Question

Balance the following oxidation–reduction reaction by the half-reaction method. (Use the lowest possible coefficients for the reaction.) Sn

Answer 1

The balanced oxidation-reduction reaction for Sn is:

`\[ \text{Sn}^(4+) + 2 \text{e}^- \rightarrow \text{Sn}^(2+) \]`.

Step-by-step explanation:

To balance the oxidation-reduction reaction for Sn, we first identify the initial and final oxidation states of tin
`(\( \text{Sn}^(4+) \) and \( \text{Sn}^(2+) \),` respectively). The half-reaction method involves balancing the oxidation and reduction half-reactions separately.

In the oxidation half-reaction, tin changes from
`\( \text{Sn}^(4+) \) to \( \text{Sn}^(2+) \)`. The electron transfer can be represented as:

`\[ \text{Sn}^(4+) \rightarrow \text{Sn}^(2+) + 2 \text{e}^- \]`

Now, in the reduction half-reaction, we balance the charge by adding electrons to the side with the higher oxidation state. Since the oxidation state of tin decreases
`\( \text{Sn}^(4+) \) to \( \text{Sn}^(2+) \)`, 2 electrons are added to the reduction half-reaction:

`\[ 2 \text{e}^- + \text{Sn}^(4+) \rightarrow \text{Sn}^(2+) \]`

To balance the number of electrons in both half-reactions, we multiply the oxidation half-reaction by 2:

`\[ 2(\text{Sn}^(4+) \rightarrow \text{Sn}^(2+) + 2 \text{e}^-) \]`

Now, we can combine the two half-reactions, ensuring that the number of electrons is the same in both, resulting in a balanced overall reaction:

`\[ \text{Sn}^(4+) + 2 \text{e}^- \rightarrow \text{Sn}^(2+) \]`.