Instrumental Analysis Fundamentals of Electrochemistry Tutorial 4

Electricity from the Ocean Floor Oceanographic instruments can be powered by making a battery between the water and sediment layers. Bacteria oxidizes organic matter using SO42- ions available in the sediment layer. As a result, HS is released and acts as one of the reactant in the electrochemical cell. The other reactant is dissolved O2 molecules available at the sediment-water interface.

Summary of Symbols, Units, and Relationships F is the Faraday’s constant (96500 C/mol). z is the total no. of moles of electrons passes in the circuit q = z . F Charge moles C/mol (coulombs, C) of e Relation between charge and moles E is the electric potential difference, E is the work (J) needed to move a coulomb of positive charge from one point to the other. Work = E . q Joules, J Volts, V coulombs Relation between work and voltage n is the moles of charge moved under a potential difference, E. ** G = ─ W = ─ n F E Relation between free energy and potential difference I is the electric current, measured in ampere (A). It is the coulombs per second moving pas a point in the circuit. I = E / R Current Volts resistance A V Ohm,  Ohm’s Law P is the work per unit time done by electricity moving through a circuit. P = work/s = E . I Power J/s V A (Watt, W) Electric Power J/mol ** the minus sign in the equation indicates that the free energy of a system decreases when the work is done on the surrounding

n is the number of moles of e- per 1 mole of reactant and rate is the rate of electrochemical reaction in mol/s I = n . F . Rate A C/mol mol/s (C/s) Relation between Current and rate of chemical reaction E is the reduction electrode potential and E is the standard reduction electrode potential when the activities of all reactants and products are unity. n is the number of electrons in the half-cell reaction. a is the activity of reactant or product of the electrochemical reaction of the half-cell when written as reduction Note: Nernst Equation can be applied to one half cell as well as to the reaction of the whole cell. Nernst equation Relation between potential of the electrode and activity of ions a is the activity of a chemical species,  is the activity coefficient, c is the concentration of chemical species Relation between activity and concentration

Example (Charge) If 3.2 g of O2 were reduced in the overall reaction with HS: how many coulombs have been transferred from HS to O2 or how many charges pass in the circuit? Solution

Example (Current) If electrons are forced into a wire which acts as anode to oxidize HS at a rate of 4.24 mmol/hr, how much current passes through? Note: The electrode at which oxidation occurs is the anode while the electrode at which reduction occurs is the cathode Solution n Generally, the current may be related to the rate of electrochemical reaction by: Rate of consumption of reactants in a unit of mol/s n is the number of moles of e- per 1 mole of reactant

Example (Work) How much work can be done if 2.4 mmol of electrons go through a potential difference of 0.70 V in the ocean-floor battery? Solution The greater the difference in potential (V), the stronger the e will be pushed around the circuit 12V battery “pushes” electrons through a circuit eight times harder than a 1.5V battery

Example (E from free energy change) Calculate the voltage that will be measured by the potentiometer in the figure, Knowing that the free energy change for the net reaction is 150 kJ/mol of Cd. Solution A spontaneous chemical reaction of negative G produces a positive voltage.

Example (Ohm’s Law) In the following circuit, a battery generates a potential difference of 3 V and the resistor has a 100  resistance. How much current and power are delivered by the battery? Solution

Nernst Equation The Nernst Equation relates the potential of the half cell to the activities of the chemical species (their concentrations) For the reaction (written as reduction) We usually calculate half-reactions at 25º C, substituting that in with the gas constant and to base 10 log gives: n is the no. of electrons in either the electrode or cell reaction Where a is the activities of species A and B.

Example (Nernst Equation) Write the Nernst equation for the reduction of O2 to water: Note that multiplying the reaction by any factor does not affect either Eº or the calculated E: Note that: asolid =1, agas = pressure aH2O =1, aion = molarity

Example (Ecell) Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if the right cell contained 0.50 M AgNO3(aq) and the left contained 0.010 M Cd(NO3)2(aq). For Ag/Ag+ electrode For Cd/Cd2+ electrode For the Cell

Note that you will get the same value of cell potential if you apply the Nernst equation directly to the overall cell reaction. Since Ecell is found to be +ve quantity, the reaction as written is spontaneous

3- Ecell = ECu/Cu2+  ESCE = 0.309 V  0.241 V = 0.068 V Example (Ecell) A cell was prepared by dipping a Cu wire and a saturated calomel electrode (ESCE = 0.241 V) into 0.10 M CuSO4 solution (ECu/Cu2+ = 0.339 V). The Cu wire was attached to the positive terminal and the calomel to the negative terminal of the potentiometer. 1- Write the half-cell reaction of Cu electrode. 2- Write the Nernst equation for the Cu electrode. 3- Calculate the cell voltage. 4- What would happen if the [Cu2+] were 4.864x10-4 M? Electrode connected to the positive terminal of the potentiometer is the cathode and the other is the anode solution 1- Cu2+(aq) + 2e-  Cu(s) 2- 3- Ecell = ECu/Cu2+  ESCE = 0.309 V  0.241 V = 0.068 V

Example (Ecell) A 100.0 ml solution containing 0.100 M NaCl was titrated with 0.100 M AgNO3 and the voltage of the cell shown in the figure was monitored. Calculate the voltage after the addition of 65.0 ml of AgNO3. (EAg/Ag+ = 0.799 V, ESCE = 0.241 V) Explain why a silver electrode can be used as an indicator electrode for both Ag+ and for halides. Solution: a) The titration reaction is: Ag+(aq) + Cl(aq)  AgCl(s). Equivalence point is reached upon addition of 100.0 ml titrant (in this case). After addition of 65.0 ml of AgNO3, we can say: (M V)reacted Cl- = (M V)Ag+ added (note that Ag+ and Cl react in the ratio of 1:1) (M V)Cl- remains unreacted = (M V)total Cl-  (M V)Ag+ = (0.1 x 100)  (0.1 x 65) (M V)Cl- remains unreacted = 3.5 mmole MCl- = 3.5 mmol/165 = 0.0212 mol/L NaCl

To find the cell voltage we need to know Ag+: where Ksp is the solubility product of insoluble AgCl so  Ag+  = Ksp / [Cl] [Ag+] depends on the concentration of Cl ion = 1.8 x 10-10 M / 0.0212 M = 8.5 x 10-9 M The potential of Ag electrode is given by: Ag electrode reaction: Ag+ + e- Ag+ EAg/Ag+ = 0.799 + 0.05916 log (8.5  10-9) = 0.322 V The cell potential: Ecell = ESCE  EAg/Ag+ = 0.241  0.322 =  0.081 V b) We can see from the example that the silver electrode can act indirectly as halide indicator electrode if solid insoluble silver halide is present. The solubility of silver halide will be affected by whatever halide ion is present and in turn the concentration of Ag+ ion as well as the potential of the Ag electrode.

Zn(s) + HgO(s) ZnO(s) + Hg( ) E = 1.35 V Exercise 1 A mercury cell used to power heart pacemaker runs on the following reaction: Zn(s) + HgO(s) ZnO(s) + Hg( ) E = 1.35 V If the power required to operate the pace maker is 0.010 W, how many kilogram of HgO (molar mass = 216.59) will be consumed in 365 days? Exercise 2 Calculate the voltage of each of the following cells: (a) Fe(s) / FeBr2(0.010 M) // NaBr(0.050 M) / Br2( ) / Pt(s) (b) Hg( ) / Hg2Cl2(s) / KCl(0.060 M) // KCl(0.040 M) / Cl2(g, 0.50 bar) / Pt(s) (EFe/Fe2+ = 0.44 V, EBr2/Br = 1.078 V, EHg/Hg2Cl2 = 0.268 V, ECl2/Cl = 1.360 V) Try to solve problems 14-3, 14-4, 14-5, 14-7, 14-12, 14-14, 14-17, 14-18, and 14-19 Harris text book, p308-310 If you are unable to solve these problems or need to revise your answer, please refer to the “Solution Manual for Quantitative Chemical Analysis”, by D.C. Harris (GUC library)