$, which is the sum of the conductivities of $\mathrm{H}^+$ and $\mathrm{OAc}^-$, we must subtract the unwanted $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions, making $\Lambda^{\circ}_{\mathrm{NaCl}}$ the necessary additional value.

The solubility product constant $K_{sp}$ represents the equilibrium state of a dissolution process, which can be linked to standard cell potential via the Nernst equation. To determine the potential for the dissolution reaction $\text{AgI} \rightarrow \text{Ag}^+ + \text{I}^-$, combine the provided half-reactions by reversing the second one (\text{AgI} + e^- \rightarrow \text{Ag} + \text{I}^-$with$E^\circ = -0.152,\text{V}$) and adding it to the first (\text{Ag} \rightarrow \text{Ag}^+ + e^-$ with $E^\circ = -0.800\,\text{V}$). This results in a net cell potential of $E^\circ_{cell} = -0.952\,\text{V}$. Using the relationship $E^\circ_{cell} = \frac{0.059}{n} \log K_{sp}$ with $n=1$, we substitute the values to obtain $-0.952 = 0.059 \log K_{sp}$. Dividing by $0.059$ yields $\log K_{sp} = -16.13$.

Kohlrausch's law of independent migration of ions states that the molar conductance of an electrolyte at infinite dilution is the sum of the molar conductances of its constituent ions. This principle allows the molar conductivity of a weak electrolyte, like acetic acid ($\mathrm{HOAc}$), to be determined by summing the ionic contributions of strong electrolytes and removing the values of ions that are not present in the target compound.

To find the molar conductance of $\mathrm{HOAc}$ at infinite dilution, we identify the necessary ions as $\mathrm{H}^+$ and $\mathrm{OAc}^-$. By adding the molar conductances of $\mathrm{HCl}$ and $\mathrm{NaOAc}$ and then subtracting the conductance of $\mathrm{NaCl}$, we effectively isolate the target ions by canceling out the contributions of $\mathrm{Na}^+$ and $\mathrm{Cl}^-$.

$\Lambda_{\mathrm{HOAc}}^{\infty } = \Lambda_{\mathrm{HCl}}^{\infty } + \Lambda_{\mathrm{NaOAc}}^{\infty } - \Lambda_{\mathrm{NaCl}}^{\infty }$

Substituting the provided values 426.2, 91.0, and 126.5 into the equation yields the final calculation:

$\Lambda_{\mathrm{HOAc}}^{\infty } = 426.2 + 91.0 - 126.5 = 390.7$

Faraday's laws of electrolysis describe the quantitative relationship between the quantity of electricity passed through an electrolyte and the mass of the substance deposited at the electrodes. The cathode reaction $Al^{3+} + 3e^{-} \rightarrow Al$ signifies that reducing one mole of aluminium ions requires three moles of electrons, which corresponds to a charge of 3 times 96,500 Coulombs for every 27 g of aluminium produced.

Preparing 5.12 kg of aluminium, which is equivalent to 5,120 g, requires calculating the total charge by multiplying the number of moles of aluminium by the charge necessary per mole of metal atoms. The total electricity needed is determined by the following expression:

$\frac{3 \times 96,500 \times 5,120}{27}$

This calculation yields a total of $5.49 \times 10^7$ Coulombs of electricity.

Electrical conductivity in an aqueous solution of a weak acid depends directly on its extent of dissociation, as a higher concentration of ions facilitates greater current flow. Among the carboxylic acids provided, the relative acidic strength follows the trend: $\text{difluoroacetic acid} > \text{fluoroacetic acid} > \text{chloroacetic acid} > \text{acetic acid}$. Because difluoroacetic acid is the strongest acid, it undergoes the greatest degree of ionization in water and generates the largest number of charge carriers, resulting in the highest electrical conductivity.

Spontaneity in a chemical reaction is fundamentally defined by a negative change in Gibbs free energy, which serves as the driving force for a process to occur without external intervention. The mathematical link between Gibbs free energy and the equilibrium constant, expressed as $\Delta G = -2.303 RT \log K_c$, dictates that for a negative $\Delta G$, the equilibrium constant $K_c$ must be greater than one, signaling that the reaction favors the formation of products. Furthermore, the relationship $\Delta G = -n F E_{\text{cell}}^\circ$ confirms that for a spontaneous reaction to generate electrical work, the standard cell potential $E_{\text{cell}}^\circ$ must be positive, as this ensures the energy change remains negative.

The total standard cell potential is determined by summing the reduction potential of the cathode and the oxidation potential of the anode. In the reaction $\mathrm{Sn}(\mathrm{s}) + 2\mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow 2\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{Sn}^{2+}(\mathrm{aq})$, tin is oxidized to $\mathrm{Sn}^{2+}$, while iron is reduced from $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$. The standard reduction potential given for the tin couple $\mathrm{Sn}^{2+}/\mathrm{Sn}$ is -0.14 V, which means the oxidation potential for the reaction $\mathrm{Sn} \rightarrow \mathrm{Sn}^{2+}$ is +0.14 V. Combining this with the reduction potential for the iron half-reaction of 0.77 V provides the overall standard cell potential
$E^{\circ}_{\mathrm{cell}} = 0.77 + 0.14 = 0.91 \text{ V}$

At equilibrium, the net cell potential $E_{\text{cell}}$ reaches zero, which links the standard cell potential $E_{\text{cell}}^\circ$ directly to the equilibrium constant $K_c$ via the Nernst equation. Because the temperature is $25^\circ \text{C}$, the constant term in the equation simplifies to $0.0591$, enabling a straightforward calculation based on the number of electrons transferred during the reaction.

With the standard cell potential at $0.591 \text{ V}$ and a single electron change represented by $n=1$, the relationship is defined by:
$E_{\text{cell}}^\circ = \frac{0.0591}{n} \log K_c$
Substituting the known values leads to:
$0.591 = \frac{0.0591}{1} \log K_c$
Solving this expression for the logarithm results in $\log K_c = 10$, which implies that the equilibrium constant is $K_c = 1 \times 10^{10}$.

The Nernst equation, E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.059}{2} \log \frac{[\mathrm{Zn}^{2+}]\ [\mathrm{H}2]}{[\mathrm{H}^{+}]^2}$, describes how changes \in ion concentrations directly impact the electrochemical potential of the cell. Adding sulfuric acid to the cathode compartment increases the$[\mathrm{H}^+]$concentration, which effectively decreases the value of the logarithmic expression \in the Nernst equation, resulting \in an overall increase \in$E{\text{cell}}$. Applying Le Chatelier's principle, an increase \in the concentration of the reactant$[\mathrm{H}^+]$ exerts stress on the system, which is relieved by shifting the reaction equilibrium to the right to produce more products.

The ease with which a metal ion undergoes oxidation is determined by its oxidation potential, which is the negative of the given standard reduction potential $E_{\mathrm{M}^{3+}/\mathrm{M}^{2+}}^o$. While Mn, Fe, and Co exhibit positive reduction potentials of +1.57 V, +0.77 V, and +1.97 V respectively, Cr stands out with a negative reduction potential of -0.41 V. Converting these to oxidation potentials reveals that Cr possesses the highest value of +0.41 V, indicating that it most easily loses an electron to transition from the +2 to the +3 oxidation state.

Kohlrausch's Law of Independent Migration of Ions states that the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. We can determine the conductivity of NaBr by combining the given electrolytes such that the desired ions, Na+ and Br-, remain and the unwanted ions, K+ and Cl-, cancel out. This leads to the following relationship:

$\Lambda^{\circ}(\mathrm{NaBr}) = \Lambda^{\circ}(\mathrm{NaCl}) + \Lambda^{\circ}(\mathrm{KBr}) - \Lambda^{\circ}(\mathrm{KCl})$

Substituting the specified values into this equation results in 126 + 152 - 150, which yields 128 S cm2 mol-1.

A hydrogen-oxygen fuel cell functions by converting chemical energy directly into electrical energy through the controlled redox reaction $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$. This combustion process facilitates the transfer of electrons from the anode to the cathode, which is the mechanism used to establish a steady electrical potential difference between the two electrodes.

Faraday's laws of electrolysis state that the mass of a substance deposited at an electrode is directly proportional to the total charge passed through the solution. Because silver ions gain one electron to form metallic silver, passing 96500 C of electricity-the charge required for one mole of electrons-deposits one mole of silver, which has a mass of 108 g.

To find the mass deposited by 9650 C, we calculate the ratio of the supplied charge to the standard Faraday constant and multiply by the molar mass of silver. The resulting equation is $\frac{108}{96500} \times 9650 = 10.8$, which shows that 10.8 g of silver is deposited on the cathode.

Determining the equilibrium constant for a cell reaction relies on the relationship between the standard cell potential and equilibrium, which at 25°C is described by the formula $\log K = \frac{n E^{\circ}_{\text{cell}}}{0.059}$. In this equation, $n$ represents the number of electrons involved in the redox change and $E^{\circ}_{\text{cell}}$ is the standard e.m.f. of the cell. Substituting the given values of $n = 2$ and $E^{\circ}_{\text{cell}} = 0.295 \;\text{V}$ into this expression gives $\log K = \frac{2 \times 0.295}{0.059}$. Simplifying the numerator leads to $\log K = \frac{0.590}{0.059}$, which results in 10. Consequently, the equilibrium constant is determined to be $K = 1 \times 10^{10}$.

Ships are typically constructed from iron, a metal that inevitably undergoes oxidation, or rusting, when continuously exposed to the harsh, salty environment of the ocean. Attaching blocks of magnesium creates a system of cathodic protection, as magnesium is more reactive than iron and serves as a sacrificial anode. Consequently, the magnesium corrodes preferentially in this electrochemical process, effectively shielding the hull from the damaging action of water and salt.

Reducing power is determined by the ease with which a metal loses electrons to act as a reducing agent. A more negative standard reduction potential indicates a stronger thermodynamic tendency for a metal to undergo oxidation, thereby increasing its effectiveness in reduction reactions. Among the provided values, metal $B$ has the most negative potential at -3.0 V, followed by metal $C$ at -1.2 V, and finally metal $A$ at +0.5 V. Because the reducing capacity scales with the magnitude of the negative potential, the metals follow the trend $B > C > A$.

The Nernst equation allows us to determine the non-standard cell potential by adjusting the standard cell potential based on the actual concentrations of the reactants and products. For the reaction $\mathrm{Zn}(\mathrm{s}) + \mathrm{Cu}^{2+}(0.1\,\mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(1\,\mathrm{M}) + \mathrm{Cu}(\mathrm{s})$, two electrons are transferred between the species, so we use $n=2$ in the formula
$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.059}{n} \log \frac{[\mathrm{Zn}^{2+}]\$}{[\mathrm{Cu}^{2+}]\$}$
By substituting the given values, the calculation becomes $E_{\text{cell}} = 1.10 - \frac{0.059}{2} \log \frac{1}{0.1}$. Because $\log 10$ equals $1$, this simplifies to $1.10 - 0.0295$, yielding a final value of $1.07\,\mathrm{V}$.