Question

A current is passed through three electrlyctic cells connected in series containing slotions if silver trioxonitrate (v), copper (ii) trioxosulphate (vi) brine respectively. if 14.5g of copper are deposited in the second electrolytic cell, calculate (a) mass of silver deposited in the first cell (b) the volume of chlorine librated in the third cell at 180c and 810mm hg pressure

Answer 1

AFTER ELECTROLYSIS Reaction Is

Cu²⁺ + 2e -------> Cu .

2Ag⁺ + 2e ------------->2Ag .

2Cl^- + 2e --------- > Cl₂ .

If we pass 2F charge all in series So , same charge supply then 1 mole of Cu , 2 mole of Ag And 1 mole of Cl₂ .

14.5g of copper are deposited in the second electrolytic cell,

So , 14.5 / 63.5 mole of Cu deposited that is 0.2283 mole of Cu deposited .

For 1 mole of Cu 2F Charge passed

So , 0.2283 mole of Cu ( 2 * 0.2283 )F charge Passes. That is 0.4566 F charge Passes.

ALL the three cell are in series so same charge passes from each cell .

(a) FOR Ag Due to 2F Charge 2 mol of Ag Form .

So , 0.4566 F form 0.4566 mole of Ag .

So , Mass of Ag = Mole of Ag * Atomic Mass of Ag = ( 0.4566 * 108 ) g . = 49.312 g . ANSWER ..

( b ) in the third cell  2F Charge gives 1mole of Cl₂ ( g) .

So , 0.4566 F charge gives(0.4566 / 2) mole of Cl₂ ( g) . = 0.2283 mole .

At . Temperature = 180 degree =( 273 + 180 ) k = 453 K .

Pressure = 810 mm of Hg = 810 / 760 atm . = 1.0657 atm .

So , Volume = mole * 0.0821 * temperature / pressure. [ by ideal gas equation ]

So , Volume = 0.2283 * 0.0821 * 453 / 1.0657 = 7.96 L . Answer .