Metalic tin (Sn) is oxidised to its maximum oxidation state by `KMnO_4` and `K_2Cr_2O_7` separately in the presence of HCl. Calculate the ratios of the volumes of decimolar solutions of `KMnO_4` and `K_2Cr_2O_7` that would be reduced by 1.0 g of Sn (Atomic weight of `Sn=118.6`).

Explanation

Sn is oxidised to `+4` oxidation state (maximum oxidation state of `Sn=+4`)
`SntoSn^(+4)+4e^(-)(n=4)`
`[5e^(-)+MnO_4^(ɵ)toMn^(2+)](n=6)`
Equivalent od `Sn=("Weight")/(Ew)=(1.0)/((118.6)/(4))=(4)/(118.6)Eq`
`Sn-=MnO_4^(ɵ)`
`Eq-=Eq`
`(4)/(118.6)-=0.1xx5` (n-factor)`xxV(L)`
`V_(MnO_4^(ɵ))=(4)/(118xx0.1xx5)=(4)/(59.3)=0.067L`
`Sn-=Cr_2O_7^(2-)`
`Eq-=Eq`
`(4)/(118.6)-=0.1xx6` (n-factor)`xxV(L)`
`V_(Cr_2O_7^(2-))=(4)/(118.6xx0.1xx6)=(4)/(71.16)=0.057L`
Alternatively
`(V_(MnO_4^(ɵ)))/(V_(Cr_2^(2-)))=(6)/(5)=6:5(("f-factor " Cr_2O_7^(2-))/("n-factor of "MnO_4^(ɵ)))`
Since the same equivalent of Sn is reating with 0.1 M of `KMnO_4` and `K_2Cr_2O_7`, the ratios of the volumes of `KMnO_4` and `K_2Cr_2O_7` reacting with the same equivalent of Sn is inversely proportional to their n-factors.