A current of 1.93 A is passed through an aqueous solution of silver nitrate ( AgNO ₃ ) for 500 seconds . What is the mass of silver deposited at the cathode? (Given: Molar mass of Ag = 108 g mol ⁻¹ , 1 F = 96500 C mol ⁻¹ )

Step-by-step solution

The total charge Q passed through the solution is given by Q = I t . Substituting the given values: Q = 1.93 A 500 s = 965 C The reduction reaction at the cathode is: Ag ^+ + e ^- Ag This indicates that 1 mole of electrons ( 96500 C ) deposits 1 mole of silver ( 108 g ). Using Faraday's first law of electrolysis, the mass m of silver deposited is: m = M nF Q Here, n = 1 for Ag ^+ . Substituting the values: m = 108 1 96500 965 m = 108 100 = 1.08 g Answer: 1.08 g