To understand the concept of half reactions for oxidation and reduction, let's break down the given chemical equation step by step.
Step 1: Identify the elements and their oxidation states on both sides of the equation.
In this equation, we have Sn (tin) and Fe (iron) atoms. On the left side, Sn has an oxidation state of +2, and Fe has an oxidation state of +3. On the right side, Sn has an oxidation state of +4, and Fe has an oxidation state of +2.
Step 2: Determine which element is undergoing oxidation and which is undergoing reduction.
In this equation, Sn is being oxidized because its oxidation state increases from +2 to +4. Fe, on the other hand, is being reduced because its oxidation state decreases from +3 to +2.
Step 3: Write the half-reactions for oxidation and reduction.
For the oxidation half-reaction, we start with Sn2+ on the left side and Sn4+ on the right side. To balance the equation, we need to add two electrons (2e-) to the left side to balance the charge. Therefore, the oxidation half-reaction is:
Sn2+ ---> Sn4+ + 2e-
For the reduction half-reaction, we start with Fe3+ on the left side and Fe2+ on the right side. Since the oxidation state of Fe decreases from +3 to +2, we can determine that two electrons are gained during this reduction process. Therefore, the reduction half-reaction is:
2Fe3+ + 2e- ---> 2Fe2+
In summary:
Oxidation half-reaction: Sn2+ ---> Sn4+ + 2e-
Reduction half-reaction: 2Fe3+ + 2e- ---> 2Fe2+
The two electrons in both half-reactions are required to balance the charges and conserve the number of electrons between the reactants and the products.