I haven't learned half reactions for oxidation and reduction so please teach me by leading me from step to step....
I don't get this..
Sn2+ +2Fe3+---> Sn4+ +2Fe2+
The answer key says
Oxidation: Sn2+--->Sn4+ +2e-
Reduction: 2Fe3+ + 2e----->2Fe2+
**where did the 2 electrons come from???
You write down the half equation.
Sn^2+ ==> Sn^4+.
The charge on the left is 2+. On the right it is 4+ . You add 2e (each with a - charge to the right so the charge balances.
That is one of the simple ones and the Fe is the same way. Others may be a little more complicated but you follow the rules. Here is a site that gives the rules along with a lot of other stuff about redox. Read this and you will be able to do all of them.
http://www.chemteam.info/Redox/Redox.html
In order to understand the half-reactions for oxidation and reduction, we need to determine the changes in oxidation states for each element in the reaction. Let's break it down step-by-step:
Step 1: Determine the initial and final oxidation states for each element.
- In the reactant Sn2+, Sn has an oxidation state of +2.
- In the reactant Fe3+, Fe has an oxidation state of +3.
- In the product Sn4+, Sn has an oxidation state of +4.
- In the product Fe2+, Fe has an oxidation state of +2.
Step 2: Compare the initial and final oxidation states for each element.
- Sn increased its oxidation state from +2 to +4, which means it was oxidized.
- Fe decreased its oxidation state from +3 to +2, which means it was reduced.
Step 3: Create the half-reactions for oxidation and reduction.
- Oxidation half-reaction: Sn2+ -> Sn4+ + 2e- (Sn is oxidized and gains 2 electrons)
- Reduction half-reaction: 2Fe3+ + 2e- -> 2Fe2+ (Fe is reduced by gaining 2 electrons)
Step 4: Balancing the half-reactions.
- The oxidation half-reaction is already balanced.
- The reduction half-reaction is already balanced.
The 2 electrons in the reduction half-reaction came from the Sn2+ ion in the oxidation half-reaction. This is because the total number of electrons gained in the reduction half-reaction must be equal to the total number of electrons lost in the oxidation half-reaction in order to maintain charge neutrality in the overall reaction.
Therefore, in the reaction Sn2+ + 2Fe3+ -> Sn4+ + 2Fe2+, Sn2+ gets oxidized by losing 2 electrons, while Fe3+ gets reduced by gaining 2 electrons.
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.