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Variable Oxidation States

0.108 g of fi...

Question

$0.108 g$ of finely divided copper was treated with an excess of ferric sulphate solution until copper was completely dissolved. The solution after the addition of excess dilute sulphuric acid required $33.7$ mL of $0.1 N K M n O_{4}$ for complete oxidation. Find the equation which represents the reaction between metallic copper and ferric sulphate solution (Atomic mass of $C u = 63.6$ ; $F e = 56$ ).

A

C u + F e_{2} (S O_{4})_{3} ⟶ C u S O_{4} + F e S O_{4}

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B

C u + F e_{2} (S O_{4})_{3} ⟶ C u_{2} S O_{4} + F e S O_{4}

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C

C u + F e_{2} (S O_{4})_{3} ⟶ C u S O_{4} + F e S O_{4} + F e S

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D

C u + F e (S O_{4}) ⟶ C u_{2} S O_{4} + F e_{2} (S O_{4})_{3}

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Solution

The correct option is A $C u + F e_{2} (S O_{4})_{3} ⟶ C u S O_{4} + F e S O_{4}$
Since, $C u$ will react with ferric sulphate to reduce $F e^{3 +}$ to $F e^{2 +}$ , the reduced state of iron is further oxidized by $K M n O_{4}$ .
Thus, Meq. of $K M n O_{4}$ used $=$ Meq. of iron sulphate oxidized $=$ Meq. of ferric sulphate used by $C u$ $=$ Meq. of $C u$
$∴$ Meq. of $C u =$ Meq. of $K M n O_{4}$ used
$\frac{0.108}{(\frac{63.6}{n})} \times = 33.7 \times 0.1$
$∴ n = 2$
It is thus, clear that during reduction of $F e^{3 +}$ , $C u$ is oxidized to $C u^{2 +}$ .
Thus, reaction is as follows:
$C u + F e_{2} (S O_{4})_{3} ⟶ C u S O_{4} + F e S O_{4}$

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Similar questions

100

mL solution of ferric alum,

F e_{2} (S O_{4})_{3} . (N H_{4})_{2} S O_{4} .24 H_{2} O

(molecular weight

= 964

g/mol) containing

2.41

g of salt was boiled with

F e

when the reaction,

F e + F e_{2} (S O_{4})_{3} \to 3 F e S O_{4}

, takes place. The unreacted iron was filtered off and the solutions was titrated with

\frac{M}{60} K_{2} C r_{2} O_{7}

in acidic medium.

F e S O_{4}

C u

F e_{2} (S O_{4})_{3}

5.5 g

of a mixture of

F e S O_{4} \cdot 7 H_{2} O

F e_{2} (S O_{4})_{3} \cdot 9 H_{2} O

5.4 m L

0.1 N K M n O_{4}

solution for complete oxidation. Calculate the gram mole of hydrated ferric sulphate in the mixture.
(At. mass

H = 1, O = 16, S = 32, F e = 56

Why is CuSO₄the expected formula of Copper Sulphate and not Cu₂SO₄ when Copper has variable valency of Cu⁺ and Cu²⁺ ; Here the higher valency is used for the salt.

However in the case of Iron Sulphate the expected formula is FeSO₄ and not Fe₂(SO₄)₃ eventhough Iron also has variable valency of Fe²⁺ and Fe³⁺ ; Here the lesser valency is used for the salt.

K M n O_{4} + H_{2} S O_{4} + F e S O_{4} \to K_{2} S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

Coefficients of sulphuric acid and ferric sulphate in the balanced equation of above reaction are______and ______respectively.

Q. Chemical equation for the reaction between

C u S O_{4}

solution and an iron nail is given below.

C u S O_{4} + F e \to F e S O_{4} + C u

(a) Write the reduction reaction taking place here.
(b) Give reason for the displacement of Cu by Fe from

C u S O_{4}

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