Read the passage carefully and answer the questions.
Sodium dichromate \( Na_2Cr_2O_7 \) and Potassium dichromate \( K_2Cr_2O_7 \), are strong oxidizing agents; in acidic solution, its oxidising action can be represented as follows:
\( Cr_2{O_7}^{2-}+ 14H^++ 6e^-→2Cr^{3+}+ 7H_2O \) \( (E° = 1.33V) \)
The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr-O-Cr bond angle of 126°
Potassium permanganate is prepared by fusion of \( MnO_2 \) with an alkali metal hydroxide and an oxidising agent like \( KNO_3 \). This produces the dark green \( K_2MnO_4 \) which disproportionates in a neutral or acidic solution to give permanganate. It is a strong oxidizing agent. Hydrogen ion concentration of the solution plays an important part in the reduction of permanganate to manganate, manganese dioxide and manganese(II) salt.
Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of \( KClO_4 \). The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K.
The number of moles of \( KMnO_4 \) required to react with one mole of sulphide ion in acidic solution is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2/5
The mole ratio in a redox reaction comes from the electron balance: the electrons given out by the reducing agent must equal the electrons taken up by the oxidising agent. In acidic solution permanganate is reduced from Mn(+7) to \( \ce{Mn^2+} \), taking up 5 electrons, while the sulphide ion is oxidised from S(−2) to sulphur, giving out 2 electrons:
\( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)
\( \displaystyle \ce{S^2- -> S + 2e^-} \)
To make the electrons equal (10 each), the first half-reaction is multiplied by 2 and the second by 5. Adding them gives the balanced ionic equation:
\( \displaystyle \ce{2MnO4^- + 16H^+ + 5S^2- -> 2Mn^2+ + 8H2O + 5S v} \)
So 5 mol of sulphide ion react with 2 mol of \( \ce{KMnO4} \), and 1 mol of sulphide ion needs:
\( \displaystyle n(\ce{KMnO4}) = \frac{2}{5}\ \mathrm{mol} \)
The answer is 2/5 mol.
Shortcut: Equate the electrons directly. Each \( \ce{KMnO4} \) takes 5 electrons and each sulphide ion gives 2, so \( 5x = 1 \times 2 \) and \( x = 2/5 \).