JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

Electricity is passed through an acidic solution of $C u^{2 +}$ till all the $C u^{2 +}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)

[Given :

$C u^{2 +} (a q) + 2 e^{-} \to C u (s) E_{r e d}^{0} = + 0.34$ V

$O_{2} (g) + 4 H^{+} + 4 e^{-} \to 2 H_{2} O E_{r e d}^{0} = + 1.23$ V

Molar mass of Cu = 63.54 g mol $^{- 1}$

Molar mass of $O_{2}$ = 32 g mol $^{- 1}$

Faraday Constant = 96500 C mol $^{- 1}$

Molar volume at STP = 22.4 L ]

Your answer

Solution

Phase 1 — while $C u^{2 +}$ is present: Cu deposition and $O_{2}$ evolution at the anode occur with equal equivalents.

Equivalents of Cu deposited $= \frac{300 \times 1 0 ^{- 3} \times 2}{63.54} = 9.44 \times 1 0^{- 3}$ eq.

Equivalents of $O_{2}$ in this phase = same. Moles of $O_{2} = \frac{9.44 \times 1 0 ^{- 3}}{4} = 2.36 \times 1 0^{- 3}$ mol.

Phase 2 — after $C u^{2 +}$ is exhausted: All current goes to $O_{2}$ evolution.

Charge passed $= I \cdot t = 0.6 \times 28 \times 60 = 1008$ C.

Moles of $O_{2} = \frac{1008}{4 \times 96500} = 2.611 \times 1 0^{- 3}$ mol.

Total volume at STP:

$V = (2.36 + 2.611) \times 1 0^{- 3} \times 22400 = 4.971 \times 1 0^{- 3} \times 22400 \approx 111.35$ mL.

Nearest integer: $111$ .

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