The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions is

Options

A. `1//3`
B. `1//6`
C. `2//3`
D. `1`

Answer

A

Explanation

`undersetulbar(Cr(2)O_(7)^(2-)+14H^(+)2Sn^(2+) rarr 3Sn^(4+)+2Cr^(3+)+7H_(2)O)(Cr_(2)O_(7)^(2-)+14H^(+)+underset((Sn^(2+) rarr Sn^(4+)+2e^(-))xx3)(6e^(-)to2Cr^(3+)+7H_(2)O)`
It is clear form this equation that `3` moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence `1` mol. Of `Sn^(2+)` will reduce `(1)/(3)`moles of `Cr_(2)O_(7)^(2-)`.