Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M+ (M arrow M+ + e-). The cation M+ is present in two different concentrations c1 and c2 as shown above. Which of the following statement is correct for generating a positive cell potential?

For a concentration cell, E = 0, so: For a concentration cell to have a positive cell potential (Ecell > 0), the ratio [Anode][Cathode] must be less than 1. This implies that [Anode] 0, we need c1 c2 (Incorrect). **Case 2: If c1 is at the cathode.** Then c2 is at the anode. Cell reaction: M(s) + M+(c1) arrow M(s) + M+(c2) For Ecell > 0, we need (c2/c1) c2. Option 2 says c1 c2 (Correct). In any spontaneous concentration cell, ions flow from the higher concentration compartment to the lower concentration compartment. Thus, for a positive voltage, the cathode must always have the higher concentration. Class 12 Chemistry: Electrochemistry

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M+ (M arrow M+ + e-). The cation M+ is present in two different concentrations c1 and c2 as shown above. Which of the following statement is correct for generating a positive cell potential?

For a concentration cell, E = 0, so: For a concentration cell to have a positive cell potential (Ecell > 0), the ratio [Anode][Cathode] must be less than 1. This implies that [Anode] 0, we need c1 c2 (Incorrect). **Case 2: If c1 is at the cathode.** Then c2 is at the anode. Cell reaction: M(s) + M+(c1) arrow M(s) + M+(c2) For Ecell > 0, we need (c2/c1) c2. Option 2 says c1 c2 (Correct). In any spontaneous concentration cell, ions flow from the higher concentration compartment to the lower concentration compartment. Thus, for a positive voltage, the cathode must always have the higher concentration. Class 12 Chemistry: Electrochemistry

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M+ (M arrow…

Year	2026	2025	2024	Total
Questions	13	19	8	40

Q72 jee_main_2026_28_january_morning Nernst Equation and Electron Transfer

Consider the following redox reaction taking place in acidic medium

BH₄⁻(aq) + ClO₃⁻(aq) arrow H₂BO₃⁻(aq) + Cl⁻(aq)

If the Nernst equation for the above balanced reaction is

Ecell = Ecell° - (RT)/(nF) ln Q,

Then the value of n is _____.

Numerical Answer. Answer: 24 to 24

Solution

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 $BH₄^-$ $BH_4^-$, Hydrogen is $-1$ $-1$ (hydride), so Boron is $+3$ $+3$. In $H₂BO₃^-$ $H_2BO_3^-$, Boron is $+3$ $+3$. The oxidation state of B doesn't change, but $H$ $H$ oxidizes from $-1$ $-1$ to $+1$ $+1$ (in water/acid). Alternatively, treat the whole moiety: Oxidation half-reaction: $BH₄^- + 3H₂O arrow H₂BO₃^- + 8H^+ + 8e^-$ In $ClO₃^-$ $ClO_3^-$, Chlorine is $+5$ $+5$. In $Cl^-$ $Cl^-$, Chlorine is $-1$ $-1$. Reduction half-reaction: $ClO₃^- + 6H^+ + 6e^- arrow Cl^- + 3H₂O$

Step 2: Balance Electrons

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4 to equalize electrons transferred: $3 \times (BH₄^- + 3H₂O arrow H₂BO₃^- + 8H^+ + 8e^-) 24e^-$ $4 \times (ClO₃^- + 6H^+ + 6e^- arrow Cl^- + 3H₂O) 24e^-$ The lowest common multiple of electrons exchanged is 24.

Final Conclusion

The balanced equation transfers 24 electrons, so $n = 24$ $n = 24$ in the Nernst equation.

Pattern Recognition

The total electrons ' $n$ $n$' in the Nernst equation is always the lowest common multiple of electrons from the balanced oxidation and reduction half-reactions.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Redox Reactions

Q73 jee_main_2026_28_january_evening Molar Conductivity And Kohlrausch Law

For strong electrolyte

Λm

increases slowly with dilution and can be represented by the equation

Λm = Λm° - Ac1/2

Molar conductivity values of the solutions of strong electrolyte AB at

18°C

are given below :

c $[mol L⁻¹]$ $[mol\ L^{-1}]$	0.04	0.09	0.16	0.25
$Λm [S cm² mol⁻¹]$ $\Lambda_{m} [S\ cm^{2} mol^{-1}]$	96.1	95.7	95.3	94.9

The value of constant A based on the above data

[in S cm² mol⁻¹/(mol/L)1/2]

unit is ____.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

$Λm = Λm° - A\sqrt(c)$ (Debye-Huckel-Onsager equation)

Core Logic

Using the equation for the first set of values ( $c = 0.04, \sqrt(c) = 0.2$ $c = 0.04, \sqrt{c} = 0.2$): $96.1 = Λm° - A(0.2)$ $96.1 = \Lambda_{m}^{\circ} - A(0.2)$ --- (1)

Using the equation for the second set of values ( $c = 0.09, \sqrt(c) = 0.3$ $c = 0.09, \sqrt{c} = 0.3$): $95.7 = Λm° - A(0.3)$ $95.7 = \Lambda_{m}^{\circ} - A(0.3)$ --- (2)

Step 1: Solve for A

Subtract eq (2) from eq (1): $96.1 - 95.7 = A(0.3) - A(0.2)$ $96.1 - 95.7 = A(0.3) - A(0.2)$ $0.4 = 0.1A$ $0.4 = 0.1A$ $A = (0.4)/(0.1) = 4$ $A = \frac{0.4}{0.1} = 4$

Pattern Recognition

Direct linear interpolation. Change in $Λm$ $\Lambda_{m}$ over change in $\sqrt(c)$ $\sqrt{c}$ gives the slope (which is A).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q74 jee_main_2026_28_january_evening Nernst Equation

A volume of x mL of

5 M NaHCO₃

solution was mixed with

10 mL

of

2 M H₂CO₃

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 mV

, then the value of x = ____ mL (nearest integer).

Sn(s) | Sn(OH)₆²⁻(0.5 M) | HSnO₂⁻(0.05 M) | OH⁻ || Bi₂O₃(s) | Bi(s)

Consider upto one place of decimal for intermediate calculations Given :

E_HSnO₂⁻|Sn(OH)₆²⁻o = -0.9 V

E_Bi₂O₃|Bio = -0.44 V

pKa_(H₂CO₃) = 6.11

(2.303RT)/(F) = 0.059 V

Antilog(1.29) = 19.5

Numerical Answer. Answer: 78 to 78

Solution

Related Formula

Ecell = Ecell° - (0.059)/(n) Q

pH = pKₐ + [Salt][Acid]

Core Logic

Note: In the question paper, $EHSnO₂^-/[Sn(OH)₆]²⁻° = -0.9 V$ is given, but standard NTA solution requires assuming $E[Sn(OH)₆]²⁻/HSnO₂^-° = -0.9 V$ for standard operation. (Our Ans. is Bonus due to this discrepancy, NTA Answer is 78). We will solve using the assumed valid logic.

$Ecell° = Ecathode° - Eanode° = -0.44 - (-0.90) = 0.46 V$

Oxidation Half: $HSnO₂^- + H₂O + 3OH^- arrow [Sn(OH)₆]²⁻ + 2e^-$ Reduction Half: $Bi₂O₃ + 3H₂O + 6e^- arrow 2Bi + 6OH^-$ Overall: $3HSnO₂^- + Bi₂O₃ + 6H₂O + 3OH^- arrow 3[Sn(OH)₆]²⁻ + 2Bi$ Here $n = 6$ $n = 6$.

Nernst Eq: $Ecell = Ecell° - (0.059)/(6) ([Sn(OH)₆]²⁻)³([HSnO₂^-]³ [OH^-]³)$ $0.2353 = 0.46 - (0.059)/(6) ((0.5)³)/((0.05)³ [OH^-]³)$ $0.2353 = 0.46 - (0.059)/(2) (10)/([OH^-])$ $( (10)/([OH^-]) ) = ((0.46 - 0.2353) \times 2)/(0.059) = (0.2247 \times 2)/(0.059) = 7.6$

Step 1: Calculate pH

$(10) - [OH^-] = 7.6$ $\log(10) - \log[OH^-] = 7.6$ $1 + pOH = 7.6 pOH = 6.6$ $1 + pOH = 7.6 \implies pOH = 6.6$ $pH = 14 - 6.6 = 7.4$ $pH = 14 - 6.6 = 7.4$

Step 2: Buffer Equation

Using Henderson-Hasselbalch equation for buffer of $NaHCO₃$ $NaHCO_3$ and $H₂CO₃$ $H_2CO_3$: $pH = pKₐ + nsaltnacid$ $7.4 = 6.11 + (5x)/(10 \times 2)$ $1.29 = (5x)/(20) = (x)/(4)$ Taking antilog: $(x)/(4) = Antilog(1.29) = 19.5$ $x = 19.5 \times 4 = 78 mL$ $x = 19.5 \times 4 = 78\text{ mL}$

Pattern Recognition

Merge Nernst equation finding unknown concentration with Buffer equations. Determine overall cell reaction to find exact stoichiometry and 'n' electrons for Nernst.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Q47 jee_main_2025_02_april_evening Conductivity and Molar Conductivity

0.2%

(w/v) solution of

NaOH

is measured to have resistivity

870.0~mΩ~m

. The molar conductivity of the solution will be

\times 10²~ mS~dm²~mol⁻¹

. (Nearest integer)

Numerical Answer. Answer: 23 to 23

Solution

Related Formula

κ = (1)/(ρ)

Λm = (κ)/(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)

$0.2%$ $0.2\%$ (w/v) $NaOH$ $\mathrm{NaOH}$ means $0.2~g$ $0.2~\mathrm{g}$ of $NaOH$ $\mathrm{NaOH}$ is present in $100~mL$ $100~\mathrm{mL}$ of solution.

Molar mass of NaOH = 23 + 16 + 1 = 40~ g~mol⁻¹

Molarity M = Mass of soluteMolar mass \times 1000VmL = (0.2)/(40) \times (1000)/(100) = 0.05~ mol~L⁻¹ = 0.05~ mol~dm⁻³

Step 2: Calculate Conductivity (kappa) in dm Units

Given resistivity $ρ = 870.0~mΩ~m = 870 \times 10⁻³~Ω~m = 0.87~Ω~m$ .

Since $1~m = 10~dm$ $1~\mathrm{m} = 10~\mathrm{dm}$:

ρ = 0.87~Ω \times (10~dm) = 8.7~Ω~dm

Now, conductivity $κ$ $\kappa$ is:

κ = (1)/(ρ) = (1)/(8.7)~Ω⁻¹~dm⁻¹

Step 3: Calculate Molar Conductivity (Lambda_m)

Λm = (κ)/(M) = (1)/(8.7)~ S~dm⁻¹0.05~ mol~dm⁻³ = (1)/(8.7 \times 0.05) = (1)/(0.435) \approx 2.29885~ S~dm²~mol⁻¹

Converting $S$ $\mathrm{S}$ to $mS$ $\mathrm{mS}$ ( $1~S = 10³~mS$ $1~\mathrm{S} = 10^3~\mathrm{mS}$):

Λm = 2.29885 \times 10³~ mS~dm²~mol⁻¹ = 22.9885 \times 10²~ mS~dm²~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 $dm⁻³$ $\mathrm{dm^{-3}}$), expressing conductivity in terms of $dm⁻¹$ $\mathrm{dm^{-1}}$ directly eliminates the need for arbitrary $1000$ $1000$ multiplication factors.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q jee_main_2025_02_april_morning Nernst Equation and Salt Hydrolysis pH

Consider the following electrochemical cell at standard condition.

Au(s) QH₂, Q NH₄X (0.01 M) Ag^+ (1 M) Ag(s)

Ecell = +0.4 V

The couple

QH₂ / Q

represents quinhydrone electrode, the half cell reaction is given below:

Quinhydrone half cell reduction equation diagram for Q47 — The diagram displays the balanced chemical equation for quinhydrone reduction, consuming two electrons and two protons to yield hydroquinone.

[ Given: EAg^+ / Ag^o = +0.8 V and (2.303 RT)/(F) = 0.06 V ]

The

pKb

value of the ammonium halide salt

(NH₄X)

used here is _____.

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

Nernst equation for the net combined redox cell expression:

E = E^° - (0.06)/(2) ( [H^+]²[Ag^+]²)

Hydrolysis equation for a salt composed of a weak base and strong acid:

pH = 7 - (1)/(2)pKb - (1)/(2)

Core Logic

Let's compute the operational values line-by-row:

Combined redox process: $QH₂ + 2Ag^+ arrow Q + 2Ag + 2H^+$ .
Standard cell potential difference: $E^°cell = E^°Ag^+/Ag - E^°Q/QH₂ = 0.8 - 0.7 = +0.1~V$ .
Apply Nernst adjustments using known concentrations ( $[Ag^+] = 1~M$ $[\mathrm{Ag}^+] = 1\mathrm{~M}$):

0.4 = 0.1 - 0.06 [H^+]

0.3 = 0.06 \times pH pH = 5

Step 1: Salt Hydrolysis Substitution

Substitute the determined $pH$ $\mathrm{pH}$ along with salt molarity ( $C = 0.01~M = 10⁻²~M$ ) into the hydrolysis equation:

5 = 7 - (1)/(2)pKb - (1)/(2) (10⁻²)

5 = 7 - (1)/(2)pKb - (1)/(2)(-2)

5 = 7 - (1)/(2)pKb + 1

5 = 8 - (1)/(2)pKb (1)/(2)pKb = 3 pKb = 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 $0.06~V$ $0.06\mathrm{~V}$ at standard ambient conditions.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Solution & Explanation

Related Formula

Core Logic

Step 1: Check Cathode Conditions

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 3

Solution

Core Logic

Step 1: Assign Oxidation States

Step 2: Balance Electrons

Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for A

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate pH

Step 2: Buffer Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Molarity (M)

Step 2: Calculate Conductivity (kappa) in dm Units

Step 3: Calculate Molar Conductivity (Lambda_m)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Salt Hydrolysis Substitution

Pattern Recognition

Chapter Mix

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