In a chlor-alkali cell, chlorine is produced at the anode by $2 Cl X^{-} Cl X_{2} + 2 e X^{-}$ . The cell operates at $3.40 V$ with a current efficiency of 92% for chlorine. The electrical energy, in kWh, consumed per kilogram of chlorine produced is closest to (Cl = 35.5, $F = 96500 C mo l^{- 1}$ , $1 kWh = 3.6 \times 1 0^{6} J$ )

Answer: C

Moles of

Cl X_{2}

per kg

= 1000/71 = 14.08 mol

, needing

2 \times 14.08 = 28.17 mol

of electrons theoretically. With 92% efficiency the charge actually passed is

28.17 \times 96500/0.92 = 2.955 \times 1 0^{6} C

. Energy

= Q V = 2.955 \times 1 0^{6} \times 3.40 = 1.005 \times 1 0^{7} J = 2.79 kWh

. Option (B) is the value obtained if the efficiency is ignored.