Electrochemistry Result Question 31
Passage
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values.
(2007, 4 $\times$ 3M = 12M)
$ \mathrm{I}_2+2 e^{-} \rightarrow 2 \mathrm{I}^{-} $$\quad \quad E^{\circ}=0.54 $
$ \mathrm{Cl}_2+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-} $$\quad \quad E^{\circ}=1.36 $
$ \mathrm{Mn}^{3+}+e^{-} \rightarrow \mathrm{Mn}^{2+} $$\quad \quad E^{\circ}=1.50 $
$ \mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} $$\quad \quad\quad E^{\circ}=0.77 $
$ \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} $$\quad \quad E^{\circ}=1.23$
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
(a) $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
(b) $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
(c) $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
(d) $\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
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Answer:
Correct Answer: 31. ( a )
Solution:
- Sodium fusion extract from aniline produces 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} $
BETA
