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Types of Redox Reactions

Identify X in...

Question

Identify X in the equation given below. $F e C l_{3} + S O_{2} + H_{2} O \to F e C l_{2} + H C l + X$

A

H_{2} S

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B

S O_{3}

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C

H_{2} S O_{4}

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D

H_{2} S O_{3}

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Solution

The correct option is C $H_{2} S O_{4}$
The balanced equation can be written as:
$2 F e C l_{3} + S O_{2} + 2 H_{2} O \to 2 F e C l_{2} + 2 H C l + H_{2} S O_{4}$
Thus X = $H_{2} S O_{4}$

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Similar questions

Q. A list of reactions are given below.

F e + 2 H C l \to F e C l_{2} + H_{2}

F e S + 2 H C l \to F e C l_{2} + H_{2} S

F e + S Δ - \to F e S

N a C l + H_{2} S O_{4} < 200^{o} C - --- \to N a H S O_{4} + H C l

C u S O_{4} + H_{2} S \to C u S + H_{2} S O_{4}

S + H_{2} S O_{4} \to 3 S O_{2} + H_{2} O

Identify redox reactions.

Q. Balance the following equation by ion electron method.

F e C l_{3} + H_{2} S \to F e C l_{2} + H C l + S

Q. Identify the oxidant and the reductant in the following reaction respectively:

(i) F e C l_{2} + H_{2} O_{2} + H C l \to F e C l_{3} + H_{2} O

S + O_{2} \to S O_{2}; Δ H = - 298.2 k J

S O_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

S O_{3} + H_{2} O \to H_{2} S O_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

the enthalpy of formation of

H_{2} S O_{4}

298

S + O_{2} \to S O_{2}

Δ H = - 298.2

S O_{2} + \frac{1}{2} O_{2} \to S O_{3}

Δ H = - 98.7

S O_{3} + H_{2} O \to H_{2} S O_{4}

Δ H = - 130.2

H_{2} + \frac{1}{2} O_{2} \to H_{2} O

Δ H = - 287.3

Then the enthalpy of formation of

H_{2} S O_{4}

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