Animated Solution for Chemistry - Redox Reactions: The reaction of sulphur in alkaline medium is given below $S_8(s)+aO{H}^{-}(aq)⟶b{S}^{2-}(aq)+c{S}_2{O}_3^{2-}(aq)+d{H}_{2}O(l)$ The value of 'a' is ....... . (Integer answer)

Imagine you are looking at a piece of elemental sulphur, $S_8$. In an alkaline medium, it doesn't just sit there; it undergoes a fascinating transformation called a disproportionation reaction. This means that the very same element is simultaneously acting as both an oxidizing and a reducing agent.

In our reaction, $S_8$ (where sulphur has an oxidation state of $0$) splits its identity. Some of it gets reduced to form sulphide ions, $S^{2-}$ (oxidation state $-2$), while the rest gets oxidized to form thiosulphate ions, $S_2{O}_3^{2-}$ (average oxidation state $+2$). Our goal is to balance this complex dance of electrons and find the exact stoichiometric coefficient for the hydroxide ion, denoted as '$a$'.

Half-Reaction Method

To tackle this, we will use the highly reliable ion-electron method. We start by splitting the main reaction into two distinct half-reactions: the reduction half and the oxidation half.

1. The Reduction Half: We write the skeletal equation: First, we balance the sulphur atoms by placing an $8$ in front of the sulphide ion: Now, we have a total charge of $-16$ on the right side. To balance this, we must add $16$ electrons to the left side:

2. The Oxidation Half: We write the skeletal equation for oxidation: Balance the sulphur atoms by placing a $4$ in front of the thiosulphate ion: Now, we have $12$ oxygen atoms on the right. To balance oxygen, we add $12$ water molecules to the left: This introduces $24$ hydrogen atoms on the left, so we add $24H^{+}$ ions to the right: Finally, we balance the charge. The right side has a net charge of $4(-2)+24(+1)=+16$. We add $16$ electrons to the right to neutralize it:

Notice how beautifully the universe aligns! Both half-reactions involve exactly $16$ electrons. We can simply add them together, and the electrons will cancel out perfectly:

But wait, there is a catch here. The problem explicitly states that the reaction occurs in an alkaline (basic) medium, but our equation currently has $H^{+}$ ions. To fix this, we add $O{H}^{-}$ ions to both sides equal to the number of $H^{+}$ ions. We add $24O{H}^{-}$ to both sides:

The $24H^{+}$ and $24O{H}^{-}$ on the right combine to form $24H_{2}O$. Now, we cancel $12H_{2}O$ from both sides to simplify:

In chemistry, we always prefer the simplest whole-number ratio for stoichiometric coefficients. We can divide the entire equation by $2$:

Now, let's compare our beautifully balanced equation with the one given in the question:

By direct comparison, we can see that the coefficient for the hydroxide ion, $a$, is exactly $12$. This method of balancing in acidic first and then converting to basic is a foolproof strategy that will save you from countless silly mistakes in the exam!