Chapter 16II. Nerstian EquationChapter 16Elements of ElectrochemistryClass Notes:ΔG = -nFEcell = -RTlnKΔG = ΔH - TΔSG = Gibbs change in free energy; a thermodynamic quantityE = internal energy of cellF = Faraday’s constantn = moles of electronsR = 8.31 constantT = temperature in Kelvinln = natural log of K or 2.303 log KK = equilibrium coefficientH = enthaplyS = entropyI. Redox Reactions- reduction – addition of electrons to an atom, as occurs during the addition of hydrogen to a molecule or the removal of oxygen from it- oxidation – loss of electron density from an atom as occurs upon addition of oxygen to a molecule or upon removal of hydrogenExample:Cu2+ + 2e- ↔ Cu E˚ = +0.335Zn2+ + 2e- ↔ Zn E˚ = -0.7990.335 - (-0.799) =I.134 VEcell = 1.134 VoltsFor an exothermic spontaneous reaction:ΔG < 0E > 0K >> 1Reduction  cathode – more positive EOxidation  anode – more negative EΔG = -nFEΔG = -(2)(96500)(1.134V)Δ= -214 kJReaction is spontaneous exothermic, produces energyII. Nerstian EquationE = E˚ - RT ln [C] c [D] d nF [A]a [B]bE˚ = 0.059 log KeqNText Notes:I. Redox Reactions- reaction in which electrons are transferred from one reactant to anotherCe4+ + Fe2+ ↔ Ce3+ + Fe3+Ce is the oxidizing agent or oxidant because it accepts electrons from iron and Fe is the reductant because it donated electrons to Ce.A redox reaction can be split into 2 half reactionsCe4+ + e- ↔ Ce3+Fe2+ ↔ Fe3+ + e-The 2 half reactions must be balanced just like any other reaction. The number of atoms of each element and the net charge on each side of the equation must be in balance so for the oxidation of Fe2+ by MnO4- the half reactions would be:MnO4- + 8H+ + 5e- ↔ Mn2+ + 4H2O5Fe2+ ↔ 5Fe3+ + 5e-The net charge for the first half reaction is (-1 – 5 + 8) = +2 which is the same on the right side. For the second half reaction it must by multiplied by 5 so that the number of electrons lost by Fe equals the number gained by MnO4. The balancednet equation for the reaction would be:MnO4- + 5Fe2+ + 8H+ ↔ Mn2+ + 5Fe3+ + 4H2OProblem 16-7a) 2Fe3+ + Sn2+  2Fe2+ + Sn4+b) Cr + 3Ag+  Cr3+ + 3Agc) 2NO3- + Cu + 4H+  2NO2 + 2H2O + Cu2+d) 2MnO4- + 5H2SO3  2Mn2+ + 5SO42- + 4H+ + 3H20e) Ti3+ + Fe(CN)63- + H2O  TiO2+ + Fe(CN)64- + 2H+f) H2O2 + 2Ce4+  O2 + 2Ce3+ + 2H+g) 2Ag + 2I- + Sn4+  2AgI + Sn2+h) UO22+ + Zn + 4H+  U4+ + Zn2+ + 2H2Oi) 5HNO2 + 2MnO4- + H+  5NO3- + 2Mn2+ + 3H2Oj) H2NNH2 + IO3- +2H+ + 2Cl  N2 + ICl2- + 3H2OProblem 16-8Oxidizing Agent Reducing Agenta) Fe3+Sn2+b) Ag+Crc) NO3- Cud) MnO4- H2SO3e) Fe(CN)63-Ti3+f) Ce4+H2O2g) Sn4+Agh) UO22+Zni) MnO4- HNO2j) IO3- H2NNH2- redox reaction may be compared to an acid/base reaction by the equation:Acid1 + base2 ↔ base1 + acid2Ared +Box ↔ Aox + BredThe oxidized B will accept electrons from reductant A and A having given up e-will become oxidizing agent A.II. Redox Reactions in Electrochemical CellsA. Types1. reaction is performed by direct contact between the oxidant and the reductant in a suitable container.2. The reactants do not come in contact with one another, salt bridge employed.- a salt bridge prevents the mixing of the contents of the 2 electrolyte solutions making up electrochemical cellsProblem 16-4The significance between the 2 standard E potentials is that the first standardpotential is for a solution that is saturated with I2, which has an I2(aq) activity muchless than 1. The second potential is for a hypothetical half cell in which the I2(aq) isunity, so it would have a greater potential since the driving force for the reductionwould be greater at the higher I2 concentration.B. Electrochemical Cell-an array consisting of 2 or 3 electrodes, each of which is in contact with an electrolyte solution. Typically, the electrolytes are in electrical contact through a salt bridge. An external metal conductor connects the electrodes.1. A cathode is an electrode where reduction occurs2. An anode is the electrode where oxidation occurs- Electrochemical Cells are either galvanic or electrolytic1. galvanic cells store and supply electrical energy (ex. Batteries)2. electrolytic cells require an external source of electrical energy for operation- Reversible versus Irreversible Cells1. in a reversible cell, reversing the current reverses the cell reaction2. in an irreversible cell, reversing the current causes a different half-reaction o occur at one or both of the electrodesProblem 16-6The potential in the presence of base would be more negative because thenickel ion activity in the basic solution would be much less than 1 molar. Sothe driving force for the reduction would be less and the electrode potentialwould be much more negative.- In a cell, electricity is carried by movement of anions toward the anode and cations toward the cathodeIII. Electrode PotentialsA. the potential difference that develops between the electrodes of the cell is a measure of the tendency for the reaction to proceed from a nonequilibrium state to the condition of equilibrium.Gibb’s Free EnergyΔG = -nFEcellIf the reactants and products are in their standard states the resulting cell potentialis called the standard cell potential and is related to Gibb’s Free Energy Equation:ΔG˚ = -nFE˚ = -RT ln KeqCell half potentials are found by:Ecell = Eright - EleftIV. Defining Electrode Potentials and Standard Electrode PotentialsA. An electrode potential is the potential of a cell consisting of the electrode in question acting as the right hand electrode and the standard hydrogen electrode acting as the left hand electrode.B. The standard electrode potential of a half reaction is defined as its electrode potential when the activities of the reactants and products are all unityV. The Nerst EquationE = E˚ - (0.0592 / n) log [C] c [D] d[A]a[B]bThe Nerst Equation relates the potential to the concentrations (activity) of the participantsin an electrochemical half-cell.- A formal potential is the electrode potential when the ratio of analytical concentrations of reactants and products of a half reaction are exactly 1.00 andthe molar concentration of any other solutes are specified.Problem 16-13E = E˚ - (0.0592 / n) log [C] c [D] d [A]a[B]ba) E = 0.337 – 0.0592/2 log 1/0.0440 = 0.297 Vb) Ksp CuCl = 1.9x10-7 = [Cu+][Cl-]E = 0.521 – 0.0592/1 log 1/[Cu+] = 0.521 – 0.0591/1 log [Cl-] / KspE = 0.521 – 0.0592/1 log 0.075/1.9x10-7 = 0.521 – 0.0592/1 log 3.95x105E = 0.521 – 0.331 = 0.190 Vc)