Which property of hydrogen is shown by the following reactions?
$F e_{3} O_{4} + 4 H_{2} \to 3 F e + H_{2} O$
$C O + H_{2} C r_{2} O_{3} - --- \to Z n O_{3} C H_{3} O H$

Reducing character

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Oxidizing character

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Combustibility

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High reactivity

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Solution

The correct option is C Reducing character
$F e_{3} O_{4} + 4 H_{2} ⟶ 3 F e + 8 H_{2} O$ $⟶ (i)$
In this reaction, $F e_{3} O_{4}$ is reduced to $F e$ which implies that $H_{2}$ acts as a reducing agent.

$C O + H_{2} C r_{2} O_{3} - --- \to Z n O_{3} C H_{3} O H$ $⟶ (i i)$

In reaction (ii) also, $C O$ is reduced to $C H_{3} O H$ which implies $H_{2}$ acts as a reducing agent.

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Similar questions

3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}

3 F e + 2 H_{2} O \to F e_{3} O_{4} + 4 H_{2}

4 H_{2} O_{(g)} + 3 F e_{(s)} ⇌ F e_{3} O_{4 (s)} + 4 H_{2 (s)}

H_{2}

3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}

3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}

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