Faraday's Law

Faraday’s Law : Faraday’s Law

Electrochemistry Review : Electrochemistry Review A coulomb (C) is the quantity of electrical charge passing any point in the circuit in 1 second when the current is 1 ampere. 1 C = 1 Ampere * 1 Second A Faraday (F) is the quantity of electricity that is capable of depositing or liberating 1 gram equivalent weight of a substance in electrolysis, approximately 9.6494 × 104 coulombs. (1 Faraday = 96494 C or 96500 approximately)

Electrochemistry Review : Electrochemistry Review Al3+ + 3 e- ? Al(s) 27g Al needs 3 mol of e-: 3F Na+ + e- ? Na(s) 23g Na needs 1 mole of e- : 1 F Ba2+ + 2 e- ? Ba(s) 137.3g Ba needs 2 mol of e- : 2F

Slide 4 : To solve stoichiometry problems: current & time --> quantity of C --> moles of e- --> moles of substance --> grams

Critical Thinking Questions : Critical Thinking Questions What is the “reduction” equation for Cu2+(aq) becoming Cu(s)? How many mole e- are required for every mol of Cu produced?

Critical Thinking Questions : Critical Thinking Questions What is the “reduction” equation for Cu2+(aq) becoming Cu(s)? How many mol e- are required for every mol of Cu produced? Cu2+ + 2 e- ----> Cu(s) 2 mol e- / mol Cu

Critical Thinking Questions : Critical Thinking Questions How many moles of electrons were required to produce 100.0 g of Ba(s)? How many coulombs were required to produce 100.0 g of Ba(s)?

Critical Thinking Questions : Critical Thinking Questions How many moles of electrons were required to produce 100.0 g of Ba? 100 g * 2 mol e- / 137.2 g = 1.46 mol e- How many coulombs were required to produce 100.0 g of Ba? 1.46 mol e- * 96500 C/ mol e- = 1.41 x 105 C

Critical Thinking Questions : Critical Thinking Questions How many grams of copper plated out when a current of 10 amperes is passed for 30.0 minutes through a solution containing Cu2+?

Critical Thinking Questions : Critical Thinking Questions How many grams of copper plated out when a current of 10 amperes is passed for 30.0 minutes through a solution containing Cu2+? #C = 10 A * 30 min * 60 sec/min = 1.80 x104 C Cu2+ + 2 e- ----> Cu(s) 2 mol e- / mol Cu = 1.80 x104 C * mol e- / 96500 C * 1 mol Cu / 2 mol e- * 63.55 g / mol = 5.93 grams Cu

Critical Thinking Questions : Critical Thinking Questions How long must a current of 5.00 ampere be applied to a solution of Ag+ to produce 10.5 grams of silver metal (Ag = 107.9 g/mol)

Critical Thinking Questions : Critical Thinking Questions How long must a current of 5.00 ampere be applied to a solution of Ag+ to produce 10.5 grams of silver metal (Ag = 107.9 g/mol) Ag+ + e- ---> Ag(s) = 10.5 g Ag * 1 mol e- / 107.87 g * 96500 C / mol e- = 9.393 x 103 C Now = 9.393 x 103 C / 5.00 A = 1.879 x 103 sec or 31.3 minutes