Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 3.
Q45 (2025)
Correct order of limiting molar conductivity for cations in water at $298\mathrm{~K}$ is:
- $\mathrm{H}^{+} > \mathrm{Na}^{+} > \mathrm{K}^{+} > \mathrm{Ca}^{2 + } > \mathrm{Mg}^{2 + }$
- $\mathrm{H}^{+} > \mathrm{Ca}^{2+} > \mathrm{Mg}^{2+} > \mathrm{K}^{+} > \mathrm{Na}^{+}$
- $\mathrm{Mg}^{2+} > \mathrm{H}^{+} > \mathrm{Ca}^{2+} > \mathrm{K}^{+} > \mathrm{Na}^{+}$
- $\mathrm{H}^{+} > \mathrm{Na}^{+} > \mathrm{Ca}^{2 + } > \mathrm{Mg}^{2 + } > \mathrm{K}^{+}$
### Related Formula
$$\lambda_m^\circ \text{ values at 298 K in S cm}^2\text{mol}^{-1}$$
### Core Logic
Standard limiting ionic molar conductivities ($\lambda^\circ$) in aqueous solution at $298\mathrm{~K}$:
- $\mathrm{H^+}: 349.8\mathrm{~S~cm^2~mol^{-1}}$ (exceptionally high due to Grotthuss proton-jumping mechanism)
- $\mathrm{Ca^{2+}}: 119.0\mathrm{~S~cm^2~mol^{-1}}$
- $\mathrm{Mg^{2+}}: 106.1\mathrm{~S~cm^2~mol^{-1}}$
- $\mathrm{K^+}: 73.5\mathrm{~S~cm^2~mol^{-1}}$
- $\mathrm{Na^+}: 50.1\mathrm{~S~cm^2~mol^{-1}}$
### Step 1: Final Order
Arranging in decreasing order:
$$\mathrm{H^+ > Ca^{2+} > Mg^{2+} > K^+ > Na^+}$$
### Pattern Recognition
Hydration effects: Smaller univalent cations ($\mathrm{Na^+}$) are more heavily hydrated than larger ones ($\mathrm{K^+}$), so $\lambda^\circ(\mathrm{K^+}) > \lambda^\circ(\mathrm{Na^+})$.
Divalent ions have higher ionic charge density and higher molar conductivities than alkali metal ions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q49 (2025)
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be $185\mathrm{~S~cm}^2\mathrm{mol}^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and $70\mathrm{~S~cm}^2\mathrm{mol}^{-1}$ , respectively. If molar conductance of $0.02\mathrm{~M}$ solution of ammonium hydroxide is $85.5\mathrm{~S~cm}^2\mathrm{mol}^{-1}$ , its degree of dissociation is given by $\mathbf{x}\times 10^{-1}$ . The value of $\mathbf{x}$ is _______. (Nearest integer)
### Related Formula
$$\Lambda_m^\circ(NH_4OH) = \lambda^\circ(NH_4^+) + \lambda^\circ(OH^-) \quad \text{(Kohlrausch's Law)}$$
$$\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}$$
### Core Logic
1. Find the limiting molar conductance of $NH_4^+$ using the $NH_4Cl$ data:
$$\Lambda_m^\circ(NH_4Cl) = \lambda^\circ(NH_4^+) + \lambda^\circ(Cl^-) = 185$$
$$\lambda^\circ(NH_4^+) = 185 - 70 = 115 \mathrm{~S \cdot cm^2 \cdot mol^{-1}}$$
2. Calculate $\Lambda_m^\circ$ for the weak electrolyte ammonium hydroxide ($NH_4OH$):
$$\Lambda_m^\circ(NH_4OH) = 115 + 170 = 285 \mathrm{~S \cdot cm^2 \cdot mol^{-1}}$$
3. Evaluate the degree of dissociation ($\alpha$):
$$\alpha = \frac{85.5}{285} = 0.3 = 3 \times 10^{-1}$$
Comparing with the format $\mathbf{x} \times 10^{-1}$, the value of $\mathbf{x}$ is **3**.
### Pattern Recognition
Kohlrausch's law allows direct algebraic recombination of ion conductances. Always construct the targeted weak base compound value by filtering out the spectator chloride contribution from the initial salt parameters.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q36 (2025)
On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively:
- $+4, +2, 0, +2$
- $+2, 0, +2, +4$
- $0, +2, +4, +2$
- $+2, 0, 0, +4$
### Related Formula
$$\text{Net Charging Reaction: } 2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$$
### Core Logic
During the **charging** cycle, the discharge chemical reactions are driven in reverse:
* **At Anode:** Lead sulfate ($PbSO_4$, where Lead is $+2$) is reduced back to metallic lead ($Pb$, oxidation state $0$):
$$x_1 = +2 \rightarrow y_1 = 0$$
* **At Cathode:** Lead sulfate ($PbSO_4$, where Lead is $+2$) is oxidized back into lead dioxide ($PbO_2$, where Lead is $+4$):
$$x_2 = +2 \rightarrow y_2 = +4$$
Thus, the values are $+2, 0, +2, +4$.
### Pattern Recognition
Be careful with wording! Discharging consumes $Pb$ and $PbO_2$ to create $PbSO_4$. **Charging** does the exact opposite, converting $PbSO_4$ ($+2$) back into its parent elements ($0$ and $+4$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q45 (2025)
Given below are two statements:
$1\text{ M}$ aqueous solution of each of $\text{Cu(NO}_3)_2$, $\text{AgNO}_3$, $\text{Hg}_2(\text{NO}_3)_2$; $\text{Mg(NO}_3)_2$ are electrolysed using inert electrodes,
Given: $E_{\text{Ag}^{+}/\text{Ag}}^{\theta} = 0.80\text{V}$ , $E_{\text{Hg}_2^{2+}/\text{Hg}}^{\theta} = 0.79\text{V}$, $E_{\text{Cu}^{2+}/\text{Cu}}^{\theta} = 0.24\text{V}$ and $E_{\text{Mg}^{2+}/\text{Mg}}^{\theta} = -2.37\text{V}$
Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be $\text{Ag}$, $\text{Hg}$ and $\text{Cu}$
Statement (II): Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statement, choose the most appropriate answer from the options given below [cite: 426, 427]
- $\text{Both statement I and statement II are incorrect}$
- $\text{Statement I is correct but statement II is incorrect}$
- $\text{Both statement I and statement II are correct}$
- $\text{Statement I is incorrect but statement II is correct}$
### Related Formula
$$\text{Ease of discharge at Cathode} \propto \text{Standard Reduction Potential } (E^0)$$
### Core Logic
- At the cathode, the metal ion with the highest standard reduction potential ($E^0$) gets reduced and deposited first. Arranging the given potentials:
$$E^0_{\text{Ag}^+/\text{Ag}} (0.80\text{V}) > E^0_{\text{Hg}_2^{2+}/\text{Hg}} (0.79\text{V}) > E^0_{\text{Cu}^{2+}/\text{Cu}} (0.24\text{V}) $$
Thus, deposition follows the order $\text{Ag}
ightarrow \text{Hg}
ightarrow \text{Cu}$ as voltage is steadily increased, confirming Statement I.
- For $\text{Mg}^{2+}$, its reduction potential is highly negative ($-2.37\text{ V}$), much lower than that of water ($-0.83\text{ V}$). Consequently, water undergoes reduction at the cathode instead of magnesium:
$$2\text{H}_2\text{O} + 2e^-
ightarrow \text{H}_2(g) + 2\text{OH}^- $$
This results in the evolution of **Hydrogen gas** at the cathode, not oxygen gas. Oxygen gas is evolved at the *anode* via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044]
### Step 1: Conclusion Match
Since Statement I is correct and Statement II is incorrect, we select option (2).
### Pattern Recognition
Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and $\text{Al}$), $\text{H}_2$ gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas ($\text{O}_2$) is an anodic product generated by water oxidation.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q26 (2025)
For the given cell:
$$Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$$
The standard cell potential of the above reaction is given by:
$$Ag^{+} + e^{-} \rightarrow Ag \quad E^{0} = x\text{ V}$$
$$Fe^{2+} + 2e^{-} \rightarrow Fe \quad E^{0} = y\text{ V}$$
$$Fe^{3+} + 3e^{-} \rightarrow Fe \quad E^{0} = z\text{ V}$$
- $x + y - z$
- $x + 2y - 3z$
- $y - 2x$
- $x + 2y$
### Related Formula
$$\Delta G^{0} = -nFE^{0}$$
### Core Logic
Using Gibbs free energy changes for individual steps to find the target reduction potential:
1. $$Ag^{+} + e^{-} \rightarrow Ag \quad \Delta G_1^0 = -1Fx$$
2. $$Fe^{2+} + 2e^{-} \rightarrow Fe \quad \Delta G_2^0 = -2Fy$$
3. $$Fe^{3+} + 3e^{-} \rightarrow Fe \quad \Delta G_3^0 = -3Fz$$
For the conversion of $$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$$, we compute the free energy change as:
$$\Delta G^0 = \Delta G_2^0 - \Delta G_3^0 = -2Fy - (-3Fz) = 3Fz - 2Fy$$
Thus, $$E^{0}_{Fe^{2+}/Fe^{3+}} = 2y - 3z$$
Combining with silver reduction:
$$E^{0}_{cell} = E^{0}_{Ag^{+}/Ag} + E^{0}_{Fe^{2+}/Fe^{3+}} = x + 2y - 3z$$
### Step 1: Final Calculation
The overall potential equals $x + 2y - 3z$. {{SOL_IMG1}}
### Pattern Recognition
Direct application of $\Delta G^0$ summation. Remember that standard cell potentials cannot be added directly unless the number of electrons involved is identical.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
