Chapter 21 Electrochemistry: Chemical Change and Electrical Work

Electrochemistry: Chemical Change and Electrical Work 21.1 Half-Reactions and Electrochemical Cells 21.2 Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy 21.3 Cell Potential: Output of a Voltaic Cell 21.4 Free Energy and Electrical Work 21.5 Electrochemical Processes in Batteries 21.6 Corrosion: A Case of Environmental Electrochemistry 21.7 Electrolytic Cells: Using Electrical Energy to Drive a Nonspontaneous Reaction

Electrochemistry Study of interchange of chemical and electrical energy. It involves oxidation –reduction reaction.

Key Points About Redox Reactions Oxidation (electron loss) always accompanies reduction (electron gain). The oxidizing agent is reduced, and the reducing agent is oxidized. The number of electrons gained by the oxidizing agent always equals the number lost by the reducing agent.

A summary of redox terminology. Figure 21.1 A summary of redox terminology. Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) OXIDATION One reactant loses electrons. Zn loses electrons. Reducing agent is oxidized. Zn is the reducing agent and becomes oxidized. Oxidation number increases. The oxidation number of Zn increases from 0 to +2. REDUCTION Other reactant gains electrons. Hydrogen ion gains electrons. Oxidizing agent is reduced. Hydrogen ion is the oxidizing agent and becomes reduced. Oxidation number decreases. The oxidation number of H decreases from +1 to 0.

Half-Reaction Method for Balancing Redox Reactions Summary: This method divides the overall redox reaction into oxidation and reduction half-reactions. Each reaction is balanced for mass (atoms) and charge. One or both are multiplied by some integer to make the number of electrons gained and lost equal. The half-reactions are then recombined to give the balanced redox equation. Advantages: The separation of half-reactions reflects actual physical separations in electrochemical cells. The half-reactions are easier to balance especially if they involve acid or base. It is usually not necessary to assign oxidation numbers to those species not undergoing change.

Balancing Redox Reactions in Acidic Solution Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) 1. Divide the reaction into half-reactions - Determine the O.N.s for the species undergoing redox. +6 -1 +3 Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(aq) Cr2O72- Cr3+ Cr is going from +6 to +3 I - I2 I is going from -1 to 0 2. Balance atoms and charges in each half-reaction - 14H+(aq) + Cr2O72- Cr3+ 2 + 7H2O(l) net: +6 Add 6e- to left. net: +12 2 Cr2O72- Cr3+ + 7H2O(l) 14H+(aq) + 6e- +

Balancing Redox Reactions in Acidic Solution continued Cr2O72- Cr3+ + 7H2O(l) 14H+(aq) + 6e- + 2 2 I - I2 + 2e- Cr(+6) is the oxidizing agent and I(-1) is the reducing agent. 3. Multiply each half-reaction by an integer, if necessary - X 3 + 2e- I - I2 2 4. Add the half-reactions together - Cr2O72- Cr3+ + 7H2O(l) 14H+ + 6e- + 2 3 I - I2 6 + 6e- Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) 14H+(aq) + Do a final check on atoms and charges.

Balancing Redox Reactions in Basic Solution Balance the reaction in acid and then add OH- so as to neutralize the H+ ions. Cr2O72-(aq) + 6 I-(aq) 2Cr3+(aq) + 3I2(s) + 7H2O(l) 14H+(aq) + + 14OH-(aq) + 14OH-(aq) Cr2O72- + 6 I- 2Cr3+ + 3I2 + 7H2O + 14OH- 14H2O + Reconcile the number of water molecules. + 14OH- Cr2O72- + 6 I- 2Cr3+ + 3I2 7H2O + Do a final check on atoms and charges.

The redox reaction between dichromate ion and iodide ion. Figure 21.2 The redox reaction between dichromate ion and iodide ion. Cr2O72- I- Cr3+ + I2 Cr2O72- Cr3+ I-

Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method PROBLEM: Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid mangaese dioxide. Balance the skeleton ionic reaction that occurs between NaMnO4 and Na2C2O4 in basic solution: MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq) PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: MnO4- MnO2 C2O42- CO32- +7 +4 +3 +4 MnO4- MnO2 C2O42- CO32- 2 4H+ + MnO4- MnO2 + 2H2O C2O42- 2CO32- + 2H2O + 4H+ +3e- +2e-

Sample Problem 21.1: Balancing Redox Reactions by the Half-Reaction Method continued: 4H+ + MnO4- +3e- MnO2+ 2H2O C2O42- + 2H2O 2CO32- + 4H+ + 2e- X 2 4H+ + MnO4- +3e- MnO2+ 2H2O X 3 C2O42- + 2H2O 2CO32- + 4H+ + 2e- 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 8H+ + 2MnO4- +6e- 2MnO2+ 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH- + 4OH- 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

Voltaic cell A cell that produces electrical energy from a spontaneous chemical reaction. (Also known as a galvanic cell) System does work on the surroundings. batteries

Electrolytic cell An electrolysis apparatus in which electrical energy from an outside source is used to produce a chemical change. Cathode: The negative electrode. Anode: The positive electrode Surroundings do work on the system. Electroplating

Cells Electrode- Conduct electricity between cell and surroundings. Electrolyte-mix of ions. Oxi-anode, e- are lost, leave cell at anode. Red-cathode, e- are gained, enter at anode.

General characteristics of voltaic and electrolytic cells. Figure 21.3 General characteristics of voltaic and electrolytic cells. VOLTAIC CELL ELECTROLYTIC CELL System does work on its surroundings Energy is released from spontaneous redox reaction Energy is absorbed to drive a nonspontaneous redox reaction Surroundings(power supply) do work on system(cell) Oxidation half-reaction X X+ + e- Oxidation half-reaction A- A + e- Reduction half-reaction Y++ e- Y Reduction half-reaction B++ e- B Overall (cell) reaction X + Y+ X+ + Y; DG < 0 Overall (cell) reaction A- + B+ A + B; DG > 0 Energy released Energy absorbed

The spontaneous reaction between zinc and copper(II) ion. Figure 21.4 The spontaneous reaction between zinc and copper(II) ion. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

A voltaic cell based on the zinc-copper reaction. Figure 21.5 A voltaic cell based on the zinc-copper reaction. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Reduction half-reaction Cu2+(aq) + 2e- Cu(s) Overall (cell) reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

A voltaic cell using inactive electrodes. Figure 21.6 A voltaic cell using inactive electrodes. Oxidation half-reaction 2I-(aq) I2(s) + 2e- Reduction half-reaction MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) Overall (cell) reaction 2MnO4-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)

Construction and operation of a Voltaic Cell Two half cells. Consists of electrode in an electrolyte. Joined by wire and a salt bridge. Voltmeter used to measure voltage. Oxi half cell-anode- left Red half cell-cathode-right

Construction and operation of a Voltaic Cell Anode conducts electrons out of the cell Cathode conducts electrons in its half cell. Electrons flow from anode to cathode thro the wire (left-right) Anode has excess of e-, -ve charge. cathode: +ve

Construction and operation of a Voltaic Cell Salt bridge: A charge imbalance results because of ions. Salt bridge helps complete the circuit and avoid such imbalances. It contains soln of nonreacting ions: Na+, sulphate in gel. To maintain neutrality ions( Na+, sulfate) diffuse thro the soln and balance the charges

Construction and operation of a Voltaic Cell Electrons move left-right thro’ wire Anions move from right-left thro’ salt bridge. Cations move from left-right thro’ salt bridge. Active electrodes- involved in reaction. Inactive electrode- graphite or platinum, conduct electrons into or out of half cell, but cannot take part in half reactions.

Line Notation Anode components listed on left, cathode components listed on right separated by double vertical lines (salt bridge) A phase difference boundary-single line. Write line notations for the following: 1. Al3+(aq) + Mg (s)→ Al (s) + Mg2+ (aq) 2. MnO4- (aq) +H+ (aq) +ClO3-(aq) → ClO4-(aq) + Mn2+ (aq) + H2O(l)

Notation for a Voltaic Cell components of anode compartment (oxidation half-cell) components of cathode compartment (reduction half-cell) phase of lower oxidation state phase of higher oxidation state phase of higher oxidation state phase of lower oxidation state phase boundary between half-cells Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s) Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s) graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite inert electrode

Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s) Sample Problem 21.2: Diagramming Voltaic Cells PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode. PLAN: Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction). e- SOLUTION: Voltmeter salt bridge Oxidation half-reaction Cr(s) Cr3+(aq) + 3e- Cr Cr3+ Ag Ag+ K+ NO3- Reduction half-reaction Ag+(aq) + e- Ag(s) Overall (cell) reaction Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)

Why Does a Voltaic Cell Work? The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit. Ecell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C)

Cell Potential: Output of a Voltaic Cell Ecell/Voltage/electromotive force(emf): Difference in the electric potential of the two cells. Electrons flow from –ve to + ve electrode Ecell > 0 for a spontaneous reaction 1 Volt (V) = 1 Joule (J)/ Coulomb (C) Ecell <0- nonspontaneous Ecell=0, equilibrium.

Table 21.1 Voltages of Some Voltaic Cells Voltage (V) Common alkaline battery 1.5 Lead-acid car battery (6 cells = 12V) 2.0 Calculator battery (mercury) 1.3 Electric eel (~5000 cells in 6-ft eel = 750V) 0.15 Nerve of giant squid (across cell membrane) 0.070

Standard cell potentials E0cell The potential measured at 298 K, with no current flowing and components in their standard states: 1 atm for gases, 1M for solutions, pure solid for electrodes.

Standard Electrode( Half-Cell) potentials E0 values corresponding to reduction half-reactions with all solutes at 1 M and 1 atm are called standard reduction potentials E 0cell= E 0cathode(reduction) – E 0anode(oxidation)

Standard Hydrogen Electrode Hydrogen gas at 1 atm is passed over a Pt electrode in contact with 1 M H+ ions. Electrode is assigned zero volts.

Figure 21.7 Determining an unknown E0half-cell with the standard reference (hydrogen) electrode. Oxidation half-reaction Zn(s) Zn2+(aq) + 2e- Reduction half-reaction 2H3O+(aq) + 2e- H2(g) + 2H2O(l) Overall (cell) reaction Zn(s) + 2H3O+(aq) Zn2+(aq) + H2(g) + 2H2O(l)

Sample Problem 21.3: Calculating an Unknown E0half-cell from E0cell PROBLEM: A voltaic cell houses the reaction between aqueous bromine and zinc metal: Br2(aq) + Zn(s) Zn2+(aq) + 2Br-(aq) E0cell = 1.83V Calculate E0bromine given E0zinc = -0.76V PLAN: The reaction is spontaneous as written since the E0cell is (+). Zinc is being oxidized and is the anode. Therefore the E0bromine can be found using E0cell = E0cathode - E0anode. SOLUTION: anode: Zn(s) Zn2+(aq) + 2e- E = +0.76 E0Zn as Zn2+(aq) + 2e- Zn(s) is -0.76V E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76) E0bromine = 1.86 - 0.76 = 1.07 V

Problems Consider a galvanic cell:Al3+(aq) + Mg (s) →Al (s) + Mg2+ Give the balanced cell reaction and calculate Eo for the cell. 3) A galvanic cell is based on the reaction ; MnO4- (aq) +H+ (aq) +ClO3-(aq) → ClO4-(aq) + Mn2+ (aq) + H2O(l) Give the balanced cell reaction and calculate Eo for the cell.

Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) F2(g) + 2e- 2F-(aq) +2.87 strength of oxidizing agent strength of reducing agent Cl2(g) + 2e- 2Cl-(aq) +1.36 MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23 NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) +0.96 Ag+(aq) + e- Ag(s) +0.80 Fe3+(g) + e- Fe2+(aq) +0.77 O2(g) + 2H2O(l) + 4e- 4OH-(aq) +0.40 Cu2+(aq) + 2e- Cu(s) +0.34 2H+(aq) + 2e- H2(g) 0.00 N2(g) + 5H+(aq) + 4e- N2H5+(aq) -0.23 Fe2+(aq) + 2e- Fe(s) -0.44 2H2O(l) + 2e- H2(g) + 2OH-(aq) -0.83 Na+(aq) + e- Na(s) -2.71 Li+(aq) + e- Li(s) -3.05

Writing Spontaneous Redox Reactions By convention, electrode potentials are written as reductions. When pairing two half-cells, you must reverse one reduction half-cell to produce an oxidation half-cell. Reverse the sign of the potential. The reduction half-cell potential and the oxidation half-cell potential are added to obtain the E0cell. When writing a spontaneous redox reaction, the left side (reactants) must contain the stronger oxidizing and reducing agents. Example: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) stronger reducing agent stronger oxidizing agent weaker oxidizing agent weaker reducing agent

Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength PROBLEM: (a) Combine the following three half-reactions into three balanced equations (A, B, and C) for spontaneous reactions, and calculate E0cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents: E0 = 0.96V (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) E0 = -0.23V (2) N2(g) + 5H+(aq) + 4e- N2H5+(aq) E0 = 1.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) PLAN: Put the equations together in varying combinations so as to produce (+) E0cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an oxidation, reverse the sign of E0. Balance the number of electrons gained and lost without changing the E0. In ranking the strengths, compare the combinations in terms of E0cell.

Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (2 of 4) SOLUTION: (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) E0 = 0.96V (a) (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- E0 = +0.23V Rev E0cell = 1.19V (1) NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) X4 (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- X3 4NO3-(aq) + 3N2H5+(aq) + H+(aq) 4NO(g) + 3N2(g) + 8H2O(l) (A) E0 = -0.96V (1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- Rev E0 = 1.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) E0cell = 0.27V (1) NO(g) + 2H2O(l) NO3-(aq) + 4H+(aq) + 3e- X2 (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) X3 2NO(g) + 3MnO2(s) + 4H+(aq) 2NO3-(aq) + 3Mn3+(aq) + 2H2O(l) (B)

Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (3 of 4) E0 = +0.23V (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- Rev E0 = 1.23V (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) E0cell = 1.46V (2) N2H5+(aq) N2(g) + 5H+(aq) + 4e- (3) MnO2(s) +4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) X2 N2H5+(aq) + 2MnO2(s) + 3H+(aq) N2(g) + 2Mn2+(aq) + 4H2O(l) (C) (b) Ranking oxidizing and reducing agents within each equation: (A): oxidizing agents: NO3- > N2 reducing agents: N2H5+ > NO (B): oxidizing agents: MnO2 > NO3- reducing agents: NO > Mn2+ (C): oxidizing agents: MnO2 > N2 reducing agents: N2H5+ > Mn2+

Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength continued (4 of 4) A comparison of the relative strengths of oxidizing and reducing agents produces the overall ranking of Oxidizing agents: MnO2 > NO3- > N2 Reducing agents: N2H5+ > NO > Mn2+

Relative Reactivities (Activities) of Metals 1. Metals that can displace H from acid: If E0 cell for the reduction of H+ is more +ve for metal A than metal B, metal A is stronger R.A than metal B and a more active metal. They are placed below H. (+ve E0) 2. Metals that cannot displace H from acid: Higher the metal in the list more –ve is the E0 cell and less active it is. 3. Metals that can displace H from water: Grop IA, II A can displace H2 from H2 O. 4. Metals that can displace other metals from solution: Lower in list can reduce higher in list.

Table 21.2 Selected Standard Electrode Potentials (298K) Half-Reaction E0(V) F2(g) + 2e- 2F-(aq) +2.87 strength of oxidizing agent strength of reducing agent Cl2(g) + 2e- 2Cl-(aq) +1.36 MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l) +1.23 NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l) +0.96 Ag+(aq) + e- Ag(s) +0.80 Fe3+(g) + e- Fe2+(aq) +0.77 O2(g) + 2H2O(l) + 4e- 4OH-(aq) +0.40 Cu2+(aq) + 2e- Cu(s) +0.34 2H+(aq) + 2e- H2(g) 0.00 N2(g) + 5H+(aq) + 4e- N2H5+(aq) -0.23 Fe2+(aq) + 2e- Fe(s) -0.44 2H2O(l) + 2e- H2(g) + 2OH-(aq) -0.83 Na+(aq) + e- Na(s) -2.71 Li+(aq) + e- Li(s) -3.05

The reaction of calcium in water. Figure 21.8 The reaction of calcium in water. Oxidation half-reaction Ca(s) Ca2+(aq) + 2e- Reduction half-reaction 2H2O(l) + 2e- H2(g) + 2OH-(aq) Overall (cell) reaction Ca(s) + 2H2O(l) Ca2+(aq) + H2(g) + 2OH-(aq)

Free Energy and Electrical Work DG a -Ecell DG = wmax = charge x (-Ecell) -Ecell = -wmax charge DG = -n F Ecell In the standard state - DG0 = -n F E0cell DG0 = - RT ln K E0cell = - (RT/n F) ln K charge = n F n = #mols e- F = Faraday constant F = 96,485 C/mol e- 1V = 1J/C F = 9.65x104J/V*mol e-

Reaction at standard-state conditions Figure 21.9 The interrelationship of DG0, E0, and K. DG0 K Reaction at standard-state conditions E0cell DG0 < 0 > 1 > 0 spontaneous at equilibrium nonspontaneous 1 > 0 < 1 < 0 DG0 = -nFEocell DG0 = -RT lnK By substituting standard state values into E0cell, we get E0cell = (0.0592V/n) log K (at 298 K) E0cell K E0cell = -RT lnK nF

Problem 5. Calculate ∆G0 for the reaction:Cu2+ (aq) + Fe(s) →Cu (s) + Fe2+ (aq) Is this reaction spontaneous?

Sample Problem 21.5: Calculating K and DG0 from E0cell PROBLEM: Lead can displace silver from solution: As a consequence, silver is a valuable by-product in the industrial extraction of lead from its ore. Calculate K and DG0 at 298 K for this reaction. Pb(s) + 2Ag+(aq) Pb2+(aq) + 2Ag(s) PLAN: Break the reaction into half-reactions, find the E0 for each half-reaction and then the E0cell. Substitute into the equations found on slide SOLUTION: Pb2+(aq) + 2e- Pb(s) E0 = -0.13V E0 = 0.13V E0 = 0.80V E0cell = 0.93V Ag+(aq) + e- Ag(s) E0 = 0.80V Ag+(aq) + e- Ag(s) Pb(s) Pb2+(aq) + 2e- 2X E0cell = log K 0.592V n DG0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V) n x E0cell 0.592V (2)(0.93V) 0.592V = DG0 = -1.8x102kJ log K = K = 2.6x1031

The Effect of Concentration on Cell Potential DG = DG0 + RT ln Q -nF Ecell = -nF Ecell + RT ln Q Ecell = E0cell - ln Q RT nF When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Ecell = E0cell - log Q 0.0592 n

Problem: 7. Consider a voltaic cell based on the reaction: Fe(s) + Cu2+(aq) →Fe2+(aq) + Cu(s) If [Cu2+] =0.30 M, what must [Fe2+] be to increase Ecell by 0.25 V above E0cell at 25 C?

Using the Nernst Equation to Calculate Ecell Sample Problem 21.6: Using the Nernst Equation to Calculate Ecell PROBLEM: In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+] = 0.010M [H+] = 2.5M P = 0.30atm H2 Calculate Ecell at 298 K. PLAN: Find E0cell and Q in order to use the Nernst equation. SOLUTION: Determining E0cell : Q = P x [Zn2+] H2 [H+]2 E0 = 0.00V 2H+(aq) + 2e- H2(g) E0 = -0.76V Zn2+(aq) + 2e- Zn(s) Q = (0.30)(0.010) (2.5)2 Zn(s) Zn2+(aq) + 2e- E0 = +0.76V Ecell = E0cell - 0.0592V n log Q Q = 4.8x10-4 Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V

Changes in potential during cell Reaction As cell discharges, current flows from anode to cathode and E cell changes. Cell will spontaneously discharge till it reaches equilibrium. Q=K. Ecell=0. Dead battery: Cell reaction has reached equilibrium. At equilibrium components in two cells have same free energy .∆G=0.

The relation between Ecell and log Q for the zinc-copper cell. Figure 21.10 The relation between Ecell and log Q for the zinc-copper cell.

Concentration cells: In a concentration cell, the half reactions are same , but the concentrations are different. Ecell depends on ratio of concs. E0cell=0.

Overall (cell) reaction Figure 21.11 A concentration cell based on the Cu/Cu2+ half-reaction. Oxidation half-reaction Cu(s) Cu2+(aq, 0.1M) + 2e- Reduction half-reaction Cu2+(aq, 1.0M) + 2e- Cu(s) Overall (cell) reaction Cu2+(aq,1.0M) Cu2+(aq, 0.1M) Identical color at end and electrode change

Sample Problem 21.7: Calculating the Potential of a Concentration Cell PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell B, electrode B dips into 4.0x10-4M AgNO3. What is the cell potential at 298 K? Which electrode has a positive charge? PLAN: E0cell will be zero since the half-cell potentials are equal. Ecell is calculated from the Nernst equation with half-cell A (higher [Ag+]) having Ag+ being reduced and plating out, and in half-cell B Ag(s) will be oxidized to Ag+. SOLUTION: Ag+(aq, 0.010M) half-cell A Ag+(aq, 4.0x10-4M) half-cell B Ecell = E0cell - 0.0592V 1 log [Ag+]dilute [Ag+]concentrated Ecell = 0 V -0.0592 log 4.0x10-2 = 0.0828V Half-cell A is the cathode and has the positive electrode.

The laboratory measurement of pH. Figure 21.12 The laboratory measurement of pH. Pt Reference (calomel) electrode Glass electrode Hg Paste of Hg2Cl2 in Hg AgCl on Ag on Pt KCl solution 1M HCl Porous ceramic plugs Thin glass membrane

Table 21.3 Some Ions Measured with Ion-Specific Electrodes Species Detected Typical Sample NH3/NH4+ Industrial wastewater, seawater CO2/HCO3- Blood, groundwater F- Drinking water, urine, soil, industrial stack gases Br- Grain, plant tissue I- Milk, pharmaceuticals NO3- Soil, fertilizer, drinking water K+ Blood serum, soil, wine H+ Laboratory solutions, soil, natural waters

batteries battery is a galvanic cell or a group of galvanic cells connected in series.

Figure 21.13 Alkaline Battery

Mercury and Silver (Button) Batteries Figure 21.14 Mercury and Silver (Button) Batteries

Figure 21.15 Lithium battery.

Figure 21.16 Lead-acid battery.

Nickel-metal hydride (Ni-MH) battery Figure 21.17 Nickel-metal hydride (Ni-MH) battery

Figure 21.18 Lithium-ion battery.

Figure 21.19 Fuel cell.

Figure 21.24 The tin-copper reaction as the basis of a voltaic and an electrolytic cell. voltaic cell electrolytic cell Oxidation half-reaction Sn(s) Sn2+(aq) + 2e- Oxidation half-reaction Cu(s) Cu2+(aq) + 2e- Reduction half-reaction Cu2+(aq) + 2e- Cu(s) Reduction half-reaction Sn2+(aq) + 2e- Sn(s) Overall (cell) reaction Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s) Overall (cell) reaction Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)

ELECTROLYTIC(recharge) Figure 21.25 The processes occurring during the discharge and recharge of a lead-acid battery. VOLTAIC(discharge) ELECTROLYTIC(recharge)

Table 21.4 Comparison of Voltaic and Electrolytic Cells Electrode Cell Type DG Ecell Name Process Sign Voltaic < 0 > 0 Anode Oxidation - Voltaic < 0 > 0 Cathode Reduction + Electrolytic > 0 < 0 Anode Oxidation + Electrolytic > 0 < 0 Cathode Reduction -

Sample Problem 21.8: Predicting the Electrolysis Products of a Molten Salt Mixture PROBLEM: A chemical engineer melts a naturally occurring mixture of NaBr and MgCl2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction. PLAN: Consider the metal and nonmetal components of each compound and then determine which will recover electrons(be reduced; strength as an oxidizing agent) better. This is the converse to which of the elements will lose electrons more easily (lower ionization energy). SOLUTION: Possible oxidizing agents: Na+, Mg2+ Possible reducing agents: Br-, Cl- Na, the element, is to the left of Mg in the periodic table, therefore the IE of Mg is higher than that of Na. So Mg2+ will more easily gain electrons and is the stronger oxidizing agent. Br, as an element, has a lower IE than does Cl, and therefore will give up electrons as Br- more easily than will Cl-. Mg2+(l) + 2Br-(l) Mg(s) + Br2(g) cathode anode

Overall (cell) reaction Figure 21.26 The electrolysis of water. Overall (cell) reaction 2H2O(l) H2(g) + O2(g) Oxidation half-reaction 2H2O(l) 4H+(aq) + O2(g) + 4e- Reduction half-reaction 2H2O(l) + 4e- 2H2(g) + 2OH-(aq)

Sample Problem 21.9: Predicting the Electrolysis Products of Aqueous Ionic Solutions PROBLEM: What products form during electrolysis of aqueous solution of the following salts: (a) KBr; (b) AgNO3; (c) MgSO4? PLAN: Compare the potentials of the reacting ions with those of water, remembering to consider the 0.4 to 0.6V overvoltage. The reduction half-reaction with the less negative potential, and the oxidation half-reaction with the less positive potential will occur at their respective electrodes. SOLUTION: E0 = -2.93V (a) K+(aq) + e- K(s) E0 = -0.42V 2H2O(l) + 2e- H2(g) + 2OH-(aq) The overvoltage would make the water reduction -0.82 to -1.02 but the reduction of K+ is still a higher potential so H2(g) is produced at the cathode. E0 = 1.07V 2Br-(aq) Br2(g) + 2e- 2H2O(l) O2(g) + 4H+(aq) + 4e- E0 = 0.82V The overvoltage would give the water half-cell more potential than the Br-, so the Br- will be oxidized. Br2(g) forms at the anode.

Sample Problem 21.9: Predicting the Electrolysis Products of Aqueous Ionic Solutions continued E0 = -0.80V (b) Ag+(aq) + e- Ag(s) E0 = -0.42V 2H2O(l) + 2e- H2(g) + 2OH-(aq) Ag+ is the cation of an inactive metal and therefore will be reduced to Ag at the cathode. Ag+(aq) + e- Ag(s) The N in NO3- is already in its most oxidized form so water will have to be oxidized to produce O2 at the anode. 2H2O(l) O2(g) + 4H+(aq) + 4e- E0 = -2.37V (c) Mg2+(aq) + 2e- Mg(s) Mg is an active metal and its cation cannot be reduced in the presence of water. So as in (a) water is reduced and H2(g) is produced at the cathode. The S in SO42- is in its highest oxidation state; therefore water must be oxidized and O2(g) will be produced at the anode.

A summary diagram for the stoichiometry of electrolysis. Figure 21.27 A summary diagram for the stoichiometry of electrolysis. MASS (g) of substance oxidized or reduced M(g/mol) AMOUNT (MOL) of substance oxidized or reduced AMOUNT (MOL) of electrons transferred Faraday constant (C/mol e-) balanced half-reaction CHARGE (C) time(s) CURRENT (A)

Sample Problem 21.10: Applying the Relationship Among Current, Time, and Amount of Substance PROBLEM: A technician is plating a faucet with 0.86 g of Cr from an electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed? PLAN: mass of Cr needed SOLUTION: Cr3+(aq) + 3e- Cr(s) divide by M mol of Cr needed 0.86g (mol Cr) (3 mol e-) (52.00 gCr) (mol Cr) = 0.050 mol e- 3mol e-/mol Cr mol of e- transferred 0.050 mol e- (9.65x104 C/mol e-) = 4.8x103 C charge (C) 9.65x104C/mol e- 4.8x103 C 12.5 min (min) (60s) = 6.4C/s = 6.4 A current (A) divide by time