The balanced reaction of oxalate with permanganate in acidic medium is :

$$5 \, \text{C}_2\text{O}_4^{2-} + 2 \, \text{MnO}_4^- + 16 \, \text{H}^+ \rightarrow 10 \, \text{CO}_2 + 2 \, \text{Mn}^{2+} + 8 \, \text{H}_2\text{O}$$

In this redox reaction, the oxalate ion $\text{C}_2\text{O}_4^{2-}$ is oxidized to carbon dioxide $\text{CO}_2$, and the permanganate ion $\text{MnO}_4^-$ is reduced to manganese(II) ion $\text{Mn}^{2+}$.

The oxidation half-reaction, representing the change for the oxalate ion, is :

$$\text{C}_2\text{O}_4^{2-} \rightarrow 2 \, \text{CO}_2 + 2 \, e^-$$

From this half-reaction, it's clear that each oxalate ion produces two CO₂ molecules and releases two electrons in the process.

So, the number of electrons involved in producing one molecule of CO₂ is :

$$\frac{2 \, e^-}{2 \, \text{CO}_2} = 1 \, e^-/\text{CO}_2$$

Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO₂.