Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 6.
Q67 (2024)
Identify the factor from the following that does not affect electrolytic conductance of a solution.
- $\text{The nature of the electrolyte added.}$
- $\text{The nature of the electrode used.}$
- $\text{Concentration of the electrolyte.}$
- $\text{The nature of solvent used.}$
### Core Logic
Conductivity of an electrolytic cell is affected by the concentration of the electrolyte, the nature of the electrolyte, and the nature of the solvent. It does not depend on the nature of the electrode used.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q90 (2024)
One Faraday of electricity liberates $x \times 10^{-1}$ gram atom of copper from copper sulphate, x is
### Core Logic
The reduction reaction for copper is:
$$Cu^{2+} + 2e^- \rightarrow Cu$$
This shows that 2 moles of electrons (2 Faraday) are required to deposit 1 mole (or 1 gram atom) of Cu.
Therefore, 1 Faraday of electricity will deposit:
$$\frac{1}{2} = 0.5 \text{ moles of Cu}$$
### Step 1: Finding x
$$0.5 \text{ mole} = 0.5 \text{ gram atom} = 5 \times 10^{-1} \text{ gram atom}$$
Hence, $x = 5$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q72 (2026)
The pH and conductance of a weak acid (HX) was found to be 5 and $4 \times 10^{-5}$ S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of $1\text{ cm}^2$ were at a distance of $15\text{ cm}$ apart. The value of the limiting molar conductivity is ..... $\text{S m}^2\text{ mol}^{-1}$. (nearest integer)
(Given: degree of dissociation of the weak acid $(\alpha) \ll 1)$
### Related Formula
$$\kappa = G \times \frac{l}{A}$$
$$\Lambda_m = \frac{\kappa \times 1000}{C}$$
$$\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}$$
$[H^+] = C\alpha$
### Core Logic
Given: $pH = 5$, so $[H^+] = 10^{-5}\text{ M}$.
Since $[H^+] = C \cdot \alpha = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$, we have $10^{-5} = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$.
First, calculate conductivity ($\kappa$):
Conductance $G = 4 \times 10^{-5}\text{ S}$.
Cell constant $G^* = \frac{l}{A} = \frac{15\text{ cm}}{1\text{ cm}^2} = 15\text{ cm}^{-1}$.
$$\kappa = G \cdot G^* = (4 \times 10^{-5}) \times 15 = 6 \times 10^{-4}\text{ S cm}^{-1}$$
Molar conductivity ($\Lambda_m$):
$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-4} \times 1000}{C} = \frac{0.6}{C}$$
Substitute $\Lambda_m$ into the proton concentration formula:
$$[H^+] = 10^{-5} = C \cdot \frac{0.6 / C}{\Lambda_m^{\infty}}$$
$$10^{-5} = \frac{0.6}{\Lambda_m^{\infty}}$$
$$\Lambda_m^{\infty} = \frac{0.6}{10^{-5}} = 60000\text{ S cm}^2\text{ mol}^{-1}$$
Convert units to $\text{S m}^2\text{ mol}^{-1}$:
Since $1\text{ m}^2 = 10^4\text{ cm}^2$,
$$\Lambda_m^{\infty} = 60000 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1} = 6\text{ S m}^2\text{ mol}^{-1}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Q67 (2026)
For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?
- $(1) \text{ Plot 1}$ {{IMG1}}
- $(2) \text{ Plot 2}$ {{IMG2}}
- $(3) \text{ Plot 3}$ {{IMG3}}
- $(4) \text{ Plot 4}$ {{IMG4}}
### Core Logic
For a closed circuit Daniell cell operating under given conditions, standard cell potential $E_{\text{cell}}^{\circ}$ remains constant with time during discharge until equilibrium or noticeable concentration shifts occur in a specific manner, or cell potential versus time plots reflect constancy of standard electromotive force terms before polarization effects take over.
### Step 1: Final Conclusion
Plot 2 accurately depicts the expected trend where $E_{\text{cell}}^{\circ}$ remains constant with time.
### Pattern Recognition
Sees: graphical representation of electrochemical cell potentials over time.
Trap: Confusing standard cell potential $E^{\circ}$ with operating cell potential $E_{\text{cell}}$.
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q75 (2026)
$\text{MX}$ is a sparingly soluble salt that follows the given solubility equilibrium at 298 K:
$$\text{MX}(\text{s}) \rightleftharpoons \text{M}^+(\text{aq}) + \text{X}^-(\text{aq}); \quad K_{\text{sp}} = 10^{-10}$$
If the standard reduction potential for $\text{M}^+(\text{aq}) \rightarrow \text{M}(\text{s})$ is $(E_{\text{M}^+/\text{M}}^{\ominus}) = 0.79 \text{ V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E_{\text{X}^-/\text{MX}(\text{s})/\text{M}}^{\ominus}$ is $\_\_\_\_ \text{ mV}$. (nearest integer)
[Given: $\frac{2.303RT}{F} = 0.059 \text{ V}$]
### Related Formula
$$E_{\text{X}^-/\text{MX}/\text{M}}^{\circ} = E_{\text{M}^+/\text{M}}^{\circ} + \frac{0.059}{n} \log K_{\text{sp}}$$r
### Core Logic
Substituting the given values:
$$E_{\text{X}^-/\text{MX}/\text{M}}^{\circ} = 0.79 + \frac{0.059}{1} \log(10^{-10})$$
$$E_{\text{X}^-/\text{MX}/\text{M}}^{\circ} = 0.79 + 0.059(-10) = 0.79 - 0.59 = 0.20 \text{ V}$$
### Step 1: Converting to Millivolts
$$0.20 \text{ V} = 200 \text{ mV}$$
### Pattern Recognition
Sees: relationship between standard reduction potential, solubility product, and metal-insoluble salt electrodes.
Trap: Sign or logarithmic base calculation errors.
### Chapter Mix
Class 12 Chemistry: Electrochemistry
