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`
Half cell potential `E_(Cr_(2)O_(7)^(-2)//Cr^(+3))^(@)` after teh reaction of `SnCI_(2)` is:

Options

A. `-1.18V`
B. `1.164V`
C. `1.18V`
D. None of these

Answer

B

Explanation

`Cr_(2)O_(7)^(2-) +3Sn^(+2) +14H^(+) rarr 3Sn^(+4) +2Cr^(+3) +7H_(2)O`
`E = (1.33 -0.15) -(0.06)/(6) log. ((0.2)^(3)(1)^(2))/((0.2)^(1)(0.1)^(3)(1)^(4))`
`= 1.164 V`.