#### How many grams of Cl2(g) can be produced by the electrolysis of molten NaCl with a current of 5.5 A for 25 min? (Atomic weight of Cl = 35.5 amu)Option: 1 3.017Option: 2 4.017Option: 3 0.085Option: 4 30.17

As we have learnt,

According to Faraday's second law, "When the same quantity of electricity is passed through different electrolytes, the amounts of the products obtained at the electrodes are directly proportional to their chemical equivalents or equivalent weights".

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Electrolysis:

$\text { Molten NaCl } \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}$
\begin{aligned} \text { Number of Faradays } &=\frac{I \times t \text { in seconds }}{96500 \mathrm{\: C}} \\ &=\frac{5.5 \mathrm{\: s} \times 25 \times 60 \mathrm{\: s}}{96500 \mathrm{\: C}}=0.085 \mathrm{\: F} \end{aligned}
$\begin{array}{l}{\text {Oxidation of } \mathrm{Cl}^{-} \text { at anode: }} \\ {2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-}} \\ {2 e^{-}=2 \mathrm{\: F}=1 \mathrm{\: mol} \mathrm{\: Cl}_{2}=71 \mathrm{\: g}}\end{array}$
$0.085 \mathrm{\: F}=\frac{71}{2} \times 0.085=3.017 \mathrm{\: g} \mathrm{\: Cl}_{2}$