An acid solution of KReO 4 sample containing 26.0 mg of combined rhenium was reduced by passage through a column of granulated zinc. The effluent solution, including the washings from the column, wa

Question

An acid solution of KReO 4 ​ sample containing 26.0 mg of combined rhenium was reduced by passage through a column of granulated zinc. The effluent solution, including the washings from the column, was then titrated with 0.1 NKMnO 4 ​ ; 11.0 mL of the standard permanganate was required for the reoxidation of all the rhenium to perrhenate ion, ReO 4 − ​ . Assuming that rhenium was the only element reduced, what is the number of electrons transferred per Re? [Assume atomic weight of Re to be 190]

TestHub · Accepted Answer

Equivalents of KMnO₄ = 0.1 N × 1000 m L / L 11.0 m L ​ = 0.0011 e q u i v a l e n t s Mass of Re = 26.0 m g = 0.026 g Atomic weight of Re = 190 g / m o l Moles of Re = 190 g / m o l 0.026 g ​ = 0.0001368 m o l Let 'n' be the number of electrons transferred per Re atom during the reoxidation by KMnO₄. This 'n' represents the change in oxidation state of Re from its reduced form back to +7. Equivalents of Re = Moles of Re × n Equivalents of Re = 0.0001368 m o l × n According to the principle of equivalence: Equivalents of Re = Equivalents of KMnO₄ 0.0001368 × n = 0.0011 n = 0.0001368 0.0011 ​ ≈ 8.04 The number of electrons transferred per Re is approximately 8. This means that the rhenium was reduced from an oxidation state of +7 to -1 by the zinc, and then reoxidized back to +7 by the permanganate. The number of electrons transferred per Re is 8.

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JEE Advanced (2028)