An acid solution of 0.2 " mol of "`KReO_(4)` was reduced with Zn and then titrated with 1.6 " Eq of "acidic `KMnO_(4)` solution for the reoxidation of the ehenium `(Re)` to the perrhenate ion `(ReO_(4)^(ɵ))`. Assuming that rhenium was the only elements reduced, what is the oxidation state to which rhenium was reduced by Zn?

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

C

Explanation

Eq of `K Re O_(4) -=` Eq of `MnO_(4)^(-)`
`0.2 xx n = 1.6` Eq, `n = (1.6)/(0.2) = 8`
Hence, ther is an `8e^(-)` reduction.
`8e^(-) +overset(+7)(ReO_(4)^(-)) rarr underset(x =- 1)(Re^(-1))`
oxidation state of `Re =- 1`