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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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A half cell is prepared by K_(2)Cr_(2)O_(7) in a buffer solution of pH =1 . Concentration of K_(2)Cr_(2)O_(7) is 1M . To 3 litre of this solution 570 gm of SnCI_(2) is added which is oxidised completely to SnCI_(4) . Given: E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06 , Atomic of mass Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15 Emf of the cell Pt|underset((0.1M))(Sn^(+2)),underset((0.2M))(Sn^(+4)),underset((1M))(H^(+)):||:underset((0.2M))(Cr_(2)O_(7)^(2-)),underset((1M))(Cr^(+3)),underset((1M))(H^(+)):|:Pt