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JEE Main 202624 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2026 · 24 January · Morning Shift · Chemistry · Electrochemistry

Electricity is passed through an acidic solution of Cu^ 2+ till all...

Electricity is passed through an acidic solution of Cu2+\mathrm{Cu}^{2+} till all the Cu2+\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: Cu2+(aq)+2e−→Cu(s)Ered o=+0.34 V\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text{red }}^{\mathrm{o}}=+0.34 \mathrm{~V}
O2( g)+4H++4e−→2H2OEred o=+1.23 V\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text{red }}^{\mathrm{o}}=+1.23 \mathrm{~V}
Molar mass of Cu=63.54 g mol−1\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}
Molar mass of O2=32 g mol−1\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}
Faraday Constant =96500Cmol−1=96500 \mathrm{C} \mathrm{mol}^{-1}
Molar volume at STP=22.4 L\mathrm{STP}=22.4 \mathrm{~L}]

Correct answer

0

Step-by-step solution

Eq of Cu = Eq of O2O_2
300×10−3×263.54=nO2×4\dfrac{300 \times 10^{-3} \times 2}{63.54} = n_{O_2} \times 4
2.36×10−3=nO22.36 \times 10^{-3} = n_{O_2}
When current is further passed
nO2×4=600×28×6096500×1000n_{O_2} \times 4 = \dfrac{600 \times 28 \times 60}{96500 \times 1000}
nO2=2.611×10−3n_{O_2} = 2.611 \times 10^{-3}
Total O2O_2 released
=[10−3×(2.36+2.611)]×22400= [10^{-3} \times (2.36 + 2.611)] \times 22400 ml
=111.35= 111.35 ml
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