Electricity is passed through an acidic solution of Cu2+ till all the Cu2+ was exhausted, leading to the deposition of…

Year	2026	2025	2024	Total
Questions	13	19	8	40

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Equivalents of metal deposited = Equivalents of gas evolved (in Phase 1)

ne^- = (Q)/(F) = (I \times t)/(96500)

Core Logic

Phase 1: Deposition of $300 mg$ $300 \text{ mg}$ Cu. By Faraday's Laws, the equivalents of copper deposited at the cathode must equal the equivalents of oxygen evolved at the anode during this period. Equivalents of $Cu = (W)/(E) = 300 \times 10⁻³(63.54)/(2)$ Equivalents of $O₂ = nO₂ \times 4$ $O_2 = n_{O_2} \times 4$ $300 \times 10⁻³ \times 263.54 = nO₂ \times 4$ $2.36 \times 10⁻³ = nO₂$ $2.36 \times 10^{-3} = n_{O_2}$ (moles of $O₂$ $O_2$ in Phase 1)

Phase 2: Current continued. $I = 600 mA = 0.6 A$ $I = 600 \text{ mA} = 0.6 \text{ A}$, $t = 28 mins = 28 \times 60 seconds$ . Moles of electrons passed $= (0.6 \times 28 \times 60)/(96500) = 0.010445 moles of e⁻$ Equivalents of $O₂$ $O_2$ evolved in Phase 2 = Moles of $e⁻$ $e^{-}$ passed = $0.010445$ $0.010445$ $nO₂ (Phase 2) = (0.010445)/(4) = 2.611 \times 10⁻³ mol$

Step 1: Calculate Total Volume of Oxygen

Total moles of $O₂$ $O_2$ evolved $= (2.36 \times 10⁻³) + (2.611 \times 10⁻³) = 4.971 \times 10⁻³ mol$

Total volume at STP: $VO₂ = ntotal \times 22400 mL$ $VO₂ = 4.971 \times 10⁻³ \times 22400 mL = 111.35 mL$ Rounding to nearest integer $111 mL$ $\implies 111 \text{ mL}$.

Pattern Recognition

Equivalents of products at cathode and anode are always equal in any given time span. Valency factor for $O₂$ $O_2$ evolution from water is 4.

Chapter Mix

Class 12 Chemistry: Electrochemistry

More Electrochemistry Previous-Year Questions

Q72 jee_main_2026_21_jan_morning Conductance of Electrolytic Solutions

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 cm²

were at a distance of

15 cm

apart. The value of the limiting molar conductivity is .....

S m² mol⁻¹

. (nearest integer) (Given: degree of dissociation of the weak acid

(α) ll 1)

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

κ = G \times (l)/(A)

Λm = (κ \times 1000)/(C)

α = ΛmΛm\infty

$[H^+] = Cα$ $[H^+] = C\alpha$

Core Logic

Given: $pH = 5$ $pH = 5$, so $[H^+] = 10⁻⁵ M$ $[H^+] = 10^{-5}\text{ M}$. Since $[H^+] = C \cdot α = C \cdot ΛmΛm\infty$ , we have $10⁻⁵ = C \cdot ΛmΛm\infty$ .

First, calculate conductivity ( $κ$ $\kappa$): Conductance $G = 4 \times 10⁻⁵ S$ $G = 4 \times 10^{-5}\text{ S}$. Cell constant $G^* = (l)/(A) = 15 cm1 cm² = 15 cm⁻¹$ .

κ = G \cdot G^* = (4 \times 10⁻⁵) \times 15 = 6 \times 10⁻⁴ S cm⁻¹

Molar conductivity ( $Λm$ $\Lambda_m$):

Λm = (κ \times 1000)/(C) = 6 \times 10⁻⁴ \times 1000C = (0.6)/(C)

Substitute $Λm$ $\Lambda_m$ into the proton concentration formula:

[H^+] = 10⁻⁵ = C \cdot 0.6 / CΛm\infty

10⁻⁵ = 0.6Λm\infty

Λm\infty = 0.610⁻⁵ = 60000 S cm² mol⁻¹

Convert units to $S m² mol⁻¹$ $\text{S m}^2\text{ mol}^{-1}$: Since $1 m² = 10⁴ cm²$ $1\text{ m}^2 = 10^4\text{ cm}^2$,

Λm\infty = 60000 \times 10⁻⁴ S m² mol⁻¹ = 6 S m² mol⁻¹

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Q67 jee_main_2026_21_jan_evening Daniell Cell and Cell Potential Variation

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?

A. $(1) Plot 1$ $(1) \text{ Plot 1}$
B. $(2) Plot 2$ $(2) \text{ Plot 2}$
C. $(3) Plot 3$ $(3) \text{ Plot 3}$
D. $(4) Plot 4$ $(4) \text{ Plot 4}$

Solution

Core Logic

For a closed circuit Daniell cell operating under given conditions, standard cell potential $Ecell°$ $E_{\text{cell}}^{\circ}$ 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 $Ecell°$ $E_{\text{cell}}^{\circ}$ remains constant with time.

Pattern Recognition

Sees: graphical representation of electrochemical cell potentials over time. Trap: Confusing standard cell potential $E°$ $E^{\circ}$ with operating cell potential $Ecell$ $E_{\text{cell}}$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q75 jee_main_2026_21_jan_evening Solubility Product and Electrode Potential

MX

is a sparingly soluble salt that follows the given solubility equilibrium at 298 K:

MX(s) leftharpoons M^+(aq) + X^-(aq); Kₛₚ = 10⁻¹⁰

If the standard reduction potential for

M^+(aq) arrow M(s)

(EM^+/M) = 0.79 V

, then the value of the standard reduction potential for the metal/metal insoluble salt electrode

EX^-/MX(s)/M

____ mV

. (nearest integer) [Given:

(2.303RT)/(F) = 0.059 V

]

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

$EX^-/MX/M° = EM^+/M° + (0.059)/(n) Kₛₚ$ r

Core Logic

Substituting the given values:

EX^-/MX/M° = 0.79 + (0.059)/(1) (10⁻¹⁰)

EX^-/MX/M° = 0.79 + 0.059(-10) = 0.79 - 0.59 = 0.20 V

Step 1: Converting to Millivolts

0.20 V = 200 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

Q72 jee_main_2026_22_january_morning Nernst Equation

Consider the following electrochemical cell at 298K

Pt | HSnO₂⁻(aq) | Sn(OH)₆²⁻(aq) || Bi₂O₃(s) | Bi(s)

If the reaction quotient at a given time is

10⁶

, then the cell EMF

(Ecell)

x \times 10⁻¹ V

(Nearest integer). Given the standard half-cell reduction potential as

E_Bi₂O₃/Bi,OH⁻⁰ = -0.44 V and E_Sn(OH)₆²⁻/HSnO₂⁻,OH⁻⁰ = -0.90 V

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Ecell = Ecell° - (0.0591)/(n) Q

(Using 0.06 for approximation if needed, as per standard practice unless specified).

Core Logic

First, calculate $E°cell$ $E^{\circ}_{cell}$:

E°cell = E°cathode - E°anode

E°cell = (-0.44 V) - (-0.90 V) = +0.46 V

Identify the number of transferred electrons ( $n$ $n$) for the overall reaction. Anode reaction: $Sn²⁺ arrow Sn⁴⁺ + 2e⁻$ $Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Tin goes from +2 in $HSnO₂^-$ $HSnO_2^-$ to +4 in $Sn(OH)₆²⁻$ $Sn(OH)_6^{2-}$). Cathode reaction: $Bi³⁺ + 3e⁻ arrow Bi$ $Bi^{3+} + 3e^{-} \rightarrow Bi$ (Bismuth goes from +3 in $Bi₂O₃$ $Bi_2O_3$ to 0). To balance the electrons, we multiply anode by 3 and cathode by 2, yielding $n = 6$ $n = 6$ total electrons transferred.

Step 1: Apply Nernst Equation

Applying the Nernst equation (approximating 0.0591 to 0.06 for simplicity):

Ecell = E°cell - (0.06)/(n) ₁₀ Q

Ecell = 0.46 - (0.06)/(6) ₁₀(10⁶)

Ecell = 0.46 - 0.01 \times 6

Ecell = 0.46 - 0.06 = 0.40 V

Step 2: Format Output

The problem asks for the answer in the form $x \times 10⁻¹ V$ $x \times 10^{-1}\text{ V}$.

0.40 = 4 \times 10⁻¹

So, $x = 4$ $x = 4$.

Pattern Recognition

Be extremely careful finding 'n'. You must balance the half-reactions. A +2 to +4 change (n=2) and a +3 to 0 change (n=3) resolves to a common multiple of 6.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q67 jee_main_2026_22_january_evening Standard Reduction Potentials and Reducing Strength

Consider the following reduction processes:

Al³⁺ + 3e^- arrow Al(s)

E⁰ = -1.66 V

Fe³⁺ + e^- arrow Fe²⁺

E⁰ = +0.77 V

Co³⁺ + e^- arrow Co²⁺

E⁰ = +1.81 V

Cr³⁺ + 3e^- arrow Cr(s)

E⁰ = -0.74 V

The tendency to act as reducing agent decreases in the order:

A. $Al > Cr > Fe²⁺ > Co²⁺$
B. $Al > Fe²⁺ > Cr > Co²⁺$
C. $Al > Cr > Co²⁺ > Fe²⁺$
D. $Cr > Fe²⁺ > Al > Co²⁺$

Solution

Related Formula

Reducing Power \propto 1Standard Reduction Potential (E⁰red)

Core Logic

Step 1: List standard reduction potential values:

$E⁰(Al³⁺/Al) = -1.66 V$
$E⁰(Cr³⁺/Cr) = -0.74 V$
$E⁰(Fe³⁺/Fe²⁺) = +0.77 V$
$E⁰(Co³⁺/Co²⁺) = +1.81 V$

Step 2: Arrange species in increasing order of $E⁰red$ $E^0_{red}$ (decreasing reducing power):

Al > Cr > Fe²⁺ > Co²⁺

Pattern Recognition

Sees: Reduction potential values for reducing strength ordering. Shortcut: More negative standard reduction potential $\implies$ stronger reducing agent.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Total Volume of Oxygen

Pattern Recognition

Chapter Mix

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Core Logic

Step 1: Final Conclusion

Pattern Recognition

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Solution

Related Formula

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Step 1: Converting to Millivolts

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Apply Nernst Equation

Step 2: Format Output

Pattern Recognition

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Related Formula

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