What is the ratio of moles of `MnO_4^(ɵ)` used per " mol of "`C_2O_4^(2-)` in acidic medium to strong basic medium?

Explanation

To solve the problem of finding the ratio of moles of \( \text{MnO}_4^{-} \) used per mole of \( \text{C}_2\text{O}_4^{2-} \) in acidic medium to strong basic medium, we can follow these steps:

### Step 1: Determine the reduction of \( \text{MnO}_4^{-} \) in acidic medium
In acidic medium, \( \text{MnO}_4^{-} \) is reduced to \( \text{Mn}^{2+} \). The half-reaction can be written as:
\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + 5 \text{e}^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
Here, the change in oxidation state is from +7 to +2, which means 5 electrons are required for the reduction.
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