Question

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation.
${{H}_{2}}S+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}\to KHS{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+S+{{H}_{2}}O$

Answer 1

Hint: First, balance all the elements except hydrogen and oxygen atoms, so in this reaction balance sulfur, chromium, potassium before balancing hydrogen and oxygen. For finding the oxidation state of the elements, take the oxidation state of hydrogen and potassium as +1 and oxidation number oxygen as -2.

Complete answer:
The given reaction is:
${{H}_{2}}S+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}\to KHS{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+S+{{H}_{2}}O$
The oxidation number of H in ${{H}_{2}}S$ is +1 and the oxidation number of S is -2.
The oxidation number of K in ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is +1, the oxidation number of Cr is +6 and the oxidation number of O is -2.
The oxidation number of H in ${{H}_{2}}S{{O}_{4}}$ is +1 and the oxidation state of $SO_{4}^{2-}$ is -2.
The oxidation number of K and H in $KHS{{O}_{4}}$ is +1, and the oxidation state of $SO_{4}^{2-}$ is -2.
The oxidation number of Cr in $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$ is +3, and the oxidation state of $SO_{4}^{2-}$ is -2.
Since the S is in the elemental form its oxidation state will be 0.
The oxidation state of H in ${{H}_{2}}O$ is +1 and the oxidation state of O is -2.
So, in the reaction, the oxidation state of chromium decreases from +6 to +3, and the oxidation state of sulfur increases from -2 to 0. Therefore, ${{H}_{2}}S$ is the reducing agent and ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is the oxidizing agent.

First, balance all the elements except hydrogen and oxygen atoms, so in this reaction balance sulfur, chromium, potassium before balancing hydrogen and oxygen.
So, the balanced equation will be:
$3{{H}_{2}}S+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+5{{H}_{2}}S{{O}_{4}}\to 2KHS{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+3S+7{{H}_{2}}O$

Note:
If the equation is in an acidic medium then to balance the hydrogen atoms, we can add hydrogen ions (${{H}^{+}}$) and if the equation is in basic medium then to balance hydrogen and oxygen we can add hydroxyl ions ($O{{H}^{-}}$).