Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 4.
Q48 (2025)
Electrolysis of $600\mathrm{~mL}$ aqueous solution of NaCl for $5\mathrm{\ \min}$ changes the $\mathrm{pH}$ of the solution to 12. The current in Amperes used for the given electrolysis is ______ (Nearest integer).
### Related Formula
Faraday's law of electrolysis equation:
$$\text{Moles of electrons (equivalents)} = \frac{I \cdot t}{F}$$
Water ion product relation:
$$\text{pH} + \text{pOH} = 14$$
### Core Logic
During the electrolysis of brine ($NaCl(aq)$), hydroxide ions ($OH^-$) are generated at the cathode:
$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$
Given metrics:
- Final $\text{pH} = 12 \implies \text{pOH} = 14 - 12 = 2$
- $[OH^-] = 10^{-2}\mathrm{\ M}$
- $\text{Volume} = 600\mathrm{\ mL} = 0.6\mathrm{\ L}$
- $\text{Time} = 5\mathrm{\ \min} = 300\mathrm{\ s}$
### Step 1: Calculate Moles of Hydroxide Produced
Find the absolute moles of $OH^-$ ions generated:
$$\text{Moles} = \text{Molarity} \times \text{Volume (L)} = 10^{-2} \times 0.6 = 6 \times 10^{-3}\text{ moles}$$
### Step 2: Relate to Electrical Current
Since $1$ mole of electrons produces $1$ mole of $OH^-$, the moles of charge equals $6 \times 10^{-3}$. Applying Faraday's equation:
$$6 \times 10^{-3} = \frac{I \times 300}{96500}$$
$$I = \frac{6 \times 10^{-3} \times 96500}{300} = 1.93\mathrm{\ A}$$
Rounding to the nearest integer gives $2$.
### Pattern Recognition
Always convert a given $\text{pH}$ value into $[OH^-]$ concentration when dealing with cathodic water reduction. Tracking the relationship where $1\ e^- \equiv 1\ OH^-$ provides a direct shortcut to link $\text{pH}$ changes to current flow.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q1302 (2025)
The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.
- $\mathrm{E}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}}^{\ominus} = +1.15\mathrm{V}$
- $\mathrm{E}_{\mathrm{T}l^{3+}/\mathrm{T}l}^{\ominus} = +1.26\mathrm{V}$
- $\mathrm{E}_{\mathrm{Al}^{3+}/\mathrm{Al}}^{\ominus} = -1.66\mathrm{V}$
- $\mathrm{E}_{\mathrm{Pb}^{4+}/\mathrm{Pb}^{2+}}^{\ominus} = +1.67\mathrm{V}$
### Related Formula
$$\text{Oxidising Capacity} \propto \text{Standard Reduction Potential } (E^{\ominus})$$
### Core Logic
A higher positive value of standard reduction potential ($E^{\ominus}$) indicates a stronger tendency to undergo reduction, hence behaving as a stronger oxidising agent. Comparing the given values:
* $\mathrm{E}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}}^{\ominus} = +1.15\mathrm{V}$
* $\mathrm{E}_{\mathrm{T}l^{3+}/\mathrm{T}l}^{\ominus} = +1.26\mathrm{V}$
* $\mathrm{E}_{\mathrm{Al}^{3+}/\mathrm{Al}}^{\ominus} = -1.66\mathrm{V}$
* $\mathrm{E}_{\mathrm{Pb}^{4+}/\mathrm{Pb}^{2+}}^{\ominus} = +1.67\mathrm{V}$
Since $+1.67\mathrm{V}$ is the highest value, $\mathrm{Pb}^{4+}$ possesses the strongest oxidising capacity.
### Pattern Recognition
Strongest oxidising agent = Most positive reduction potential. Weakest oxidising agent = Most negative reduction potential.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 12 Chemistry: p-Block Elements
Q1303 (2025)
The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
- A small decrease in molar conductivity is observed at infinite dilution.
- A small increase in molar conductivity is observed at infinite dilution.
- Molar conductivity increases sharply with increase in concentration.
- Molar conductivity decreases sharply with increase in concentration.
### Related Formula
For weak electrolytes, the degree of dissociation $\alpha$ increases sharply near infinite dilution according to Ostwald's Dilution Law:
$$\alpha = \sqrt{\frac{K_a}{C}}$$
### Core Logic
When a weak electrolyte is diluted (concentration $C \rightarrow 0$), its molar conductivity increases steeply. Conversely, when plotted against $\sqrt{C}$, as concentration increases, the degree of dissociation drops rapidly, causing a sharp decrease in molar conductivity. This matches the curve given below:
{{SOL_IMG1}}
### Pattern Recognition
Weak electrolyte plots feature a steep asymptotic exponential-like rise towards the y-axis as $C \rightarrow 0$, meaning a sharp decrease occurs with increasing concentration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q1310 (2025)
For a $\mathrm{Mg} \mid \mathrm{Mg}^{2+}$ (aq) $\parallel \mathrm{Ag}^{+}(\mathrm{aq}) \mid \mathrm{Ag}$ the correct Nernst Equation is :
- $\mathrm{E_{cell} = E_{cell}^{o} - \frac{RT}{2F}\ln\frac{[Ag^{+}]}{[Mg^{2 + }]}}$
- $\mathrm{E}_{\mathrm{cell}} = \mathrm{E}_{\mathrm{cell}}^{\circ} + \frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{[\mathrm{Ag}^{+}]^{2}}{[\mathrm{Mg}^{2+}]}$
- $\mathrm{E_{cell} = E_{cell}^{o} - \frac{RT}{2F}\ln\frac{[Mg^{2 + }]}{[Ag^{+}]}}$
- $\mathrm{E_{cell} = E_{cell}^{o} - \frac{RT}{2F}\ln\frac{[Ag^{+}]^{2}}{[Mg^{2 + }]}}$
### Related Formula
$$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{nF} \ln Q$$
### Core Logic
Let us explicitly formulate the complete chemical oxidation-reduction equations :
Anode oxidation: $\mathrm{Mg}_{(s)} \rightarrow \mathrm{Mg}^{2+}_{(aq)} + 2e^-$
Cathode reduction: $2\mathrm{Ag}^{+}_{(aq)} + 2e^- \rightarrow 2\mathrm{Ag}_{(s)}$
Net total equation :
$$\mathrm{Mg}_{(s)} + 2\mathrm{Ag}^{+}_{(aq)} \rightleftharpoons \mathrm{Mg}^{2+}_{(aq)} + 2\mathrm{Ag}_{(s)}$$
Total transferred moles of electrons $n = 2$ .
Reaction quotient :
$$Q = \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Ag}^{+}]^2}$$
Substituting into Nernst form :
$$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{RT}{2F} \ln\left( \frac{[\mathrm{Mg}^{2+}]}{[\mathrm{Ag}^{+}]^2} \right)$$
Inverting the inside quotient changes the sign of the logarithm term from negative to positive:
$$E_{\text{cell}} = E_{\text{cell}}^{\circ} + \frac{RT}{2F} \ln\left( \frac{[\mathrm{Ag}^{+}]^2}{[\mathrm{Mg}^{2+}]} \right)$$
### Pattern Recognition
A standard negative logarithmic quotient can always toggle into an addition configuration by inverting the products/reactants variables concentration ratio.
Q82 (2024)
The potential for the given half cell at 298K is
$(-)\dots\dots\dots\dots \times 10^{-2} \mathrm{~V}$.
$2\mathrm{H}^+_{\text{(aq)}} + 2e^- \rightarrow \mathrm{H}_2\mathrm{(g)}$
$[\mathrm{H}^+] = 1 \mathrm{M}, P_{\mathrm{H}_2} = 2 \mathrm{~atm}$
Given: $2.303\mathrm{RT/F} = 0.06\mathrm{V}$, $\log 2 = 0.3$
### Related Formula
$$E = E^\circ - \frac{2.303RT}{nF} \log Q$$
For the Standard Hydrogen Electrode half-reaction: $2H^+ + 2e^- \rightarrow H_2$
$$E_{H^+/H_2} = E^\circ_{H^+/H_2} - \frac{0.06}{2} \log \frac{P_{H_2}}{[H^+]^2}$$
### Step 1: Substitute the given values
$E^\circ_{H^+/H_2} = 0.00 \mathrm{~V}$ (by definition)
$[H^+] = 1 \mathrm{~M}$
$P_{H_2} = 2 \mathrm{~atm}$
$n = 2$ electrons
$$E = 0.00 - \frac{0.06}{2} \log \left( \frac{2}{1^2} \right)$$
### Step 2: Solve the calculation
$E = -0.03 \log 2$
Given $\log 2 = 0.3$
$E = -0.03 \times 0.3$
$E = -0.009 \mathrm{~V}$
$E = -0.9 \times 10^{-2} \mathrm{~V}$
### Step 3: Match the requested format
The question asks for $(-) \dots \times 10^{-2} \mathrm{~V}$.
This gives exactly $0.9$.
For NAT type with integer expected, $0.9$ can be rounded to $1$. However, exact calculation yields $0.9$. According to official JEE rounding, $0.9 \approx 1$.
### Pattern Recognition
Hydrogen electrode non-standard potential depends strictly on pressure of $H_2$ and concentration of $H^+$. If $[H^+]=1$, increasing $H_2$ pressure lowers the potential below zero (makes it negative).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
