A 4.0 M aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. (Na = 23, Hg = 200, 1 F = 96500 C)
The following examples illustrate a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
The total charge (coulombs) required for complete electrolysis is

Options

A. 24125
B. 48250
C. 96500
D. 193000

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of NaCl in the solution.
Given:
- Molarity (M) = 4.0 M
- Volume of solution = 500 mL = 0.5 L (since 1 L = 1000 mL)

Using the formula for molarity:
\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (in L)} \]
\[ \text{Moles of NaCl} = 4.0 \, \text{mol/L} \times 0.5 \, \text{L} = 2.0 \, \text{moles} \]

### Step 2: Determine the moles of chlorine gas produced.
From the electrolysis of NaCl, the reaction can be summarized as:
\[ 2 \, \text{NaCl} + 2 \, \text{H}_2\text{O} \rightarrow \text{Cl}_2 + 2 \, \text{Na}^+ + 2 \, \text{OH}^- \]
From the balanced equation, we see that:
- 2 moles of NaCl produce 1 mole of Cl2.

Thus, from 2 moles of NaCl, we get:
\[ \text{Moles of Cl}_2 = \frac{2.0 \, \text{moles NaCl}}{2} = 1.0 \, \text{mole Cl}_2 \]

### Step 3: Calculate the total charge required for the electrolysis.
The charge required to produce 1 mole of Cl2 can be calculated using Faraday's law. The charge (Q) required for 1 mole of a substance can be calculated as:
\[ Q = n \times F \]
Where:
- \( n \) = number of moles of Cl2 = 1.0 mole
- \( F \) = Faraday's constant = 96500 C/mol

Thus, the total charge required is:
\[ Q = 1.0 \, \text{mole} \times 96500 \, \text{C/mol} = 96500 \, \text{C} \]

### Step 4: Conclusion
The total charge required for complete electrolysis of 500 mL of a 4.0 M NaCl solution is **96500 C**.

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