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`
Number of moles of `Cr^(+3)` formed are

Options

A. 2
B. 6
C. 4
D. 3

Answer

A

Explanation

Cathode: `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
Moles of `Cr^(+3) =(2)/(3) xx 3 = 2`