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Suppose we had a 0.3A current producing 15mL of hydrogen gas in 6 min and 30 seconds.

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?

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Suppose we had a 0.3A current producing 15mL of hydrogen gas in 6 min and 30 seconds.

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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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