`50 cm^(3)` of 0.04 M `K_2Cr_2O_7` in acidic medium oxidizes a sample of `H_2S` gas to sulphur. Volume of 0.03 `M KMnO_4` required to oxidize the same amount of `H_2S` gas to sulphur, in acidic medium is

Options

A. `60 cm^(3)`
B. `80 cm^(3)`
C. `90 cm^(3)`
D. `120 cm^(3)`

Answer

B

Explanation

`Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_(2)O`
(`H_(2)S to S + 2e^(-) + 2H^(+)) xx 3`
`Cr_(2)O_(7)^(2-) + 8H^(+) + 3H_(2)S to 2Cr^(3+) + 3S + 7H_(2)O`
`MnO_(4)^(2-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O) xx2`
`(H_(2)S to S + 2e^(-) + 2H^(+))xx5`
`2MnO_(4)^(2-) + 6H^(+) + 5H_(2)S to 2Mn^(2+) + 8H_(2)O`
From equation given above.
1 mol of `k_(2)Cr_(2)O_(7)` oxidises 3 mol of `H_(2)S` or 1 mol of `H_(2)S` si oxidised by `(1)/(3)` mol of `K_(2)Cr_(2)O_(7)`
SImilary . 2mol of `KMnO_(4)` oxidise 5 mol of `H_(2)S` or 1 mol of `H_(2)S` is oxidised by `(2)/(5)` mol of `KMnO_(4)`.
Therefore, the ratio of moles of `K_(2)Cr_(2)O_(7)` and `KMnO_(4)` which oxidise same number of
moles of `H_(2)S` is `(1)/(3) : (2)/(5) or 5 : 6`.
MOles of `K_(2)Cr_(2)O_(7)` in 50 `cm^(3)` of 0.04 M
Solution ` = 2 xx 10^(-3)` moles
`therefore` Moles of `KMnO_(4)` that would oxidse same amont of `H_(2)S = 2xx 10^(-3) xx (6)/(5)`
` = 2.4 xx 10^(-3)` moles
`therefore` Volume of `KMnO_(4)` required
` = (1000)/(0.03) xx 2.4 xx 10 `
` = 80 cm^(3)`