How many hours does it take to reduce $3$ moles of ${F e}^{3 +}$ to ${F e}^{2 +}$ with $2.0 A$ current intensity?

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Solution

Given, Current $= 2.0$ Ampere
$3 {F e}^{3 +} + 3 e^{-} ⟶ 3 {F e}^{2 +}$
Charge required $= F a r a d a y = 3 \times 96500$ coulomb
$= 289500$ coulomb
We know, $C h a r g e = C u r r e n t \times T i m e$
or Time $= \frac{289500}{2}$
$= 144750$ sec
$= 40.20$ hrs.

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