12/1/2014 CH1101 Tutorial 4 2014 Basic Electrochemistry Professor Mike Lyons melyons@tcd.ie 1 12/1/2014 2 12/1/2014 Annual Examination 2010 3 12/1/2014 4 12/1/2014 5 12/1/2014 Use the following standard reduction potentials to calculate the equilibrium Constant for the formation of the Zn(NH3)42+ ion. Zn(NH3)42+(aq) + 2e- Zn(s) + 4NH3(aq) E0 = -1.04 V Zn2+ (aq) + 2e- Zn(s) E0 = - 0.76 V. The reaction of interest is: Zn2+(aq) + 4NH3(aq) Zn(NH3)42+(aq). This can be obtained by reversing the first half reaction and adding it to the second. Zn(s) + 4NH3(aq) + Zn2+(aq) + 2e- Zn(s) + Zn(NH3)42+(aq) + 2eEcell0 = 1.04 V – 0.76 V = + 0.28 V Now we recall K = exp[-G0/RT] = exp[nFEcell0/RT], lnK = nFE0cell/RT lnK = (2 x 96485 Cmol-1x0.28 V)/(8.314 Jmol-1K-1x298K) = 21.8 K = e21.8 = 2.9x109. K is large so reaction is very product favoured. 6 12/1/2014 Faraday’s Laws of Electrolysis. The quantity (moles) of product formed by the passage of an electric current is stoichiometrically equivalent to the amount (moles) of electrons supplied. The amount of product formed during an electrolysis process is calculated from the stoichiometry of the reaction, the magnitude of the current flowing, and the time during which the current flows. • • Michael Faraday : 1791-1867. We will focus a lot on metal plating or metal electrodeposition reactions. Coulombs C Mn+ + ne- M Amperes A Seconds s Charge = current flowing x time taken Charge passed by 1 mole electrons = 1F = 96, 500 C. n mol e- = 1 mol M 13 Interfacial electron transfer at electrode/solution interfaces: oxidation and reduction processes. Electron sink electrode (Anode). Electron source electrode (Cathode). P ne- Q Oxidation or de-electronation. P = reductant (electron donor) Q = Product A ne- B Reduction or electronation. A = oxidant (electron acceptor) B = Product • The greater the applied voltage, the larger the resulting current flow, and the greater the rate of the chemical reaction. • The rate at which charge is moved across the M/S interface = the rate at which chemistry is accomplished at the M/S interface. # electrons transferred Current (A) Electrode area (cm2) dq • In electrolysis we use an applied voltage i dt to perform chemistry at a M/S interface. • The applied voltage drives the chemical reaction which does not occur spontaneously. Charge (C) • The current flowing across the M/S interface is a measure of the rate of Time (s) the chemical transformation at the interface. nFA dN nFAf dt Amount of Faraday Material (mol) Constant (Cmol-1) Reaction flux (rate) mol cm-2s-1 14 7 12/1/2014 8 12/1/2014 Electrolysis of a molten chromium salt for 1.5 hr with a constant Current of 5.0 A resulted in the deposition of 4.835 g of chromium Metal. Determine the charge on the chromium ion in the salt. Our strategy is as follows. From the mass of Cr deposited, the amount in mole of Cr deposited can be evaluated. The current passed and deposition time can be used to evaluate the total quantity of charge passed during electrolysis. The valency z of the Crz+ ion can be evaluated by relating amount of Cr metal formed to amount of electrons used. Amount (mol) Cr: nCr = mCr/Acr = 4.835 g/51.996 gmol-1 = 0.093 mol Amount charge passed Q = Ixt = 5.0 Cs-1 x 1.5 hr x 60 minhr-1 x 60smin-1 Q = 27,000 C. 1 mol electrons equivalent to charge of 96, 485 C (1 Faraday F) Hence, 27,000 C equivalent to 27,000 C/96485 Cmol-1 = 0.28 mol Hence 0.28 mol electrons produces 0.093 mol Cr metal Hence we conclude that 1 mol Cr required 0.28 mol/0.093 mol = 3 electrons Crz+ + ze- Cr, 1 mol Cr = z mol eHence z = 3 And we write: Cr3+ + 3e- Cr A summary diagram for the stoichiometry of electrolysis MASS (g) of substance oxidized or reduced M(g/mol) AMOUNT (MOL) of substance oxidized or reduced AMOUNT (MOL) of electrons transferred balanced half‐ reaction Faraday constant (C/mol e‐) CHARGE (C) time(s) CURRENT (A) 18 9