Question

How much time in hours will be required to reduce 3moles of $F {e^ {3 +}} \to F {e^ {2 +}} $ with current of $2A$?

Answer 1

Hint: Faraday’ s first law of electrolysis states that amount of substance deposited or liberated at an electrode is directly proportional to amount of charge passed through the solution.
$W \propto Q$
where $W$ is the amount of substance deposited and $Q$ is charged in the coulomb.

Complete step by step answer:
-We know that when $I$ ampere current is passed through a solution for $t$ seconds then,
$Q = I \times T$
-We have $Fe^{3+} + e^ - \to Fe^{2 +} $
from the above equation it is clear that 1 mole of $F{e^{3 + }}$ will be reduced to $F{e^{2 + }}$ by gaining $1mole$ of electrons and we know that charge on one mole of electrons is $96500$ coulomb .
$96500$ coulomb = 1 Faraday
-One Faraday is the charge required to liberate or deposit one-gram equivalent of a substance at corresponding electrode.
-Therefore for $3mole$ we have $3F {e^ {3 +}} + 3{e^ -} \to 3F {e^ {2 +}} $
$ \Rightarrow $$Q = 3 \times 96500C = 289500C$
 -Now we have both $Q$ and $I$, so we can easily calculate the time.
$Q = IT$
$\therefore T = \dfrac{Q}{I} = \dfrac{{289500}}{2}$
 $ \Rightarrow $$T = 144750\sec $
-Dividing $T$ by $3600$ in order to convert it into hours.
$T = \dfrac{{144750}}{{3600}} = 40.20$$ \Rightarrow $ $T = \dfrac{{144750}}{{3600}} = 40.20hrs$
So, the time required to reduce $3mole$ of $F {e^ {3 +}} + {e^ -} \to F {e^ {2 +}} $ is $40.20hrs$.

Note:
The second law of faraday states that when the same amount of charge is passed through different electrolyte solutions connected in series then the weight of substances deposited at electrodes are in the ratio of their equivalent weights.
 For electrolysis, $\Delta G = positive $ so it is a non-spontaneous process.
In electrolytic cells, D.C current is used.