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An acid solution of 0.2 " mol of "KReO(4...

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?

A

1

B

2

C

`-1`

D

`-2`

Text Solution

Verified by Experts

The correct Answer is:
C
" Eq of "`KReO_(4)-=" Eq of "MnO_(4)^(ɵ)`
`0.2xxn=1.6Eq`
`n=(1.6)/(0.2)=8`
Hence, there is an `8e^(-)` reduction.
`8e^(-)+underset(x=7)Reoverset(+7)(O_(4)^(ɵ))tounderset(x=-1)(Re^(-1))`
Oxidation state of `Re=-1`
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