Chapter 19: Electrochemistry Overview of the Chapter review oxidation-reduction chemistry basics galvanic cells spontaneous chemical reaction generates a voltage set-up of galvanic cell & identification of: anode (and half-reaction) cathode (and half-reaction) net cell reaction cell potential (E or E ) relationship between cell potential and work, free energy, Q and K Electrolytic cells electric current drives a non-spontaneous chemical rxn stoichiometric calculations based on cell rxn, current, time Redox Review Oxidation-Reduction Reactions aka Redox Reactions oxidation: loss of electrons reduction: gain of electrons Consider the reaction of calcium metal with oxygen: 2 Ca (s) + O2 (g)! 2 CaO (s) What happens to each element during the course of this reaction? Oxidation and Reduction Half Reactions for the reaction of calcium with oxygen: oxidation! reaction: Ca! Ca2+ + 2 e note: in oxidation! rxn e s are products reduction! reaction: O2 + 4 e! 2 O2 note: in reduciton! rxn e s are reactants net redox reaction: 2 Ca + O2! 2 Ca2+ + 2 O2 note: add together the!-reactions; multiply as necessary to have same number of electrons in each half reaction before adding
Oxidation Numbers (Nox) An oxidation number indicates the amount of electropositive or electronegative character of an atom - particularly as part of a polyatomic species. Determining Oxidation Numbers: ex. Determine Nox of sulfur in SO3. Nox S + 3(Nox O) = 0 Nox S + 3( 2) = 0 Nox S = +6 ex. Determine Nox of sulfur in SO3 2. Nox S + 3(Nox O) = 2 Nox S + 3( 2) = 2 Nox S = +4 ex. Determine Nox N, H, P, and O in (NH4)3PO4. consider NH4 + and PO4 3 separately: Nox N + 4(Nox H) = +1 Nox P + 4(Nox O) = 3 Nox N + 4(+1) = +1 Nox P + 4( 2) = 3 Nox N = 3 Nox P = +5 Recognizing Oxidation and Reduction Compare oxidation numbers of elements in reactants and products: ex: if Nox increases - the element is oxidized if Nox decreases - the element is reduced 2 Al (s) + Cr2O3 (s)! Al2O3 (s) + 2 Cr (s) Recognizing Oxidation and Reduction Identify the oxidizing and reducing agents: oxidizing agent reactant that facilitates oxidation by taking on (i.e. gaining) electrons lost during oxidation process oxidizing agent is the reactant that contains the element that is reduced reducing agent reactant that facilitates reduction by providing (i.e. losing) electrons that are gained during reduction process reducing agent is the reactant that contains the element that is oxidized
Recognizing Oxidation and Reduction Consider the following reaction: 2 Ca3(PO4)2 + 6 SiO2 + 10 C! P4 + 6 CaSiO3 + 10 CO Identify the following: element oxidized element reduced oxidizing agent reducing agent Balancing Redox Equations: the Half-Reaction Method in acidic solution 1. Identify the element oxidized and the element reduced. 2. Write the skeletal oxidation & reduction!-reactions. 3. Balance elements other than H and O. 4. Balance O s by adding H2O. 5. Balance H s by adding H +. 6. Balance charge by adding e s. *at this point the individual!-reactions are balanced* 7. Prepare to add the!-reactions together; multiply as necessary so that the # of e s in oxidation!-reaction equals # of e s in the reduction!-reaction. 8. Add!-reactions together; clean up. Balance the following redox reaction that occurs in acidic solution using the!-reaction method: I2 + NO3! IO3 + NO2 step 2: ox! rxn: I2! IO3 red! rxn: NO3! NO2 step 3: ox! rxn: I2! 2 IO3 red! rxn: NO3! NO2 step 4: ox! rxn: 6 H2O + I2! 2 IO3 red! rxn: NO3! NO2 + H2O step 5: ox! rxn: 6 H2O + I2! 2 IO3 + 12 H + red! rxn: 2 H + + NO3! NO2 + H2O Balance the following redox reaction that occurs in acidic solution using the!-reaction method (continued): I2 + NO3! IO3 + NO2 step 6: ox! rxn: 6 H2O + I2! 2 IO3 + 12 H + + 10 e red! rxn: 1 e + 2 H + + NO3! NO2 + H2O step 7: ox! rxn: 6 H2O + I2! 2 IO3 + 12 H + + 10 e red! rxn: 10"(1 e + 2 H + + NO3! NO2 + H2O) step 8: 6 H2O + I2 + 20 H + + 10 NO3! 2 IO3 + 12 H + + 10 NO2 + 10 H2O net rxn: I2 + 8 H + + 10 NO3! 2 IO3 + 10 NO2 + 4 H2O
Redox Reaction in an Electrochemical Cell Cu (s) + 2 AgNO3 (aq) Cu(NO3)2 (aq) + 2 Ag (s) net ionic equation: Cu + 2 Ag+ Cu2+ + 2 Ag copper metal in solution of silver nitrate With this set-up, any energy released by the reaction goes into sol n causing T. If we want the energy produced to do something, we need to change the experimental set-up. consider the rxn: Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) salt bridge anode compartment oxidation occurs here electrons produced anode! reaction: Zn Zn2+ + 2 e designated on battery cathode compartment reduction occurs here electrons consumed cathode! reaction: Cu2+ + 2 e Cu designated + on battery galvanic (or voltaic) cell: a spontaneous chemical reaction generates an electric current separate the oxidation and reduction processes each!-reaction occurs in its own compartment oxidation occurs at the anode reduction occurs at the cathode The electrons produced in the oxidation process are transferred across an electrically conducting wire to compartment where reduction will occur. consider the rxn: Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) anode compartment cathode compartment salt bridge or porous barrier allows migration of spectator ions between compartments maintain charge neutrality anions migrate toward the anode cations migrate toward the cathode
the actual, physical electrodes in this cell are a strip of Zn (the anode) and a strip of Cu (the cathode) the electrodes must be connected by an electrically conducting wire to complete the circuit need a complete circuit for: e movement chemical reaction A galvanic cell is constructed based on the following redox reaction: Fe (s) + 2 Fe3+ (aq) 3 Fe2+ (aq) the e transfer generates a potential difference (measured in volts,v) the specific voltage generated is dependent on the anode & cathode used, and [ ] s or P s Line Notation for Galvanic Cells in a galvanic cell, the direction of e flow is always: anode cathode short-hand representation of galvanic cells use single vertical lines to indicate/separate species of different phases within same compartment use double vertical lines to indicate separation of anode and cathode compartments Write the balanced!-reactions that occur at the anode and the cathode. Sketch this galvanic cell indicating the following: the anode and the cathode anode anode compartment cathode compartment cathode the species in the anode & cathode compartments the direction of electron flow the direction of cation and anion migration examples: Zn (s) Zn2+ (aq) Cu2+ (aq) Cu (s) Fe (s) Fe2+ (aq) Fe2+ (aq), Fe3+ (aq) Pt (s)
Cell Potential, E and E oxidation-reduction reaction results in e transfer as e s move from anode to cathode, a voltage is generated cell potential, E is measured in volts 1 volt = 1 joule of work done as 1 Coulomb of charge moves from a point of higher potential to a point of lower potential 1 V = 1 J/C standard cell potential, E voltage generated when: solids and liquids are in their pure form solutions at 1 M concentration gases at 1 atm pressure Cell Potential, E and E E cell = E cathode E anode standard half-cell designated as reference electrode E s of other half-cells determined and tabulated relative to this reference electrode standard hydrogen electrode (SHE): 2 H + (aq) + 2 e H2 (g) [H + ] = 1 M PH2 = 1 atm T = 298 K E = 0.000 V Standard Hydrogen Electrode Table of Standard Reduction Potentials! reactions written as reductions oxidizing agent + e reducing agent top to bottom, increasing E increasing tendency for reduction decreasing tendency for oxidation relative strengths of oxidizing and reducing agents strongest reducing agents are higher in the table stronger oxidizing agents are lower in the table correlation between sign of E and spontaneity + E, reduction is spontaneous E, oxidation is spontaneous
Spontaneous Reactions and Cell Potential in a galvanic cell, a spontaneous chemical reaction generates a + potential E cell must be positive E cathode > E anode comparing any two! reactions in the Table of Standard Reduction Potentials: the! reaction higher in the table (smaller E ) will be the anode, oxidation! reaction the! reaction lower in the table (larger E ) will be the cathode, reduction! reaction Consider a galvanic cell based on the following half-cells: Ni 2+ (aq) + 2 e Ni (s); E = 0.23 V Cd 2+ (aq) + 2 e Cd (s); E = 0.40 V Write the anode! reaction, the cathode! reaction, and the overall cell reaction. Then determine E cell.
Consider a galvanic cell based on the following half-cells: Can nitrate ion in an acidic solution oxidize iron (II) ions? Pb 2+ + 2 e Pb; E = 0.13 V PbO2 + SO4 2 + 4 H + + 2 e PbSO4 + 2 H2O; E = 1.69 V Write the anode! reaction, the cathode! reaction, and the overall cell reaction. Then determine E cell. Can MnO4 in an acidic solution oxidize Ni? Ag? Is Cr or Al a stronger reducing agent?
Complete Description of a Galvanic Cell At this point for a galvanic cell, you should be able to: write the balanced anode and cathode! reactions write the balanced cell reaction determine the standard cell potential (E ) in volts line notation sketch or draw the set-up of the cell showing: anode and cathode species in anode and cathode compartments salt bridge direction of e flow direction of ion migration Work, Cell Potential, and Efficiency the maximum work that can be done by a cell is given by: wmax = nfe BUT... this maximum work is never achieved to calculate the actual work done by a cell: wactual = nfe can also determine the cell s efficiency: wactual efficiency = wmax x 100 E x 100 efficiency = E Work, Free Energy, and Electrochemistry work is done in a galvanic cell as e s are transferred from anode to cathode work is done by the cell exothermic ; w is maximum work that can be done by a galvanic cell is equal to the free energy change for the cell reaction wmax = G wmax = (quantity of charge transferred)(operating voltage of cell) wmax = G = nfe n = mol e s transferred F = Faraday constant = 96485 C/mol e note: When E is +, G is. This confirms that the chemical reaction in a galvanic cell is spontaneous. Consider a galvanic cell based on these!-reactions: Cu2+ + 2 e Cu; E =0.34V Ag+ + e Ag; E =0.80V Calculate G (in kj) for this galvanic cell s reaction. If the cell operates at a potential of 0.35 V, determine its efficiency.
The Nernst Equation: Dependence of E on Composition we have discussed the relationship between G and E : G = nfe at standard conditions: s and l in pure form sol ns at 1M concentration gases at 1 atm pressure galvanic cells can operate at conditions other than standard: G = nfe relationship between G, G, E, E, composition? RT Nernst equation: E = E ln Q nf Cell Potential and the Equilibrium Constant starting with the Nernst equation: E = E RT ln Q nf for a system at equilibrium, E = 0 and Q = K: E = RT ln K nf rework the equations a bit: at T = 25 C combine R, T, and F convert ln to log E = E.0592 V log Q; E =.0592 V log K n n Consider a galvanic cell based on the following cell rxn: Cu (s) + 2 Fe 3+ (aq) Cu 2+ (aq) + 2 Fe 2+ (aq); E cell = 0.43 V Determine G and E at 25 C if the cell is run with the following ion concentrations: [Fe 2+ ] = 0.20 M [Fe 3+ ] = 1.0 x 10 4 M [Cu 2+ ] = 0.25 M Determine the equilibrium constant, K, for this reaction at 25 C. Consider the following galvanic cell: anode: Tl (s) Tl + (aq) + e ; E anode =.34V cathode: Ni 2+ (aq) + 2 e Ni (s); E cathode =.24V Determine the concentration of Tl + (aq) if the cell operates at E = 0.15V at 25 C when [Ni 2+ ] = 3.3 M.
Electrolysis an electrical current is used to drive a nonspontaneous chemical reaction in an electrolytic cell: w is + G is + Ecell is we will use the relationships between: mol e transferred and mol product formed stoichiometry of balanced! reactions mol e transferred and quantity of charge Faraday constant, F quantity of charge and time current, amperes (1A = 1C/s) Determine the time required to plate 85.5 g Zn onto a frame if 23.0 A current is passed through a solution of ZnSO4 (aq). strategy? Zn 2+ (aq) + 2 e Zn (s) g Zn mol Zn mol e C time Determine the current required to plate 2.86 g Cr in 2.5 min onto a bathroom fixture from Cr2(SO4)3 (aq) strategy? Cr 3+ (aq) + 3 e Cr (s) g Cr mol Cr mol e C current, A Determine the mass of nickel that can be produced in an experiment when 13.7 A passed through a solution of Ni(NO3)2 (aq) for 5.0 min. strategy? Ni 2+ (aq) + 2 e Ni (s) time C mol e mol Ni g Ni