1. Write the oxidation half-reaction for oxalate in acidic medium

Oxalate ion is $\ce{C2O4^{2-}}$. It gets oxidized to carbon dioxide:

$\ce{C2O4^{2-} -> 2CO2 + 2e^-}$

2. Interpret the half-reaction

From the equation above:

• $1$ oxalate ion produces $2$ molecules of $\ce{CO2}$
• and in doing so, it loses $2$ electrons.

So, electrons involved per molecule of $\ce{CO2}$ are:

$\frac{2\ e^-}{2\ \text{molecules of } CO_2} = 1\ e^-$

3. Check via oxidation number

In $\ce{C2O4^{2-}}$, let oxidation state of each carbon be $x$:

$2x + 4(-2) = -2$ $2x - 8 = -2$ $2x = 6 \Rightarrow x = +3$

In $\ce{CO2}$, carbon is $+4$.

Thus each carbon goes from $+3$ to $+4$, meaning loss of:

$1\ e^- \text{ per carbon atom}$

Since one molecule of $\ce{CO2}$ contains one carbon atom, producing one molecule of $\ce{CO2}$ corresponds to 1 electron.

4. Evaluate options

• A: $10$ $\rightarrow$ incorrect
• B: $2$ $\rightarrow$ incorrect
• C: $1$ $\rightarrow$ correct
• D: $5$ $\rightarrow$ incorrect

Therefore, the correct answer is C.