Question

Calculate the volume of gas liberated at the anode at STP during the electrolysis of a $CuS{O}_4$ solution by a current of $1A$ passed for $16$ minutes and $5$ seconds.

• A. $224mL$
• B. $56mL$
• C. $112mL$
• D. $448mL$

Answer 1

Answer: (B)

$56mL$

Given:-

Current, $I=1A$

Time, $t$=$16$ min $5$ sec= $965$ seconds

$n=2$

$C{u}^{+2}+2e^{-}⟶Cu$

$4O{H}^{-1}⟶O_2+2H_{2}O+4e^{-1}$

$W=\frac{I\times t\times M}{nF}$

$\Rightarrow W=\frac{1\times 965\times 63.5}{2\times 96500}$

$\Rightarrow W=0.3175g$

$2\times 63.5g⟶22.4litre{O}_2$

$\Rightarrow 0.3175g⟶\frac{0.3175}{63.5\times 2}\times 22.4$

Volume of gas liberated during electrolysis = $56ml$