Number of moles of `K_2Cr_2O_7` reduced by one mole of `Sn^(2+)` ion is

Options

A. `1/3`
B. 3
C. `1/6`
D. 6

Answer

A

Explanation

`underset"1 M"(Cr_2O_7^(2-)) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_2O`
`underset"1 M"(Sn^(2+)) to Sn^(4+) + 2e^-`
1 M of `Sn^(2+)` on oxidation gives `2e^-`
`6e^-` reduce 1 M of `Cr_2O_7^(2-)` completely.
Then `2e^-` will reduce `M/6xx2=M/6` i.e.`1/3` moles of `K_2Cr_2O_7`