A current of $5.0 A$ is passed for $1930 s$ through concentrated brine (chlor-alkali cell), with chlorine liberated at the anode at $100%$ current efficiency. Using $F = 96500 C mo l^{- 1}$ and $22.4 L mo l^{- 1}$ at STP, the volume of $Cl X_{2}$ at STP is

Answer: B

Q = 5.0 \times 1930 = 9650 C

, so electrons

= 9650/96500 = 0.10 mol

. The anode reaction

2 Cl X^{-} Cl X_{2} + 2 e X^{-}

needs 2 mol of electrons per mole of

Cl X_{2}

, giving

0.05

mol

Cl X_{2}

V = 0.05 \times 22.4 = 1.12 L