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The ratio of the mass discharged at anode on passing the same charge through three electrolytic cells A, B, and C containing an aqueous solution of NaCl, MgCl2 and AlCl3 respectively is

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\\\text{Na metal deposited as }\mathrm{Na}^{+}\text{ ion, Mg metal deposited as }\mathrm{Mg}^{2+} ion \\ \text{Al metal deposited as }\mathrm{Al}^{3+} ion \\\\ \therefore \text{to form 1 atom of }\mathrm{Na}, \mathrm{Mg}$ and \mathrm{Al}

                                                \\\downarrow & \ \ \ \ \ \ \downarrow \ \ \ \ \ \downarrow \\1 \mathrm{e}^{-} \ \ & 2 \mathrm{e}^{-} \ \ \ & 3 \mathrm{e}^{-}

\\\text{if 3 mol Faraday of electricity is passed, then } \\ 3 \ \mathrm{mol} \text{ of } \mathrm{Na}, \frac{3}{2} \ \mathrm{mol}$ \text{ of } $\mathrm{Mg}$ and $1\ \mathrm{mol}$ \text{ of } $\mathrm{Al}$

Ratio will be 6 : 3 : 1

 

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lovekush

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