What mass of Cu would be produced by the cathodic reduction of Cu\(^{2+}\) when 1.60A of current passes through a solution of CuSO\(_4\) for 1 hour. (F=96500Cmol\(^{-1}\) , Cu=64)

  • A 7.64g
  • B 3.82g
  • C 1.91g
  • D 0.96g
jamb 2018chemistry

The correct answer is C. 1.91g

M = \(\frac{MmIT}{96500n}\)

  Where

  M=mass

  Mm = Molar mass = 64g/mol

  I = current = 1.6A

  T= Time  =1h r=3600s

  N= No of Charge = +2

  M = \(\frac{64x \times 1.6 \times 3600}{96500 \times 2}\)

  M=1.91g

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