Question

When calomel react with $N{H_4}OH$ solution the compound formed is:
(A) $N{H_2} - Hg - Cl$
(B) $H{g_2}C{l_2}N{H_3}$
(C) $Hg{(N{H_3})_2}C{l_2}$
(D) $HgC{l_2}N{H_3}$

Answer 1

To answer the above question we first need to understand what is the reaction that occurs in the calomel and it consists of what kind of compounds which will react with what to form different compounds.

Complete step by step answer:
We first start understanding what calomel is, it is a mineral which has a composition of mercury and chloride with having formula $H{g_2}C{l_2}$ , it has a very specific property of turning black when it is made to react with ammonia $N{H_3}$ .
Now calomel is used to measure the $pH$ of a substance as it acts as an electrode between the metallic mercury and chloride solution filled in a calomel electrode. It is believed to make a half cell made up of mercury which is coated with calomel.it also consists of a platinum wire, the platinum wire is used because platinum is not believed to react easily as it does not donates its electron normally and therefore it can be used for a very long time in the electrode. Also with the platinum wire the calomel also consists of potassium chloride $KCl$ which is the saturated solution, also it consists of ammonium hydroxide with chemical formula $N{H_4}OH$ the reaction in the calomel is given below
$H{g_2}C{l_2} + 2N{H_4}OH \to Hg + Hg(N{H_2})Cl + N{H_4}Cl + 2{H_2}O$
Calomel reacts with ammonium hydroxide to form mercury, ammonium chloride and a water molecule.
$Hg(N{H_2})Cl$ is the compound which I black in colour because calomel reacts with ammonia to produce black mass.
Option (A) is correct.

Note:
In the electrochemistry platinum electrode is also used as standard because it does not react with each other and therefore it can be used to calculate the potential of other compounds or elements.