First slide

Faraday's Laws

Question

The electricity required in coulombs for the oxidation of 0.1 mol of FeO to Fe₂O₃ is

Easy

A

96500 C
B

2 x 96500 C
C

9650 C
D

96.5 C

Solution

$H_{2} O + 2 FeO \to {Fe}_{2} O_{3} + 2 H^{+} + 2 e^{-}$
For 2 moles → 2 × 96500 coulombs
So for 0.1 mol → 9650 coulombs

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