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The charge required for the reduction of...

The charge required for the reduction of 1 mol of `Cr_(2)O_(7)^(2-)` to `Cr^(3+)` is :

A

`96500C`

B

`2xx96500C`

C

`3xx96500C`

D

`6xx96500C`

Text Solution

Verified by Experts

The correct Answer is:
d
`6e^(c-)+Cr_(2)O_(7)^(2-) rarr 2Cr^(3+)`
So charge `=6xxF=6xx96500C`
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