In basic medium \(\mathrm{CrO}_{4}{ }^{2-}\) oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\) to form \(\mathrm{SO}_{4}{ }^{2-}\) and itself changes into \(\mathrm{Cr}(\mathrm{OH})_{4}{ }^{-}\). The volume of \(0.154 \mathrm{M} \mathrm{CrO}_{4}{ }^{2-}\) required to react with 40 mL of \(0.25 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\) is \_\_\_\_ mL. (Rounded - off to the nearest integer)

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Redox Reactions and Electrochemistry

In basic medium

CrO_{4}^{2 -}

oxidises

S_{2} O_{3}^{2 -}

to form

SO_{4}^{2 -}

and itself changes into

Cr (OH)_{4}^{-}

.
The volume of

0.154 M CrO_{4}^{2 -}

required to react with 40 mL of

0.25 M S_{2} O_{3}^{2 -}

is ____ mL. (Rounded - off to the nearest integer)

Answer: 173

S_{2} O_{3}^{2 -}

the two sulfur atoms are at an average oxidation state of

+ 2

and both finish at

+ 6

S O_{4}^{2 -}

, so each thiosulfate ion gives up 8 electrons. Chromium falls from

+ 6

Cr O_{4}^{2 -}

+ 3

Cr (OH)_{4}^{-}

, so each chromate ion accepts 3 electrons. Equating the electrons lost and gained,

8 n (S_{2} O_{3}^{2 -}) = 3 n (Cr O_{4}^{2 -})

The thiosulfate taken is

0.25 \times 0.040 = 0.010

mol, so

n (Cr O_{4}^{2 -}) = \frac{8}{3} \times 0.010 = 2.67 \times 1 0^{- 2} mol

V = \frac{2.67 \times 1 0 ^{- 2}}{0.154} = 0.1732 L

The volume of chromate solution required is 173 mL.

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Integer answer