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Discovering the Quantitative Aspects of Electrochemistry

Dive into the world of electrochemistry, exploring concepts like Faraday's constant, charge calculations, and electrode reactions. Learn about reaction kinetics, equilibrium constants, and the Nernst equation, all essential in understanding electrochemical processes.

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Discovering the Quantitative Aspects of Electrochemistry

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  1. Lets look at the hydrolysis of water… • Suppose we had a 0.3A current producing 15mL of hydrogen gas in 6 min and 30 seconds. • How much charge did we have per one mole of electrons? • We need to find the temp, atmospheric pressure and calculate moles… But since the gas was collected over water… What else do we need? • Lets turn to A-24 for the chart.

  2. Don’t Hesitate… Let’s Calculate!

  3. Quantitative Aspects of Electrochemistry How close are we to the “correct answer?” = 96,500 C/mol e- = 1 Faraday

  4. Well that is swell, but what is the use of this? Consider electrolysis of aqueous silver ion. Ag+ (aq) + e- ---> Ag(s) 1 mol e- ---> 1 mol Ag If we could measure the moles of e-, we could know the quantity of Ag formed. But how to measure moles of e-?

  5. Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Calc. charge Charge (C) = current (A) x time (t) = (1.5 amps)(15.0 min)(60 s/min) = 1350 C

  6. Quantitative Aspects of Electrochemistry 1.50 amps flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution (a) Charge = 1350 C (b) Calculate moles of e- used (c) Calc. quantity of Ag

  7. Your turn to try! The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Calculate moles of e- c) Calculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C

  8. Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last? Solution a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol c) Charge = 423,000 C d) Calculate time About 78 hours

  9. Where did this Faraday stuff come from anyways Michael Faraday1791-1867 Originated the terms anode, cathode, anion, cation, electrode. Discoverer of • electrolysis • magnetic props. of matter • electromagnetic induction • benzene and other organic chemicals Was a popular lecturer…. Like your teacher!!!! I would love to hear an oxidation Haiku!

  10. Oxidation Haiku! • Lost an electron • But now feeling positive • Oxidized is cool! Got any more Haiku’s?

  11. Reduction Haiku!!! • Gained some electrons • Gave me a negative mood! • Now I can say Ger! I really got a coulomb out of those!

  12. I know you are craving to know the answer to one question… are Eo and ∆Go related??? YES! Eo is related to ∆Go, the free energy change for the reaction. ∆Go = - n F Eo where F = Faraday constant = 96,500 J/V•mol (C/mol) and n is the number of moles of electrons transferred Hey! What Gibbs? Why am I stuck in the middle? Michael Faraday 1791-1867

  13. Eo and ∆Go Hey Kids! Have a Great Day.. Not just a Faraday! ∆Go = - n F Eo For a product-favored reaction Reactants ----> Products ∆Go < 0 and so Eo > 0 Eo is positive For a reactant-favored reaction Reactants <---- Products ∆Go > 0 and so Eo < 0 Eo is negative

  14. Walther Nernst, the famous German physical chemist, developed an electric lamp, known as the "Nernst lamp", which he sold for a very large sum of money. A colleague of his, not without spite asked him whether his next project will be making diamonds. At 25oC E at Nonstandard Conditions • The NERNST EQUATION • E = potential under nonstandard conditions • n = no. of electrons exchanged • ln = “natural log” • If [P] and [R] = 1 mol/L, then E = E˚ • If [R] > [P], then E is ______________ than E˚ • If [R] < [P], then E is ______________ than E˚ more positive less positive

  15. Nernst answered, "No, I can afford to buy them now, so I don't need to make them". Other forms of the Nernst Equation At 25oC

  16. E and the Equilibrium Constant • The cell potential changes as the concentrations change. As reactants are converted to products, the value of Enet must decline from initially positive value to zero • A potential of zero means that no net reaction occurs. The cell is at equilibrium. • Therefore: • E= 0 = Eo – (0.0592 V/n) log K • log K = nEo/0.0592 (at 25C) • pg 980 example 20.10 I am getting bored, make this your last example!

  17. In the following reaction…. Fe(s) + Cd2+(aq) Fe2+(aq) + Cd (s) Eonet = +0.04 V a.) What is the value of the equilibrium constant? b.) What are the equilibrium concentrations of Fe2+ and Cd2+ ions if each began with a concentration of 1.0M? a.) log K = (2.00) (0.04 V)/ 0.0592 = 1.35 K = 22.45 b.) K = 22.45= [Fe2+]/ [Cd2+] = (1.0 +X)/(1.0 –X) X= 0.9147 Therefore Fe2+= 1.9 M and Cd2+ = 0.10 M Wow! What an exciting way to end your lectures for the year!!!

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