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$K_4[Fe(CN)_6]$ can be used to detect one or more out of $Fe^{2+}, Fe^{3+}, Zn^{2+}, Cu^{2+}, Cd^{2+}$

(a) $Fe^{2+}, Fe^{3+}$
(b) $Fe^{3+}, Zn^{2+}, Cu^{2+}$
(c) All but $Fe^{3+}$
(d) All but $Fe^{2+}$

1 Answer

Answer: All but $Fe^{2+}$
$2ZnCl_2 + K_4[Fe(CN)_6] \xrightarrow{CH_3COOH} Zn_2[Fe(CN)_6]_3 \downarrow (\text{Bluish white}) + 4KCl$
$2FeCl_3 + 2K_4[Fe(CN)_6] \xrightarrow{CH_3COOH} Fe_2[Fe(CN)_6]_3 \downarrow (\text{Prussian blue}) + 12KCl$
$2Cu(NO_3)_2 + K_4[Fe(CN)_6] \xrightarrow{CH_3COOH} Cu_2[Fe(CN)_6] \downarrow (\text{Reddish brown})+ 4KNO_3$
$2Cd(NO_3)_2 + K_4[Fe(CN)_6] \xrightarrow{CH_3COOH} Cd_2[Fe(CN)_6] \downarrow (\text{Bluish white}) + 4KNO_3$
Thus except $Fe^{2+}$, all other ions gives $Fe^{3+}, Zn^{2+}, Cu^{2+}$ and $Cd^{2+}$ can be detected using $K_4[Fe(CN)_6]$


answered Jan 28, 2014 by mosymeow_1
edited Jan 28, 2014 by mosymeow_1