Calculate volume of the gases liberated at STP if L of $0.2$ molar solution of $C u S O_{4}$ is electrolysed by $5.79 A$ Current for $10000 s e c o n d s .$
$5.79 \times 10000$

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Solution

At Cathode :- ${C u}^{2 +} + {2 e}^{-} \to C u$
at anode : $4 {O H}^{(-)} + {4 e}^{-} \to 2 H_{2} O + O_{2}$
Mol. of Faraday $= \frac{I \times t}{F} = \frac{5.79 \times 10000}{96500}$
$= 0.6 m o l$
4 mol Faraday = 1 mol of $O_{2}$ = 22.4 lit of $O_{2}$
Volume of gas liberated $= \frac{1}{4} \times 22.4 \times 0.6$
$= 3.36 l i t$

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Calculate volume of the gases liberated at STP if L of 0.2 molar solution of CuSO4 is electrolysed by 5.79A Current for 10000seconds.5.79×10000

At Cathode :- Cu2++2e−→Cuat anode : 4OH(−)+4e−→2H2O+O2Mol. of Faraday =I×tF=5.79×1000096500 =0.6mol4 mol Faraday = 1 mol of O2 = 22.4 lit of O2Volume of gas liberated =14×22.4×0.6 =3.36lit

Calculate volume of the gases liberated at STP if L of $0.2$ molar solution of $C u S O_{4}$ is electrolysed by $5.79 A$ Current for $10000 s e c o n d s .$
$5.79 \times 10000$