A certain quantity of electricity is passed through an aqueous solutions of AgNO₃ and CuSO₄ in series. If the mass of Ag deposited is 1.08 g, What is the mass deposited of Cu?

0.3175 g

According to Faraday's Second Law of Electrolysis, When same quantity of charge is passed through the solution of different electrolytes, the mass of substance deposited or liberated at electrode is directly proportional to its equivalent weight.

m α E ;     E = Equivalent weight

\(\frac{m_1}{m_2}\) = \(\frac{E_1}{E_2}\)

\(\frac{1.08}{m_2}\) = \(\frac{108}{31.75}\)

m₂ = \(\frac{1.08 × 31.75}{108}\) = 0.3175 g