A current of 3.0A is passed through 750 ml of 0.45 M solution of CuSO₄ for 2 hours with a current efficiency of 90%. If the volume of the solution is assumed to remain constant, what would be the final molarity of CuSO₄ solution?

Step-by-step solution

The reaction for the electrolysis of CuS O₄ is Cu ²⁺ + 2e ⁻ Cu(s). The total charge passed through the solution is Q = I t efficiency . Q = 3.0 A (2 3600 s ) 0.90 = 19440 C . The number of moles of electrons passed is n_ e = Q F = 19440 96500 0.20145 mol . Since 2 moles of electrons are required to reduce 1 mole of Cu ²⁺ , the moles of Cu ²⁺ reduced is n_ reduced = n_ e 2 = 0.20145 2 = 0.100725 mol . The initial moles of CuSO₄ is n_ initial = M V = 0.45 M 0.750 L = 0.3375 mol . The final moles of CuSO₄ is n_ final