A charge of $0.4 F$ is passed through concentrated aqueous sodium acetate between inert electrodes. Assume that only acetate is oxidised at the anode (Kolbe reaction), that water is reduced at the cathode, and that current efficiency is 100 %. Using molar volume $22.4 L mo l^{- 1}$ at STP, the total volume of gases evolved at both electrodes is

Answer: B

Anode:

2 CH X_{3} COO X^{-} C X_{2} H X_{6} + 2 CO X_{2} + 2 e X^{-}

, so 0.4 mol of electrons gives 0.2 mol

C X_{2} H X_{6}

and 0.4 mol

CO X_{2}

. Cathode:

2 H X_{2} O + 2 e X^{-} H X_{2} + 2 OH X^{-}

gives 0.2 mol

H X_{2}

. Total gas

= 0.8

mol, volume

= 0.8 \times 22.4 = 17.92 L