Electrochemistry 2 Question 32
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31. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $\mathrm{H}{2} \mathrm{SO}{4}$ in the presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of
======= ####31. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $\mathrm{H}{2} \mathrm{SO}{4}$ in the presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $\mathrm{Fe}{4}\left[\mathrm{Fe}(\mathrm{CN}){6}\right]_{3}$
(c) $\mathrm{Fe}{4}\left[\mathrm{Fe}(\mathrm{CN}){6}\right]_{2}$
(b) $\mathrm{Fe}{3}\left[\mathrm{Fe}(\mathrm{CN}){6}\right]_{2}$
(d) $\mathrm{Fe}{3}\left[\mathrm{Fe}(\mathrm{CN}){6}\right]_{3}$
Passage IV
Tollen’s reagent is used for the detection of aldehydes. When a solution of $\mathrm{AgNO}{3}$ is added to glucose with $\mathrm{NH}{4} \mathrm{OH}$, then gluconic acid is formed.
$$ \begin{aligned} & \mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; \quad E_{\text {red }}^{\circ}=0.80 \mathrm{~V} \ & \mathrm{C}{6} \mathrm{H}{12} \mathrm{O}{6}+\mathrm{H}{2} \mathrm{O} \longrightarrow \underset{\text { Gluconic acid }}{\mathrm{C}{6} \mathrm{H}{12} \mathrm{O}{7}+2 \mathrm{H}^{+}+2 e^{-} ;} \quad E{\text {oxi }}^{\circ}=-0.05 \mathrm{~V} \ & \mathrm{Ag}\left(\mathrm{NH}{3}\right){2}^{+}+e^{-} \longrightarrow \mathrm{Ag}(s)+2 \mathrm{NH}{3} ; \quad E{\text {red }}^{\circ}=0.337 \mathrm{~V} \ & {\left[\text { Use } 2.303 \times \frac{R T}{F}=0.0592 \text { and } \frac{F}{R T}=38.92 \text { at } 298 \mathrm{~K}\right]} \end{aligned} $$
$(2006,3 \times 4 \mathrm{M}=12 \mathrm{M})$
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Solution:
- Sodium fusion extract from aniline produces $\mathrm{NaCN}$ which reacts with $\mathrm{Fe}^{2+}$ to form $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$. The complex ion then reacts with $\mathrm{Fe}^{3+}$ to give blue precipitate of prussian blue.
$$ \mathrm{Fe}^{3+}+\left[\mathrm{Fe}(\mathrm{CN}){6}\right]^{4-} \rightleftharpoons \underset{\text { Prussian blue }}{\mathrm{Fe}{4}\left[\mathrm{Fe}(\mathrm{CN}){6}\right]{3}} $$
