Ch. 20: Electrochemistry Lecture 4: Electrolytic Cells & Faraday’s Law.

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Presentation transcript:

Ch. 20: Electrochemistry Lecture 4: Electrolytic Cells & Faraday’s Law

Electrolytic Processes Electrolytic processes are NOT spontaneous. An external power source is required for the cell to work. They have: A negative cell potential, (-E0) A positive free energy change, (+G)

Electrolysis of Water Anode rxn: -1.23 V Cathode rxn: -0.83 V -2.06 V

Electroplating of Silver Anode reaction: Ag(s)  Ag+ + e- Cathode reaction: Ag+ + e-  Ag(s) Electroplating requirements: 1. Solution of the plating metal 2. Anode made of the plating metal 3. Cathode with the object to be plated 4. Source of current

Current, I I = q/t Current: flow of charge. e– through wire, and Ions through solutions and salt bridge I = q/t

Relating I, q, and t I = q/t I = current (amperes or amp or coulombs/seconds, C/s) q = charge (coulombs, C) = n*F t = time (seconds, s)

Relating I, q, and t I = q/t Zn(s)+ Cu2+  Zn2+ + Cu(s) E0 = + 1.10 V Calculate the mass of copper deposited on the copper electrode in a galvanic cell in 30 seconds if the current, I, is 30 amperes (C/s). Calculate q: b) Use Faraday’s constant, 96,485 coulombs/mol e-, determine number of moles of electrons: q = nF c) Calculate the mass of copper deposited: q = I * t = (30 C/s) * (30 s) = 900 C

Solving a Faraday’s Law Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current? I = 20.0 C/s Ag+ + e-  Ag I = q/t 5.0 g 1 mol Ag 1 mol e- 96 485 C 1 s 1 mol e- 20.0 C 107.87 g 1 mol Ag = 2.2 x 102 s

I = q/t Problem 1 on HW I = current (amperes or coulombs/seconds, C/s) q = charge (coulombs, C) = nxF t = time (seconds, s)