The pH and conductance of a weak acid (HX) was found to be 5 and 4 times 10^-5$4 \times 10^{-5}$ S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1text cm^2$1\text{ cm}^2$ were at a distance of 15text cm$15\text{ cm}$ apart. The value of the limiting molar conductivity is ..... textS m^2text mol^-1$\text{S m}^2\text{ mol}^{-1}$. (nearest integer)
(Given: degree of dissociation of the weak acid (alpha) ll 1)$(\alpha) \ll 1)$
Numerical Answer Type:
Enter a numerical valueAnswer: 6 to 6+4 marks
Solution & Explanation
### Related Formula
kappa = G times fraclA$$\kappa = G \times \frac{l}{A}$$Lambda_m = frackappa times 1000C$$\Lambda_m = \frac{\kappa \times 1000}{C}$$alpha = fracLambda_mLambda_m^infty$$\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}$$[H^+] = Calpha$[H^+] = C\alpha$
### Core Logic
Given: pH = 5$pH = 5$, so [H^+] = 10^-5text M$[H^+] = 10^{-5}\text{ M}$.
Since [H^+] = C cdot alpha = C cdot fracLambda_mLambda_m^infty$[H^+] = C \cdot \alpha = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$, we have 10^-5 = C cdot fracLambda_mLambda_m^infty$10^{-5} = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$.
First, calculate conductivity (kappa$\kappa$):
Conductance G = 4 times 10^-5text S$G = 4 \times 10^{-5}\text{ S}$.
Cell constant G^* = fraclA = frac15text cm1text cm^2 = 15text cm^-1$G^* = \frac{l}{A} = \frac{15\text{ cm}}{1\text{ cm}^2} = 15\text{ cm}^{-1}$.
kappa = G cdot G^* = (4 times 10^-5) times 15 = 6 times 10^-4text S cm^-1$$\kappa = G \cdot G^* = (4 \times 10^{-5}) \times 15 = 6 \times 10^{-4}\text{ S cm}^{-1}$$
Molar conductivity (Lambda_m$\Lambda_m$):
Lambda_m = frackappa times 1000C = frac6 times 10^-4 times 1000C = frac0.6C$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-4} \times 1000}{C} = \frac{0.6}{C}$$
Substitute Lambda_m$\Lambda_m$ into the proton concentration formula:
[H^+] = 10^-5 = C cdot frac0.6 / CLambda_m^infty$$[H^+] = 10^{-5} = C \cdot \frac{0.6 / C}{\Lambda_m^{\infty}}$$10^-5 = frac0.6Lambda_m^infty$$10^{-5} = \frac{0.6}{\Lambda_m^{\infty}}$$Lambda_m^infty = frac0.610^-5 = 60000text S cm^2text mol^-1$$\Lambda_m^{\infty} = \frac{0.6}{10^{-5}} = 60000\text{ S cm}^2\text{ mol}^{-1}$$
Convert units to textS m^2text mol^-1$\text{S m}^2\text{ mol}^{-1}$:
Since 1text m^2 = 10^4text cm^2$1\text{ m}^2 = 10^4\text{ cm}^2$,
Lambda_m^infty = 60000 times 10^-4text S m^2text mol^-1 = 6text S m^2text mol^-1$$\Lambda_m^{\infty} = 60000 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1} = 6\text{ S m}^2\text{ mol}^{-1}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Equilibrium
Keywords:#limiting molar conductivity#weak acid conductance#JEE Main 2026 Morning Q72#Electrochemistry JEE Main 2026
More Electrochemistry Previous-Year Questions — Page 2
Q35jee_main_2025_07_april_morningKohlrausch's Law
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 mathbfx_1$\mathbf{x}_1$, mathbfx_2$\mathbf{x}_2$ and mathbfx_3$\mathbf{x}_3$mathrmS\ cm^2\ mathrmmol^-1$\mathrm{S\ cm}^2\ \mathrm{mol}^{-1}$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by mathbfx_1 + mathbfx_2 + 2mathbfx_3$\mathbf{x}_1 + \mathbf{x}_2 + 2\mathbf{x}_3$.
In the light of the given statements, choose the correct answer from the options given below:
A.textBoth Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
B.textStatement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
C.textStatement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$
D.textBoth Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
Solution
### Related Formula
lambda_m^infty = nu_+ lambda_+^infty + nu_- lambda_-^infty$$\lambda_m^{\infty} = \nu_+ \lambda_+^{\infty} + \nu_- \lambda_-^{\infty}$$
### Core Logic
Statement I: Mohr's salt is a double salt with chemical formula:
mathrmFeSO_4 cdot (NH_4)_2SO_4 cdot 6H_2O$$\mathrm{FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O}$$
When dissolved in water, it completely dissociates into three distinct ionic species:
mathrmFe^2+ text (ferrous), quad mathrmNH_4^+ text (ammonium), quad textand mathrmSO_4^2- text (sulphate)$$\mathrm{Fe}^{2+} \text{ (ferrous)}, \quad \mathrm{NH}_4^+ \text{ (ammonium)}, \quad \text{and } \mathrm{SO}_4^{2-} \text{ (sulphate)}$$
Thus, Statement I is true.
Statement II: According to Kohlrausch's law of independent migration of ions:
lambda_m^infty(textMohr's Salt) = 1 cdot lambda_m^infty(mathrmFe^2+) + 2 cdot lambda_m^infty(mathrmNH_4^+) + 2 cdot lambda_m^infty(mathrmSO_4^2-)$$\lambda_m^{\infty}(\text{Mohr's Salt}) = 1 \cdot \lambda_m^{\infty}(\mathrm{Fe}^{2+}) + 2 \cdot \lambda_m^{\infty}(\mathrm{NH}_4^+) + 2 \cdot \lambda_m^{\infty}(\mathrm{SO}_4^{2-})$$lambda_m^infty = x_1 + 2x_2 + 2x_3$$\lambda_m^{\infty} = x_1 + 2x_2 + 2x_3$$
Statement II claims the expression is x_1 + x_2 + 2x_3$x_1 + x_2 + 2x_3$ (missing the coefficient 2$2$ 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$(NH_4)_2$, requiring a multiplier of 2$2$ for ammonium ion conductance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 12 Chemistry: d- and f-Block Elements
Q48jee_main_2025_07_april_morningNernst Equation
1 Faraday electricity was passed through mathrmCu^2+$\mathrm{Cu}^{2+}$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through mathrmAg^+$\mathrm{Ag}^{+}$ (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.
The cell assembly combines Cu and Ag half cells after individual initial electrolysis modifications.
Given:
mathrmE_mathrmCu^2+/mathrmCu^circ = 0.34 mathrm~V$\mathrm{E}_{\mathrm{Cu}^{2+}/\mathrm{Cu}}^{\circ} = 0.34 \mathrm{~V}$mathrmE_mathrmAg^+/mathrmAg^circ = 0.8 mathrm~V$\mathrm{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}^{\circ} = 0.8 \mathrm{~V}$frac2.303RTF = 0.06 mathrm~V$\frac{2.303RT}{F} = 0.06 \mathrm{~V}$
Numerical Answer.Answer: 0.4 to 0.4
Solution
### Related Formula
E_textcell = E^circ_textcell - frac0.06n log Q$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{n} \log Q$$
### Core Logic
First, analyze the electrolysis step to determine final ionic concentrations:
1. **For mathrmCu^2+/mathrmCu$\mathrm{Cu}^{2+}/\mathrm{Cu}$ half-cell**:
- Initial moles of mathrmCu^2+ = 1.5 text M times 1 text L = 1.5 text mol$\mathrm{Cu}^{2+} = 1.5 \text{ M} \times 1 \text{ L} = 1.5 \text{ mol}$.
- Reductive half-reaction: mathrmCu^2+ + 2mathrme^- rightarrow mathrmCu$\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}$.
- Passing 1 text Faraday$1 \text{ Faraday}$ converts: frac12 = 0.5 text mol$\frac{1}{2} = 0.5 \text{ mol}$ of mathrmCu^2+$\mathrm{Cu}^{2+}$.
- Remaining moles of mathrmCu^2+ = 1.5 - 0.5 = 1.0 text mol$\mathrm{Cu}^{2+} = 1.5 - 0.5 = 1.0 \text{ mol}$.
- Final concentration [mathrmCu^2+] = 1.0 text M$[\mathrm{Cu}^{2+}] = 1.0 \text{ M}$.
2. **For mathrmAg^+/mathrmAg$\mathrm{Ag}^{+}/\mathrm{Ag}$ half-cell**:
- Initial moles of mathrmAg^+ = 0.2 text M times 1 text L = 0.2 text mol$\mathrm{Ag}^+ = 0.2 \text{ M} \times 1 \text{ L} = 0.2 \text{ mol}$.
- Reductive half-reaction: mathrmAg^+ + mathrme^- rightarrow mathrmAg$\mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag}$.
- Passing 0.1 text Faraday$0.1 \text{ Faraday}$ converts: 0.1 text mol$0.1 \text{ mol}$ of mathrmAg^+$\mathrm{Ag}^+$.
- Remaining moles of mathrmAg^+ = 0.2 - 0.1 = 0.1 text mol$\mathrm{Ag}^+ = 0.2 - 0.1 = 0.1 \text{ mol}$.
- Final concentration [mathrmAg^+] = 0.1 text M$[\mathrm{Ag}^+] = 0.1 \text{ M}$.
Now, connect the two components into a galvanic cell:
- Anode reaction: mathrmCu(s) rightarrow mathrmCu^2+mathrm(aq) + 2mathrme^-$\mathrm{Cu(s)} \rightarrow \mathrm{Cu}^{2+}\mathrm{(aq)} + 2\mathrm{e}^-$
- Cathode reaction: 2mathrmAg^+mathrm(aq) + 2mathrme^- rightarrow 2mathrmAg(s)$2\mathrm{Ag}^{+}\mathrm{(aq)} + 2\mathrm{e}^- \rightarrow 2\mathrm{Ag(s)}$
- Net cell reaction: mathrmCu(s) + 2mathrmAg^+mathrm(aq) rightarrow mathrmCu^2+mathrm(aq) + 2mathrmAg(s)$\mathrm{Cu(s)} + 2\mathrm{Ag}^{+}\mathrm{(aq)} \rightarrow \mathrm{Cu}^{2+}\mathrm{(aq)} + 2\mathrm{Ag(s)}$
- n = 2$n = 2$
Calculate standard cell potential:
E^circ_textcell = E^circ_mathrmAg^+/mathrmAg - E^circ_mathrmCu^2+/mathrmCu = 0.80 - 0.34 = 0.46 text V$$E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag}^+/\mathrm{Ag}} - E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.80 - 0.34 = 0.46 \text{ V}$$
Applying Nernst Equation:
E_textcell = E^circ_textcell - frac0.062 log left( frac[mathrmCu^2+][mathrmAg^+]^2 right)$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{2} \log \left( \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2} \right)$$E_textcell = 0.46 - 0.03 log left( frac1(0.1)^2 right) = 0.46 - 0.03 log(100)$$E_{\text{cell}} = 0.46 - 0.03 \log \left( \frac{1}{(0.1)^2} \right) = 0.46 - 0.03 \log(100)$$E_textcell = 0.46 - 0.03(2) = 0.46 - 0.06 = 0.40 text V$$E_{\text{cell}} = 0.46 - 0.03(2) = 0.46 - 0.06 = 0.40 \text{ V}$$
(Note: The potential is 0.4text V$0.4\text{ V}$ or 400text mV$400\text{ mV}$).
### Pattern Recognition
Electrolysis modifies the bulk concentrations. First, use Faraday's laws to get the new concentration values ([Cu^2+] = 1.0text M$[Cu^{2+}] = 1.0\text{ M}$, [Ag^+] = 0.1text 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
Q50jee_main_2025_08_april_eveningNernst Equation
Consider the following half-cell reduction reaction:
textCr_2textO_7^2-text(aq) + 6e^- + 14textH^+text(aq) longrightarrow 2textCr^3+text(aq) + 7textH_2textO(l)$$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 6e^- + 14\text{H}^+\text{(aq)} \longrightarrow 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$$
The process is conducted with a concentration ratio of frac[textCr^3+]^2[textCr_2textO_7^2-] = 10^-6$\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$. The specific pH value at which the EMF (E$E$) of this reduction half-cell becomes exactly zero is _________ (as the nearest integer value).
Given parameters: E^circ_textCr_2textO_7^2-/textCr^3+ = 1.33 text V$E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \text{ V}$ and frac2.303RTF = 0.059 text V$\frac{2.303RT}{F} = 0.059 \text{ V}$.
Numerical Answer.Answer: 10 to 10
Solution
### Related Formula
The Nernst equation for a reduction half-cell is:
E = E^circ - frac2.303RTnF log Q$$E = E^\circ - \frac{2.303RT}{nF} \log Q$$
For this reaction, the reaction quotient Q$Q$ is:
Q = frac[textCr^3+]^2[textCr_2textO_7^2-] cdot [textH^+]^14$$Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}] \cdot [\text{H}^+]^{14}}$$
### Execution
Step 1: Identify the number of transferred electrons (n = 6$n = 6$) and substitute the condition E = 0$E = 0$:
0 = 1.33 - frac0.0596 log left( frac10^-6[textH^+]^14 right)$$0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\text{H}^+]^{14}} \right)$$
Step 2: Isolate the logarithmic term:
1.33 = frac0.0596 left[ log(10^-6) - log([textH^+]^14) right]$$1.33 = \frac{0.059}{6} \left[ \log(10^{-6}) - \log([\text{H}^+]^{14}) \right]$$frac1.33 times 60.059 = -6 - 14 log[textH^+]$$\frac{1.33 \times 6}{0.059} = -6 - 14 \log[\text{H}^+]$$
Step 3: Perform the arithmetic division:
135.254 = -6 - 14 log[textH^+]$$135.254 = -6 - 14 \log[\text{H}^+]$$
Step 4: Rearrange the terms using the definition of pH (-log[textH^+] = textpH$-\log[\text{H}^+] = \text{pH}$):
135.254 + 6 = 14 cdot textpH$$135.254 + 6 = 14 \cdot \text{pH}$$141.254 = 14 cdot textpH$$141.254 = 14 \cdot \text{pH}$$textpH = frac141.25414 = 10.089$$\text{pH} = \frac{141.254}{14} = 10.089$$
Rounding to the nearest integer value gives **10**.
### Pattern Recognition
The exponent of the hydrogen ion concentration ([textH^+]^14$[\text{H}^+]^{14}$) 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 (textpH approx 10$\text{pH} \approx 10$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Ionic Equilibrium
Q35jee_main_2025_29_jan_eveningBatteries and Commercial Cells
Match List-I with List-II:
List-I (Applications)
List-II (Batteries/Cell)
(A) Transistors
(I) Anode - Zn/Hg; Cathode - HgO + C
(B) Hearing aids
(II) Hydrogen fuel cell
(C) Invertors
(III) Anode - Zn; Cathode - Carbon
(D) Apollo space ship
(IV) Anode - Pb; Cathode - Pb | PbO_2$PbO_2$
Choose the correct answer from the options given below:
A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
B. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
C. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
D. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Solution
### 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$MnO_2$
ightarrow$
ightarrow$ (III).
* Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C
ightarrow$
ightarrow$ (I).
* Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO_2$PbO_2$
ightarrow$
ightarrow$ (IV).
* Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells
ightarrow$
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
Q36jee_main_2025_29_jan_eveningProducts of Electrolysis
O_2$O_{2}$ gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO_3$AgNO_{3}$ using silver electrodes.
(B) an aqueous solution of AgNO_3$AgNO_{3}$ using platinum electrodes.
(C) a dilute solution of H_2SO_4$H_{2}SO_{4}$ using platinum electrodes.
(D) a high concentration solution of H_2SO_4$H_{2}SO_{4}$ using platinum electrodes.
Choose the correct answer from the options given below:
A. (B) and (C) only
B. (A) and (D) only
C. (B) and (D) only
D. (A) and (C) only
Solution
### Core Logic
Analyzing anodic reactions during electrolysis:
* Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag
ightarrow Ag^+ + e^-$Ag
ightarrow Ag^{+} + e^{-}$). No oxygen is evolved.
* Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO_3^-$NO_3^-$ ions:2H_2O
ightarrow O_2 + 4H^+ + 4e^-$$2H_2O
ightarrow O_2 + 4H^{+} + 4e^{-}$$
* Case (C): In dilute H_2SO_4$H_2SO_4$, water oxidation takes place, releasing O_2$O_2$ gas at the anode.
* Case (D): In concentrated H_2SO_4$H_2SO_4$, oxidation of SO_4^2-$SO_4^{2-}$ creates peroxodisulphate ions (S_2O_8^2-$S_2O_8^{2-}$), 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^-$NO_3^-$ or dilute SO_4^2-$SO_4^{2-}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
More Electrochemistry Questions — jee_main_2026_21_jan_morning
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