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Chemistry 1011 Chemistry 1011 Introductory Chemistry II TOPIC http://www.mi.mun.ca/~pfisher/chemistry.html Electrochemistry Password for final exams Midgley Chemistry 1011 Slot 5 TEXT REFERENCE Masterton and Hurley Chapter 18 1 18.2 Standard Voltages 2 Cell Voltage YOU ARE EXPECTED TO BE ABLE TO: • The force that pushes the electrons through the external circuit of a cell is known as the • Define the standard electrode potential of a half cell • Order species according to their ease of oxidation or reduction based on a table of standard reduction potentials • Calculate the net cell voltage, Eo, of a combination of half cells from standard electrode potential data • Determine whether a given redox reaction will be spontaneous or non-spontaneous Chemistry 1011 Slot 5 Chemistry 1011 Slot 5 3 – Potential difference, or – Electromotive force (emf), or – Voltage • It is measured in volts • The magnitude of the voltage depends on – The nature of the redox reaction – The concentrations of the ions in solution, (or pressures of any gases) Chemistry 1011 Slot 5 4 The Zinc – Hydrogen Voltaic Cell Standard Voltage • In order to compare the voltages of different cells, or to calculate the expected voltage of a given cell, measurements are taken under standard conditions: – Current flow is almost zero – All ions and molecules in solution are at a concentration of 1.0 mol/L – All gases are at a pressure of 1.0 atm Chemistry 1011 Slot 5 5 Chemistry 1011 Slot 5 6 1 The Standard Voltage of the Zinc – Hydrogen Voltaic Cell Zn(s) + 2H+(aq) → 1.0 mol/L Standard Half Reaction Voltages Zn2+(aq) + H2(g) 1.0 mol/L 1.0 atm Zn | Zn2+ || H+ | H2 | Pt • Cell voltage with no current flowing is +0.762V • This is the standard voltage for this cell Zn(s) + 2H+(aq, 1.0M) → Zn2+(aq, 1.0M) + H2(g, 1.0atm) Eo = +0.762V Chemistry 1011 Slot 5 7 Obtaining Values for Standard Half Reaction Voltages 2H+(aq,1.0M) + 2e− → H2(g,1.0atm) Eored (H+ → H2) = 0.000V 8 • Once one half reaction standard voltage is established, others can be deduced: • For: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) the standard cell voltage is +1.101V Zn(s) → Zn2+(aq,1.0M) + 2e− Cu2+(aq,1.0M) + 2e− → Cu(s) Eoox (Zn → Zn2+) = +0.762V Eored (Cu2+→ Cu) = ??V Since Eo = Eoox + Eored +1.101V = +0.762V + Eored Since Eo = Eoox + Eored Eoox (Zn → Zn2+) = +0.762V Eored = +0.339V 9 Standard Reduction Potentials Chemistry 1011 Slot 5 10 Standard Reduction Potentials Oxidizing Agent Li+(aq) + e− Na+(aq) + e− Zn2+(aq) + 2e− Ni2+(aq) + 2e− 2H+(aq) + 2e− Cu2+(aq) + 2e− Ag+(aq) + e− NO3−(aq) + 4H+(aq) + 3e− MnO4−(aq) + 8H+(aq) + 5e− F2(g) + 2e− • Standard half cell voltages are found in tables of standard potentials • These are the values for reduction half reactions based upon the convention that 2H+(aq,1.0M) + 2e− → H2(g,1.0atm) Eored (H+ → H2) = 0.000V • Standard reduction potential = Eored Chemistry 1011 Slot 5 Chemistry 1011 Slot 5 Obtaining Values for Standard Half Reaction Voltages • Standard half reaction voltages are determined by arbitrarily assigning the value of zero to the standard reduction half reaction for hydrogen ions to give hydrogen gas Chemistry 1011 Slot 5 • Each half reaction has a standard voltage • Eoox (standard oxidation voltage) • Eored (standard reduction voltage) Eo = Eoox + Eored o • Only E can be measured - the standard voltage of a half reaction cannot be measured directly 11 Reducing Agent → Li(s) → Na(s) → Zn(s) → Ni(s) → H2(g) → Cu(s) → Ag(s) → NO(g) + 2H2O → Mn2+(aq) + 4H2O → 2F−(aq) Chemistry 1011 Slot 5 Eored (V) -3.040 -2.714 -0.762 -0.236 0.000 +0.339 +0.799 +0.964 +1.512 +2.889 12 2 Standard Reduction Potentials • Elements above hydrogen in the table of standard reduction potentials will react with a solution of hydrogen ions to produce hydrogen gas M(s) + 2H+(aq) → M2+(aq) + H2(g) M2+(aq) + 2e− → M(s) Eored = negative M(s) → M2+(aq) + 2e− Eoox = positive 0.000V 2H+(aq) + 2e− → H2(g) Eored = M | M2+ || H+ | H2 | Pt Eocell = positive • Elements below hydrogen in the table of standard reduction potentials will NOT react with a solution of hydrogen ions to produce hydrogen gas Chemistry 1011 Slot 5 Standard Voltages for Voltaic Cells • The table of standard reduction potentials gives standard voltages for reduction half reactions • Standard voltages for oxidation half reactions are obtained by reversing these reactions and changing the sign of the Eored value → Zn(s) Zn2+(aq) + 2e− Eored = -0.762 • If: • Then: Zn(s) → Zn2+(aq) + 2e− Eoox = +0.762 13 Chemistry 1011 Slot 5 Oxidizing Agents Computing Standard Cell Potential • An oxidizing agent is a species that can gain electrons • The standard voltage of a cell is the sum of the standard potentials for the two half reactions • For the cell: Zn | Zn2+ || Cu2+ | Cu Eoox = +0.762V Zn(s) → Zn2+(aq) + 2e− Eored = +0.339V Cu2+(aq) + 2e− → Cu(s) Zn(s) + Cu2+(aq) → – The strongest oxidizing agents are the species that gain electrons most readily – They have the largest positive Eored values – Oxidizing strength increases moving down the left column of the table of standard reduction potentials – Oxidizing agents in the table of standard reduction potentials can oxidize any species above Zn2+(aq) + Cu(s) • Eocell = Eoox + Eored = + 0.762 + 0.339 = 1.101V Chemistry 1011 Slot 5 14 15 Chemistry 1011 Slot 5 16 Strong Reducing and Oxidizing Agents Reducing Agents • A reducing agent is a species that readily loses electrons – The strongest reducing agents are the species that lose electrons most readily – They have the largest negative Eored values (The largest positive Eoox values) – Reducing strength increases moving up the right column of the table of standard reduction potentials – Reducing agents in the table of standard reduction potentials can reduce any species below Reducing agent causes another species to be reduced - it is oxidized Li(s) → Li+(aq) + e− Eoox = +3.040V Oxidizing agent causes another species to be oxidized - it is reduced F2(g) + 2e− R → 2F−(aq) Eored = +2.889V Table of Standard Reduction Potentials R = strongest reducing agent O = strongest oxidizing agent O Chemistry 1011 Slot 5 17 Chemistry 1011 Slot 5 18 3 Spontaneity of Redox Reactions Reaction of Copper with Dilute Hydrochloric Acid?? • In order for a redox reaction to occur spontaneously, the calculated cell potential MUST BE POSITIVE • Questions: • Possible oxidation half reaction: Cu(s) → Cu2+(aq) + 2e− Eoox = -0.339V + • Possible reduction half reaction (H and Cl− ions are present - Cl− ions cannot be reduced): 2H+(aq) + 2e− → H2(g) Eored = 0.000 – Will copper metal be oxidized to Cu2+ ions by dilute hydrochloric acid? – Will copper metal be oxidized to Cu2+ ions by dilute nitric acid? Chemistry 1011 Slot 5 • Net possible reaction: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) • Net calculated cell voltage Eocell = Eoox + Eored = - 0.339 + 0.000 = - 0.339 V • Reaction will not be spontaneous i.e no reaction 19 Reaction of Copper with Dilute Nitric Acid?? 20 Voltaic Cells with Inert Electrodes • Possible oxidation half reaction: Cu(s) → Cu2+(aq) + 2e− Chemistry 1011 Slot 5 Eoox = -0.339V • Possible reduction half reactions (H+ and NO3− ions are present): 2H+(aq) + 2e− → H2(g) Eored = 0.000V NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O Eored = +0.964V • Net spontaneous reaction (Add multiples of the two half reactions so that same #electrons (6) in each half): 3Cu(s) + 2NO3−(aq) + 8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O • Half cells will frequently be constructed with inert electrodes (often carbon or platinum) • The Hydrogen half cell is one example: H+ | H2 | Pt • A cell with two inert electrodes might be: Pt | Fe2+(aq) | Fe3+(aq) || Cl −(aq) | Cl2(g) | Pt • Net calculated cell voltage: Eocell = Eoox + Eored = - 0.339 + 0.964 = + 0.629 V • Reaction will be spontaneous i.e reaction takes place Chemistry 1011 Slot 5 21 Chemistry 1011 Slot 5 22 Chemistry 1011 Slot 5 24 The Leclanché Cell • The Leclanché cell is the ordinary commercial flashlight battery Zn | Zn2+ ||MnO2 | Mn2O3 | C • Anode half reaction: Zn(s) → Zn2+(aq) + 2e− Eoox = +0.762V • Cathode half reaction (complex): 2MnO2(s) + 2NH4+(aq) + 2e− → Mn2O3(s) + 2NH3(aq) + H2O Eored = +0.7 V • Net cell voltage Eocell = 1.5V Chemistry 1011 Slot 5 23 4