A current of 2.5 amperes is passed through 800 ml of 0.48 M solution of CuSO ₄ for 1.0 hour with a current efficiency of 80 % . If the volume of the solution remains unchanged, what is the final molarity of the solution?

Step-by-step solution

The reaction at the cathode is Cu ²⁺ + 2e⁻ Cu (s) . The total charge passed through the solution is Q = I t efficiency = 2.5 A (1.0 3600 s ) 0.80 = 7200 C . The number of moles of electrons transferred is n_ e = Q F = 7200 96500 0.07461 mol . Since 2 moles of electrons are required to reduce 1 mole of Cu ²⁺ ions, the moles of Cu ²⁺ consumed is n_ Cu ²⁺ = n_ e 2 = 0.07461 2 = 0.037305 mol . The initial moles of Cu ²⁺ in 800 ml (0.8 L) of 0.48 M solution is n_ initial = 0.48 M 0.8 L = 0.384 mol . The final moles of C