Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 2.
Q48 (2025)
1 Faraday electricity was passed through $\mathrm{Cu}^{2+}$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through $\mathrm{Ag}^{+}$ (0.2 M, 1 L)/Ag electrolytic cells. After this, the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is ______ V.
{{IMG1}}
Given:
$\mathrm{E}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^{\circ} = 0.34 \mathrm{~V}$
$\mathrm{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}^{\circ} = 0.8 \mathrm{~V}$
$\frac{2.303RT}{F} = 0.06 \mathrm{~V}$
### Related Formula
$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{n} \log Q$$
### Core Logic
First, analyze the electrolysis step to determine final ionic concentrations:
1. **For $\mathrm{Cu}^{2+}/\mathrm{Cu}$ half-cell**:
- Initial moles of $\mathrm{Cu}^{2+} = 1.5 \text{ M} \times 1 \text{ L} = 1.5 \text{ mol}$.
- Reductive half-reaction: $\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}$.
- Passing $1 \text{ Faraday}$ converts: $\frac{1}{2} = 0.5 \text{ mol}$ of $\mathrm{Cu}^{2+}$.
- Remaining moles of $\mathrm{Cu}^{2+} = 1.5 - 0.5 = 1.0 \text{ mol}$.
- Final concentration $[\mathrm{Cu}^{2+}] = 1.0 \text{ M}$.
2. **For $\mathrm{Ag}^{+}/\mathrm{Ag}$ half-cell**:
- Initial moles of $\mathrm{Ag}^+ = 0.2 \text{ M} \times 1 \text{ L} = 0.2 \text{ mol}$.
- Reductive half-reaction: $\mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag}$.
- Passing $0.1 \text{ Faraday}$ converts: $0.1 \text{ mol}$ of $\mathrm{Ag}^+$.
- Remaining moles of $\mathrm{Ag}^+ = 0.2 - 0.1 = 0.1 \text{ mol}$.
- Final concentration $[\mathrm{Ag}^+] = 0.1 \text{ M}$.
Now, connect the two components into a galvanic cell:
- Anode reaction: $\mathrm{Cu(s)} \rightarrow \mathrm{Cu}^{2+}\mathrm{(aq)} + 2\mathrm{e}^-$
- Cathode reaction: $2\mathrm{Ag}^{+}\mathrm{(aq)} + 2\mathrm{e}^- \rightarrow 2\mathrm{Ag(s)}$
- Net cell reaction: $\mathrm{Cu(s)} + 2\mathrm{Ag}^{+}\mathrm{(aq)} \rightarrow \mathrm{Cu}^{2+}\mathrm{(aq)} + 2\mathrm{Ag(s)}$
- $n = 2$
Calculate standard cell potential:
$$E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag}^+/\mathrm{Ag}} - E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.80 - 0.34 = 0.46 \text{ V}$$
Applying Nernst Equation:
$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{2} \log \left( \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2} \right)$$
$$E_{\text{cell}} = 0.46 - 0.03 \log \left( \frac{1}{(0.1)^2} \right) = 0.46 - 0.03 \log(100)$$
$$E_{\text{cell}} = 0.46 - 0.03(2) = 0.46 - 0.06 = 0.40 \text{ V}$$
(Note: The potential is $0.4\text{ V}$ or $400\text{ mV}$).
### Pattern Recognition
Electrolysis modifies the bulk concentrations. First, use Faraday's laws to get the new concentration values ($[Cu^{2+}] = 1.0\text{ M}$, $[Ag^+] = 0.1\text{ M}$). Then plug these straight into standard Nernst equations.
### Evaluation Rubric / Model Answer
Requires complete calculations showing concentrations updated by electrolysis, followed by a double-transfer Nernst equation calculation.
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q50 (2025)
Consider the following half-cell reduction reaction:
$$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 6e^- + 14\text{H}^+\text{(aq)} \longrightarrow 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$$
The process is conducted with a concentration ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$. The specific pH value at which the EMF ($E$) of this reduction half-cell becomes exactly zero is _________ (as the nearest integer value).
Given parameters: $E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \text{ V}$ and $\frac{2.303RT}{F} = 0.059 \text{ V}$.
### Related Formula
The Nernst equation for a reduction half-cell is:
$$E = E^\circ - \frac{2.303RT}{nF} \log Q$$
For this reaction, the reaction quotient $Q$ is:
$$Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}] \cdot [\text{H}^+]^{14}}$$
### Execution
Step 1: Identify the number of transferred electrons ($n = 6$) and substitute the condition $E = 0$:
$$0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\text{H}^+]^{14}} \right)$$
Step 2: Isolate the logarithmic term:
$$1.33 = \frac{0.059}{6} \left[ \log(10^{-6}) - \log([\text{H}^+]^{14}) \right]$$
$$\frac{1.33 \times 6}{0.059} = -6 - 14 \log[\text{H}^+]$$
Step 3: Perform the arithmetic division:
$$135.254 = -6 - 14 \log[\text{H}^+]$$
Step 4: Rearrange the terms using the definition of pH ($-\log[\text{H}^+] = \text{pH}$):
$$135.254 + 6 = 14 \cdot \text{pH}$$
$$141.254 = 14 \cdot \text{pH}$$
$$\text{pH} = \frac{141.254}{14} = 10.089$$
Rounding to the nearest integer value gives **10**.
### Pattern Recognition
The exponent of the hydrogen ion concentration ($[\text{H}^+]^{14}$) heavily influences the cell potential. A small shift in pH causes a large change in EMF due to this factor of 14, which explains why the potential drops to zero even in a highly basic environment ($\text{pH} \approx 10$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Ionic Equilibrium
Q35 (2025)
Match List-I with List-II:
<div style="margin: 16px 0; overflow-x: auto;"><table style="width: 100%; border-collapse: collapse; font-size: 14px; text-align: left; background-color: rgba(0, 0, 0, 0.2); border-radius: 8px; overflow: hidden; border: 1px solid rgba(255, 255, 255, 0.1);"><thead style="background-color: rgba(255, 255, 255, 0.05); border-bottom: 1px solid rgba(255, 255, 255, 0.1);"><tr><th style="padding: 12px 16px; font-weight: 600; color: #e4e4e7;">List-I (Applications)</th><th style="padding: 12px 16px; font-weight: 600; color: #e4e4e7;">List-II (Batteries/Cell)</th></tr></thead><tbody><tr style="border-bottom: 1px solid rgba(255, 255, 255, 0.05);"><td style="padding: 12px 16px; color: #a1a1aa;">(A) Transistors</td><td style="padding: 12px 16px; color: #a1a1aa;">(I) Anode - Zn/Hg; Cathode - HgO + C</td></tr><tr style="border-bottom: 1px solid rgba(255, 255, 255, 0.05);"><td style="padding: 12px 16px; color: #a1a1aa;">(B) Hearing aids</td><td style="padding: 12px 16px; color: #a1a1aa;">(II) Hydrogen fuel cell</td></tr><tr style="border-bottom: 1px solid rgba(255, 255, 255, 0.05);"><td style="padding: 12px 16px; color: #a1a1aa;">(C) Invertors</td><td style="padding: 12px 16px; color: #a1a1aa;">(III) Anode - Zn; Cathode - Carbon</td></tr><tr><td style="padding: 12px 16px; color: #a1a1aa;">(D) Apollo space ship</td><td style="padding: 12px 16px; color: #a1a1aa;">(IV) Anode - Pb; Cathode - Pb | $PbO_2$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
- (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
### Core Logic
Matching applications to their respective electrochemical cells:
* Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with $MnO_2$ $
ightarrow$ (III).
* Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C $
ightarrow$ (I).
* Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | $PbO_2$ $
ightarrow$ (IV).
* Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells $
ightarrow$ (II).
### Pattern Recognition
Space missions universally trigger fuel cell pairs in standard test patterns due to the secondary requirement of gathering pure drinking water byproduct.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q36 (2025)
$O_{2}$ gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of $AgNO_{3}$ using silver electrodes.
(B) an aqueous solution of $AgNO_{3}$ using platinum electrodes.
(C) a dilute solution of $H_{2}SO_{4}$ using platinum electrodes.
(D) a high concentration solution of $H_{2}SO_{4}$ using platinum electrodes.
Choose the correct answer from the options given below:
- (B) and (C) only
- (A) and (D) only
- (B) and (D) only
- (A) and (C) only
### Core Logic
Analyzing anodic reactions during electrolysis:
* Case (A): With active Ag electrodes, silver oxidation occurs at the anode ($Ag
ightarrow Ag^{+} + e^{-}$). No oxygen is evolved.
* Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over $NO_3^-$ ions:$$2H_2O
ightarrow O_2 + 4H^{+} + 4e^{-}$$
* Case (C): In dilute $H_2SO_4$, water oxidation takes place, releasing $O_2$ gas at the anode.
* Case (D): In concentrated $H_2SO_4$, oxidation of $SO_4^{2-}$ creates peroxodisulphate ions ($S_2O_8^{2-}$), inhibiting oxygen evolution.
### Pattern Recognition
Remember that active electrodes participate directly in redox reactions, whereas inert electrodes (Pt, Graphite) yield oxygen gas when water is oxidized in the presence of oxoanions like $NO_3^-$ or dilute $SO_4^{2-}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q46 (2025)
Given below is the plot of the molar conductivity vs $\sqrt{\text{concentration}}$ for KCl in aqueous solution.
{{IMG1}}
If, for the higher concentration of KCl solution, the resistance of the conductivity cell is $100\Omega$ then the resistance of the same cell with the dilute solution is $\mathrm{x}\Omega$
The value of $\mathbf{x}$ is (Nearest integer)
### Related Formula
Conductivity relationship with cell parameters:
$$\kappa = G \cdot G^* = \frac{G^*}{R}$$
$$\lambda_m = \frac{\kappa \times 1000}{C}$$
where $G^*$ represents the static cell constant.
### Step 1: Setting Up Ratios
Using concentration subscripts $c$ (concentrated) and $d$ (dilute):
$$\frac{\kappa_c}{\kappa_d} = \frac{R_d}{R_c}$$
Expressing conductivity through molar conductivity values:
$$\kappa = \frac{\lambda_m \cdot C}{1000}$$
$$\frac{(\lambda_m \cdot C)_c}{(\lambda_m \cdot C)_d} = \frac{R_d}{R_c}$$
Substituting the graphical read coordinates ($C_c = 0.15^2, C_d = 0.1^2$ with scaled $\lambda_m$ parameters):
$$\frac{100 \cdot (0.15)^2}{150 \cdot (0.1)^2} = \frac{R_d}{100}$$
$$R_d = 150\,\Omega$$
### Pattern Recognition
Sees: Resistance correlation across specific graph coordinates.
Shortcut: Equate cell parameters through $\kappa \propto \frac{1}{R}$ and solve for the target resistance directly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
