Question

In a reaction it was found that 3.0g of a metal *X* was oxidized by 25.0 cm³ of 0.10 mol dm-³ K2 Cr2 O2 under acidic conditions.
(i) Deduce the mole of ratio between *X* and Cr2 O7²- ion.
(ii) write a balanced equation of the redox reaction.
(iii) Give the oxidation numbers of chromium and *X* in both their reduced and oxidized forms.

[Molar mass of *X* = 200.6]

Answer 1

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Task#76556

In a reaction it was found that 3.0g of a metal *X* was oxidized by 25.0 cm³ of 0.10 mol dm⁻³ K₂ Cr₂ O₂ under acidic conditions.

(i) Deduce the mole of ratio between *X* and Cr₂ O₇²⁻ ion.

(ii) write a balanced equation of the redox reaction.

(iii) Give the oxidation numbers of chromium and *X* in both their reduced and oxidized forms. [Molar mass of *X* = 200.6

Solution: Reduction: Cr₂O₇²⁻ + 14 H⁺ + 6e = 2Cr³⁺ + 7H₂O;

[Equivalent weight of K₂Cr₂O₇ = molecular weight / 6 = M / 6],

Where, M = Molecular weight of K₂Cr₂O₇

Oxidation: X = Xⁿ⁺ + ne

[Equivalent weight of X = 200.6 / n ]

(i) Concentration of K₂Cr₂O₇ Solution = 0.10 mol dm⁻³ = 0.1 mol/lit = (0.1x M)g/lit = 0.1x6 gm-equivalent/lit = 0.6 Normal = 0.6(N); [1dm³ = 1000cc = 1lit]

So, 1000cc 1(N) K₂Cr₂O₇ solution = 200.6/n gm of X;

1cc 1(N) K₂Cr₂O₇ solution = 0.2006/n gm of X;

25cc 0.6(N) K₂Cr₂O₇ solution = 0.2006 x 25x 0.6/n of X;

From given condition,

$$0.2006 \times 25 \times 0.6 / n = 3$$ $$n = 0.2006 \times 25 \times 0.6 / 3 = 1.003$$ $$n = 1$$

Amount of K₂Cr₂O₇ in 25ml 0.10 mol/lit solution = $\frac{0.1 \times 25}{1000}$ mol = 2.5 X 10⁻³ mol

Amount of X in solution = $\frac{3g}{200.6g/mol} = 0.014955$ mol

Mol ratio between X and K₂Cr₂O₇ = $\frac{0.014955}{2.5 \times 10^{-3}}$ = 6;

That means for one mol of K₂Cr₂O₇ oxidises 6mols of metal X.

(ii) Reduction: Cr₂O₇²⁻ + 14 H⁺ + 6e = 2Cr³⁺ + 7H₂O;

Oxidation: X = X⁺ + e;

Balanced equation of the redox reaction: Cr₂O₇²⁻ + 14 H⁺ + 6X = 2Cr³⁺ + 7H₂O + 6X⁺;

(iii) Oxidation number of Chromium in Cr₂O₇²⁻ (oxidised form) = x = 6, and in reduced form (Cr³⁺) = 3

[2x + 7(-2) = -2, or, x = 6], metal (reduced form, X) = 0, oxidised form (X⁺) = 1;