Consider the following redox reaction taking place in acidic medium
$$BH_{4}^{-}(aq)+ClO_{3}^{-}(aq)\rightarrow H_{2}BO_{3}^{-}(aq)+Cl^{-}(aq)$$
If the Nerst equation for the above balanced reaction is $$E_{cell}=E_{cell}^{\circ}-\frac{RT}{nF}ln Q$$, then the value of n is______.(Nearest integer)

Find the value of $$n$$ in the Nernst equation for the balanced redox reaction:

$$BH_4^-(aq) + ClO_3^-(aq) \to H_2BO_3^-(aq) + Cl^-(aq)$$ (acidic medium)

We assign oxidation states. In $$BH_4^-$$, hydrogen is $$-1$$, so boron must be $$+3$$ since $$+3 + 4(-1) = -1$$. In $$H_2BO_3^-$$, hydrogen is $$+1$$ and oxygen is $$-2$$, giving $$2(+1) + B + 3(-2) = -1$$ and hence $$B = +3$$. Thus boron remains at +3, while each hydrogen atom is oxidized from $$-1$$ to $$+1$$, losing two electrons apiece.

In $$ClO_3^-$$, chlorine is $$+5$$, and in $$Cl^-$$ it is $$-1$$, indicating that each chlorine atom gains six electrons during reduction.

The oxidation half-reaction $$BH_4^- \to H_2BO_3^-$$ involves four hydrogen atoms each going from $$-1$$ to $$+1$$, resulting in a loss of 4 × 2 = 8 electrons per $$BH_4^-$$. The reduction half-reaction $$ClO_3^- \to Cl^-$$ involves a gain of 6 electrons per $$ClO_3^-$$. To equalize electron transfer, LCM of 8 and 6 = 24.

Accordingly, combining three $$BH_4^-$$ (lose 24 e) with four $$ClO_3^-$$ (gain 24 e) balances the overall reaction and yields $$n = 24$$.

The correct answer is $$\boxed{24}$$.