Number of moles of `K_(2)Cr_(2)O_(7)` can be reduced by 1 mole of `Sn^(2+)` ions is:

Options

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

Answer

A

Explanation

`{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2 Cr^(3+)+7H_(2)O),((SN^(2+)rarr Sn^(4+)+2e^(-))xx3),(bar(Cr_(2)O_(7)^(2-)+14 H^(+)+3 Sn^(2+) rarr 3Sn^(4+) +2 Cr^(3+)+7 H_(2)O)):}`
It is clear from 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-)`.