Electrochemistry – JEE Main ChemistryPractice Questions & PYQs

Standard electrode potentials are intensive properties that cannot be added directly, whereas Gibbs free energy is an extensive property that allows summation across consecutive reaction steps using the relationship $\Delta G^\circ = -nFE^\circ$. The reduction of $\mathrm{FeO}_4^{2-}$ to $\mathrm{Fe}^{3+}$ involves a three-electron transfer with a potential of 2.2 V, and the subsequent reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$ involves a single-electron transfer with a potential of 0.70 V. Combining these results in an overall four-electron reduction of $\mathrm{FeO}_4^{2-}$ to $\mathrm{Fe}^{2+}$.

Representing the total energy change as the sum of the individual steps gives:

$-n_3 F E_3^\circ = (-n_1 F E_1^\circ) + (-n_2 F E_2^\circ)$

Substituting the values for the electrons and potentials into the equation:

$-4 F E_3^\circ = -(3 \times 2.2 F) - (1 \times 0.70 F)$

Dividing both sides by $-F$ simplifies the expression to:

$4 E_3^\circ = 3(2.2) + 0.70$

The value of the right side equals 7.3, so $4 E_3^\circ = 7.3$. Dividing by 4 gives an electrode potential of $E_3^\circ = 1.825$ V. Comparing this to the format $x \times 10^{-3}$ V, the value of $x$ is 1825.

Molar conductivity is defined by the expression $\Lambda_m = \frac{\kappa \times 1000}{M}$, where $\kappa$ represents the conductivity and $M$ denotes the molarity of the solution. Since the conductivity $\kappa$ is identical for both cells, the relationship between the two molar conductivities depends entirely on their respective molar concentrations.

Calculating the molarity for the first cell gives $M_1 = \frac{10}{0.02} = 500 \text{ M}$, which results in $\Lambda_{m1} = \frac{\kappa \times 1000}{500} = 2\kappa$. For the second cell, the molarity is $M_2 = \frac{20}{0.08} = 250 \text{ M}$, yielding $\Lambda_{m2} = \frac{\kappa \times 1000}{250} = 4\kappa$. By comparing these two values, it is evident that $\Lambda_{m2} = 2\Lambda_{m1}$.

Electrolysis of dilute sulphuric acid results in the decomposition of water, liberating hydrogen at the cathode and oxygen at the anode. First, determine the total charge passed using the relation $Q = I \times t$. With a current of 0.10 A flowing for 7200 seconds, which is equivalent to 2 hours, the total charge accumulated is 720 C. Dividing this total charge by the Faraday constant, 96500 C mol-1, gives approximately 0.00746 mol of electrons.

The stoichiometric relationships indicate that 4 moles of electrons are required to produce 1 mole of oxygen gas, whereas 2 moles of electrons are required to produce 1 mole of hydrogen gas. Applying these ratios, the oxygen produced is 0.001865 mol and the hydrogen produced is 0.00373 mol. Summing these values gives a total gas amount of approximately 0.005595 mol. Multiplying this total moles by the molar volume of 22.7 L mol-1 yields 0.127 L. Finally, converting this volume into cubic centimeters results in 127 cm3.

The Nernst equation provides the essential relationship to determine electron transfer in an electrochemical cell by linking the measured cell potential to standard conditions and the reaction quotient. In this specific configuration, the cell operates with a zinc anode undergoing oxidation and a tin cathode undergoing reduction, allowing us to establish the standard cell potential as the difference between the reduction potential of the cathode and the anode: $E^{\circ}_{cell} = E^{\circ}(\mathrm{Sn}^{x+}/\mathrm{Sn}) - E^{\circ}(\mathrm{Zn}^{2+}/\mathrm{Zn}) = 0.008\ \mathrm{V} - (-0.763\ \mathrm{V}) = 0.771\ \mathrm{V}$.

Substituting the known values into the Nernst equation, $E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log Q$, we get $0.801\ \mathrm{V} = 0.771\ \mathrm{V} - \frac{0.06}{n} \log(10^{-2})$. Simplifying the voltage difference results in $0.03\ \mathrm{V} = -\frac{0.06}{n} \times (-2)$, which effectively becomes $0.03 = \frac{0.12}{n}$. Solving for the variable $n$ reveals that the number of electrons involved in the reaction is 4.

The cell constant of a conductivity cell is a characteristic parameter defined by the geometric properties of the electrodes, specifically given by the product of the solution's conductivity $\kappa$ and its measured resistance $R$, expressed as $G^* = \kappa \times R$. For a 0.01 M KCl solution with a conductivity of 0.152 $\times 10^{-3}$ S cm$^{-1}$ and a resistance of 1750 $\Omega$, the cell constant is determined by substituting these values directly into the relationship. Performing the multiplication $0.152 \times 10^{-3} \times 1750 = 266 \times 10^{-3}$ yields a cell constant of 266 $\times 10^{-3}$ cm$^{-1}$.

The Nernst equation connects the measured cell potential to the ion concentrations in an electrochemical cell. For the reaction $\mathrm{H}_2(\mathrm{g}) + \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow 2\mathrm{H}^{+}(\mathrm{aq}) + \mathrm{Cu}(\mathrm{s})$, two electrons are transferred, and the standard cell potential is 0.34 V. Utilizing the equation E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{2} \log \frac{[\mathrm{H}^{+}]^2}{[\mathrm{Cu}^{2+}]\}$, we substitute the known values 0.576 V and 0.01 M into the expression:

$0.576 = 0.34 - 0.03 \log \frac{[\mathrm{H}^{+}]^2}{0.01}$

Rearranging the logarithmic term gives 0.236 = -0.03(2 \log \[\mathrm{H}^{+}]$ + 2)$, which simplifies to$0.236 = 0.06 \mathrm{pH} - 0.06$after applying the definition$\mathrm{pH} = -\log [\mathrm{H}^{+}]$. Solving this yields 4.93, which rounds to the nearest integer, 5.

The Nernst equation relates the actual cell potential to the standard cell potential by accounting for the concentrations of the participating ions. For the reaction $\mathrm{Cu}(s) + 2\mathrm{Ag}^{+}(aq) \rightarrow \mathrm{Cu}^{2+}(aq) + 2\mathrm{Ag}(s)$, the potential is given by:

$E_{\mathrm{cell}} = E_{\mathrm{cell}}^0 - \frac{0.06}{2} \log \frac{[\mathrm{Cu}^{2+}]\$}{[\mathrm{Ag}^{+}]^2}$

Inserting the provided concentration values, the expression becomes $0.43 = E_{\mathrm{cell}}^0 - 0.03 \log \frac{10^{-3}}{(10^{-2})^2}$, which simplifies to $0.43 = E_{\mathrm{cell}}^0 - 0.03 \log 10$. This calculation results in a standard cell potential $E_{\mathrm{cell}}^0$ of $0.46\ \mathrm{V}$.

The standard cell potential is determined by subtracting the reduction potential of the anode from the reduction potential of the cathode, expressed as:

$E_{\mathrm{cell}}^0 = E_{\mathrm{Ag}^{+}/\mathrm{Ag}}^0 - E_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^0$

By substituting the known value $0.80\ \mathrm{V}$ for silver and the calculated $0.46\ \mathrm{V}$ for the cell, the standard electrode potential for copper is $0.80\ \mathrm{V} - 0.46\ \mathrm{V} = 0.34\ \mathrm{V}$. This value corresponds to $34 \times 10^{-2}\ \mathrm{V}$.

A reliable reference electrode must maintain a consistent electromotive force (EMF) despite changes in ambient conditions, specifically temperature. The term $\left(\frac{\partial E}{\partial T}\right)_P$ represents the temperature coefficient of the cell, which indicates how much the potential shifts as the temperature varies. To minimize error, we prioritize an electrode that exhibits the least sensitivity to these thermal changes. When evaluating the coefficients provided for the cells-$1 \times 10^{-4}$ for A, $2 \times 10^{-4}$ for B, $0.1 \times 10^{-4}$ for C, and $0.2 \times 10^{-4}$ for D-cell C stands out as having the smallest temperature coefficient, making it the best choice for a stable reference electrode.

Determining the Gibbs free energy change for an electrochemical reaction requires finding the actual cell potential under non-standard conditions using the Nernst equation and subsequently relating this potential to free energy. For the given reaction, copper acts as the anode since it undergoes oxidation, and tin acts as the cathode since it undergoes reduction. The standard cell potential is determined by subtracting the standard oxidation potential of the anode from the cathode potential:
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = -0.14 - 0.34 = -0.48 \text{ V}$
Once the standard cell potential is established, the Nernst equation allows for the calculation of the cell potential at the specified concentrations. Given that the number of electrons transferred is $n=2$, the expression is
$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\mathrm{Cu}^{2+}]\$}{[\mathrm{Sn}^{2+}]\$} \right)$
Substituting the known concentration values where the ratio of copper ions to tin ions is 10 yields $\log 10$, which is 1. This results in an actual potential of $-0.48 - 0.0295$, or $-0.5095$ V.

The Gibbs free energy change is then derived from the formula $\Delta G = -nFE_{\text{cell}}$. Substituting the values for $n$, $F$, and the calculated potential gives
$\Delta G = -2 \times 96500 \times (-0.5095) = 98333.5 \text{ J mol}^{-1}$
Converting this value to kilojoules per mole provides 98.3335 kJ mol$^{-1}$. To match the target format of $x \times 10^{-1}$ kJ mol$^{-1}$, we express this as $983.335 \times 10^{-1}$ kJ mol$^{-1}$, making the nearest integer for $x$ equal to 983.

The standard electrode potential, denoted as $E^\circ_{M^{2+}/M}$, measures the thermodynamic tendency of a metal ion to gain two electrons and reduce to its metallic state. Most 3d transition metals exhibit negative potentials, indicating they are more stable as ions than as neutral solids, but these values vary significantly across the series. Specifically, the reduction potentials are -0.90 V for Chromium, -0.44 V for Iron, -0.76 V for Zinc, and +0.34 V for Copper. Since Copper is the only element among these with a positive reduction potential, it possesses the highest value in the series.

Electrochemical cells are classified based on their ability to be recharged, where primary batteries are irreversible and secondary batteries can be restored by reversing the redox reactions. The reaction $\mathrm{Cd}(s)+2\mathrm{Ni}(\mathrm{OH})_3(s) \rightarrow \mathrm{CdO}(s)+2\mathrm{Ni}(\mathrm{OH})_2(s)+\mathrm{H}_2\mathrm{O}(l)$ represents the discharge of a secondary nickel-cadmium battery, while the $\mathrm{Zn(Hg)}+\mathrm{HgO}(s) \rightarrow \mathrm{ZnO}(s)+\mathrm{Hg}(l)$ reaction identifies the mercury cell as a primary battery. Conversely, the conversion of lead sulfate back into lead, lead dioxide, and sulfuric acid via $2\mathrm{PbSO}_4(s)+2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{Pb}(s)+\mathrm{PbO}_2(s)+2\mathrm{H}_2\mathrm{SO}_4(aq)$ constitutes the charging of a secondary lead-acid battery, and the $2\mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2\mathrm{H}_2\mathrm{O}(l)$ equation characterizes the operation of a fuel cell.

The behavior of molar conductivity ($\Lambda_m$) upon dilution is fundamentally governed by the nature of the electrolyte. Strong electrolytes like KI are already fully dissociated, so their molar conductivity increases only marginally as the solution is diluted, rendering the claim of a steep increase incorrect. Conversely, weak electrolytes such as carbonic acid undergo increased dissociation upon dilution, leading to a sharp rise in molar conductivity, which invalidates the claim that it increases slowly.

Kohlrausch's law of independent migration of ions establishes that the limiting molar conductivity of an electrolyte is the sum of the conductivities of its constituent ions. To determine the limiting molar conductivity of silver iodide, one must sum the conductivities of sodium iodide and silver nitrate, as this combination introduces the desired silver and iodide ions while also including sodium and nitrate ions. Subtracting the limiting molar conductivity of sodium nitrate removes the unwanted contributions from the sodium and nitrate ions, effectively isolating the value for silver iodide. This arithmetic operation is represented by the formula:

$\Lambda_m^{\infty }(\mathrm{AgI}) = \Lambda_m^{\infty }(\mathrm{NaI}) + \Lambda_m^{\infty }(\mathrm{AgNO}_3) - \Lambda_m^{\infty }(\mathrm{NaNO}_3)$

Substituting the given figures leads to the following calculation:

$12.7 + 13.3 - 12.0 = 14.0$

The final limiting molar conductivity of silver iodide is 14.0 mS m^2 mol^-1.

Faraday's laws of electrolysis establish that the amount of substance deposited is proportional to the quantity of electricity passed through the electrolyte. The cathodic reduction process, $\mathrm{Fe}^{3+} + 3\mathrm{e}^{-} \rightarrow \mathrm{Fe}$, confirms that 3 moles of electrons are required to produce 1 mole of metallic iron. Given that 1 mole of iron has a mass of 56 g and 1 mole of electrons carries 96500 C, the total charge needed to deposit 0.3482 g of iron is calculated as $\frac{0.3482}{56} \times 3 \times 96500$, which results in 1800.06 C. Substituting this charge and the current of 1.5 A into the equation $Q = I \times t$, we get $1.5 \times t = 1800.06$. Solving for $t$ gives 1200 s, which is equivalent to 20 min.

The standard Gibbs free energy change, represented as $\Delta_f G^{\circ}$, is determined by the equation $\Delta_f G^{\circ} = -nFE_{\text{cell}}^{\circ}$, which correlates the electrochemical cell's potential with its thermodynamic spontaneity. In the specified cell, hydrogen gas undergoes oxidation at the anode, $\mathrm{H}_2 \rightarrow 2\mathrm{H}^{+} + 2e^{-}$, while silver ions are reduced at the cathode, $\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}$. Balancing the electron transfer between these half-reactions confirms that two electrons are exchanged, so $n=2$.

Substituting the given values into the formula yields the expression $-2 \times 96500 \times 0.5332$. Calculating this product results in -102907.6 Joules per mole. Converting this energy to kilojoules per mole requires dividing by 1000, which results in -102.9 kilojoules per mole. Rounding to the nearest integer provides a final value of -103 kilojoules per mole.

Standard electrode potential is not an extensive property and is not directly additive, but the Gibbs free energy change, defined as $\Delta G^o = -nFE^o$, is a state function that allows for the algebraic combination of reactions. To determine the potential for the reduction of $Sn^{4+}$ to $Sn^{2+}$, the Gibbs energy of the target reaction is derived from the sum of the free energy changes of the known half-reactions involving the transfer of electrons.

Using the relationship for standard electrode potentials, the value is calculated by taking the weighted sum of the potentials divided by the number of electrons in the target reaction. Given the oxidation potential of $0.140 \, \text{V}$ for $Sn \rightarrow Sn^{2+} + 2e^{-}$ and the reduction potential of $0.010 \, \text{V}$ for $Sn^{4+} + 4e^{-} \rightarrow Sn$, the calculation is as follows:
$E^o = \frac{4(0.010) + 2(0.140)}{2} = 0.16 \, \text{V}$
This result is equal to $16 \times 10^{-2} \, \text{V}$.

A spontaneous electrochemical reaction is characterized by a positive standard cell potential, which is determined by subtracting the reduction potential of the anode from the reduction potential of the cathode. In this cell, the reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$ functions as the cathodic process, while the oxidation of $\mathrm{I}^{-}$ to $\mathrm{I}_2$ occurs at the anode. Substituting the given values into the formula $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$, we calculate $E^\circ_{\text{cell}} = 0.77 - 0.54 = 0.23$ V. Given that the potential is represented as $x \times 10^{-2}$ V, we equate $0.23 = x \times 10^{-2}$ to find that $x = 23$.

A half-cell reaction is considered pH dependent if its balanced redox equation requires the participation of hydrogen ions ($\mathrm{H}^{+}$) or hydroxide ions ($\mathrm{OH}^{-}$) to maintain charge and mass balance. For the reduction of the permanganate ion, the reaction stoichiometry changes significantly depending on the acidity of the medium, as illustrated by the following reduction reaction in an acidic solution:
$\mathrm{MnO}_4^{-} + 8\mathrm{H}^{+} + 5\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$
Because this process directly consumes $\mathrm{H}^{+}$ ions, the electrode potential varies as the pH of the solution changes, whereas the other provided half-cells rely solely on electron transfer without involving protons or hydroxide ions.

The Nernst equation, $E_{\text{cell}} = E^{\ominus}_{\text{cell}} - \frac{0.06}{n} \log Q$, relates the cell potential to the concentration of species in the reaction $\mathrm{H}_2(\mathrm{g}) + \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow 2\mathrm{H}^+(\mathrm{aq}) + \mathrm{Cu}(\mathrm{s})$, where $n = 2$ electrons are transferred and the standard cell potential is $E^{\ominus}_{\text{cell}} = 0.34\ \mathrm{V}$. Substituting the known values for the $\mathrm{pH} = 3$, which implies $[\mathrm{H}^+] = 10^{-3}\ \mathrm{M}$, the measured cell potential $E_{\text{cell}} = 0.31\ \mathrm{V}$, and the reaction quotient Q = \frac{[\mathrm{H}^+]^2}{[\mathrm{Cu}^{2+}]\ P_{\mathrm{H}_2}}$into the Nernst expression gives$0.31 = 0.34 - \frac{0.06}{2} \log \frac{10^{-6}}{10^{-x}}$. Simplifying this expression leads to$-0.03 = -0.03(x - 6)$, which yields$1 = x - 6$, confirming that$x = 7$.

The quantity of electricity required for an electrochemical reduction is determined by the number of moles of electrons involved, as one mole of electrons provides one Faraday of charge. Reducing the dichromate ion involves shifting the oxidation state of chromium from +6 to +3. Since there are two chromium atoms in 1 mole of $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, the total transfer involves 6 electrons, as represented by the half-reaction $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 6\mathrm{e}^- \to 2\mathrm{Cr}^{3+}$. Therefore, 6 Faraday of electricity are necessary to fully reduce 1 mole of the reactant.