Give equation and describe the process for the developing of black and white photographic film. When sodium thiosulphate solution is treated with acidic solution turns milky white. Give the half reaction of the above described process.

Options

A. `AgBr` solution
B. Hypo solution
C. `Na_(2)3_(4)O_(6)` solution
D. `FeC_(2)O_(4)`, solution

Answer

To develop black and white photographic film, a series of chemical reactions take place involving silver bromide (AgBr) and hydroquinone. Below is a step-by-step solution to the question, including the relevant chemical equations and half-reactions.

### Step 1: Understanding the Development Process
The development of black and white photographic film primarily involves the reduction of silver ions from silver bromide (AgBr) to metallic silver (Ag) using a reducing agent like hydroquinone.

### Step 2: Reaction of Silver Bromide with Hydroquinone
The first reaction involves silver bromide reacting with hydroquinone. The equation for this reaction can be written as:

\[ 2 \text{AgBr} + \text{C}_6\text{H}_6\text{O}_2 \rightarrow 2 \text{Ag} + 2 \text{HBr} + \text{C}_6\text{H}_4\text{O}_2 \]

Here, hydroquinone (C₆H₆O₂) reduces Ag⁺ ions to Ag (metallic silver), while itself being oxidized to quinone (C₆H₄O₂).

### Step 3: Formation of Soluble Complex with Sodium Thiosulfate
Next, the developed silver can form a soluble complex with sodium thiosulfate (Na₂S₂O₃). The reaction is as follows:

\[ \text{AgBr} + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_3\text{AgS}_2\text{O}_3 + \text{NaBr} \]

This reaction produces a soluble complex known as sodium argentothiosulfate.

### Step 4: Acidification of Sodium Thiosulfate
When sodium thiosulfate solution is treated with an acidic solution, it turns milky white due to the formation of colloidal sulfur. The reaction can be represented as:

\[ \text{Na}_2\text{S}_2\text{O}_3 + 2 \text{H}^+ \rightarrow 2 \text{Na}^+ + \text{H}_2\text{SO}_3 + \text{Colloidal S} \]

Here, the acid (H⁺) reacts with sodium thiosulfate to produce sulfur dioxide (H₂SO₃) and colloidal sulfur, which causes the solution to appear milky.

### Step 5: Half-Reaction of the Process
The half-reaction for the reduction of silver ions can be written as:

\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

This half-reaction shows the reduction of silver ions to metallic silver during the development process.

### Summary of Equations
1. Development of film:
\[ 2 \text{AgBr} + \text{C}_6\text{H}_6\text{O}_2 \rightarrow 2 \text{Ag} + 2 \text{HBr} + \text{C}_6\text{H}_4\text{O}_2 \]
2. Formation of soluble complex:
\[ \text{AgBr} + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_3\text{AgS}_2\text{O}_3 + \text{NaBr} \]
3. Acidification reaction:
\[ \text{Na}_2\text{S}_2\text{O}_3 + 2 \text{H}^+ \rightarrow 2 \text{Na}^+ + \text{H}_2\text{SO}_3 + \text{Colloidal S} \]
4. Half-reaction for silver reduction:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]