Question

In basic medium CrO4 2- oxidises S2O3 2- to form SO​42- ​ and itself changes to Cr (OH)4-. How many ml of 1.54 M CrO4 2-are required to react with 40 ml of 0.246 MS2O3 2- ?

Answer 1

Here,
CrO₄²⁻ ⇔ Cr(OH)₄⁻
Change in oxidation number of Cr = oxidation number of Cr in CrO₄²⁻ - oxidation number of Cr in Cr(OH)₄⁻
= +6 - (+3) = +3
∴ n - factor = 3

Similarly, S₂O₃²⁻ ⇔ 2SO₄²⁻
change in oxidation number of S = oxidation number of S in S₂O₃²⁻ - oxidation number of S in SO₄²⁻
= +4 - (+12) = -8
∴ n- factor = 8

Now, equivalent of CrO₄²⁻ = equivalent of S₂O₃²⁻
N₁V₁ = N₂V₂
∵ normality = n-factor × molarity
so, n₁M₁V₁ = n₂M₂V₂
⇒3 × 1.54 × V₁ = 8 × 0.246 × 40
⇒V₁ = 320 × 0.246/(3 × 1.54) = 17.038 ml

Hence, volume of CrO₄²⁻ = 17.0.38 mL

Answer 2

Answer:

Here,

CrO₄²⁻ ⇔ Cr(OH)₄⁻

Change in oxidation number of Cr = oxidation number of Cr in CrO₄²⁻ - oxidation number of Cr in Cr(OH)₄⁻

= +6 - (+3) = +3

∴ n - factor = 3

Similarly, S₂O₃²⁻ ⇔ 2SO₄²⁻

change in oxidation number of S = oxidation number of S in S₂O₃²⁻ - oxidation number of S in SO₄²⁻

= +4 - (+12) = -8

∴ n- factor = 8

Now, equivalent of CrO₄²⁻ = equivalent of S₂O₃²⁻

N₁V₁ = N₂V₂

∵ normality = n-factor × molarity

so, n₁M₁V₁ = n₂M₂V₂

⇒3 × 1.54 × V₁ = 8 × 0.246 × 40

⇒V₁ = 320 × 0.246/(3 × 1.54) = 17.038 ml

Hence, volume of CrO₄²⁻ = 17.0.38 mL