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Ion Electron Method

The equivalen...

Question

The equivalent weight of ${F e}_{3} O_{4}$ in the reaction, would be :

${F e}_{3} O_{4} + {K M n O}_{4} \to {F e}_{2} O_{3} + {M n O}_{2}$

M / 6

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M

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2 M

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M / 3

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Solution

The correct option is B $M$
N factor for ions in redox reactions is equal to the number of electrons transferred.

Here n = 1

$F e^{+ 2} \to F e^{+ 3} + e^{-}$

so, equivalent $w e i g h t = M / 1 = M$

So option (B) is correct.

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Similar questions

F e C_{2} O_{4}

F e C_{2} O_{4} ⟶ {F e}^{3 +} + C O_{2}

{F e}_{2} S_{3}

{F e}_{2} S_{3}

{F e}_{3} S_{3} + O_{2} ⟶ {F e S O}_{4} + {S O}_{2}

F e O (s) + {F e}_{2} O_{3} (s) ⟶ {F e}_{3} O_{4} (s)

2 F e (s) + O_{2} (g) ⟶ 2 F e O (s); Δ H^{o} = - 544 k J

4 F e (s) + 3 O_{2} (g) ⟶ 2 {F e}_{2} O_{; 3} (s); Δ H^{o} = - 1648.4 k J

{F e}_{3} O_{4} (s) ⟶ 3 F e (s) + 2 O_{2} (g); Δ H^{o} = + 1118.4 k J

Δ G = - 177 K c a l

(1) 2 F e (s) + \frac{3}{2} O_{2} (g) ⟶ {F e}_{2} O_{3} (s)

Δ G = - 19 K c a l f o r (2) {4 F e}_{2} O_{3} (s) ⟶ {3 F e}_{3} O_{4} (s)

{F e}_{3} O_{4}

F e

F e O

{F e}_{2} O_{3}

{F e}_{3} O_{4}

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