CuSO_4 solution is electrolysed for 15 minutes to deposit 0.4725 g of copper at the cathode. The current in amperes required is (Faraday…

Question

CuSO_4 solution is electrolysed for 15 minutes to deposit 0.4725 g of copper at the cathode. The current in amperes required is (Faraday =96,500Cmol^(-1), atomic weight of copper =63)

MARKS Content Team · Accepted Answer

Correct answer: (B) 1.608. The reaction at cathode takes place as below: Cu^(2+ )+ 2e^(-) → Cus For 63 g of Cu = 2F electricity requiredSo, for 0.475 g, electricity required= 2F/63×0.4725=(2×96500×0.4725)/63=1447.5 C According to Faradays's law: Q=It→I=Q/tI=1447.5/(15×60)=1.608 C

${CuSO}_{4}$ solution is electrolysed for $15$ minutes to deposit $0.4725 g$ of copper at the cathode. The current in amperes required is $($ Faraday $= 96, 500 {Cmol}^{- 1}$ , atomic weight of copper $= 63)$

Solution

Relevant Questions

CuSO4 solution is electrolysed for 15 minutes to deposit 0.4725 g of copper at the cathode. The current in amperes required is (Faraday =96,500Cmol-1, atomic weight of copper =63)

Solution

Relevant Questions

${CuSO}_{4}$ solution is electrolysed for $15$ minutes to deposit $0.4725 g$ of copper at the cathode. The current in amperes required is $($ Faraday $= 96, 500 {Cmol}^{- 1}$ , atomic weight of copper $= 63)$