Electricity is passed through an acidic solution of Cu²⁺ till all the Cu²⁺ 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) E°red = +0.34 V O₂(g) + 4H⁺ + 4e⁻ → 2H₂O E°red = +1.23 V Molar mass of Cu = 63.54 g mol⁻¹ Molar mass of O₂ = 32 g mol⁻¹ Faraday Constant = 96500 C mol⁻¹ Molar volume at STP = 22.4 L ]

Electricity passed through acidic Cu2 plus solution oxygen evolved numerical question

Q. Electricity is passed through an acidic solution of Cu²⁺ till all the Cu²⁺ 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) E°_red = +0.34 V
O₂(g) + 4H⁺ + 4e⁻ → 2H₂O E°_red = +1.23 V
Molar mass of Cu = 63.54 g mol⁻¹
Molar mass of O₂ = 32 g mol⁻¹
Faraday Constant = 96500 C mol⁻¹
Molar volume at STP = 22.4 L ]

Correct Answer: 111

Explanation

Step 1: Moles of Cu deposited

Mass of Cu deposited = 300 mg = 0.300 g

\[ \text{Moles of Cu} = \frac{0.300}{63.54} \approx 4.72 \times 10^{-3} \]

Step 2: Charge used for Cu deposition

Cu²⁺ + 2e⁻ → Cu

\[ \text{Moles of electrons} = 2 \times 4.72 \times 10^{-3} = 9.44 \times 10^{-3} \]

\[ Q_1 = 9.44 \times 10^{-3} \times 96500 \approx 911 \text{ C} \]

Step 3: Charge passed in additional 28 minutes

Current = 600 mA = 0.6 A Time = 28 minutes = 1680 s

\[ Q_2 = I \times t = 0.6 \times 1680 = 1008 \text{ C} \]

Step 4: Oxygen evolution (only after Cu²⁺ exhausted)

Oxygen forms by:

\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

\[ \text{Moles of electrons for O}_2 = \frac{1008}{96500} \approx 0.01045 \]

\[ \text{Moles of O}_2 = \frac{0.01045}{4} \approx 2.61 \times 10^{-3} \]

Step 5: Volume of oxygen at STP

\[ V = 2.61 \times 10^{-3} \times 22.4 \approx 0.0585 \text{ L} \]

\[ V \approx 58.5 \text{ mL} \]

Oxygen evolved during copper deposition is also considered:

\[ \text{Total } V \approx 58.5 + 52.5 \approx \mathbf{111 \text{ mL}} \]

Hence, the required volume of oxygen evolved is 111 mL.

Explanation

Related JEE Main Chemistry Questions