Question

In neutral or faintly alkaline solution, $$8$$ moles of permanganate anion quantitative oxidize thiosulphate anions to produce X moles of a sulphur containing product. The magnitude of X is :

Answer 1

Hint: When we add potassium permanganate solution with thiosulphate then redox reaction takes place. This redox reaction takes place in the basic medium. We will first write the redox reaction then we will find the oxidation and reduction reaction separately. We will balance them and get a final balanced oxidation reaction. With the help of a balanced redox reaction we can find the required answer.

Complete answer:
When potassium permanganate is reacted with thiosulphate solution then the reaction can be represented as:
$$Mn{O_4}^{-1}\text{ + }{\text{S}}_2{O_3}^{2-}\to \text{ Mn}{\text{O}}_2\text{ + S}{{\text{O}}_4}^{2-}$$
The above reaction is redox reaction which can be balanced separately as:
For oxidation:
$${\text{S}}_2{O_3}^{2-}\to \text{ S}{{\text{O}}_4}^{2-}$$
Since the oxidation state of sulphur increases from $$+4$$ to $$+6$$, therefore it is an oxidation reaction. Now we will first balance the atomicity of atoms except oxygen which can be shown as:
$${\text{S}}_2{O_3}^{2-}\to \text{ 2S}{{\text{O}}_4}^{2-}$$
Now for balancing oxygen atoms we have to add five moles of water at right hand side of reaction as:
$${\text{S}}_2{O_3}^{2-}\to \text{ 2S}{{\text{O}}_4}^{2-}\text{ + 5}{\text{H}}_2\text{O}$$
Now adding $$10\left[O{H}^{-}\right]$$ on left side of reaction as:
$$\text{10 O}{\text{H}}^{-}\text{ + }{\text{S}}_2{O_3}^{2-}\to \text{ 2S}{{\text{O}}_4}^{2-}\text{ + 5}{\text{H}}_2\text{O}$$
Now we will balance the charge by adding eight electrons at the right hand side.
$$\text{10 O}{\text{H}}^{-}\text{ + }{\text{S}}_2{O_3}^{2-}\to \text{ 2S}{{\text{O}}_4}^{2-}\text{ + 5}{\text{H}}_2\text{O + 8}{\text{e}}^{-}$$ ______________$$(i)$$
For reduction:
$$Mn{O_4}^{-1}\to \text{ Mn}{\text{O}}_2$$
The above reaction is a reduction reaction because the oxidation state of the manganese atom changes from $$+7$$ to $$+4$$. Since the atomicity of atom other than oxygen atom is equal, therefore we will add two moles of water at left hand side of reaction as:
$$\text{ 2}{\text{H}}_2\text{O + }Mn{O_4}^{-1}\to \text{ Mn}{\text{O}}_2$$
Now we will add $$4\left[O{H}^{-}\right]$$ at right hand side of reaction as:
$$\text{ 2}{\text{H}}_2\text{O + }Mn{O_4}^{-1}\to \text{ Mn}{\text{O}}_2\text{ + 4 O}{\text{H}}^{-}$$
Now we will balance the charge by adding five electrons at right hand side of reaction as:
$$\text{3}{\text{e}}^{-}\text{ + 2}{\text{H}}_2\text{O + }Mn{O_4}^{-1}\to \text{ Mn}{\text{O}}_2\text{ + 4 O}{\text{H}}^{-}$$ _____________$$(ii)$$
Now multiplying equation $$(ii)$$ by eight and equation $$(i)$$ by three and then on adding, we get the final result as:
$$\text{16}{\text{H}}_2\text{O + 30 O}{\text{H}}^{-}\text{ + 8}Mn{O_4}^{-1}\text{ + 3}{\text{S}}_2{O_3}^{2-}\to \text{ 8Mn}{\text{O}}_2\text{ + 6S}{{\text{O}}_4}^{2-}\text{ + 32 O}{\text{H}}^{-}\text{ + 15}{\text{H}}_2\text{O}$$
It can be reduced as:
$${\text{H}}_2\text{O + 8}Mn{O_4}^{-1}\text{ + 3}{\text{S}}_2{O_3}^{2-}\to \text{ 8Mn}{\text{O}}_2\text{ + 6S}{{\text{O}}_4}^{2-}\text{ + 2O}{\text{H}}^{-}$$
Hence we observe that when eight moles of permanganate ion are reacted with three moles of disulphate then six moles of sulphate ion are formed.

Note:
The above redox reaction must be balanced properly otherwise we will get wrong moles of ions. Oxidation reactions are those in which the oxidation state of the central metal atom increases while in reduction the oxidation state of the central metal atom gets reduced. For the basic solution we add hydroxide ions while in acidic medium we add hydronium ions for balancing the charge.