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Standard X

Chemistry

Electrolysis and Electrolytes

An electric c...

Question

An electric current is passed through two electrolytic cells connected in series, one containing $A g {N O}_{3}$ (aq) and other $H_{2} {S O}_{4}$ (aq). The volume of $O_{2}$ (in litres) measured at $25^{o} C$ and 750 mm Hg would be liberated in litre from $H_{2} {S O}_{4}$ , if 1 mole of $A g^{+}$ are deposited from $A g {N O}_{3}$ .
(Round off to integer)

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Solution

$∵$ Equivalent of $O_{2}$ = Equivalent of $A g$
or $m_{O_{2}}$ /8 = 1 ( $∵$ 1 mole of $A g$ =1 eq. of $A g$ )
$∴$ $m_{O_{2}}$ = 8g
Using gas equation for volume of $O_{2}$ , at $T = 298 K, P = \frac{750}{760}$ atm

$\frac{750}{760} \times V_{O_{2}} = \frac{8}{32} \times 0.0821 \times 298$

$∴$ $V_{O_{2}}$ $= 6.20$ litres.
Therefore the answer is 6.

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An electric current is passed through two electrolytic cells connected in series, one containing AgNO3(aq) and other H2SO4(aq). The volume of O2 (in litres) measured at 25oC and 750 mm Hg would be liberated in litre from H2SO4, if 1 mole of Ag+ are deposited from AgNO3.(Round off to integer)

∵ Equivalent of O2 = Equivalent of Ag or mO2/8 = 1 (∵ 1 mole of Ag =1 eq. of Ag)∴ mO2 = 8gUsing gas equation for volume of O2, at T=298 K,P=750760 atm750760×VO2=832×0.0821×298∴ VO2 =6.20 litres.Therefore the answer is 6.