Question

Which of the following is not formed when ${{H}_{2}}S$ reacts with acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ solution.
(A) $CrS{{O}_{4}}$
(B) $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$
(C) $S$
(D) ${{H}_{2}}O$

Answer 1

Hint: In the reaction between the hydrogen sulphide and acidic potassium dichromate, an oxidation-reduction reaction takes place. With the ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ as the oxidising agent and the ${{H}_{2}}S$ as the reducing agent.

Complete step by step solution:
In the reaction of potassium dichromate with hydrogen sulphide in presence of an acidic medium of sulphuric acid, a redox reaction takes place as follows:
During the reaction, the potassium chromate acts as the oxidizing agent, due to the presence of the hexavalent chromium in $(+6)$ oxidation state. It is thus reduced to $(+3)$oxidation state.
Then, the reduction half-cell reaction will be:
\[C{{r}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\]
Also, the sulphide ${{S}^{2-}}$ acting as the reducing agent, is oxidised to sulfur (S) in the reaction. Then the oxidation half-cell reaction will be:
${{S}^{2-}}\to S+2{{e}^{-}}$
Thus, combining the oxidation and reduction half-cell reaction, the balanced equation obtained is:
\[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+3{{H}_{2}}S+4{{H}_{2}}S{{O}_{4}}\to 3S+{{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+7{{H}_{2}}O\]
The products formed are sulfur, potassium sulphate, chromium sulphate and water.

Therefore, it is seen that the option (A)- $CrS{{O}_{4}}$ is not formed during the reaction of ${{H}_{2}}S$ with acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ solution.

Additional information:
During this reaction, a colour change in the solution is observed, from reddish orange to green. And, the sulphide gas is precipitated to sulfur which is pale yellow in colour.
Also, the potassium dichromate solution is used in volumetric analysis to determine the unknown concentration of secondary solution, thus acting as the primary standard.

Note: In the dichromate, the chromium being present in a higher oxidation state, which increases its electronegativity. Thus, it gains electrons during the reaction and gets reduced to lower oxidation state. Also, the sulphuric acid is only present to create an acidic medium for the reaction to proceed.