Calculate the total n-factor for the oxidation half reaction given below:
$F e C_{2} O_{4} (a q) \to F e^{3 +} (a q) + C O_{2} (g)$

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Solution

The correct option is C 3
For the given reaction :
$+ 2 F e {+ 3 C}_{2} O_{4} (a q) ⟶ F e^{3 +} (a q) + C O_{2} (g)$
Oxidation :
$F e^{2 +} ⟶ F e^{3 +}$
$n_{f} = | 2 - 3 | \times 1 = 1$
Oxidation :
$_{2} O_{4}^{2 -} ⟶ + 4 C O_{2}$
$n_{f} = | 3 - 4 | \times 2 = 2$
$(n_{f})_{T} = (n_{f})_{F e} + (n_{f})_{C} = 2 + 1 = 3$

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Similar questions

F e C_{2} O_{4} (a q) \to F e^{3 +} (a q) + C O_{2} (g)

The oxidation half reaction for following reaction is

Fe²⁺(aq) + Cr₂O₇^2-(aq) → Fe³⁺ (aq) + Cr³⁺(aq)

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

E^{0}

F e^{3 +} (a q) + 3 e^{-} \to F e (s)

E^{0}

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

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

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