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.
$I_2+2e^{-}\to 2I^{-}{E}^{\circ }=0.54$
$C{l}_2+2e^{-}\to 2C{l}^{-}{E}^{\circ }=1.36$
$M{n}^{3+}+e^{-}\to M{n}^{2+}{E}^{\circ }=1.50$
$F{e}^{3+}+e^{-}\to F{e}^{2+}{E}^{\circ }=0.77$
$O_2+4H^{+}+4e^{-}\to 2H_{2}O{E}^{\circ }=1.23$
Sodium fusion extract, obtained from aniline, on treatment with iron(II) sulphate and $H_{2}S{O}_4$ in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of

• A. $F{e}_4{\left[Fe(CN{)}_6\right]}_3$
• B. $F{e}_3{\left[Fe(CN{)}_6\right]}_2$
• C. $F{e}_4{\left[Fe(CN{)}_6\right]}_2$
• D. $F{e}_3{\left[Fe(CN{)}_6\right]}_3$

Answer 1

Answer: (A)

The sodium extract obtained from aniline is $Na+C+N\to NaCN$ On reaction with iron sulphate: $F{e}^{2+}+C{N}^{-}\to {\left[Fe(CN{)}_6\right]}^{4-}$ $4F{e}^{3+}+3{\left[Fe(CN{)}_6\right]}^{4-}\to \underset{\text{Prussian blue ppt.}}{F{e}_4{\left[Fe(CN{)}_6\right]}_3}$