Question

Redox reactions 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 Daniel 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.
$ I _{2}+2 e^{-} \rightarrow 2 I ^{-} E ^{\circ}=0.54 $
$Cl _{2}+2 e^{-} \rightarrow 2 Cl ^{-} E ^{\circ}=1.36$
$ Mn ^{3+}+e^{-} \rightarrow Mn ^{2+} E ^{\circ}=1.50$
$Fe ^{3+}+e^{-} \rightarrow Fe ^{2+} E ^{\circ}=0.77$
$ O _{2}+4 H ^{+}+4 e^{-} \rightarrow 2 H _{2} O E ^{\circ}=1.23$
Sodium fusion extract, obtained from aniline, on treatment with iron(II) sulphate and $H _{2} SO _{4}$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of

• A. $F e_{4}\left[F e(C N)_{6}\right]_{3}$
• B. $F e_{3}\left[F e(C N)_{6}\right]_{2}$
• C. $F e_{4}\left[F e(C N)_{6}\right]_{2}$
• D. $F e_{3}\left[F e(C N)_{6}\right]_{3}$

Answer 1

Answer: (A)

The sodium extract obtained from aniline is $Na + C + N \rightarrow NaCN$ On reaction with iron sulphate: $F e^{2+}+C N^{-} \rightarrow\left[F e(C N)_{6}\right]^{4-}$ $4 F e^{3+}+3\left[F e(C N)_{6}\right]^{4-} \rightarrow \underset{\text{Prussian blue ppt.}}{F e_{4}\left[F e(C N)_{6}\right]_{3}}$