Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry).
Q47 (2025)
$0.2\%$ (w/v) solution of $\mathrm{NaOH}$ is measured to have resistivity $870.0~\mathrm{m\Omega~m}$. The molar conductivity of the solution will be $\times 10^{2}~\mathrm{mS~dm^{2}~mol^{-1}}$. (Nearest integer)
### Related Formula
$$\kappa = \frac{1}{\rho}$$
$$\Lambda_m = \frac{\kappa}{M}$$
### Core Logic
To compute the molar conductivity, we first calculate the molarity of the solution and the conductivity of the electrolyte from the given resistivity.
### Step 1: Calculate Molarity (M)
$0.2\%$ (w/v) $\mathrm{NaOH}$ means $0.2~\mathrm{g}$ of $\mathrm{NaOH}$ is present in $100~\mathrm{mL}$ of solution.
$$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40~\mathrm{g~mol^{-1}}$$
$$\text{Molarity } M = \frac{\text{Mass of solute}}{\text{Molar mass}} \times \frac{1000}{V_{\mathrm{mL}}} = \frac{0.2}{40} \times \frac{1000}{100} = 0.05~\mathrm{mol~L^{-1}} = 0.05~\mathrm{mol~dm^{-3}}$$
### Step 2: Calculate Conductivity (kappa) in dm Units
Given resistivity $\rho = 870.0~\mathrm{m\Omega~m} = 870 \times 10^{-3}~\Omega~m = 0.87~\Omega~m$.
Since $1~\mathrm{m} = 10~\mathrm{dm}$:
$$\rho = 0.87~\Omega \times (10~\mathrm{dm}) = 8.7~\Omega~dm$$
Now, conductivity $\kappa$ is:
$$\kappa = \frac{1}{\rho} = \frac{1}{8.7}~\Omega^{-1}~dm^{-1}$$
### Step 3: Calculate Molar Conductivity (Lambda_m)
$$\Lambda_m = \frac{\kappa}{M} = \frac{\frac{1}{8.7}~\mathrm{S~dm^{-1}}}{0.05~\mathrm{mol~dm^{-3}}} = \frac{1}{8.7 \times 0.05} = \frac{1}{0.435} \approx 2.29885~\mathrm{S~dm^2~mol^{-1}}$$
Converting $\mathrm{S}$ to $\mathrm{mS}$ ($1~\mathrm{S} = 10^3~\mathrm{mS}$):
$$\Lambda_m = 2.29885 \times 10^3~\mathrm{mS~dm^2~mol^{-1}} = 22.9885 \times 10^2~\mathrm{mS~dm^2~mol^{-1}}$$
Rounding off to the nearest integer gives **23**.
### Pattern Recognition
Ensure careful handling of volumetric conversions. Since concentration is expressed in moles per liter (equivalent to $\mathrm{dm^{-3}}$), expressing conductivity in terms of $\mathrm{dm^{-1}}$ directly eliminates the need for arbitrary $1000$ multiplication factors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q472 (2025)
Consider the following electrochemical cell at standard condition.
$$\mathrm{Au(s)} \vert \mathrm{QH}_2, \mathrm{Q} \vert \mathrm{NH}_4\mathrm{X} (0.01 \mathrm{M}) \vert \vert \mathrm{Ag}^+ (1 \mathrm{M}) \vert \mathrm{Ag(s)}$$
$$\mathrm{E}_{\text{cell}} = +0.4 \mathrm{V}$$
The couple $\mathrm{QH}_2 / \mathrm{Q}$ represents quinhydrone electrode, the half cell reaction is given below:
{{IMG1}}
$$\left[ \text{Given}: E_{Ag^+ / Ag}^o = +0.8 \mathrm{V} \text{ and } \frac{2.303 RT}{F} = 0.06 \mathrm{V} \right]$$
The $\mathrm{pK_b}$ value of the ammonium halide salt $(\mathrm{NH}_4\mathrm{X})$ used here is _____.
### Related Formula
Nernst equation for the net combined redox cell expression:
$$E = E^\circ - \frac{0.06}{2}\log\left(\frac{[\mathrm{H}^+]^2}{[\mathrm{Ag}^+]^2}\right)$$
Hydrolysis equation for a salt composed of a weak base and strong acid:
$$\mathrm{pH} = 7 - \frac{1}{2}\mathrm{pK_b} - \frac{1}{2}\log\mathrm{C}$$
### Core Logic
Let's compute the operational values line-by-row:
* Combined redox process: $\mathrm{QH_2 + 2Ag^+ \rightarrow Q + 2Ag + 2H^+}$.
* Standard cell potential difference: $E^\circ_{\text{cell}} = E^\circ_{\mathrm{Ag^+/Ag}} - E^\circ_{\mathrm{Q/QH_2}} = 0.8 - 0.7 = +0.1\mathrm{~V}$.
* Apply Nernst adjustments using known concentrations ($[\mathrm{Ag}^+] = 1\mathrm{~M}$):
$$0.4 = 0.1 - 0.06 \log [\mathrm{H}^+]$$
$$0.3 = 0.06 \times \mathrm{pH} \implies \mathrm{pH} = 5$$
### Step 1: Salt Hydrolysis Substitution
Substitute the determined $\mathrm{pH}$ along with salt molarity ($C = 0.01\mathrm{~M} = 10^{-2}\mathrm{~M}$) into the hydrolysis equation:
$$5 = 7 - \frac{1}{2}\mathrm{pK_b} - \frac{1}{2}\log(10^{-2})$$
$$5 = 7 - \frac{1}{2}\mathrm{pK_b} - \frac{1}{2}(-2)$$
$$5 = 7 - \frac{1}{2}\mathrm{pK_b} + 1$$
$$5 = 8 - \frac{1}{2}\mathrm{pK_b} \implies \frac{1}{2}\mathrm{pK_b} = 3 \implies \mathrm{pK_b} = 6$$
### Pattern Recognition
Quinhydrone electrodes act as excellent pH indicators in electrochemical cells. Note that each change of 1 pH unit shifts the cell output potential by exactly $0.06\mathrm{~V}$ at standard ambient conditions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Q26 (2025)
$40\mathrm{~mL}$ of a mixture of $\mathrm{CH}_{3}\mathrm{COOH}$ and $\mathrm{HCl}$ (aqueous solution) is titrated against $0.1\mathrm{~M~NaOH}$ solution conductometrically. Which of the following statements is correct? {{IMG1}}
- The concentration of $\mathrm{CH}_3\mathrm{COOH}$ in the original mixture is $0.005\mathrm{~M}$
- The concentration of $\mathrm{HCl}$ in the original mixture is $0.005\mathrm{~M}$
- $\mathrm{CH}_3\mathrm{COOH}$ is neutralised first followed by neutralisation of $\mathrm{HCl}$
- Point 'C' indicates the complete neutralisation of $\mathrm{HCl}$
### Related Formula
At the equivalence point during titration:
$$M_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}$$
### Core Logic
In a mixture of a strong acid ($\mathrm{HCl}$) and a weak acid ($\mathrm{CH}_3\mathrm{COOH}$):
1. $\mathrm{HCl}$ is a strong acid and is completely ionized. When $\mathrm{NaOH}$ is added, highly mobile $\mathrm{H}^+$ ions are replaced by less mobile $\mathrm{Na}^+$ ions, causing a sharp drop in conductance (segment AB).
2. At point B ($2.0\mathrm{~mL}$), $\mathrm{HCl}$ is completely neutralized.
3. Segment BC represents the neutralization of the weak acid $\mathrm{CH}_3\mathrm{COOH}$ to form highly conducting sodium acetate, causing a moderate rise in conductance up to point C ($5.0\mathrm{~mL}$).
4. Beyond point C, excess $\mathrm{OH}^-$ ions cause a rapid rise in conductance (segment CD).
### Step 1: Calculate concentration of $\mathrm{HCl}$
Volume of $\mathrm{NaOH}$ used to neutralize $\mathrm{HCl}$ is $V_1 = 2.0\mathrm{~mL}$:
$$M_{\mathrm{HCl}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 2.0\mathrm{~mL}$$
$$M_{\mathrm{HCl}} = \frac{0.2}{40} = 0.005\mathrm{~M}$$
### Step 2: Calculate concentration of $\mathrm{CH}_3\mathrm{COOH}$
Volume of $\mathrm{NaOH}$ used to neutralize $\mathrm{CH}_3\mathrm{COOH}$ is $V_2 = 5.0\mathrm{~mL} - 2.0\mathrm{~mL} = 3.0\mathrm{~mL}$:
$$M_{\mathrm{CH}_3\mathrm{COOH}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 3.0\mathrm{~mL}$$
$$M_{\mathrm{CH}_3\mathrm{COOH}} = \frac{0.3}{40} = 0.0075\mathrm{~M}$$
### Pattern Recognition
Conductometric titration curves are analyzed sequentially: the strongest electrolyte is always neutralized first. A steep drop in conductance always signals the neutralization of a strong acid ($\mathrm{H}^+$ depletion). A weak acid titration shows a gentle upward slope due to salt formation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Q34 (2025)
The standard cell potential $\left(E_{\mathrm{cell}}^{\ominus}\right)$ of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as $1.21\mathrm{~V}$. The standard half cell reduction potential for $\mathrm{O}_2$ $\left(E_{\mathrm{O}_2/\mathrm{H}_2\mathrm{O}}^{\circ}\right)$ is $1.229\mathrm{~V}$. Choose the correct statement:
- The standard half cell reduction potential for the reduction of $\mathrm{CO}_2$ $\left(E_{\mathrm{CO}_2/\mathrm{CH}_3\mathrm{OH}}^{\circ}\right)$ is $19\mathrm{~mV}$
- Oxygen is formed at the anode.
- Reactants are fed at one go to each electrode.
- Reduction of methanol takes place at the cathode.
### Related Formula
Standard cell EMF is related to standard reduction potentials:
$$E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}$$
### Core Logic
In a methanol-oxygen fuel cell:
- Anode reaction (Oxidation): Methanol is oxidized to carbon dioxide:
$$\mathrm{CH}_3\mathrm{OH} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CO}_2 + 6\mathrm{H}^+ + 6e^-$$
- Cathode reaction (Reduction): Oxygen is reduced to water:
$$\mathrm{O}_2 + 4\mathrm{H}^+ + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O}$$
Hence, cathode is the oxygen electrode, and anode is the methanol electrode.
### Step 1: Calculate Standard Reduction Potential of Anode
Using the EMF equation:
$$1.21\mathrm{~V} = 1.229\mathrm{~V} - E_{\mathrm{anode}}^{\circ}$$
$$E_{\mathrm{anode}}^{\circ} = 1.229 - 1.21 = 0.019\mathrm{~V} = 19\mathrm{~mV}$$
The standard half-cell reduction potential for the $\mathrm{CO}_2/\mathrm{CH}_3\mathrm{OH}$ couple is $19\mathrm{~mV}$, matching Option (1).
### Pattern Recognition
Fuel cells are galvanic cells where reactants (like fuels and oxidants) are fed continuously to the electrodes, not at one go. Oxidation always occurs at the anode (methanol) and reduction at the cathode (oxygen).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q35 (2025)
Given below are two statements:
Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate.
Statement II: If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are $\mathbf{x}_1$, $\mathbf{x}_2$ and $\mathbf{x}_3$ $\mathrm{S\ cm}^2\ \mathrm{mol}^{-1}$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by $\mathbf{x}_1 + \mathbf{x}_2 + 2\mathbf{x}_3$.
In the light of the given statements, choose the correct answer from the options given below:
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is false but Statement II is true}$
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are true}$
### Related Formula
$$\lambda_m^{\infty} = \nu_+ \lambda_+^{\infty} + \nu_- \lambda_-^{\infty}$$
### Core Logic
Statement I: Mohr's salt is a double salt with chemical formula:
$$\mathrm{FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O}$$
When dissolved in water, it completely dissociates into three distinct ionic species:
$$\mathrm{Fe}^{2+} \text{ (ferrous)}, \quad \mathrm{NH}_4^+ \text{ (ammonium)}, \quad \text{and } \mathrm{SO}_4^{2-} \text{ (sulphate)}$$
Thus, Statement I is true.
Statement II: According to Kohlrausch's law of independent migration of ions:
$$\lambda_m^{\infty}(\text{Mohr's Salt}) = 1 \cdot \lambda_m^{\infty}(\mathrm{Fe}^{2+}) + 2 \cdot \lambda_m^{\infty}(\mathrm{NH}_4^+) + 2 \cdot \lambda_m^{\infty}(\mathrm{SO}_4^{2-})$$
$$\lambda_m^{\infty} = x_1 + 2x_2 + 2x_3$$
Statement II claims the expression is $x_1 + x_2 + 2x_3$ (missing the coefficient $2$ for ammonium). Thus, Statement II is false.
### Pattern Recognition
Kohlrausch's law matches stoichiometric coefficients directly to the ion quantities released. Mohr's salt formula contains $(NH_4)_2$, requiring a multiplier of $2$ for ammonium ion conductance.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 12 Chemistry: d- and f-Block Elements
