1. We need to balance the redox equation:

$2\mathrm{IO_3^-} + x\mathrm{I^-} + 12\mathrm{H^+} \rightarrow 6\mathrm{I_2} + 6\mathrm{H_2O}$

We must find the value of $x$.

2. Check iodine atoms on the right side:

$6\mathrm{I_2}$ contains $6 \times 2 = 12$ iodine atoms.

3. Count iodine atoms on the left side:

• From $2\mathrm{IO_3^-}$, iodine atoms = $2$
• From $x\mathrm{I^-}$, iodine atoms = $x$

So total iodine atoms on the left =

$2 + x$

4. Equate iodine atoms on both sides:

$2 + x = 12$

$x = 10$

5. Quick charge check: Left side charge:

• $2\mathrm{IO_3^-} \Rightarrow -2$
• $10\mathrm{I^-} \Rightarrow -10$
• $12\mathrm{H^+} \Rightarrow +12$

Net charge:

$-2 -10 +12 = 0$

Right side is neutral: $6\mathrm{I_2} + 6\mathrm{H_2O}$ So the equation is charge balanced as well.

6. Therefore, the correct value is:

$\boxed{10}$

So the correct option is A.