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. `1.187`
B. `1.191`
C. `1.285`
D. None of these

Answer

B

Explanation

Cathode: `E_(Cr_(2))O_(7)^(-2) +6e^(-) +14H^(+) rarr 2Cr^(+3) +7H_(2)O`
Anode: `[Sn^(2+) rarr Sn^(+4) +2e^(-)] xx 3`
Net cell reaction:
`Cr_(2)O_(7)^(2-) +3Sn^(+2) +14H^(+) rarr 3Sn^(+4) +2Cr^(+3) +7H_(2)O`
`{:("Initial conc.",1M,1M,10^(-1)M,-,-,-),(At.,(2)/(3)M,-,10^(-1),1M,(2)/(3)M,-):}`
completion
`E_(Cr_(2)O_(7)^(-2)//cr^(+3)) = E_(Cr_(2)O_(7)^(-2)//Cr^(+3))^(@) (-0.06)/(6)log.([Cr^(+3)]^(2))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))`
`= 1.33 -0.01 log.((2//3)^(2))/((2//3)(10^(-1))^(14)) =1.191`