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When 0.1 mole Cr(2)O(7)^(-2) is oxidised...

When 0.1 mole `Cr_(2)O_(7)^(-2)` is oxidised then quantity of elecricity required to completely oxidise `Cr_(2)O_(7_^(-2))` to `Cr^(+3)` is :-

A

9650 C

B

96500 C

C

57900 C

D

54900 C

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(Cr_(2)O_(7)^(-2),to,2Cr^(+3),+6e^(-),),(0.1 "mole",,,0.6"mole",):}`
Change required = 0.6 F `= 0.6xx96500`
=57900C
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