Question

During the Kolbe electrolysis of aqueous solution of sodium acetate, 45 ml of $H_2$ gas is liberated at the cathode. Then, the volume of the gases liberated at the anode is:

• A. 15 ml
• B. 90 ml
• C. 135 ml
• D. 70 ml

Answer 1

Answer: (C)

135 ml

The reactions which occur at anode are:

$2C{H}_{3}CO{O}^{\bullet }\to 2C{H}_3^{\bullet }+2C{O}_2\left(g\right)$

$2C{H}_3^{\bullet }\to C_2{H}_6\left(g\right)$

The reaction which occurs at cathode is:

$2N{a}^{+}+2H_{2}O\to 2NaOH+H_2$

For every single mole of gaseous product formed at the cathode, three moles of gaseous products are formed at the anode.

So, for 45 ml of $H_2$ gas produced at the cathode, $45\times 3=135ml$ of gaseous product is formed at the anode.

Option C is correct.