500 mL 4 (M) of `NaCl` solution is electrolysed liberating `Cl_(2)(g)` at an electrode. Calculate (i) the total number of moles of `Cl_(2)` (ii) the maximum amount of amalgam, if Hg- cathode is used. (Hg = 200) (iii) the amount of electricity required for complete electrolysis.

Explanation

`"500 mL (M) NaCl"-="2 mol "Cl^(-) and "2 mol "Na^(+)`
(i) `Cl^(-)rarr(1)/(2)Cl_(2)+e" So, 2 mol "Cl^(-)-="1 mol "Cl_(2)`
(ii) `Na^(+)+e rarr Na" So, 2 mol "Na^(+)-="2 mol Na"-="46 g Na"`
`therefore" Maximum amount of amalgam "=2xx200+46=446g`
(iii) `"2 mol "Cl_(2)="2 mol Na"-=2xx96500-=193000C`