Question

What product is formed by mixing the solution of ${K_4}[Fe{(CN)_6}]$ with the solution of $FeC{l_3}$ :
(A) $Ferro - ferricyanide$
(B) $Ferric - ferrocyanide$
(C) $Ferri - ferricyanide$
(D) $Potassium \ ferricyanide$

Answer 1

Hint: In this question there are two reactants are given as per which we've to find the product by using both originally we've to know that how to mix any 2 reactants also make a product and balance the both side of equation. As well as knowing the special names of numerous rudiments they're veritably useful in numerous matters of work.

Complete Step by Step Solution:
As per when we see both reactants the oxidation state of $Fe$ by ${K_4}[Fe{(CN)_6}]$ is $ + 2$ whereas the oxidation state of $Fe$ by $FeC{l_3}$ is $ + 3$ .

As from which we can balance the response from the following oxidation state grounded,
As by writing complete response we get that,$3{K_4}[Fe{(CN)_6}] + 4FeC{l_3}\xrightarrow{{}}F{e_4}{[Fe{(CN)_6}]_3} + 12KCl$

As we get data from the above equation, the product formed from the response of the reactants will be $F{e_4}{[Fe{(CN)_6}]_3}$ .
And we know that the emulsion formed is named as by the name of Prussian blue coloured emulsion.

As well as it is also said by the name of $Ferric - Ferrocyanide$ .
As a result we get the answer as $Ferric - Ferrocyanide$ .
Hence, the correct option is (B).

Note: When we complete any response there are the main factors which we've to follow originally we've to check that all the nature of rudiments are inversely put to the right position. Always we've to know that when a response takes place with Potassium Hexacyanoferrate (II) results, Iron ions always give a violent blue precipitate of Hexacyanoferrate (Prussian blue).