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)
$ \begin{gathered} \downarrow Na _2 S _2 O _3 \\ \text { Colourless solution } \end{gathered} $
(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) $\longrightarrow$
$$ K _4\left[\stackrel{+2}{Fe}(CN) _6\right]+\underset{\substack{\text { Brownish yellow } \\ \text { solution }}}{KI _3(\text { redox })} $$
$$ \begin{aligned} & K _4\left[Fe(CN) _6\right]+ZnSO _4 \rightarrow K _2 Zn _3\left[Fe(CN) _6\right] _2 \\ & \text { or } K _2 Zn\left[Fe(CN) _6\right] \\ & \text { White ppt } \end{aligned} $$
$K _2 Zn\left[Fe(CN) _6\right]$ reacts with $NaOH$ as
$K _2 Zn\left[Fe(CN) _6\right]+NaOH \longrightarrow\left[Zn(OH) _4\right]^{2-}+\left[Fe(CN) _6\right]^{4-}$