In the development of photographic film, the black and white photographic film is developed as follows:

The silver bromide reacts with hydroquinone producing black coloured silver particles along with HBr and quinone. During this process, silver bromide gets reduced to silver and hydroquinone is oxidised to quinone.

IIT-JEE 2005 Mains Chemistry - d and f Block Elements Question 3 English Explanation 1

The hydroquinone acts as a developer during this process.

The unreacted silver bromide then reacts with sodium thiosulphate to produce soluble complex and sodium bromide.

$$\mathrm{AgBr + \mathop {2N{a_2}{S_2}{O_3}}\limits_{hypo\,solution} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{so{\mathop{\rm lub}} le} + NaBr}$$

The reaction occurs in acidic medium. In acidic medium, sulphur from sodium thiosulphate will precipitate out. The colloidal sulphur is obtained which gives milky white turbidity.

$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{SO}_3+\underset{\substack{\text { colloidal } \\ \text { sulphur }}}{\mathrm{S} \downarrow}$

Final Answer :

IIT-JEE 2005 Mains Chemistry - d and f Block Elements Question 3 English Explanation 2

$$\mathrm{AgBr + \mathop {2N{a_2}{S_2}{O_3}}\limits_{hypo\,solution} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{so{\mathop{\rm lub}} le} + NaBr}$$

D And F Block Elements

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