Question

How long should aqueous \(\text{NaCl}\) be electrolysed by passing 100 ampere current, so that 0.5 mol chlorine is released at anode?

• A. 96500 seconds
• B. 9650 seconds
• C. 965 seconds
• D. 96.5 seconds

Hint:

Use \(Q = It\) and Faraday constant \(= 96500 C\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Faraday's laws of electrolysis, the amount of substance produced at an electrode is proportional to the quantity of electricity passed through the electrolyte.
Step 2: Key Formula or Approach:
\[ Q = I \times t = n \times F \times (\text{moles of product}) \] Where $n$ is the number of electrons per molecule of product, and $F \approx 96500\text{ C/mol}$.
Step 3: Detailed Explanation:
The anodic reaction for chlorine release is: \[ 2\text{Cl}^-(\text{aq}) \longrightarrow \text{Cl}_2(\text{g}) + 2\text{e}^- \] From this, $1$ mole of \(\text{Cl}_2\) requires $2$ moles of electrons (\(n = 2\)).
Total charge required for $0.5\text{ mol Cl}_2$: \[ Q = 0.5 \times 2 \times 96500\text{ C} = 96500\text{ C} \] We are given current \(I = 100\text{ A}\).
\[ t = \frac{Q}{I} = \frac{96500\text{ C}}{100\text{ A}} = 965\text{ seconds} \] Step 4: Final Answer:
The time required is $965$ seconds.