A half cell is prepared by `K_(2)Cr_(2)O_(7)` in a buffer solutoin of pH = 1. Concentration of `K_(2)Cr_(2)O_(7)` is 1 M. To 3 litre of this solution 570 gm of `SnCl_(2)` is added which is oxidised completely to `SnCl_(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

`40.K_(2)Cr_(2)O_(7)+3SnCl_(2)+H^(+)rarr SnCl_(4)+2Cr+3`
mole of `SnCl_(2)=3 " " therefore` mole of `Cr^(+3)=2`