Question Number 227417 by Spillover last updated on 24/Jan/26
Answered by master_brainly last updated on 31/Jan/26
$$\:\:\:\left(\boldsymbol{\mathrm{a}}\right)\: \\ $$Faraday's first law states that the mass of a substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. Faraday's second law states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances produced at the electrodes are proportional to their equivalent masses.
Answered by Spillover last updated on 01/Feb/26
$${Q}={It}\:\:\:\: \\ $$$$\:{t}=\mathrm{32}{min}+\mathrm{10}\:{sec}=\mathrm{32}×\mathrm{60}+\mathrm{10}=\mathrm{1930}{C} \\ $$$${t}=\mathrm{1930}{C} \\ $$$${Q}=\mathrm{0}.\mathrm{50}×\mathrm{1930}=\mathrm{965}{C} \\ $$$$\mathrm{0}.\mathrm{44}{g}\rightarrow\mathrm{965}{C} \\ $$$$\mathrm{88}{g}\rightarrow?\:{x} \\ $$$${x}=\mathrm{193000}\:{coulombs} \\ $$$${from} \\ $$$$\:\mathrm{1}{F}=\mathrm{96500}\:{coulombs} \\ $$$${number}\:{of}\:{Faradays}=\frac{\mathrm{193000}}{\mathrm{96500}}=\mathrm{2}{F} \\ $$$$ \\ $$
Answered by Spillover last updated on 01/Feb/26
$$\left({ii}\right){From}\:{Faradays}\:{second}\:{law}\:{of}\:{electroysis} \\ $$$$\mathrm{1}{F}=\mathrm{1}{e}=\mathrm{1}{mole}=\mathrm{1}{volume} \\ $$$$\mathrm{2}{F}=\mathrm{2}\overset{−} {{e}}= \\ $$$${X}\:{ions}\:{X}^{\mathrm{2}+} \\ $$$$\left({iii}\right){Hydroxy}\:{of}\:{X}\:\:\:\:\:\:\:\:\:\:\:{X}\left({OH}\right)_{\mathrm{2}} \\ $$