Electrochem: It s Got Potential! Presented by: Denise DeMartino Westlake High School, Eanes ISD Pre-AP, AP, and Advanced Placement are registered trademarks of the College Board, which was not involved in the production of, and does not endorse, this product. The presenter(s) are solely responsible for the required permissions and content of the lessons. LTF and Laying the Foundation are registered trademarks of Laying the Foundation, Inc. Visit: www.layingthefoundation.org for more information.
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ELECTROCHEMISTRY Terms to Know: the study of the interchange of chemical and electrical energy OIL RIG oxidation is loss, reduction is gain (of electrons) Oxidation the loss of electrons, increase in charge Reduction the gain of electrons, reduction of charge Oxidation number the assigned charge on an atom Oxidizing agent (OA) the species that is reduced and thus causes oxidation Reducing agent (RA) the species that is oxidized and thus causes reduction How to balance using the ½ reaction method: 1.Write the two half-reactions separating the oxidized and reduced species. Acidic Solution: 2. Balance all atoms, except hydrogen and oxygen, in each half-reaction. 3. Balance oxygen atoms by adding water to the side with the fewest number of oxygen atoms. 4. Balance hydrogen atoms by adding H + to the side with the fewest number of hydrogen atoms. 5. Balance the charge by adding electrons to the side with the most positive charge. 6. If needed, multiply each half-reaction by a whole number which will produce an equal number of electrons transferred in the two half-reactions. 7. Add the half-reactions and simplify by canceling species common to both sides. Basic Solution: Follow steps 1-7; add a number of OH - ions equal to the number of H + ions to both sides of the equation. On the side of the equation with the H + ions, H + ions and OH - ions combine to form H2O. Water which appears on both sides of the equation can be canceled out. NOTE: No electrons should appear in the final balanced equation! Balance this redox reaction: MnO 4 - + Fe 2+ Mn 2+ + Fe 3+ [acidic solution] BrO 3 (aq) + Fe 2+ (aq) Br (aq) + Fe 3+ (aq) (basic solution) Page 3 1
STANDARD REDUCTION POTENTIALS The electromotive force (abbreviated emf), denoted E cell, is the 'driving force' or 'electrical pressure' which is responsible for the movement of electrons from the anode towards the cathode in a voltaic cell. The volt is the standard unit of measuring the electromotive force in an electrochemical cell. By definition, it requires 1 joule of energy to transport 1 coulomb of electrical charge across a potential of 1 volt. The emf of an electrochemical cell can be measured using a voltmeter. A voltmeter is used to measure the potential difference, or the difference in electrical potential between the two half-reactions occurring in the electrochemical cell. The greater the difference between the electrical potential of the two half-reactions, the greater the voltage. Calculating Standard Cell Potential Symbolized by E cell 1. Decide which element is oxidized or reduced using the table of reduction potentials. Remember: THE MORE POSITIVE REDUCTION POTENITAL GETS TO BE REDUCED. 2. Write both equations AS IS from the chart with their voltages. 3. Reverse the equation that will be oxidized and change the sign of the voltage [this is now E oxidation ] 4. Balance the two half reactions **do not multiply voltage values** 5. Add the two half reactions and the voltages together. 6. E cell = E oxidation + E reduction means standard conditions: 1atm, 1M, 25 C 7. A positive E cell means that the reaction is SPONTANEOUS. Standard Reduction Potentials at 25 C Half-Reaction E Li + (aq) + 1e - Li(s) -3.05 v K + (aq) + 1e - K(s) -2.93 v Ca 2+ (aq) + 2e - Ca(s) -2.87 v Na + (aq) + 1e - Na(s) -2.71 v Mg 2+ (aq) + 2e - Mg(s) -2.37 v Zn 2+ (aq) + 2e - Zn(s) -0.76 v Ni 2+ (aq) + 2e - Ni(s) -0.25 v Sn 2+ (aq) + 2e - Sn(s) -0.136 v Pb 2+ (aq) + 2e - Pb(s) -0.126 v 2H + (aq) + 2e - H 2 (s) 0.00 v AgCl(s) + 1e - Ag(s) + Cl - (aq) Cu 2+ (aq) + 2e - Cu(s) I 2 (s) + 2e - 2I - (aq) Ag + (aq) + 1e - Ag(s) NO - 3 (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) Br 2 (l) + 2e - 2Br - (aq) O 2 (g) + 4H + (aq) + 4e - 2H 2 O (l) Cr 2 O 2-7 (aq) + 14H + (aq) + 6e - 2Cr 3+ (aq) + 7H 2 O(l) Cl 2 (g) + 2e - 2Cl - (aq) MnO 4 - (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) H 2 O 2 (aq) + 2H + (aq) + 2e - 2H 2 O(l) F 2 (g) + 2e - 2F - (aq) 2H 2 O(l) + 2e- H 2 (g) + 2OH- (aq) +0.22 v +0.34 v +0.53 v +0.80 v +0.96 v +1.07 v +1.23 v +1.33 v +1.36 v +1.52 v +1.77 v +2.87 v -0.83 v Page 4 2
Examples: 1) Consider a galvanic cell based on the reaction Al 3+ (aq) + Mg(s) Al(s) + Mg 2+ (aq) Give the balanced cell reaction and calculate E for the cell. 2) Which of the following species is the strongest oxidizing agent, MnO 4 (in acidic solution), Br2 (l), or Ca 2+ (aq)? 3. Will aluminum displace Cu 2+ ion from an aqueous solution of Cu(NO 3 ) 2? ELECTROCHEMISTRY INVOLVES TWO MAIN TYPES OF PROCESSES: A. Galvanic (voltaic) cells which are spontaneous chemical reactions (battery) B. Electrolytic cells which are non-spontaneous and require external e - source (DC power source) C. BOTH of these fit into the category entitled Electrochemical cells GALVANIC CELLS Parts of the voltaic or galvanic cell: o Anode--the electrode where oxidation occurs. After a period of time, the anode may appear to become smaller as it falls into solution. o Cathode-- the anode where reduction occurs. After a period of time it may appear larger, due to ions from solution plating onto it. o inert electrodes used when a gas is involved OR ion to ion involved such as Fe 3+ being reduced to Fe 2+ rather than Fe 0. Made of Pt or graphite. o Salt bridge -- a device used to maintain electrical neutrality in a galvanic cell. This may be filled with agar which contains a neutral salt or it may be replaced with a porous cup. It connects the two compartments, ions flow from it, AND it keeps each cell neutral. Use KNO 3 as the salt when constructing your own diagram so that no precipitation occurs! o Electron flow -- always from anode to cathode. (through the wire) o Standard cell notation (line notation) - anode/solution// cathode solution/ cathode Ex. Zn/Zn 2+ (1.0 M) // Cu 2+ (1.0M) / Cu o Voltmeter - measures the cell potential (emf). Usually is measured in volts. Page 5 3
Terms to know in order to construct a spontaneous cell one that can act as a battery: AN OX oxidation occurs at the anode (may show mass decrease) RED CAT reduction occurs at the cathode (may show mass increase) FAT CAT The electrons in a voltaic or galvanic cell ALWAYS flow From the Anode To the CAThode ANIONS from the salt move to the anode while CATIONS from the salt move to the cathode! Ca+hode the cathode is + in galvanic cells Salt Bridge bridge between cells whose purpose is to provide ions to balance the charge. Usually made of a salt filled agar (KNO 3 ) or a porous cup. Example: Draw a diagram of the cells in which the following reactions occur. In each case, label the anode and cathode, the anode and cathode electrode material, the half-reaction at each electrode, the ions in the anode and cathode compartments and salt bridge, the direction of electron flow, and the direction of ion movement. Also calculate E of the cell 1) Co(s) + Cu 2+ (aq) Co 2+ (aq) + Cu(s) 2) cell is made from Cu/Cu +2 (1M) and Fe/Fe +3 (1M) Page 6 4
When cell is not at standard conditions, use Nernst Equation to calculate the cell potential: RT E = E o - ------- ln Q nf R = Gas constant 8.315 J/K mol F = Faraday constant Q = reaction quotient [products coefficient ]/[reactants coefficient ] E = Energy produced by reaction T = Temperature in Kelvins n = # of electrons exchanged in BALANCED redox equation Rearranged, another useful form 0.0592 NERNST EQUATION: E = E o - ---------- log Q @ 25ΕC (298K) n As E declines with reactants converting to products, E eventually reaches zero. Zero potential means reaction is at equilibrium [dead battery]. Also, Q =K AND ΔG = 0 as well. LeChatlier s principle can also be applied. An increase in the concentration of a reactant will favor the forward reaction and the cell potential will increase. The converse is also true! Example: 1) For the cell reaction 2Al(s) + 3Mn 2+ (aq) 2Al 3+ (aq) + 3Mn(s) E cell = predict whether E cell is larger or smaller than E cell if [Al 3+ ] = 2.0 M, [Mn 2+ ] = 1.0 M. 2) Determine E o cell and E cell based on the following half-reactions: VO + 2 + 2H + + e - VO 2+ + H 2 O Zn 2+ + 2e - Zn E = 1.00 V E = -0.76V Where [VO 2 + ] = 2.0 M, [H + ] = 0.50 M, [VO 2+ ] = 1.0 x 10-2 M, [Zn 2+ ] = 1.0 x 10-1 M at 25 C. Page 7 5
CELL POTENTIAL, ELECTRICAL WORK & FREE ENERGY The work that can be accomplished when electrons are transferred through a wire depends on the push or emf which is defined in terms of a potential difference [in volts] between two points in the circuit. work( J ) emf ( V ) = ch arg e( C) thus one joule of work is produced [or required] when one coulomb of charge is transferred between two points in the circuit that differ by a potential of one volt G = Gibb s free energy. n = number of moles of electrons. F = Faraday constant 96,500 C/mole e- ΔG o = -nfe o SUMMARY OF GIBB S FREE ENERGY AND CELLS -E o implies NONspontaneous + E o implies spontaneous (would be a good battery!) E = 0, equilibrium reached (dead battery) larger the voltage, more spontaneous the reaction ΔG will be negative in spontaneous reactions K>1 are favored Two important equations: ΔG = - nfe [ minus nunfe ] ΔG = - RTlnK [ ratlink ] G = Gibbs free energy [Reaction is spontaneous if ΔG is negative] n = number of moles of electrons. F = Faraday constant 9.6485309 x 10 4 J/V (1 mol of electrons carries 96,500C ) E = cell potential R = 8.31 J/mol K T = Kelvin temperature K = equilibrium constant [products] coeff. /[reactants] coeff **Favored conditions: E cell > 0 ΔG < 0 K>1** Examples: 1) Using the table of standard reduction potentials, calculate G for the reaction. Is this reaction spontaneous? Cu 2+ (aq) + Fe(s) Cu(s) + Fe 3+ (aq) Page 8 6
2) For the oxidation-reduction reaction S 4 O 6 2- (aq) + Cr 2+ (aq) Cr 3+ (aq) + S 2 O 3 2- (aq) The appropriate half-reactions are 2- S 4 O 6 + 2e - 2-2S 2 O 3 Cr 3+ + e - Cr 2+ E = 0.17V E = -0.50 V Balance the redox reaction, and calculate E and K (at 25 C). Electrolysis and Nonspontaneous Cells A voltaic cell is an electrochemical cell in which the reaction occurs spontaneously. An electrolytic cell is an electrochemical cell in which the reaction occurs only after adding electrical energy. The calculated E for a voltaic cell is always positive. It is negative for an electrolytic cell. To make an electrolytic cell work, you must add energy (this is usually done with a voltaic cell - a battery). This battery forces the electrons to travel from the postive electrode to the negative electrode, the exact opposite of what you would expect. This means in an electrolytic cell, the postive electrode is the anode, and the negative electrode is the cathode. EPA Electrolytic Postive Anode There are two rules that you should know that are exceptions in an aqueous solution. (1) No alkali or alkaline earth metal can be reduced in an aqueous solution - water is more easily reduced. (2) No polyatomic can be oxidized in an aqueous solution - water is more easily oxidized. Since you are running an electrical current through a solution to cause electrolysis, you can measure that current and tell how much substance is going to be produced. To do this you need to know a few units and definitions: The amount of electrical charge which flows through an electrochemical cell is measured in units of coulombs. The rate at which the charge flows is a current and is measured in amperes, or amps. By definition, or 1 C = 1 amp. 1 sec 1 amp = 1 C 1 sec A Faraday is the amount of charge associated with 1 mol of electrons. A Faraday has the value of 96,500 C. Using these constants and dimensional analysis, you can work any electrochemical calculation problem. Page 9 7
Example: 1) Show a sketch of an electrolytic cell of water. 2) Show sketches of electrolytic cells of (a) molten sodium chloride, (b) aqueous sodium chloride, (c) aqueous iron(iii) sulfate, and (d) aqueous sodium sulfate. 3) Calculate the mass of copper metal produced during the passage of 2.50 amps of current through a solution of copper (II) sulfate for 50.0 minutes. 4) What volume of oxygen gas (measure at STP) is produced by the oxidation of water that occurs concurrently with the above problem. Page 10 8
Problems 1. Answer parts (a) through (c) below, which relate to reactions involving copper, Cu and the copper(ii) ion, Cu 2+. A standard voltaic cell is constructed using the half reactions represented below. Cu 2+ (aq) + 2e Cu(s) E o = 0.34 V X + (aq) + e X(s) E o =??? V (a) The value of the standard potential, E o for this cell is 0.46 V. After several minutes it was noted that small flakes were adhering to the X electrode. (i) Which metal, Cu or X, is the cathode? Justify your answer. (ii) Determine the standard reduction potential, E o, for the X + /X electrode and identify metal X. (iii) (iv) Write a balanced net ionic equation for this electrochemical cell. The cell was changed so that [Cu 2+ ] is 0.01 M and [X + ] is 0.10 M. Calculate the new potential, E, for the cell at 25 C. In another experiment, current was passed through a solution of Cu(NO 3 ) 2 for 30 minutes. (a) Calculate the amount of current passed through the solution if 1.19 grams of Cu metal formed. (b) The 1.19 grams of Cu metal produced in Part (b) was filtered, rinsed, and added to 250. ml of 0.25 M nitric acid, HNO 3. The reaction represented below occurs. 3 Cu + 8 HNO 3 3 Cu(NO 3 ) 2 + 2NO + 4 H 2 O (i) Identify the limiting reactant. Show work to support your answer. (ii) On the basic of the limiting reactant identified in Part (i) calculate the value of the [Cu 2+ ] ions after the reaction is complete. 2. What is the sum of the coefficients on the reactants side in this balanced equation? H + + MnO 4 + C 2 O 4 2 + H 2 O + CO 2 + Mn 2+ (A) 5 (B) 8 (C) 10 (D) 14 (E) 23 Page 11 9
Questions 3-4: Mn(s) + Cu +2 (aq) Mn 2+ (aq) + Cu(s) E = 1.52 volts 3. Which expression gives the value of ΔG in kj/mol for this reaction? (A) -2 x 8.31 x 1.52 x 1,000 (B) -2 x 96,500 x 1.52 8.31 (C) -2 x 96,500 x 1.52 1,000 (D) - 2 x 96,500 1.52 x 8.31 (E) -2 x 8.31 x 1.52 1,000 4) Which expression gives the voltage for such a cell at non-standard conditions where [Cu 2+ ] is 1.00 M and [Mn +2 ] is 0.010 M? (A) 1.52 + 0.0591 (B) 1.52 + (0.0591 x 2) (C) 1.52 x 0.0591 2 (D) 1.52 + 0.0591 2 (E) 1.52 x 2 0.0591 5) Zn(s) + Cu +2 (aq) Zn 2+ (aq) + Cu(s) If the equilibrium constant for the reaction above is 1.66 x 10 37, which of the following correctly describes the standard voltage, E, and the standard free energy change, ΔG, for this reaction? (A) E is positive and ΔG is negative. (B) E is negative and ΔG is positive. (C) ΔG and E are both positive. (D) ΔG and E and are both negative. (E) ΔG and E and are both zero. Page 12 10