An acidic solution of Cu 2+ salt containing 0.4 g of Cu 2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 mL and the current at 1.2 A. Calculate the volume of gases evolved at NTP during the entire electrolysis.

Question

An acidic solution of Cu 2+ salt containing 0.4 g of Cu 2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 mL and the current at 1.2 A. Calculate the volume of gases evolved at NTP during the entire electrolysis. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

If the salt is CuSO 4 During deposition of Cu at cathode, O 2( g ) will evolve at anode gram-equivalent of Cu deposited =(0.4 × 2/63.5)=0.0126 Volume of O 2 liberated at NTP at anode =0.0126 × 5600 mL =70.56 mL In the next 7 min , H 2 at cathode and O 2 at anode would be produced. Faraday's passed =(1.2 × 7 × 60/96500)=5.22 × 10-3 ⇒ Volume of H 2( at NTP )=5.22 × 10-3 × 11200 mL =58.46 mL Volume of O 2( at NTP )=5.22 × 10-3 × 5600 mL =29.23 mL Therefore, O 2(g) at NTP =70.56+29.23=99.79 mL H 2(g) at NTP =58.46 mL