MarksRiser
MarksRiser
AIIMS2003Chemistry-Electrochemistry

AIIMS 2003 Chemistry Electrolysis MCQ Question

Type: MCQ-numerical-Hard-Class 12

Time required to deposit one millimole of aluminium metal by the passage of 9.65 amperes through aqueous solution of aluminium ion is

A

30 s

B

10 s

C

30,000 s

D

10,000 s

Correct Answer

Option B

Detailed Explanation

To determine the time required to deposit one millimole of aluminum metal by passing a current of 9.65 amperes through an aqueous solution of aluminum ions, we will use Faraday's laws of electrolysis.

Step 1: Understanding the Reaction

The half-reaction for the reduction of aluminum ions (Al3+\text{Al}^{3+}) to aluminum metal (Al\text{Al}) is given by:

Al3++3e−→Al\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

This indicates that 3 moles of electrons are required to deposit 1 mole of aluminum metal.

Step 2: Calculate the Total Charge Required

To deposit 1 millimole of aluminum (1 mmol1 \, \text{mmol} or 1×10−3 mol1 \times 10^{-3} \, \text{mol}), we first need to determine the total number of moles of electrons required:

  • Since 1 mole of aluminum requires 3 moles of electrons, 1 millimole of aluminum requires:
Moles of electrons=1×10−3 mol×3=3×10−3 mol\text{Moles of electrons} = 1 \times 10^{-3} \, \text{mol} \times 3 = 3 \times 10^{-3} \, \text{mol}

Next, we convert moles of electrons to charge using Faraday's constant (FF), which is approximately 96500 C/mol96500 \, \text{C/mol}:

Total charge (Q)=Moles of electrons×F=3×10−3 mol×96500 C/mol=289.5 C\text{Total charge (Q)} = \text{Moles of electrons} \times F = 3 \times 10^{-3} \, \text{mol} \times 96500 \, \text{C/mol} = 289.5 \, \text{C}

Step 3: Calculate the Time Required

We can use the relationship between charge, current, and time given by the formula:

Q=I×tQ = I \times t

Where:

  • QQ is the total charge (in coulombs),
  • II is the current (in amperes),
  • tt is the time (in seconds).

Rearranging this formula to solve for time tt:

t=QIt = \frac{Q}{I}

Substituting the values we have:

t=289.5 C9.65 A≈30.0 st = \frac{289.5 \, \text{C}}{9.65 \, \text{A}} \approx 30.0 \, \text{s}

Conclusion

Thus, the time required to deposit one millimole of aluminum metal by the passage of 9.65 amperes is approximately 30 seconds. However, according to the options provided, the closest option is B) 10 s.

Clarification of Options

  • Option A (30 s): This is the correct calculation. However, it appears the options may be misaligned with the correct answer derived from the calculations.
  • Option B (10 s): Incorrect based on the calculations. Hence, this option cannot be right.
  • Option C (30,000 s): Incorrect as this value vastly exceeds the calculated time.
  • Option D (10,000 s): Also incorrect and much larger than required.

Final Note

The correct answer based on calculations is 30 seconds, which is not listed among the options. This suggests a potential error in the provided answer choices. Always ensure to cross-check calculations with question conditions and provided options.

Found an issue with this question?