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CHAPTER 18 Electrochemistry OXIDATION–REDUCTION Reactions where electrons are transferred from one atom to another are called oxidation–reduction reactions redox reactions for short Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction Oil Rig OXIDATION & REDUCTION Oxidation is the process that occurs when the oxidation number of an element increases an element loses electrons a compound adds oxygen a compound loses hydrogen a half-reaction has electrons as products Reduction is the process that occurs when the oxidation number of an element decreases an element gains electrons a compound loses oxygen a compound gains hydrogen a half-reaction has electrons as reactants RULES FOR ASSIGNING OXIDATION STATES Rules are in order of priority 1. Free elements have an oxidation state = 0 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 2. Monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = −1 in NaCl 3. (a) The sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0 RULES FOR ASSIGNING OXIDATION STATES 3. (b) The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl2 RULES FOR ASSIGNING OXIDATION STATES 5. In their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority Nonmetal Oxidation State Example F −1 CF4 H +1 CH4 O −2 CO2 Group 7A −1 CCl4 Group 6A −2 CS2 Group 5A −3 NH3 EXAMPLE: DETERMINE THE OXIDATION STATES OF ALL THE ATOMS IN A PROPANOATE ION, C3H5O2– There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b (C3) + (H5) + (O2) = −1 Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order H = +1 O = −2 Note: unlike (C3) + 5(+1) + 2(−2) = −1 charges, oxidation (C3) = −2 states can be C = −⅔ fractions! PRACTICE – ASSIGN AN OXIDATION STATE TO EACH ELEMENT IN THE FOLLOWING • Br2 Br = 0 (Rule 1) • K+ K = +1 (Rule 2) • LiF Li = +1 (Rule 4a) & F = −1 (Rule 5) • CO2 O = −2 (Rule 5) & C = +4 (Rule 3a) • SO42− O = −2 (Rule 5) & S = +6 (Rule 3b) • Na2O2 Na = +1 (Rule 4a) & O = −1 (Rule 3a) Oxidation and Reduction • Oxidation occurs when an atom’s oxidation • state increases during a reaction Reduction occurs when an atom’s oxidation state decreases during a reaction CH4 + 2 O2 → CO2 + 2 H2O −4 +1 0 +4 –2 +1 −2 oxidation reduction Copyright © 2011 Pearson Education, Inc. Oxidation–Reduction • Oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them • The reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized • The reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent Copyright © 2011 Pearson Education, Inc. Common Oxidizing Agents Copyright © 2011 Pearson Education, Inc. Common Reducing Agents Copyright © 2011 Pearson Education, Inc. Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reaction Reducing Agent Oxidizing Agent Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O 0 +7 −2 +1 +3 +4 −2 +1 −2 Reduction Oxidation Copyright © 2011 Pearson Education, Inc. Practice – Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions Sn4+ + Ca → Sn2+ + Ca2+ +4 0 +2 +2 Ca is oxidized, Sn4+ is reduced Ca is the reducing agent, Sn4+ is the oxidizing agent F2 + S → SF4 0 0 +4−1 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent Copyright © 2011 Pearson Education, Inc. 18.2 Balancing Redox Reactions • Some redox reactions can be balanced by the method we previously used, but many are hard to balance using that method many are written as net ionic equations many have elements in multiple compounds • The main principle is that electrons are transferred – so if we can find a method to keep track of the electrons it will allow us to balance the equation Copyright © 2011 Pearson Education, Inc. Half-Reactions • We generally split the redox reaction into two separate half-reactions – a reaction just involving oxidation or reduction the oxidation half-reaction has electrons as products the reduction half-reaction has electrons as reactants 3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+ 0 −1 +1 −2 −1 +5 −2 +1 oxidation: I− → IO3− + 6 e− reduction: Cl2 + 2 e− → 2 Cl− Tro: Chemistry: A Molecular Approach, 2/e 16 Copyright © 2011 Pearson Education, Inc. Balancing Redox Reactions by the Half-Reaction Method • In this method, the reaction is broken down into • two half-reactions, one for oxidation and another for reduction Each half-reaction includes electrons electrons go on the product side of the oxidation halfreaction – loss of electrons electrons go on the reactant side of the reduction halfreaction – gain of electrons • Each half-reaction is balanced for its atoms • Then the two half-reactions are adjusted so that the electrons lost and gained will be equal when combined Copyright © 2011 Pearson Education, Inc. Balancing Redox Reactions 1. assign oxidation states a) determine element oxidized and element reduced 2. write ox. & red. half-reactions, including electrons a) ox. electrons on right, red. electrons on left of arrow 3. balance half-reactions by mass a) b) c) d) first balance elements other than H and O add H2O where need O add H+ where need H if reaction done in Base, neutralize H+ with OH− 4. balance half-reactions by charge a) balance charge by adjusting electrons 5. balance electrons between half-reactions 6. add half-reactions 7. check by counting atoms and total charge Copyright © 2011 Pearson Education, Inc. Example 18.2: Balancing redox reactions in acidic solution 1. assign oxidation states and determine element oxidized and element reduced Fe2+ + MnO4– → Fe3+ + Mn2+ 2. separate into oxidation & reduction halfreactions ox: Fe2+ → Fe3+ +2 +7 −2 +3 +2 oxidation reduction red: MnO4– → Mn2+ Copyright © 2011 Pearson Education, Inc. Example 18.2: Balancing redox reactions in acidic solution 3. balance halfreactions by mass a) first balance atoms other than O and H b) balance O by adding H2O to side that lacks O c) balance H by adding H+ to side that lacks H Fe2+ → Fe3+ MnO4– → Mn2+ MnO4– → Mn2+ + 4H2O MnO4– + 8H+ → Mn2+ + 4H2O Copyright © 2011 Pearson Education, Inc. Example 18.2: Balancing redox reactions in acidic solution 4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) b) electrons on product side for oxidation electrons on reactant side for reduction Fe2+ → Fe3+ + 1 e− MnO4– + 8H+ → Mn2+ + 4H2O +7 +2 MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O Copyright © 2011 Pearson Education, Inc. Example 18.2: Balancing redox reactions in acidic solution 5. balance electrons Fe2+ → Fe3+ + 1 e− } x 5 between halfMnO4– + 8H+ + 5 e− → Mn2+ + 4H2O reactions 6. add half2+ → 5 Fe3+ + 5 e− 5 Fe reactions, canceling MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O electrons and common 5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+ species reactant side Element product side 7. Check that 5 Fe 5 numbers of 1 Mn 1 4 O 4 atoms and 8 H 8 total charge +17 charge +17 are equal Copyright © 2011 Pearson Education, Inc. Practice – Balance the following equation in acidic solution I– + Cr2O72− → Cr3+ + I2 Copyright © 2011 Pearson Education, Inc. Practice – Balancing redox reactions 1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction halfreactions I− + Cr2O72– → I2 + −1 +6 −2 0 Cr3+ +3 oxidation reduction ox: I− → I2 red: Cr2O72– → Cr3+ Copyright © 2011 Pearson Education, Inc. Practice – Balancing redox reactions 3. balance halfreactions by mass a) first balance atoms other than O and H b) balance O by adding H2O to side that lacks O c) balance H by adding H+ to side that lacks H ox: I − → I2 ox: 2 I − → I2 red: Cr2O72– → Cr3+ red: Cr2O72– → 2 Cr3+ red: Cr2O72– → 2Cr3+ +7H2O Cr2O72– +14H+ → 2Cr3+ +7H2O Copyright © 2011 Pearson Education, Inc. Practice – Balancing redox reactions 4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) b) electrons on product side for oxidation electrons on reactant side for reduction 2 I− → I2 + 2e− Cr2O72– +14H+ → 2Cr3+ +7H2O Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O Copyright © 2011 Pearson Education, Inc. Practice – Balancing redox reactions 5. 6. 7. balance 2 I− → I2 + 2e− electrons between 6 I− → 3 I2 + 6e−} x 3 halfreactions 2– +14H+ + 6e−→ 2Cr3+ +7H O Cr O 2 7 2 add halfreactions, canceling Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2 electrons reactant side Element product side and 2 Cr 2 common 6 I 6 species 7 O 7 check 14 H 14 +6 charge +6 Copyright © 2011 Pearson Education, Inc. EXAMPLE 18.3 Balancing Redox Reactions Occurring in Basic Solution Balance the equation occurring in basic solution: SOLUTION To balance redox reactions occurring in basic solution, follow the half-reaction method outlined above, but add an extra step to neutralize the acid with OH– as shown in step 3 below. 1. Assign oxidation states. 2. Separate the overall reaction into two halfreactions. Oxidation: Reduction: 3. Balance each half-reaction with respect to mass. • Balance all elements other than H and O. • Balance O by adding H2O. • Balance H by adding H+. • Neutralize H+ by adding enough OH– to neutralize each H+. Add the same number of OH– ions to each side of the equation. © 2011 Pearson Education, Inc. continued… Copyright © 2011 Pearson Education, Inc. 4. Balance each half-reaction with respect to charge. 5. Make the number of electrons in both halfreactions equal. 6. Add the half-reactions together. 7. Verify that the reaction is balanced. FOR PRACTICE 18.3 Balance the following redox reaction occurring in basic solution. © 2011 Pearson Education, Inc. Copyright © 2011 Pearson Education, Inc. 18.3 VOLTAIC (OR GALVANIC) CELLS: GENERATING ELECTRICITY FROM SPONTANEOUS CHEMICAL REACTIONS Electrical Current When we talk about the current of a liquid in a stream, we are discussing the amount of water that passes by in a given period of time. When we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time. whether as electrons flowing through a wire, or ions flowing through a solution. REDOX REACTIONS GENERATING CURRENT Redox reactions involve the transfer of electrons from one substance to another. Therefore, redox reactions have the potential to generate an electric current. To use that current, we need to separate the place where oxidation is occurring from the place where reduction is occurring. A VOLTAIC (GALVANIC) CELL ELECTROCHEMISTRY Electrochemistry is the study of redox reactions that produce or require an electric current. The conversion between chemical energy and electrical energy is carried out in an electrochemical cell. Spontaneous redox reactions take place in a voltaic cell. aka galvanic cells Nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy . COMPONENTS OF AN ELECTROCHEMICAL CELL • Oxidation and reduction half-reactions are kept separate in halfcells. • Electron flow through a wire along with ion flow through a solution constitutes an electric circuit. • Requires a conductive solid electrode to allow the transfer of electrons. through external circuit metal or graphite • requires ion exchange between the two half-cells of the system electrolyte (salt bridge for galvanic cell) THE ELECTRODES Anode Oxidation occurs at anode In a voltaic cell, anode is labelled (-) with a negative sign because the anode releases the electrons. Cathode Reduction occurs at cathode In a voltaic cell, cathode is labelled (+) with a positive sign because cathode (attracts) receives the electrons. Electrons are (-) and will like to travel to anything with a (+) (cathode in this case). Electrons are released from the oxidation process at the anode. A VOLTAIC (GALVANIC) CELL The salt bridge is required to complete the circuit and maintain charge balance. CURRENT Current is the number of electrons that flow through the system per second unit = Ampere 1 A of current = 1 Coulomb of charge flowing by each second 1 A = 6.242 x 1018 electrons per second Electrode surface area dictates the number of electrons that can flow larger batteries produce larger currents VOLTAGE The difference in potential energy between the reactants and products is the potential difference unit = Volt 1 V = 1 J of energy per Coulomb of charge the voltage needed to drive electrons through the external circuit Amount of force pushing the electrons through the wire is called the electromotive force, emf CELL POTENTIAL The difference in potential energy between the anode and the cathode in a voltaic cell is called the cell potential. The cell potential depends on the relative ease with which the oxidation and reduction happens. The cell potential under standard conditions is called the standard emf, E°cell 25 °C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions ELECTROCHEMICAL CELL NOTATION Shorthand description of a voltaic cell Electrode | electrolyte || electrolyte | electrode Oxidation half-cell on the left, reduction half-cell on the right Single | = phase barrier if multiple electrolytes in same phase, a comma is used rather than | often use an inert electrode Double line || = salt bridge VOLTAIC CELL Anode = Zn(s) the anode is oxidized to Zn2+ Cathode = Cu(s) Cu2+ ions are reduced at the cathode Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) ELECTRODES Many times the anode is made of the metal that is oxidized and the cathode is made of the same metal as is produced by the reduction. However, if the redox reaction we are running involves the oxidation or reduction of an ion to a different oxidation state, or the oxidation or reduction of a gas, we may use an inert electrode. an inert electrode is one that not does participate in the reaction, but just provides a surface for the transfer of electrons to take place on Because the half-reaction involves reducing the Mn oxidation state from +7 to +2, we use an electrode that will provide a surface for the electron transfer without reacting with the MnO4−. Platinum works well because it is extremely nonreactive and conducts electricity. Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s) 18.4 STANDARD REDUCTION POTENTIALS A half-reaction with a strong tendency to occur has a large positive half-cell potential. When two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur. We cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction. We select as a standard halfreaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 V standard hydrogen electrode, SHE Zn(s) | Zn2+(1 M) || H+(1 M) | H2(g) | Pt(s) HALF-CELL POTENTIALS SHE reduction potential is defined to be exactly 0 V. Standard Reduction Potentials compare the tendency for a particular reduction half-reaction to occur relative to the reduction of H+ to H2 under standard conditions Half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red Half-reactions with a weaker tendency toward reduction than the SHE have a value for E°red For an oxidation half-reaction, E°oxidation = − E°reduction CALCULATING CELL POTENTIALS UNDER STANDARD CONDITIONS E°cell = E°oxidation + E°reduction When adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the halfreactions to balance the equation This is due to the fact that Volt = Joules/Coulomb (potential energy per unit charge) (It is an intensive property like density) Example 18.4: Calculate Ecell for the reaction at 25 C Al(s) + NO3−(aq) + 4 H+(aq) Al3+(aq) + NO(g) + 2 H2O(l) separate the ox (anode): Al(s) Al3+(aq) + 3 e− reaction into the oxidation red: NO3−(aq) + 4 H+(aq) + 3 e− NO(g) + 2 H2O(l) and reduction half-reactions find the E for Eox of Al = −Ered of Al3+ = +1.66 V each halfEred of NO3− = +0.96 V reaction and E = (+1.66 V) + (+0.96 V) = +2.62 V cell sum to get Ecell Copyright © 2011 Pearson Education, Inc. Practice – Calculate Ecell for the reaction at 25 C IO3– (aq) + 6 H+(aq) + 5 I−(aq) → 3 I2(s) + 3 H2O(l) Reduction Half-Reaction Ered, V F2(g) + 2e− 2 F−(aq) +2.87 IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e− Ag(s) +0.80 I2(s) + 2e− 2 I−(aq) +0.54 Cu2+(aq) + 2e− Cu(s) +0.34 Cr3+(aq) + 1e− Cr2+(aq) −0.50 Mg2+(aq) + 2e− Mg(s) −2.37 Copyright © 2011 Pearson Education, Inc. Practice – Calculate Ecell for the reaction at 25 C 2 IO3– + 12 H+ + 10 I−→ 6 I2 + 6 H2O separate the ox (anode): 2 I−(s) I2(aq) + 2 e− reaction into the oxidation red: IO3−(aq) + 6 H+(aq) + 5 e− ½ I2(s) + 3 H2O(l) and reduction half-reactions find the E for Eox of I− = −Ered of I2 = −0.54 V each halfEred of IO3− = +1.20 V reaction and E = (−0.54 V) + (+1.20 V) = +0.66 V cell sum to get Ecell Copyright © 2011 Pearson Education, Inc. PREDICTING THE SPONTANEOUS DIRECTION OF AN OXIDATION-REDUCTION REACTION A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table. If paired the other way, the reverse reaction is spontaneous. Cu2+(aq) + 2 e− Cu(s) Ered = +0.34 V Zn2+(aq) + 2 e− Zn(s) Ered = −0.76 V Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) spontaneous Cu(s) + Zn2+(aq) Cu2+(aq) + Zn(s) nonspontaneous Example 18.5: Predict if the following reaction is spontaneous under standard conditions Fe(s) + Mg2+(aq) Fe2+(aq) + Mg(s) separate the ox: reaction into the oxidation red: and reduction half-reactions look up the relative positions of the reduction half-reactions red: red: Fe(s) Fe2+(aq) + 2 e− Mg2+(aq) + 2 e− Mg(s) Fe2+(aq) + 2 e− Fe(s) Mg2+(aq) + 2 e− Mg(s) because Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written Copyright © 2011 Pearson Education, Inc. the reaction is Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s) spontaneous in ox: Mg(s) Mg2+(aq) + 2 e− the reverse red: Fe2+(aq) + 2 e− Fe(s) direction sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode Copyright © 2011 Pearson Education, Inc. Practice – Decide whether each of the following will be spontaneous as written or in the reverse direction F2(g) + 2 I−(aq) I2(s) + 2 F−(aq) Reduction Half-Reaction F2(g) + 2e− 2 F−(aq) spontaneous as written Mg(s) + 2 Ag+(aq) Mg2+(aq) + 2 Ag(s) spontaneous as written Cu2+(aq) + 2e− Cu(s) Cr3+(aq) + 1e− Cr2+(aq) spontaneous in reverse +2 Cr2+(aq) Cu(s) + 2 Ag+(aq) + 1e− Ag(s) I2(s) + 2e− 2 I−(aq) Cu2+(aq) + 2 I−(aq) I2(s) + Cu(s) Cu2+(aq) IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l) Cr3+(aq) Mg2+(aq) + 2e− Mg(s) spontaneous as written Copyright © 2011 Pearson Education, Inc. Practice – Which of the following materials can be used to oxidize Cu without oxidizing Ag? a) F− b) I− c) I2 d) Cr3+ Reduction Half-Reaction F2(g) + 2e− 2 F−(aq) IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l) Ag+(aq) + 1e− Ag(s) I2(s) + 2e− 2 I−(aq) Cu2+(aq) + 2e− Cu(s) Cr3+(aq) + 1e− Cr2+(aq) Mg2+(aq) + 2e− Mg(s) Copyright © 2011 Pearson Education, Inc. Practice – Sketch and label a voltaic cell in which one half-cell has Ag(s) immersed in 1 M AgNO3, and the other half-cell has a Pt electrode immersed in 1 M Cr(NO3)2 and 1 M Cr(NO3)3 Write the half-reactions and overall reaction, and determine the cell potential under standard conditions. Reduction Half-Reaction Ered, V F2(g) + 2e− 2 F−(aq) +2.87 IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e− Ag(s) +0.80 I2(s) + 2e− 2 I−(aq) +0.54 Cu2+(aq) + 2e− Cu(s) +0.34 Cr3+(aq) + 1e− Cr2+(aq) −0.50 Mg2+(aq) + 2e− Mg(s) −2.37 Copyright © 2011 Pearson Education, Inc. e− → e− → e− → e− → salt bridge anode = Pt cathode = Ag Cr2+ Ag+ Cr3+ ox: Cr2+(aq) Cr3+(aq) + 1 e− red: Ag+(aq) + 1 e− Ag(s) E = +0.50 V E tot: Cr2+(aq) + Ag+(aq) Cr3+(aq) + Ag(s) = +0.80 V E = +1.30 V Copyright © 2011 Pearson Education, Inc. PREDICTING WHETHER A METAL WILL DISSOLVE IN AN ACID Metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid as a single displacement reaction Almost all metals will dissolve in HNO3 having N reduced rather than H Au and Pt dissolve in HNO3 + HCl NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l) Practice – Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction. Reduction Half-Reaction Au3+(aq) + 3e− Au(s) a) Ag b) Cu Cu2+(aq) + 2e− Cu(s) c) Fe 2H+(aq) + 2e− H2(g) d) Cr Cr3+(aq) + 3e− Cr(s) Ag+(aq) + 1e− Ag(s) Fe3+(aq) + 3e− Fe(s) Mg2+(aq) + 2e− Mg(s) c) 2 Fe(s) + 6 HC2H3O2(aq) → 2 Fe(C2H3O2)3(aq) + 3 H2(g) d) 2 Cr(s) + 6 HC2H3O2(aq) → 2 Cr(C2H3O2)3(aq) + 3 H2(g) Copyright © 2011 Pearson Education, Inc. 18.5 CELL POTENTIAL, FREE ENERGY AND K (EQUILIBRIUM CONSTANT) For a spontaneous reaction one that proceeds in the forward direction with the chemicals in their standard states DG° < 1 (negative) E° > 1 (positive) K>1 DG° = −RTlnK = −nFE°cell n is the number of electrons F = Faraday’s Constant = 96,485 C/mol e− EXAMPLE 18.6 Relating ∆G and Ecell Use the tabulated electrode potentials to calculate ∆G for the reaction: Is the reaction spontaneous? SORT You are given a redox reaction and asked to find ∆G. STRATEGIZE Use the tabulated values of electrode potentials to calculate Ecell. Then use Equation 18.3 to calculate ∆G from Ecell. CONCEPTUAL PLAN SOLVE Break the reaction up into oxidation and reduction half-reactions and find the standard electrode potentials for each. Find Ecell by subtracting Ean from Ecat. SOLUTION Oxidation (Anode) : Reduction (Cathode) : 2 Br (aq) Br22 (l ) 2 e I 22 ( s) 2 e 2 I (aq) I 22 ( s) 2 Br (aq) 2 I ( aq) Br22 (l ) E 1.09 V E 0.54 V Ecell cell Ecat cat Ean an 0.55 V Calculate ∆G from Ecell. The value of n (the number of moles of electrons) corresponds to the number of electrons that are canceled in the half-reactions. Remember that 1 V = 1 J/C. Since ∆G is positive, the reaction is not spontaneous under standard conditions. continued… © 2011 Pearson Education, Inc. EXAMPLE 18.7 Relating Ecell and K Use the tabulated electrode potentials to calculate K for the oxidation of copper by H+ (at 25 C). SORT You are given a redox reaction and asked to find K. GIVEN: FIND: K STRATEGIZE Use the tabulated values of electrode potentials to calculate Ecell. Then use Equation 18.6 to calculate ∆G from Ecell. CONCEPTUAL PLAN SOLVE Break the reaction up into oxidation and reduction half-reactions and find the standard electrode potentials for each. Find Ecell by subtracting Ean from Ecat. SOLUTION Calculate K from Ecell. The value of n (the number of moles of electrons) corresponds to the number of electrons that are canceled in the half-reactions. continued… © 2011 Pearson Education, Inc. 18.6 CELL POTENTIAL AND CONCENTRATION We know there is a relationship between the reaction quotient, Q, the equilibrium constant, K, and the free energy change, DGº. Changing the concentrations of the reactants and products so they are not 1 M will affect the standard free energy change, DGº . Because DGº determines the cell potential, Ecell, the voltage for the cell will be different when the ion concentrations are not 1 M. NERNST EQUATION EXAMPLE 18.8 Calculating Ecell Under Nonstandard Conditions Determine the cell potential for an electrochemical cell based on the following two half-reactions: Oxidation: Reduction: SORT You are given the half-reactions for a redox reaction and the concentrations of the aqueous reactants and products. You are asked to find the cell potential. GIVEN: FIND: Ecell STRATEGIZE Use the tabulated values of electrode potentials to calculate Ecell. Then use Equation 18.9 to calculate Ecell. CONCEPTUAL PLAN SOLVE Write the oxidation and reduction half-reactions, multiplying by the appropriate coefficients to cancel the electrons. Find the standard electrode potentials for each. Find Ecell. SOLUTION continued… © 2011 Pearson Education, Inc. Calculate Ecell from Ecell.The value of n (the number of moles of electrons) corresponds to the number of electrons (6 in this case) canceled in the half-reactions. Determine Q based on the overall balanced equation and the given concentrations of the reactants and products. (Note that pure liquid water, solid MnO2, and solid copper are omitted from the expression for Q.) CHECK The answer has the correct units (V). The value of Ecell is larger than , as expected based on Le Châtelier’s principle because one of the aqueous reactants has a concentration greater than standard conditions and the one aqueous product has a concentration less than standard conditions. Therefore the reaction has a greater tendency to proceed toward products and has a greater cell potential. © 2011 Pearson Education, Inc. SUMMARIZING ELECTROCHEMICAL CELLS In all electrochemical cells, Anode is where OXIDATION happens Cathode is where REDUCTION happens In voltaic cells, the source of electrons is anode, and anode is – In voltaic cells, the cathode draws electrons and has a + sign. In electrolytic cells, anode = oxidation, but an external power source is hooked to anode with the + terminal connected to anode (to pull electrons away from anode) In electrolytic cells, cathode = reduction, but an external power source is hooked to cathode with the – terminal connected to cathode (to force electrons towards cathode) ELECTROLYSIS Electrolysis is the process of using electrical energy to break a compound apart Electrolysis is done in an electrolytic cell Electrolytic cells can be used to separate elements from their compounds ELECTROLYSIS In electrolysis we use electrical energy to overcome the energy barrier of a non-spontaneous reaction, allowing it to occur The reaction that takes place is the opposite of the spontaneous process 2 H2(g) + O2(g) 2 H2O(l) spontaneous 2 H2O(l) 2 H2(g) + O2(g) electrolysis Some applications are (1) metal extraction from minerals and purification, (2) production of H2 for fuel cells, (3) metal plating ELECTROLYTIC CELLS The electrical energy is supplied by a Direct Current power supply AC alternates the flow of electrons so the reaction won’t be able to proceed Some electrolysis reactions require more voltage than Ecell predicts. This is called the overvoltage ELECTROLYTIC CELLS The source of energy is a battery or DC power supply The + terminal of the source is attached to the anode The − terminal of the source is attached to the cathode Electrolyte can be either an aqueous salt solution or a molten ionic salt Cations in the electrolyte are attracted to the cathode and anions are attracted to the anode Cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized ELECTROLYSIS OF PURE COMPOUNDS The compound must be in molten (liquid) state Electrodes normally graphite Cations are reduced at the cathode to metal element Anions oxidized at anode to nonmetal element ELECTROLYSIS OF NACL(L) ELECTROLYSIS OF WATER ELECTROPLATING In electroplating, the work piece is the cathode Cations are reduced at cathode and plate to the surface of the work piece The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution. 80 MIXTURES OF IONS When more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode most positive E°red (on top) When more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode most negative E°red (on bottom) EXAMPLE 18.9A: PREDICT THE HALF-REACTION OCCURRING AT THE ANODE AND CATHODE FOR THE ELECTROLYSIS OF A MIXTURE OF MOLTEN ALBR3(L) AND MGBR2(L) in the electrolysis of a molten salt, the anion is oxidized and the cation is reduced ox: 2 Br−(l) → Br2(g) + 2 e− red: Al3+(aq) + 3 e− → Al(s) Mg2+(aq) + 2 e− → Mg(s) use the values of E° of the half-reactions as a guide to determine which halfreaction is easiest red: Al3+(aq) + 3 e− → Al(s) Mg2+(aq) + 2 e− → Mg(s) E°= −1.66 V E°= −2.37 V Top reduction is preferred, Al 3+ will reduce to Al. oxidation occurs at the anode: anode and reduction cathode: at the cathode 2 Br−(l) → Br2(g) + 2 e− Al3+(aq) + 3 e− → Al(s) PRACTICE – PREDICT THE HALF-REACTION OCCURRING AT THE ANODE AND CATHODE FOR THE ELECTROLYSIS OF A MIXTURE OF MOLTEN MGCL2(L) AND MGBR2(L) in the electrolysis of a molten salt, the anion is oxidized and the cation is reduced ox: use the values of E° of the half-reactions as a guide to determine which halfreaction is easiest ox: red: 2 Cl−(l) Cl2(g) + 2 e− 2 Br−(l) Br2(g) + 2 e− Mg2+(aq) + 2 e− Mg(s) Cl2(g) + 2 e− 2Cl−(l) E°= 1.36 V Br2(g) + 2 e− 2 Br−(l) E°= 1.09 V Br is on bottom, Br - is preferred for oxidation to Br. oxidation occurs at the anode: anode and reduction cathode: at the cathode 2 Br−(l) Br2(g) + 2 e− Mg2+(aq) + 2 e− Mg(s) ELECTROLYSIS OF AQUEOUS SOLUTIONS Possible cathode reactions reduction of cation to metal reduction of water to H2 2 H2O + 2 e− H2 + 2 OH− Possible anode reactions oxidation of anion to element oxidation of H2O to O2 O2 + 4 e− + 4H+ 2 H2O Ered° = −0.83 v @ stand. cond. Ered° = −0.41 v @ pH 7 Ered° = 1.23 v @ stand. cond. Ered° = 0.82 v @ pH 7 EXAMPLE 18.9B: PREDICT THE HALF-REACTION OCCURRING AT THE ANODE AND CATHODE FOR THE ELECTROLYSIS LII(AQ) in the electrolysis of an aqueous salt, the anion or water (or electrode) is oxidized, and the cation or water is reduced ox: 2 I−(aq) I2(s) + 2 e− 2 H2O(l) O2(g) + 4e− + 4H+(aq) red: Li+(aq) + 1 e− Li(s) 2 H2O(l) + 2 e− H2(g) + 2 OH−(aq) use the values of E° of the half-reactions as a guide to determine which halfreaction is easiest ox: O2 + 4e− + 4H+2 H2O I2(s) + 2 e− 2 I−(aq) E°= 0.82 V E°= 0.54 V red: 2 H2O + 2 e− H2 + 2 OH− E°= −0.41 V Li+(aq) + 1 e− Li(s) E°= −3.04 V oxidation occurs at the anode: anode and reduction cathode: at the cathode 2 I−(aq) I2(s) + 2 e− 2 H2O(l) + 2 e− H2(g) + 2 OH−(aq) PRACTICE – PREDICT THE HALF-REACTION OCCURRING AT THE ANODE AND CATHODE FOR THE ELECTROLYSIS NABR(AQ) in the electrolysis of an aqueous salt, the anion or water (or electrode) is oxidized, and the cation or water is reduced ox: 2 Br−(aq) Br2(l) + 2 e− 2 H2O(l) O2(g) + 4 e− + 4H+(aq) red: Na+(aq) + 1 e− Na(s) 2 H2O(l) + 2 e− H2(g) + 2 OH−(aq) use the values of E° of the half-reactions as a guide to determine which halfreaction is easiest ox: Br2(s) + 2 e− 2 Br−(aq) O2 + 4e− + 4H+2 H2O red: 2 H2O + 2 e− H2 + 2 OH− Na+(aq) + 1 e− Na(s) oxidation occurs at the anode: anode and reduction cathode: at the cathode Tro: Chemistry: A Molecular Approach, 2/e E°= 1.09 V E°= 0.82 V E°= −0.41 V E°= −3.04 V 2 H2O(l) O2(g) + 4 e− + 4H+(aq) 2 H2O(l) + 2 e− H2(g) + 2 OH−(aq) 86 STOICHIOMETRY OF ELECTROLYSIS In an electrolytic cell, the amount of product made is related to the number of electrons transferred essentially, the electrons are a reactant The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time 1 Amp = 1 Coulomb of charge/second 1 mole of e− = 96,485 Coulombs of charge Faraday’s constant EXAMPLE 18.10: CALCULATE THE MASS OF AU THAT CAN BE PLATED IN 25 MIN USING 5.5 A FOR THE HALF-REACTION AU3+(AQ) + 3 E− → AU(S) Given: Find: 3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g Conceptual t(s), amp Plan: charge (C) mol e− mol Au g Au Relationships: Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e− PRACTICE – CALCULATE THE AMPERAGE REQUIRED TO PLATE 2.5 G OF AU IN 1 HOUR (3600 SEC) AU3+(AQ) + 3 E− → AU(S) Given: Find: Conceptual Plan: 3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s current, A g Au mol Au mol e− charge amps Relationships: Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−