Qualitative Analysis - Result Question 27
27. For the given aqueous reaction which of the statement(s) is (are) true?
Excess $KI+K _3\left[Fe(CN) _6\right] \xrightarrow{\text { Dilute } H _2 SO _4}$
Brownish-yellow solution
$\downarrow ZnSO _4$
(White precipitate + Brownish-yellow filtrate)
$\downarrow Na _2 S _2 O _3 $
Colourless solution
(2012)
(a) The first reaction is a redox reaction
(b) White precipitate is $Zn _3\left[Fe(CN) _6\right] _2$
(c) Addition of filtrate to starch solution gives blue colour
(d) White precipitate is soluble in $NaOH$ solution
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Answer:
Correct Answer: 27. (a,c,d)
Solution:
$K _3\left[\stackrel{+3}{Fe}(CN) _6\right]+KI$ (excess) $\longrightarrow K _4\left[\stackrel{+2}{Fe}(CN) _6\right]+\underset{\substack{\text { Brownish yellow } \\ \text { solution }}}{KI _3(\text { redox })}$
$\begin{aligned} & K _4\left[Fe(CN) _6\right]+ZnSO _4 \rightarrow K _2 Zn _3\left[Fe(CN) _6\right] _2 \\ & \text { or } K _2 Zn\left[Fe(CN) _6\right] \\ & \text { White ppt } \end{aligned}$
$ \underset{\begin{array}{c} \text { Brownish } \\ \text { yellow } \\ \text { filtrate } \end{array}}{\mathrm{I}_3^{-}}+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \longrightarrow \underset{\substack{\text { Clear } \\ \text { solution }}}{\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6}+2 \mathrm{NaI}+\underset{\begin{array}{c} \text { (Turns starch } \\ \text { solution } \\ \text { blue) } \end{array}}{\mathrm{I}_2} $
$K _2 Zn\left[Fe(CN) _6\right]$ reacts with $NaOH$ as
$K _2 Zn\left[Fe(CN) _6\right]+ NaOH \longrightarrow\left[Zn(OH) _4\right]^{2-}+\left[Fe(CN) _6\right]^{4-}$
BETA
