Question

Balance the following equation by oxidation number method.
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+HCl\to KCl+CrC{{l}_{3}}+{{H}_{2}}O+C{{l}_{2}}$

Answer 1

Hint As we know that oxidation number method is basically used to balance the chemical reaction based on their oxidation state. It is basically found that the idea behind this method is that electrons are transferred between the charged atoms or molecules.

Complete Step by step solution:
- We will firstly write the oxidation numbers of all of all the atoms.
${{K}_{2}}^{+1}C{{r}_{2}}^{+6}{{O}_{7}}^{-2}+{{H}^{+1}}C{{l}^{-1}}\to {{K}^{+1}}C{{l}^{-1}}+C{{r}^{+3}}C{{l}^{-1}}_{3}+{{H}^{+1}}_{2}{{O}^{-2}}+C{{l}^{0}}_{2}$
- We can see here that the oxidation number of Cr Is decreased and of chlorine is increased.

& {{K}_{2}}C{{r}_{2}}^{+6}{{O}_{7}}\to 2C{{r}^{+3}}C{{l}_{3}}--(1) \\\ & H\overset{-1}{\mathop{Cl}}\,\to \overset{0}{\mathop{Cl}}\,---(2) \\\ \end{aligned}$$ \- We can say that the decrease in the oxidation number of Cr is equal to 6 units per molecule ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ The decrease in the oxidation number of Cl is equal to 1 unit per molecule HCl. \- By multiplying equation 2 with 6 we get: $${{K}_{2}}C{{r}_{2}}{{O}_{7}}+6HCl\to 2CrC{{l}_{3}}+3C{{l}_{2}}$$ \- As we can see here to balance potassium and chlorine, there are 14 molecules of HCl required. $${{K}_{2}}C{{r}_{2}}{{O}_{7}}+14HCl\to 2CrC{{l}_{3}}+3C{{l}_{2}}+2KCl$$ \- And in order to balance oxygen and hydrogen, we will add $7{{H}_{2}}O$ to RHS. $${{K}_{2}}C{{r}_{2}}{{O}_{7}}+14HCl\to 2CrC{{l}_{3}}+3C{{l}_{2}}+2KCl+7{{H}_{2}}O$$ In this way we can balance the equation by oxidation number method. **Note:** \- We must make sure that the sum of the signs on the right-hand side is equal to the sum of the signs on the left-hand side. If both are equal, then we can say that the reaction is being balanced. \- We must also understand that if the reaction takes place in an acidic or basic medium. As the ions that are to be added will differ on the product side, that is it is basic then we will see the hydroxide ions on the reactant side. Whereas, if it is acidic then we will see hydrogen ions on the reactant side.