At anode, in 1 mole NaCl 71.0 gm Cl− ion is present.
2Cl−→Cl2 + 2e−
71.0 g 71.0 g 2×96500 C
Charge =Q=i×t×efficiency=50×10−3×12×3600×14
Moles of Cl2 liberated =12×96500×50×10−3×12×3600×14=2.8milimoles
At cathode:
2e−+2H2O→H2(g)+2OH−(aq)
Moles of H2 liberated = moles of Cl2 liberated =2.8 milimoles
Total moles =5.6 milimoles (Cl2+H2)
Total volume of gas liberated as STP =5.61000×22.4×1000=125.44 ml