Presentation on theme: "1 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."— Presentation transcript:
1 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. Lets look at the hydrolysis of water…
2 Dont Hesitate… Lets Calculate!
3 How close are we to the correct answer? Quantitative Aspects of Electrochemistry = 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 Solution (a)Charge = 1350 C (b)Calculate moles of e- used 1.50 amps flow thru a Ag + (aq) solution for 15.0 min. What mass of Ag metal is deposited? (c)Calc. quantity of Ag
7 Your turn to try! The anode reaction in a lead storage battery is Pb(s) + HSO 4 - (aq) ---> PbSO 4 (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 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) + HSO 4 - (aq) ---> PbSO 4 (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 About 78 hours d)Calculate time
9 Michael Faraday Originated the terms anode, cathode, anion, cation, electrode. Discoverer of electrolysiselectrolysis magnetic props. of mattermagnetic props. of matter electromagnetic inductionelectromagnetic induction benzene and other organic chemicalsbenzene and other organic chemicals Was a popular lecturer…. Like your teacher!!!! I would love to hear an oxidation Haiku! Where did this Faraday stuff come from anyways
10 Oxidation Haiku! Lost an electron But now feeling positive Oxidized is cool! Got any more Haikus?
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 YES! E o is related to G o, the free energy change for the reaction. G o = - n F E o where F = Faraday constant = 96,500 J/Vmol (C/mol) and n is the number of moles of electrons transferred Michael Faraday I know you are craving to know the answer to one question… are E o and G o related??? Hey! What Gibbs? Why am I stuck in the middle?
13 E o and G o G o = - n F E o For a product-favored reaction Reactants ----> Products Reactants ----> Products G o 0 E o is positive For a reactant-favored reaction Reactants <---- Products Reactants <---- Products G o > 0 and so E o 0 and so E o < 0 E o is negative Hey Kids! Have a Great Day.. Not just a Faraday!
14 E at Nonstandard Conditions The NERNST EQUATIONThe NERNST EQUATION E = potential under nonstandard conditionsE = potential under nonstandard conditions n = no. of electrons exchangedn = no. of electrons exchanged ln = natural logln = natural log If [P] and [R] = 1 mol/L, then E = E˚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˚ If [R] < [P], then E is ______________ than E˚If [R] < [P], then E is ______________ than E˚ more positive less positive At 25 o C 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.
15 Other forms of the Nernst Equation At 25 o C Nernst answered, "No, I can afford to buy them now, so I don't need to make them".
16 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 = E o – ( V/n) log K log K = nE o / (at 25C) pg 980 example E and the Equilibrium Constant I am getting bored, make this your last example!
17 In the following reaction…. Fe (s) + Cd 2+ (aq) Fe 2+ (aq) + Cd (s) E o net = 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)/ = 1.35 K = b.) K = 22.45= [Fe 2+ ]/ [Cd 2+ ] = (1.0 +X)/(1.0 –X) X= Therefore Fe 2+ = 1.9 M and Cd 2+ = 0.10 M a.) log K = (2.00) (0.04 V)/ = 1.35 K = b.) K = 22.45= [Fe 2+ ]/ [Cd 2+ ] = (1.0 +X)/(1.0 –X) X= Therefore Fe 2+ = 1.9 M and Cd 2+ = 0.10 M Wow! What an exciting way to end your lectures for the year!!!