Calculate the total n-factor for the reduction half reaction given below:
$F e_{2} (S O_{4})_{3} ⟶ F e O + S O_{2}$

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Solution

The correct option is B 8
For the given reaction :
$+ 3 F e_{2} (+ 6 S O_{4})_{3} (s) ⟶ + 2 F e O (s) + + 4 S O_{2} (g)$
Reduction :
$F e^{+ 3} ⟶ F e^{+ 2}$
$n_{f} = | 3 - 2 | \times 2 = 2$
Reduction :
$S^{+ 6} ⟶ S^{+ 4}$
$n_{f} = | 6 - 4 | \times 3 = 6$
$(n_{f})_{T} = (n_{f})_{F e} + (n_{f})_{S} = 2 + 6 = 8$

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