How much time is required to deposit 6.48 g of silver at the cathode when a current of 9.65 A is passed through an aqueous solution of silver nitrate? (Given: Molar mass of Ag = 108 ~g ~mol ⁻¹, 1 ~F = 96500 ~C )

Step-by-step solution

The reduction reaction at the cathode for silver nitrate is: Ag ⁺ + e ⁻ Ag ( s ) Here, the number of moles of electrons required per mole of Ag is n = 1 . According to Faraday's First Law of Electrolysis: w = M I t n F Rearranging for time ( t ): t = w n F M I Given: w = 6.48 ~g M = 108 ~g ~mol ⁻¹ I = 9.65 ~A F = 96500 ~C Substituting the values into the equation: t = 6.48 1 96500 108 9.65 t = 6.48 108 10000 = 0.06 10000 = 600 ~s Converting the time into minutes: Time in minutes = 600 60 = 10 ~min . Answer: 10 min