It requires `40.0mL` of `0.50M Ce^(4+)` to titrate `10.0 mL` of `1.0MSn^(2+) "to" Sn^(4+)`. What is the oxidation state of cerium in the reduced product?

Explanation

To determine the oxidation state of cerium in the reduced product, we can follow these steps:

### Step 1: Write the half-reaction for the oxidation of Sn²⁺ to Sn⁴⁺.
The oxidation half-reaction can be written as:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]
This indicates that each Sn²⁺ ion loses 2 electrons to become Sn⁴⁺.

### Step 2: Write the half-reaction for the reduction of Ce⁴⁺.
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