Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{cu}^{2+}$ 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 the solution fixed at 200 mL . The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer) [Given : $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text{red}}^{0}=+0.34 \mathrm{~V}$ $\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text{red}}^{0}=+1.23 \mathrm{~V}$ Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$ Faraday Constant $=96500 \mathrm{C} \mathrm{~mol}^{-1}$ Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$ ]

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Redox Reactions and Electrochemistry

Electricity is passed through an acidic solution of

Cu^{2 +}

till all the

cu^{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 the solution fixed at 200 mL . The total volume of oxygen evolved at STP during the entire process is ______ mL.
(Nearest integer)
[Given :

Cu^{2 +} (aq) + 2 e^{-} \to Cu (s) E_{red}^{0} = + 0.34 V

O_{2} (g) + 4 H^{+} + 4 e^{-} \to 2 H_{2} O E_{red}^{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

]

Answer: 111

Solution

While

C u^{2 +}

is being deposited, an equal quantity of charge liberates oxygen at the anode. The copper deposited is

\frac{0.300}{63.54} = 4.72 \times 1 0^{- 3} mol

, needing

9.44 \times 1 0^{- 3} mol

of electrons, which give

\frac{9.44 \times 1 0 ^{- 3}}{4} = 2.36 \times 1 0^{- 3} mol

O_{2}

. The further charge passed is

Q = 0.600 \times 28 \times 60 = 1008 C

, that is

\frac{1008}{96500} = 1.045 \times 1 0^{- 2} mol

of electrons, yielding another

\frac{1.045 \times 1 0 ^{- 2}}{4} = 2.61 \times 1 0^{- 3} mol

O_{2}

. The total oxygen is

4.97 \times 1 0^{- 3} mol

, which at STP occupies

4.97 \times 1 0^{- 3} \times 22400 = 111 mL

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Integer answer

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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Redox Reactions and Electrochemistry

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