Find the n-factor the half equation for the oxidation of sulphite ion :SO2−3(aq)+H2O→SO2−4(aq)+2H+(aq)+2e−

To find the n-factor for the half-equation involving the oxidation of the sulfite ion ($\text{SO}_3^{2-}$) to sulfate ion ($\text{SO}_4^{2-}$), we can follow these steps:

1. ### Break Down the Problem

We need to identify the change in oxidation state of sulfur in the reaction:

$\text{SO}_3^{2-}(aq) + \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-}(aq) + 2\text{H}^+(aq) + 2e^-$

2. ### Relevant Concepts

• The oxidation state of sulfur in sulfite ($\text{SO}_3^{2-}$) is +4.
• The oxidation state of sulfur in sulfate ($\text{SO}_4^{2-}$) is +6.
• The n-factor is defined as the total number of electrons gained or lost per formula unit of the substance in the reaction.

3. ### Analysis and Detail

1. Determine the change in oxidation state of sulfur:

   • In $\text{SO}_3^{2-}$, S has an oxidation state of +4.
   • In $\text{SO}_4^{2-}$, S has an oxidation state of +6.
   • Change in oxidation state = +6 - (+4) = +2.

2. Determine the number of electrons involved:

   • The half-equation shows that there are 2 electrons ($2e^-$) lost during this oxidation process.

3. Determine the n-factor:

   • Since 2 electrons are transferred, the n-factor is equal to the number of electrons transferred per ion.

4. ### Verify and Summarize

The n-factor for the oxidation of the sulfite ion ($\text{SO}_3^{2-}$) to the sulfate ion ($\text{SO}_4^{2-}$) is confirmed as 2.

Final Answer

The n-factor for the half-equation is 2.