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

Question

In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. The magnitude of X is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

3e- + MnO4- + 2H2O → MnO2 + 4OH-] × 8 S2O32- + 10OH- → 2SO42- + 5H2O + 8e-]× 3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/62cca1897c146660f898561a6e96031e-.png" /> 8 moles of MnO 4-ion gives 6 moles of SO 42- ion

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

$3 e^{-} + M n O_{4}^{-} + 2 H_{2} O \to M n O_{2} + 4 O H^{-}] \times 8$
$S_{2} O_{3}^{2 -} + 10 O H^{-} \to 2 S O_{4}^{2 -} + 5 H_{2} O + 8 e^{-}] \times 3$

8 moles of $M n O_{4}^{-}$ ion gives 6 moles of $S O_{4}^{2 -}$ ion

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

3e−+MnO4−​+2H2​O→MnO2​+4OH−]×8 S2​O32−​+10OH−→2SO42−​+5H2​O+8e−]×3 8 moles of MnO4−​ion gives 6 moles of SO42−​ ion

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

$3 e^{-} + M n O_{4}^{-} + 2 H_{2} O \to M n O_{2} + 4 O H^{-}] \times 8$
$S_{2} O_{3}^{2 -} + 10 O H^{-} \to 2 S O_{4}^{2 -} + 5 H_{2} O + 8 e^{-}] \times 3$

8 moles of $M n O_{4}^{-}$ ion gives 6 moles of $S O_{4}^{2 -}$ ion