Question

For the given aqueous reactions, which of the statement(s) is (are) true?

excess $KI+K_3[Fe(CN{)}_6]\stackrel{\text{dilute}\mathrm{H}{}_2\mathrm{SO}{}_4}{\to }$ brownish - yellow solution $\stackrel{\mathrm{ZnSO}{}_4}{\to }$ white precipitate $+\underbrace{\text{brownish - yellow filtrate}}\stackrel{\mathrm{Na}{}_2\mathrm{S}{}_2\mathrm{O}{}_3}{\to }$ colourless solution

• A. The first reaction is a redox reaction.
• B. White precipitate is $Z{n}_3{\left[Fe(CN{)}_6\right]}_2$
• C. Addition of filtrate to starch solution gives blue colour.
• D. White precipitate is soluble in $NaOH$ solution.

Answer 1

Answer: (A), (C), (D)

$K_3[\stackrel{+3}{\mathrm{Fe}}(CN{)}_6]+KI$(excess)$\to K_4[\stackrel{+2}{\mathrm{Fe}}(CN{)}_6]+\underset{\text{Brownish yellow solution}}{K{I}_3}$(redox reaction )

$\underset{\text{(Brownish yellow filtrate)}}{I_3^{-}}+2N{a}_2{S}_2{O}_3\to \underset{\text{Clear solution}}{N{a}_2{S}_4{O}_6}++2NaI+I^{-}$

$K_4[Fe(CN{)}_6]+ZnS{O}_4\to \underset{\text{(white ppt.}}{K_{3}Z{n}_3[Fe(CN{)}_6{]}_3}\stackrel{\mathrm{NaOH}}{\to }\underset{\text{soluble}}{N{a}_2[Zn(OH{)}_4]}$