Electrochemistry Chapter 7

Electrical Conductance Conductors: Metallic Conductors/Electronic conductors: Which allow electricity to pass through it by movements of electrons. Electrolytic conductors/ electrolyte: Which allow electricity to pass through it by movements of ions.

Types of Electrolytes Based on the extent of dissociation, electrolytes are classified into two types: Strong electrolytes: The elctrolytes which completely dissociate into ions in aqueous solution. For example, HCl, HNO3, NaOH, KOH, NaCl and KCl. Weak electrolytes: The elctrolytes which do not undergoes complete dissociation into its ions in aqueous solution. For example, CH3COOH, HCOOH and NH4OH.

Mechanism involved in typical electrolytic conduction. Conductivity of Solutions of Electrolytes Mechanism involved in typical electrolytic conduction. Anode Cathode Engineering Chemistry Copyright  2011 Wiley India Pvt. Ltd. All rights reserved.

Resistance: Obstruction to flow of current Here, ρ = specific resistance. It is the resistance offered by conductor having 1cm length and 1cm2 area of cross section.

Conductance (C): Ease of flow of current Here Here k is specific conductance. It is a conductance of conductor having length 1cm and area of cross section 1cm2. Specific conductance = conductance × cell constant

Measurement of specific conductance Specific conductance = conductance × cell constant Cell constant of the conductivity cell is determined by measuring the resistance of solution (using conductivity meter) of an electrolyte of known specific conductance at given temperature. Generally a solution of KCl at 298K is used for which specific conductance values are reported at different concentration and temperature in literature. Once the cell constant of conductivity cell is determined, that cell can be used for determination of specific conductance of any solution by measuring resistance.

Equivalent conductance: It is the conductance given by all the ions produced by 1 gm equivalent of electrolyte solution. Here k ohm-1cm-1, C= geq , Λ = ohm cm2(geq)-1. Molar conductance: It is the conductance given by all the ions produced by 1 mole of electrolyte solution. Here k ohm-1cm-1, C= mol , Λ = ohm cm2mol-1.

Oxidation half-reaction (lose e-) Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2- 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-)

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1.

HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3- Oxidation numbers of all the atoms in HCO3- ? O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4

Reactions at electrodes Left: Galvanic cell. Electrons are deposited on the anode (so it is neg) and collected from the cathode (so it is positive) Right: Electro-lytic cell. Elec-trons are forced out of the anode (positive) and into the cathode (negative) Electrochemical/ Galvanic cell Electrolytic cell

Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction E0 = Ecathode - Eanode cell = 0.34 – (-0.76) = 1.10 V

Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode cathode

Origin of electrode potential (reduction) Origin of electrode potential (oxidation) Engineering Chemistry Copyright  2011 Wiley India Pvt. Ltd. All rights reserved.

Measurement of EMF of the Cell EMF and Potential Difference Engineering Chemistry Copyright  2011 Wiley India Pvt. Ltd. All rights reserved.

Standard Reduction Potentials Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm and temperature 298K. Reduction Reaction 2e- + 2H+ (aq) H2 (g) Oxidation Reaction H2 (g) 2H+ (aq) +2e- E0 = 0 V Standard hydrogen electrode (SHE)

Standard Reduction Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (g, 1 atm), Pt Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e- Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

Standard Reduction Potentials E0 = 0.76 V cell Standard emf (E0 ) cell E0 = Ecathode - Eanode cell Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (g, 1 atm), Pt E0 = EH /H - EZn /Zn cell + 2+ 2 0.76 V = 0 - EZn /Zn 2+ EZn /Zn = -0.76 V 2+ Zn2+ (1 M) + 2e- Zn E0 = -0.76 V

Standard Reduction Potentials E0 = 0.34 V cell E0 = Ecathode - Eanode cell Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 2+ ECu /Cu = 0.34 V 2+ Pt , H2 (g, 1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): H2 (1 atm) 2H+ (1 M) + 2e- Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

E0 is for the reaction as written The more positive E0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

Cd is the stronger oxidizer What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V Cd is the stronger oxidizer Cd will oxidize Cr Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V Anode (oxidation): Cr (s) Cr3+ (1 M) + 3e- x 2 Cathode (reduction): 2e- + Cd2+ (1 M) Cd (s) x 3 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M) E0 = -0.40 – (-0.74) cell E0 = 0.34 V cell

Spontaneity of Redox Reactions DG = -nFEcell n = number of moles of electrons in reaction F = 96,500 J V • mol DG0 = -nFEcell = 96,500 C/mol DG0 = -RT ln K = -nFEcell Ecell = RT nF ln K (8.314 J/K•mol)(298 K) n (96,500 C/mol) ln K = = 0.0257 V n ln K Ecell = 0.0592 V n log K Ecell

The Effect of Concentration on Cell Emf DG = DG0 + RT ln Q DG = -nFEcell DG0 = -nFEcell -nFE = -nFE0 + RT ln Q Nernst equation E = E0 - ln Q RT nF Q = reaction quotient R= 8.314JK-1Mol-1 F=96500 C/mol At 298 K

At equilibrium: Ecell =0 K= equilibrium constant At 298 K = 0.0257 n ln K Ecell

Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq) Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq) E0Fe2+/Fe = - 0.44 V; E0cd2+/cd = -0.40V Oxidation: Cd Cd2+ + 2e- n = 2 Reduction: 2e- + Fe2+ 2Fe E0 = EFe /Fe – ECd /Cd 2+ E0 = -0.44 – (-0.40) E0 = -0.04 V - 0.0591 2 log -0.04 E = 0.010 0.60 E = 0.0125 E > 0 Spontaneous

What is the emf of a cell consisting of a Fe/Fe2+ half cell and a Pt/H2H+ half cell if [Fe2+] = 0.12M, [H+] = 0.250 and PH2 = 1.2 atm? Anode : Fe (s) → Fe2+ (0.12M) + 2e- Cathode : 2H+ (0.250M) + 2e- → H2 (g) (1.2 atm) Fe (s) + 2H+ (0.250) → Fe2+ ( 0.12M) + H2 (g) (1.2 atm) = 0 – (-0.44) = 0.44 V Ecell = 0.429 V

What is the equilibrium constant for the following reaction at 250C What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) = 0.0591 V n log K Ecell E0Fe2+/Fe = - 0.44 V E0Ag+/Ag = 0.80V Oxidation: 2Ag 2Ag+ + 2e- n = 2 Reduction: 2e- + Fe2+ Fe E0 = EFe /Fe – EAg /Ag 2+ + = 0.0591 n log K Ecell E0 = -0.44 – (0.80) E0 = -1.24 V 0.0591 x n E0 cell K = antilog10 0.0591 x 2 -1.24 K = antilog10 K = 1.09 x 10-42

Concentration Cells A concentration cell is a galvanic cell which generates electrical energy at the expense of chemical energy. A concentration cell: When the circuit is completed, reactions occur that tend to make the concentrations of Cu2+ the same in the two half-cells. Oxidation occurs in half-cell having lower concentration of electrolyte, and reduction occurs in half-cell having higher concentration of electrolyte. Engineering Chemistry Copyright  2011 Wiley India Pvt. Ltd. All rights reserved.

Practice Problems An iron wire is immersed in a solution containing 1M ZnSO4 and 1M NiSO4. Predict which of the following reactions is likely to proceed? (i) Iron reduces Zn2+ ions (ii) Iron reduces Ni2+ ions 2. Can a solution of 1M AgNO3 be stored in lead container? 3. What is the emf of a cell consisting of a Pb/Pb2+ half cell and a Pt/H2H+ half cell if [Pb2+] = 0.10M, [H+] = 0.050 and PH2 = 1.0 atm?