the number fo electrons required to ...

the number fo electrons required to reduce chromium completely in `Cr_(2) O_(7) ^(2-) ` to `Cr_(3+)` in acidic meddium , is

A

5

B

3

C

6

D

2

Verified by Experts

The correct Answer is:

C

`Cr_(2) O_(7) ^(-2) +14H^(+)+6e^(-) to 2Cr^(+3) +7H_(2) O`
`therefore ` no of electron = 6

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