Question

The volume of gases evolved by passing $0.965$ ampere current for $10$ hours through an aqueous solution of $C{H}_{3}COONa$ at STP.

• A. $8.06L$
• B. $16.13L$
• C. $32.24L$
• D. $1.76L$

Answer 1

The correct option is B $16.13L$

Reaction
At Cathode : $2H_{2}O+2e^{-}\to H_2+2O{H}^{-}$
At Anode : $\frac{2C{H}_{3}CO{O}^{-}\to C_2{H}_6+2C{O}_2+2e^{-}}{2H_{2}O+C{H}_{3}CO{O}^{-}\to H_{2\left(g\right)}+C_2{H}_{2\left(g\right)}+2C{O}_2\left(g\right)+2O{H}^{-}}$
$\text{No. of Faraday}=\frac{0.965\times 10\times 60\times 60}{96500}=0.36$
If $2$ moles $e^{-}$ was transferred, the moles of gases produced $=4$
$\therefore \text{For 0.36 moles gases produced}=\frac{4}{2}\times 0.36mole$
$=0.72\text{mole}$
$\therefore \text{Volume}=16.13L$