Balance the following equations
a. `Fe^(2+) + Sn^(+2) rarr Sn^(4+) + fe^(2+)`

Explanation

To balance the redox reaction given in the question, we will follow these steps:

### Step 1: Identify the oxidation and reduction half-reactions
The reaction given is:
\[ \text{Fe}^{2+} + \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + \text{Fe}^{2+} \]

From the equation, we can see that:
- **Tin (Sn)** is being oxidized from \( \text{Sn}^{2+} \) to \( \text{Sn}^{4+} \).
- **Iron (Fe)** is being reduced from \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \).

### Step 2: Write the half-reactions
1. **Oxidation half-reaction** (for Sn):
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]
(Here, Sn loses 2 electrons)

2. **Reduction half-reaction** (for Fe):
\[ \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \]
(Here, Fe gains 1 electron)

### Step 3: Equalize the number of electrons transferred
To balance the electrons, we need to make sure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

- The oxidation half-reaction involves 2 electrons.
- The reduction half-reaction involves 1 electron.

To equalize, we can multiply the reduction half-reaction by 2:
\[ 2\left(\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}\right) \]
This gives us:
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \]

### Step 4: Combine the half-reactions
Now we can combine the two half-reactions:
1. Oxidation:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]
2. Reduction:
\[ 2\text{Fe}^{3+} + 2e^- \rightarrow 2\text{Fe}^{2+} \]

Adding these together:
\[ \text{Sn}^{2+} + 2\text{Fe}^{3+} \rightarrow \text{Sn}^{4+} + 2\text{Fe}^{2+} \]

### Step 5: Final balanced equation
The final balanced equation is:
\[ \text{Sn}^{2+} + 2\text{Fe}^{3+} \rightarrow \text{Sn}^{4+} + 2\text{Fe}^{2+} \]