Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 8.
Q71 (2026)
Electricity is passed through an acidic solution of $Cu^{2+}$ till all the $Cu^{2+}$ was exhausted, leading to the deposition of $300 \text{ mg}$ of Cu metal. However, a current of $600 \text{ mA}$ was continued to pass through the same solution for another $28 \text{ minutes}$ by keeping the total volume of the solution fixed at $200 \text{ mL}$. The total volume of oxygen evolved at STP during the entire process is ____ $\text{mL}$. (Nearest integer)
[Given :
$Cu^{2+}(aq)+2e^{-} \rightarrow Cu(s) \quad E_{red}^0 = +0.34 \text{ V}$
$O_2(g)+4H^++4e^{-} \rightarrow 2H_2O \quad E_{red}^0 = +1.23 \text{ V}$
Molar mass of $Cu = 63.54 \text{ g mol}^{-1}$
Molar mass of $O_2 = 32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = $22.4 \text{ L}$ ]
### Related Formula
$$\text{Equivalents of metal deposited} = \text{Equivalents of gas evolved (in Phase 1)}$$
$$n_{e^-} = \frac{Q}{F} = \frac{I \times t}{96500}$$
### Core Logic
Phase 1: Deposition of $300 \text{ mg}$ Cu.
By Faraday's Laws, the equivalents of copper deposited at the cathode must equal the equivalents of oxygen evolved at the anode during this period.
Equivalents of $Cu = \frac{W}{E} = \frac{300 \times 10^{-3}}{\frac{63.54}{2}}$
Equivalents of $O_2 = n_{O_2} \times 4$
$\frac{300 \times 10^{-3} \times 2}{63.54} = n_{O_2} \times 4$
$2.36 \times 10^{-3} = n_{O_2}$ (moles of $O_2$ in Phase 1)
Phase 2: Current continued.
$I = 600 \text{ mA} = 0.6 \text{ A}$, $t = 28 \text{ mins} = 28 \times 60 \text{ seconds}$.
Moles of electrons passed $= \frac{0.6 \times 28 \times 60}{96500} = 0.010445 \text{ moles of } e^{-}$
Equivalents of $O_2$ evolved in Phase 2 = Moles of $e^{-}$ passed = $0.010445$
$n_{O_2} \text{ (Phase 2)} = \frac{0.010445}{4} = 2.611 \times 10^{-3} \text{ mol}$
### Step 1: Calculate Total Volume of Oxygen
Total moles of $O_2$ evolved $= (2.36 \times 10^{-3}) + (2.611 \times 10^{-3}) = 4.971 \times 10^{-3} \text{ mol}$
Total volume at STP:
$V_{O_2} = n_{total} \times 22400 \text{ mL}$
$V_{O_2} = 4.971 \times 10^{-3} \times 22400 \text{ mL} = 111.35 \text{ mL}$
Rounding to nearest integer $\implies 111 \text{ mL}$.
### Pattern Recognition
Equivalents of products at cathode and anode are always equal in any given time span. Valency factor for $O_2$ evolution from water is 4.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q74 (2026)
Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1/30 of the molar conductivity of another weak acid HZ with concentration of 0.02 of M. If $\lambda_{\mathrm{Q}^{-}}^{0}$ happened to be equal with $\lambda_{\mathrm{Z}^{-}}^{0}$ , then the difference of the $\mathrm{pK}_{\mathrm{a}}$ values of the two weak acids ( $\mathrm{pK}_{\mathrm{a}}(\mathrm{HQ}) - \mathrm{pK}_{\mathrm{a}}(\mathrm{HZ})$ ) is ____ (Nearest integer).
[Given : degree of dissociation ( $\alpha$ ) << 1 for both weak acids, $\lambda^{\circ}$ : limiting molar conductivity of ions]
### Related Formula
$$\alpha = \frac{\lambda_m}{\lambda_m^{\infty}}$$
$$K_a \simeq C \alpha^2 \quad (\text{for } \alpha \ll 1)$$
### Core Logic
For weak acid HQ:
$\alpha_1 = \frac{\lambda_m(\mathrm{HQ})}{\lambda_m^{\infty}(\mathrm{HQ})}$
$K_a(\mathrm{HQ}) = C_1 \alpha_1^2 = 0.18 \left(\frac{\lambda_m(\mathrm{HQ})}{\lambda_m^{\infty}(\mathrm{HQ})}\right)^2$
For weak acid HZ:
$\alpha_2 = \frac{\lambda_m(\mathrm{HZ})}{\lambda_m^{\infty}(\mathrm{HZ})}$
$K_a(\mathrm{HZ}) = C_2 \alpha_2^2 = 0.02 \left(\frac{\lambda_m(\mathrm{HZ})}{\lambda_m^{\infty}(\mathrm{HZ})}\right)^2$
### Step 1: Evaluate Ratios
Since limiting molar conductivities of anions $\lambda_{\mathrm{Q}^-}^0$ and $\lambda_{\mathrm{Z}^-}^0$ are equal, and both have $H^+$ as the cation:
$\lambda_m^{\infty}(\mathrm{HQ}) = \lambda_m^{\infty}(\mathrm{HZ})$
Take the ratio of their ionization constants:
$$\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})} = \frac{C_1}{C_2} \cdot \left[ \frac{\lambda_m(\mathrm{HQ})}{\lambda_m(\mathrm{HZ})} \right]^2$$
We are given $\lambda_m(\mathrm{HQ}) = \frac{1}{30} \lambda_m(\mathrm{HZ})$, so the ratio inside the bracket is $\frac{1}{30}$.
$$\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})} = \frac{0.18}{0.02} \times \left(\frac{1}{30}\right)^2$$
$$\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})} = 9 \times \frac{1}{900} = \frac{1}{100}$$
### Step 2: Logarithmic Difference
Taking the negative logarithm on both sides:
$$-\log\left(\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})}\right) = -\log(10^{-2})$$
$$pK_a(\mathrm{HQ}) - pK_a(\mathrm{HZ}) = 2$$
### Pattern Recognition
When $\lambda_{\text{anion}}^0$ is identical for both acids, $\lambda_m^\infty$ cancels out entirely in comparative ratios. Use $K_a = C \cdot (\lambda_m / \lambda_m^\infty)^2$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Q72 (2026)
Consider the following redox reaction taking place in acidic medium
$BH_{4}^{-}(aq) + ClO_{3}^{-}(aq) \rightarrow H_{2}BO_{3}^{-}(aq) + Cl^{-}(aq)$
If the Nernst equation for the above balanced reaction is
$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{nF} \ln Q,$
Then the value of n is _____.
### Core Logic
To find 'n', the number of moles of electrons transferred, we must balance the overall redox equation and calculate the net electron exchange.
### Step 1: Assign Oxidation States
In $BH_4^-$, Hydrogen is $-1$ (hydride), so Boron is $+3$. In $H_2BO_3^-$, Boron is $+3$. The oxidation state of B doesn't change, but $H$ oxidizes from $-1$ to $+1$ (in water/acid).
Alternatively, treat the whole moiety:
Oxidation half-reaction: $BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-$
In $ClO_3^-$, Chlorine is $+5$. In $Cl^-$, Chlorine is $-1$.
Reduction half-reaction: $ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O$
### Step 2: Balance Electrons
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4 to equalize electrons transferred:
$3 \times (BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-) \implies 24e^-$
$4 \times (ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O) \implies 24e^-$
The lowest common multiple of electrons exchanged is 24.
### Final Conclusion
The balanced equation transfers 24 electrons, so $n = 24$ in the Nernst equation.
### Pattern Recognition
The total electrons '$n$' in the Nernst equation is always the lowest common multiple of electrons from the balanced oxidation and reduction half-reactions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Redox Reactions
Q73 (2026)
For strong electrolyte $\Lambda_{m}$ increases slowly with dilution and can be represented by the equation
$$\Lambda_{m} = \Lambda_{m}^{\circ} - Ac^{1/2}$$
Molar conductivity values of the solutions of strong electrolyte AB at $18^{\circ}C$ are given below :
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">c $[mol\ L^{-1}]$</th><th style="border: 1px solid #888; padding: 8px;">0.04</th><th style="border: 1px solid #888; padding: 8px;">0.09</th><th style="border: 1px solid #888; padding: 8px;">0.16</th><th style="border: 1px solid #888; padding: 8px;">0.25</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">$\Lambda_{m} [S\ cm^{2} mol^{-1}]$</td><td style="border: 1px solid #888; padding: 8px;">96.1</td><td style="border: 1px solid #888; padding: 8px;">95.7</td><td style="border: 1px solid #888; padding: 8px;">95.3</td><td style="border: 1px solid #888; padding: 8px;">94.9</td></tr></tbody></table></div>
The value of constant A based on the above data $[in\ S\ cm^{2} mol^{-1}/(mol/L)^{1/2}]$ unit is ____.
### Related Formula
$$\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}$$ (Debye-Huckel-Onsager equation)
### Core Logic
Using the equation for the first set of values ($c = 0.04, \sqrt{c} = 0.2$):
$96.1 = \Lambda_{m}^{\circ} - A(0.2)$ --- (1)
Using the equation for the second set of values ($c = 0.09, \sqrt{c} = 0.3$):
$95.7 = \Lambda_{m}^{\circ} - A(0.3)$ --- (2)
### Step 1: Solve for A
Subtract eq (2) from eq (1):
$96.1 - 95.7 = A(0.3) - A(0.2)$
$0.4 = 0.1A$
$A = \frac{0.4}{0.1} = 4$
### Pattern Recognition
Direct linear interpolation. Change in $\Lambda_{m}$ over change in $\sqrt{c}$ gives the slope (which is A).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q74 (2026)
A volume of x mL of $5\text{ M } NaHCO_{3}$ solution was mixed with $10\text{ mL}$ of $2\text{ M } H_{2}CO_{3}$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $235.3\text{ mV}$, then the value of x = ____ mL (nearest integer).
$Sn(s) | Sn(OH)_{6}^{2-}(0.5\text{ M}) | HSnO_{2}^{-}(0.05\text{ M}) | OH^{-} || Bi_{2}O_{3}(s) | Bi(s)$
Consider upto one place of decimal for intermediate calculations
Given :
$E_{HSnO_{2}^{-}|Sn(OH)_{6}^{2-}}^{o} = -0.9\text{ V}$
$E_{Bi_{2}O_{3}|Bi}^{o} = -0.44\text{ V}$
$pKa_{(H_{2}CO_{3})} = 6.11$
$\frac{2.303RT}{F} = 0.059\text{ V}$
$\text{Antilog(1.29)} = 19.5$
### Related Formula
$$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log Q$$
$$pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$$
### Core Logic
Note: In the question paper, $E_{HSnO_2^-/[Sn(OH)_6]^{2-}}^{\circ} = -0.9\text{ V}$ is given, but standard NTA solution requires assuming $E_{[Sn(OH)_6]^{2-}/HSnO_2^-}^{\circ} = -0.9\text{ V}$ for standard operation. (Our Ans. is Bonus due to this discrepancy, NTA Answer is 78). We will solve using the assumed valid logic.
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = -0.44 - (-0.90) = 0.46\text{ V}$
Oxidation Half: $HSnO_2^- + H_2O + 3OH^- \rightarrow [Sn(OH)_6]^{2-} + 2e^-$
Reduction Half: $Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$
Overall: $3HSnO_2^- + Bi_2O_3 + 6H_2O + 3OH^- \rightarrow 3[Sn(OH)_6]^{2-} + 2Bi$
Here $n = 6$.
Nernst Eq:
$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{6} \log \frac{([Sn(OH)_6]^{2-})^3}{([HSnO_2^-]^3 [OH^-]^3)}$
$0.2353 = 0.46 - \frac{0.059}{6} \log \frac{(0.5)^3}{(0.05)^3 [OH^-]^3}$
$0.2353 = 0.46 - \frac{0.059}{2} \log \frac{10}{[OH^-]}$
$\log \left( \frac{10}{[OH^-]} \right) = \frac{(0.46 - 0.2353) \times 2}{0.059} = \frac{0.2247 \times 2}{0.059} = 7.6$
### Step 1: Calculate pH
$\log(10) - \log[OH^-] = 7.6$
$1 + pOH = 7.6 \implies pOH = 6.6$
$pH = 14 - 6.6 = 7.4$
### Step 2: Buffer Equation
Using Henderson-Hasselbalch equation for buffer of $NaHCO_3$ and $H_2CO_3$:
$pH = pK_a + \log \frac{n_{\text{salt}}}{n_{\text{acid}}}$
$7.4 = 6.11 + \log \frac{5x}{10 \times 2}$
$1.29 = \log \frac{5x}{20} = \log \frac{x}{4}$
Taking antilog:
$\frac{x}{4} = \text{Antilog}(1.29) = 19.5$
$x = 19.5 \times 4 = 78\text{ mL}$
### Pattern Recognition
Merge Nernst equation finding unknown concentration with Buffer equations. Determine overall cell reaction to find exact stoichiometry and 'n' electrons for Nernst.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
