Chapter 18 Electrochemistry

Chapter 18 Electrochemistry
Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 18 Electrochemistry Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Harnessing the Power in Nature
The goal of scientific research is to understand nature Once we understand the nature of something, we can then use it efficiently The average U.S. household currently consumes 1000 kWh of electricity per month Almost all of this electricity is generated at a remote power plant by either the combustion of fossil fuels or nuclear fission Tro: Chemistry: A Molecular Approach, 2/e

Electricity from Chemistry
Many chemical reactions involve the transfer of electrons between atoms or ions electron transfer reactions all single displacement, and combustion reactions some synthesis and decomposition reactions The flow of electrons is associated with electricity Basic research into the nature of this relationship may, in the near future, lead to cheaper, more efficient ways of generating electricity fuel cell Tro: Chemistry: A Molecular Approach, 2/e

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
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 Oil Rig 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 CCl4 Group 6A CS2 Group 5A −3 NH3 Tro: Chemistry: A Molecular Approach, 2/e

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 (C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔ Note: unlike charges, oxidation states can be fractions! Tro: Chemistry: A Molecular Approach, 2/e

Practice – Assign an oxidation state to each element in the following
Br2 K+ LiF CO2 SO42− Na2O2 Br = 0 (Rule 1) K = +1 (Rule 2) Li = +1 (Rule 4a) & F = −1 (Rule 5) O = −2 (Rule 5) & C = +4 (Rule 3a) O = −2 (Rule 5) & S = +6 (Rule 3b) Na = +1 (Rule 4a) & O = −1 (Rule 3a) Tro: Chemistry: A Molecular Approach, 2/e 10

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 CH O2 → CO2 + 2 H2O − – −2 oxidation reduction Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Common Oxidizing Agents
Tro: Chemistry: A Molecular Approach, 2/e

Common Reducing Agents

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
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+ + MnO H2O +7 −2 +1 +3 +4 −2 +1 −2 Oxidation Reduction Tro: Chemistry: A Molecular Approach, 2/e

Practice – Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions Sn4+ + Ca → Sn Ca2+ F2 + S → SF4 Ca is oxidized, Sn4+ is reduced Ca is the reducing agent, Sn4+ is the oxidizing agent −1 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Half-Reactions 3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+
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+ − − − −2 +1 oxidation: I− → IO3− + 6 e− reduction: Cl2 + 2 e− → 2 Cl− Tro: Chemistry: A Molecular Approach, 2/e

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 half- reaction – loss of electrons electrons go on the reactant side of the reduction half- reaction – 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 Tro: Chemistry: A Molecular Approach, 2/e

Example 18.2: Balancing redox reactions
in acidic solution 1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction half- reactions Fe2+ + MnO4– → Fe3+ + Mn2+ +2 +7 −2 +3 +2 oxidation reduction ox: Fe2+ → Fe3+ red: MnO4– → Mn2+ Tro: Chemistry: A Molecular Approach, 2/e

in acidic solution 3. balance half-reactions 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 Tro: Chemistry: A Molecular Approach, 2/e

in acidic solution 4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction Fe2+ → Fe e− MnO4– + 8H+ → Mn2+ + 4H2O +7 +2 MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O Tro: Chemistry: A Molecular Approach, 2/e

in acidic solution 5. balance electrons between half-reactions 6. add half-reactions, canceling electrons and common species 7. Check that numbers of atoms and total charge are equal Fe2+ → Fe e− } x 5 MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O 5 Fe2+ → 5 Fe e− MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O 5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+ reactant side Element product side 5 Fe 1 Mn 4 O 8 H +17 charge Tro: Chemistry: A Molecular Approach, 2/e

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) first balance elements other than H and O b) add H2O where need O c) add H+ where need H d) 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Balance the following equation in acidic solution I– + Cr2O72− → Cr3+ + I2

Practice – Balancing redox reactions
1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction half- reactions I− + Cr2O72– → I Cr3+ −1 +6 −2 +3 oxidation reduction ox: I− → I2 red: Cr2O72– → Cr3+ Tro: Chemistry: A Molecular Approach, 2/e 26

3. balance half-reactions 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 Tro: Chemistry: A Molecular Approach, 2/e 27

4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction 2 I− → I2 + 2e− 2 I− → I2 Cr2O72– +14H+ → 2Cr3+ +7H2O Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O Tro: Chemistry: A Molecular Approach, 2/e

5. balance electrons between half-reactions 6. add half-reactions, canceling electrons and common species 7. check 6 I− → 3 I2 + 6e− 2 I− → I2 + 2e− } x 3 Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2 reactant side Element product side 2 Cr 6 I 7 O 14 H +6 charge Tro: Chemistry: A Molecular Approach, 2/e 29

assign oxidation states I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
Example 18.3: Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution assign oxidation states I(aq) + MnO4(aq)  I2(aq) + MnO2(s) separate into half-reactions ox: red: assign oxidation states separate into half-reactions ox: I(aq)  I2(aq) red: MnO4(aq)  MnO2(s) − − −2 Reduction Oxidation Tro: Chemistry: A Molecular Approach, 2/e

Example 18.3: Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
balance half-reactions by mass in base, neutralize the H+ with OH− ox: 2 I(aq)  I2(aq) red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) 4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq) 4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq) 2 H2O(l) + MnO4(aq)  MnO2(s) + 4 OH(aq) balance half-reactions by mass in base, neutralize the H+ with OH− ox: 2 I(aq)  I2(aq) red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) 4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq) 4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq) balance half-reactions by mass in base, neutralize the H+ with OH− ox: 2 I(aq)  I2(aq) red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) 4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq) balance half-reactions by mass first, elements other than H and O ox: 2 I(aq)  I2(aq) red: MnO4(aq)  MnO2(s) balance half-reactions by mass ox: I(aq)  I2(aq) red: MnO4(aq)  MnO2(s) balance half-reactions by mass then O by adding H2O ox: 2 I(aq)  I2(aq) red: MnO4(aq)  MnO2(s) + 2 H2O(l) balance half-reactions by mass then H by adding H+ ox: 2 I(aq)  I2(aq) red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 31

balance Half-reactions by charge ox: 2 I(aq)  I2(aq) + 2 e red: MnO4(aq) + 2 H2O(l) + 3 e− MnO2(s) + 4 OH(aq) Balance electrons between half-reactions ox: 2 I(aq)  I2(aq) + 2 e } x3 red: MnO4(aq) + 2 H2O(l) + 3 e−  MnO2(s) + 4 OH(aq) }x2 ox: 6 I(aq)  3 I2(aq) + 6 e red: 2 MnO4(aq) + 4 H2O(l) + 6 e− 2 MnO2(s) + 8 OH(aq) balance Half-reactions by charge ox: 2 I(aq)  I2(aq) red: MnO4(aq) + 2 H2O(l)  MnO2(s) + 4 OH(aq) Balance electrons between half-reactions balance Half-reactions by charge ox: 2 I(aq)  I2(aq) + 2 e red: MnO4(aq) + 2 H2O(l) + 3 e− MnO2(s) + 4 OH(aq) Balance electrons between half-reactions Tro: Chemistry: A Molecular Approach, 2/e

add the half-rxns ox: 6 I(aq)  3 I2(aq) + 6 e red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq) tot: 6 I(aq)+ 2 MnO4(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH(aq) check add the half-rxns ox: 6 I(aq)  3 I2(aq) + 6 e red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq) check Reactant Count Element Product 6 I 2 Mn 12 O 8 H 8 charge Tro: Chemistry: A Molecular Approach, 2/e 33

Practice – Balance the following equation in basic solution I– + CrO42− → Cr3+ + I2

1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction half- reactions I− + CrO42– → I Cr3+ −1 +6 −2 +3 oxidation reduction ox: I− → I2 red: CrO42– → Cr3+ Tro: Chemistry: A Molecular Approach, 2/e 35

3. balance half-reactions 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 d) if in basic solution, neutralize the H+ with OH− ox: I− → I2 ox: 2 I− → I2 red: CrO42– → Cr3+ red: CrO42– → Cr3+ +4H2O CrO42– +8H+ → Cr3+ +4H2O CrO42– +8H2O → Cr3+ +4H2O +8OH− CrO42– +8H+ +8OH− → Cr3+ +4H2O +8OH− CrO42– +4H2O → Cr3+ +8OH− Tro: Chemistry: A Molecular Approach, 2/e 36

4. balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction 2 I− → I2 + 2e− 2 I− → I2 CrO42– +4H2O+ 3e− → Cr3+ +8OH− CrO42– +4H2O → Cr3+ +8OH− Tro: Chemistry: A Molecular Approach, 2/e

5. balance electrons between half-reactions 6. add half-reactions, canceling electrons and common species 7. check 6 I− → 3 I2 + 6e− 2 I− → I2 + 2e− } x 3 2CrO42– +8H2O+ 6e− → 2Cr3+ +16OH− CrO42– +4H2O+ 3e− → Cr3+ +8OH− } x 2 2CrO42– +8H2O + 6 I− → 2Cr3+ +16OH− + 3 I2 reactant side Element product side 2 Cr 6 I 16 O H −10 charge Tro: Chemistry: A Molecular Approach, 2/e 38

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 Tro: Chemistry: A Molecular Approach, 2/e

Redox Reactions & 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 Tro: Chemistry: A Molecular Approach, 2/e

Electric Current Flowing Directly Between Atoms

Electric Current Flowing Indirectly Between Atoms

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 Tro: Chemistry: A Molecular Approach, 2/e

Electrochemical Cells
Oxidation and reduction half-reactions are kept separate in half-cells 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 Tro: Chemistry: A Molecular Approach, 2/e

Electrodes Anode Cathode electrode where oxidation occurs
anions are attracted to it connected to positive end of battery in an electrolytic cell loses weight in electrolytic cell Cathode electrode where reduction occurs cations are attracted to it connected to negative end of battery in an electrolytic cell gains weight in electrolytic cell electrode where plating takes place in electroplating Tro: Chemistry: A Molecular Approach, 2/e

Voltaic Cell the salt bridge is required to complete the circuit and maintain charge balance Tro: Chemistry: A Molecular Approach, 2/e

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 = x 1018 electrons per second Electrode surface area dictates the number of electrons that can flow larger batteries produce larger currents Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode 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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Voltaic Cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Cathode = Cu(s)
Cu2+ ions are reduced at the cathode Anode = Zn(s) the anode is oxidized to Zn2+ Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Fe(s) | Fe2+(aq) || MnO4(aq), Mn2+(aq), H+(aq) | Pt(s)
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) Tro: Chemistry: A Molecular Approach, 2/e

Cell Potential 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 Tro: Chemistry: A Molecular Approach, 2/e

Which Way Will Electrons flow?
Under standard conditions, zinc has a stronger tendency to oxidize than copper Zn → Zn e− Cu e− → Cu Potential Energy Electron Flow Therefore the electrons flow from zinc; making zinc the anode Zn → Zn e− E°= +0.76 Cu → Cu e− E°= −0.34 Tro: Chemistry: A Molecular Approach, 2/e

Standard Reduction Potential
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 half-reaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 v standard hydrogen electrode, SHE Tro: Chemistry: A Molecular Approach, 2/e

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 stronger tendency toward oxidation than the SHE have a  value for E°red For an oxidation half-reaction, E°oxidation = − E°reduction Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

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 half- reactions to balance the equation Tro: Chemistry: A Molecular Approach, 2/e

separate the reaction into the oxidation and reduction half-reactions
Example 18.4: Calculate Ecell for the reaction at 25 C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l) separate the reaction into the oxidation and reduction half-reactions ox (anode): Al(s)  Al3+(aq) + 3 e− red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) find the E for each half-reaction and sum to get Ecell Eox of Al = −Ered of Al3+ = V Ered of NO3− = V Ecell = (+1.66 V) + (+0.96 V) = V Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate Ecell for the reaction at 25 C IO3–(aq) + 6 H+(aq) + 5 I−(aq) → 3 I2(s) + 3 H2O(l) Reduction Half-Reaction Ered, 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate Ecell for the reaction at 25 C 2 IO3– + 12 H I−→ 6 I2 + 6 H2O separate the reaction into the oxidation and reduction half-reactions ox (anode): 2 I−(s)  I2(aq) + 2 e− red: IO3−(aq) + 6 H+(aq) + 5 e−  ½ I2(s) + 3 H2O(l) find the E for each half-reaction and sum to get Ecell Eox of I− = −Ered of I = −0.54 V Ered of IO3− = V Ecell = (−0.54 V) + (+1.20 V) = V 2 Tro: Chemistry: A Molecular Approach, 2/e

Tendencies from the Table of Standard Reduction Potentials
A redox reaction will be spontaneous when there is a strong tendency for the oxidizing agent to be reduced and the reducing agent to be oxidized higher on the table of Standard Reduction Potentials = stronger tendency for the reactant to be reduced lower on the table of Standard Reduction Potentials = stronger tendency for the product to be oxidized Tro: Chemistry: A Molecular Approach, 2/e

Predicting Spontaneity of Redox Reactions
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) Ered = V Zn2+(aq) + 2 e−  Zn(s) Ered = −0.76 V Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) spontaneous Cu(s) + Zn2+(aq)  Cu2+(aq) + Zn(s) nonspontaneous Tro: Chemistry: A Molecular Approach, 2/e

Example 18.5: Predict if the following reaction is spontaneous under standard conditions Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s) separate the reaction into the oxidation and reduction half-reactions ox: Fe(s)  Fe2+(aq) + 2 e− red: Mg2+(aq) + 2 e−  Mg(s) look up the relative positions of the reduction half-reactions red: Fe2+(aq) + 2 e−  Fe(s) because Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written Tro: Chemistry: A Molecular Approach, 2/e

the reaction is spontaneous in the reverse direction
Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s) ox: Mg(s)  Mg2+(aq) + 2 e− red: Fe2+(aq) + 2 e−  Fe(s) sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode Tro: Chemistry: A Molecular Approach, 2/e

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) 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) spontaneous as written Mg(s) + 2 Ag+(aq)  Mg2+(aq) + 2 Ag(s) spontaneous as written Cu2+(aq) + 2 I−(aq)  I2(s) + Cu(s) spontaneous in reverse Cu2+(aq) + 2 Cr2+(aq)  Cu(s) + 2 Cr3+(aq) spontaneous as written Tro: Chemistry: A Molecular Approach, 2/e

Practice – Which of the following materials can be used to oxidize Cu without oxidizing Ag?
Cr3+ F− I− I2 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) Tro: Chemistry: A Molecular Approach, 2/e

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 Ered, 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 Tro: Chemistry: A Molecular Approach, 2/e

ox: Cr2+(aq)  Cr3+(aq) + 1 e− E = +0.50 V
salt bridge anode = Pt cathode = Ag Cr2+ Ag+ Cr3+ ox: Cr2+(aq)  Cr3+(aq) + 1 e− E = V red: Ag+(aq) + 1 e−  Ag(s) E = V tot: Cr2+(aq) + Ag+(aq)  Cr3+(aq) + Ag(s) E = V Tro: Chemistry: A Molecular Approach, 2/e

Predicting Whether a Metal Will Dissolve in an Acid
Metals dissolve in acids if the reduction of the metal ion is easier than the reduction of H+(aq) 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) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Which of the following metals will dissolve in HC2H3O2(aq)
Practice – Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction. Ag Cu 2 Fe(s) + 6 HC2H3O2(aq) → 2 Fe(C2H3O2)3(aq) + 3 H2(g) 2 Cr(s) + 6 HC2H3O2(aq) → 2 Cr(C2H3O2)3(aq) + 3 H2(g) Ag Cu Fe Cr Reduction Half-Reaction Au3+(aq) + 3e− Au(s) Ag+(aq) + 1e− Ag(s) Cu2+(aq) + 2e− Cu(s) 2H+(aq) + 2e− H2(g) Fe3+(aq) + 3e− Fe(s) Cr3+(aq) + 3e− Cr(s) Mg2+(aq) + 2e− Mg(s) Tro: Chemistry: A Molecular Approach, 2/e

E°cell, DG° and K For a spontaneous reaction DG° = −RTlnK = −nFE°cell
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− Tro: Chemistry: A Molecular Approach, 2/e

Example 18.6: Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given: Find: I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) DG°, (J) Conceptual Plan: Relationships: E°ox, E°red E°cell DG° Solve: ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 V red: I2(l) + 2 e− → 2 I−(aq) E° = V tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V Answer: because DG° is +, the reaction is not spontaneous in the forward direction under standard conditions Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate DG for the reaction at 25C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l) Reduction Half-Reaction Ered, 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate DG for the reaction at 25 C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) 6 I2(s) + 6 H2O(l) Given: Find: 2IO3−(aq)+12H+(aq)+10I−(aq) → 6 I2(s) + 6 H2O(l) G°, (J) Conceptual Plan: Relationships: E°ox, E°red E°cell G° Solve: ox : 2 I−(s) → I2(aq) + 2 e− Eº = −0.54 V red: IO3−(aq) + 6 H+(aq) + 5 e− → ½ I2(s) + 3 H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V Answer: because DG° is −, the reaction is spontaneous in the forward direction under standard conditions Tro: Chemistry: A Molecular Approach, 2/e

Example 18.7: Calculate K at 25 °C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given: Find: Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K Conceptual Plan: Relationships: E°ox, E°red E°cell K Solve: ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 V red: 2 H+(aq) + 2 e− → H2(aq) E° = V tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 V Answer: because K < 1, the position of equilibrium lies far to the left under standard conditions Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate K for the reaction at 25C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Reduction Half-Reaction Ered, 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate K for the reaction at 25 C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Given: Find: 2 IO3−(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l) K Conceptual Plan: Relationships: E°ox, E°red E°cell K Solve: ox : 2 I−(s) → I2(aq) + 2e− Eº = −0.54 V red: IO3−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V Answer: because K >> 1, the position of equilibrium lies far to the right under standard conditions Tro: Chemistry: A Molecular Approach, 2/e

Cell Potential when Ion Concentrations Are Not 1 M
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 Tro: Chemistry: A Molecular Approach, 2/e

Ecell When Ion Concentrations are Not 1 M

Deriving the Nernst Equation

Example 18.8: Calculate Ecell at 25 °C for the reaction 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l) Given: Find: 3Cu(s) + 2MnO4−(aq) + 8H+(aq) → 2MnO2(s) + Cu2+(aq) + 4H2O(l) [Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M Ecell Conceptual Plan: Relationships: E°ox, E°red E°cell Ecell Solve: ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 V red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = V tot: 3Cu(s) + 2MnO4−(aq) + 8H+(aq) → 2MnO2(s) + Cu2+(aq) + 4H2O(l)) E° = V Check: units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate Ecell for the reaction at 25C if all ion concentrations are 0.10 M 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l) Reduction Half-Reaction Ered, 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 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate Ecell at 25 °C for the reaction 2 IO3−(aq)+12 H+(aq)+10 I−(aq)  6 I2(s) + 6 H2O(l) Given: Find: 2 IO3−(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l) [H+] = 0.10 M, [IO3−] = 0.10 M, [I−] = 0.10 M Ecell Conceptual Plan: Relationships: E°ox, E°red E°cell Ecell Solve: ox : 2 I−(s) → I2(aq) + 2e− Eº = −0.54 V red: IO3−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V Check: units are correct, Ecell < E°cell as expected because all the ions are reactants and < 1M Tro: Chemistry: A Molecular Approach, 2/e

Concentration Cells It is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different The difference in energy is due to the entropic difference in the solutions the more concentrated solution has lower entropy than the less concentrated Electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode Tro: Chemistry: A Molecular Approach, 2/e

Concentration Cell When the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode) When the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow Cu(s) Cu2+(aq) (0.010 M)  Cu2+(aq) (2.0 M) Cu(s) Tro: Chemistry: A Molecular Approach, 2/e

LeClanche’ Acidic Dry Cell
Electrolyte in paste form ZnCl2 + NH4Cl or MgBr2 Anode = Zn (or Mg) Zn(s) ® Zn2+(aq) + 2 e− Cathode = graphite rod MnO2 is reduced 2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e− ® 2 NH4OH(aq) + 2 Mn(O)OH(s) Cell voltage = 1.5 V Expensive, nonrechargeable, heavy, easily corroded Tro: Chemistry: A Molecular Approach, 2/e

Alkaline Dry Cell Same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste Anode = Zn (or Mg) Zn(s) ® Zn2+(aq) + 2 e− Cathode = graphite or brass rod MnO2 is reduced 2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e− ® 2 NH4OH(aq) + 2 Mn(O)OH(s) Cell voltage = 1.54 V Longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc Tro: Chemistry: A Molecular Approach, 2/e

PbO2(s) + 4 H+(aq) + SO42−(aq) + 2 e−
Lead Storage Battery Six cells in series Electrolyte = 30% H2SO4 Anode = Pb Pb(s) + SO42−(aq) ® PbSO4(s) + 2 e− Cathode = Pb coated with PbO2 PbO2 is reduced PbO2(s) + 4 H+(aq) + SO42−(aq) + 2 e− ® PbSO4(s) + 2 H2O(l) Cell voltage = 2.09 V Rechargeable, heavy Tro: Chemistry: A Molecular Approach, 2/e

NiCad Battery Electrolyte is concentrated KOH solution Anode = Cd
Cd(s) + 2 OH−(aq) → Cd(OH)2(s) + 2 e− E0 = 0.81 V Cathode = Ni coated with NiO2 NiO2 is reduced NiO2(s) + 2 H2O(l) + 2 e− → Ni(OH)2(s) + 2OH− E0 = 0.49 V Cell voltage = 1.30 v Rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown Tro: Chemistry: A Molecular Approach, 2/e

Ni-MH Battery Electrolyte is concentrated KOH solution
Anode = metal alloy with dissolved hydrogen oxidation of H from H0 to H+ M∙H(s) + OH−(aq) → M(s) + H2O(l) + e− E° = 0.89 V Cathode = Ni coated with NiO2 NiO2 is reduced NiO2(s) + 2 H2O(l) + 2 e− → Ni(OH)2(s) + 2OH− E0 = 0.49 V Cell voltage = 1.30 V Rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad Tro: Chemistry: A Molecular Approach, 2/e

Lithium Ion Battery Electrolyte is concentrated KOH solution
Anode = graphite impregnated with Li ions Cathode = Li - transition metal oxide reduction of transition metal Work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode Rechargeable, long life, very light, more environmentally friendly, greater energy density Tro: Chemistry: A Molecular Approach, 2/e

Fuel Cells Like batteries in which reactants are constantly being added so it never runs down! Anode and cathode both Pt coated metal Electrolyte is OH– solution Anode reaction: 2 H2 + 4 OH– → 4 H2O(l) + 4 e− Cathode Reaction: O2 + 4 H2O + 4 e− → 4 OH– Tro: Chemistry: A Molecular Approach, 2/e

Electrochemical Cells
In all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode In voltaic cells anode is the source of electrons and has a (−) charge cathode draws electrons and has a (+) charge In electrolytic cells electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Electrolytic Cells The source of energy is a battery or DC power supply The + terminal of the souce 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 Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

Electrolysis of Water Tro: Chemistry: A Molecular Approach, 2/e

Electrolysis of NaCl(l)

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. Tro: Chemistry: A Molecular Approach, 2/e

Mixtures of Ions When more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode least negative or most positive E°red When more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode least negative or most positive E°ox Tro: Chemistry: A Molecular Approach, 2/e

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 half-reaction is easiest red: Al3+(aq) + 3 e− → Al(s) E°= −1.66 V Mg2+(aq) + 2 e− → Mg(s) E°= −2.37 V because the reduction of Al3+ has a more positive E°, it is easiest and will occur oxidation occurs at the anode and reduction at the cathode anode: 2 Br−(l) → Br2(g) + 2 e− cathode: Al3+(aq) + 3 e− → Al(s) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Predict the half-reaction occurring at the anode and cathode for the electrolysis of a mixture of molten MgCl2(l) and MgBr2(l) Tro: Chemistry: A Molecular Approach, 2/e

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: 2 Cl−(l)  Cl2(g) + 2 e− 2 Br−(l)  Br2(g) + 2 e− red: Mg2+(aq) + 2 e−  Mg(s) use the values of E° of the half-reactions as a guide to determine which half-reaction is easiest ox: 2 Cl−(l)  Cl2(g) + 2 e− E°= −1.36 V 2 Br−(l)  Br2(g) + 2 e− E°= −1.09 V because the reduction of Br− has a more positive E°, it is easiest and will occur oxidation occurs at the anode and reduction at the cathode anode: 2 Br−(l)  Br2(g) + 2 e− cathode: Mg2+(aq) + 2 e−  Mg(s) Tro: Chemistry: A Molecular Approach, 2/e

Electrolysis of Aqueous Solutions
Possible cathode reactions reduction of cation to metal reduction of water to H2 2 H2O + 2 e− ® H2 + 2 OH− E° = −0.83 stand. cond. E° = −0.41 pH 7 Possible anode reactions oxidation of anion to element oxidation of H2O to O2 2 H2O ® O2 + 4 e− + 4H+ E° = −1.23 stand. cond. E° = −0.82 pH 7 oxidation of electrode particularly Cu graphite doesn’t oxidize Half-reactions that lead to least negative Ecell will occur unless overvoltage changes the conditions Tro: Chemistry: A Molecular Approach, 2/e

Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations 2 I− → I2 + 2 e− E° = −0.54 v 2 H2O → O2 + 4e − + 4H+ E° = −0.82 v possible oxidations 2 I− → I2 + 2 e− E° = −0.54 v 2 H2O → O2 + 4e− + 4H+ E° = −0.82 v possible reductions Na+ + 1e − → Na0 E° = −2.71 v 2 H2O + 2 e − → H2 + 2 OH− E° = −0.41 v possible reductions Na+ + 1e− → Na0 E° = −2.71 v 2 H2O + 2 e− → H2 + 2 OH− E° = −0.41 v overall reaction 2 I−(aq) + 2 H2O(l) → I2(aq) + H2(g) + 2 OH−(aq) Tro: Chemistry: A Molecular Approach, 2/e

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 half-reaction is easiest ox: 2 I−(aq)  I2(s) + 2 e− E°= −0.54 V 2 H2O ® O2 + 4e− + 4H+ E°= −0.82 V red: Li+(aq) + 1 e−  Li(s) E°= −3.04 V 2 H2O + 2 e− ® H2 + 2 OH− E°= −0.41 V oxidation occurs at the anode and reduction at the cathode anode: 2 I−(aq)  I2(s) + 2 e− cathode: 2 H2O(l) + 2 e− ® H2(g) + 2 OH−(aq) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Predict the half-reaction occurring at the anode and cathode for the electrolysis NaBr(aq) Tro: Chemistry: A Molecular Approach, 2/e

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 half-reaction is easiest ox: 2 Br−(aq)  Br2(s) + 2 e− E°= −1.09 V 2 H2O ® O2 + 4e− + 4H+ E°= −0.82 V red: Na+(aq) + 1 e−  Na(s) E°= −3.04 V 2 H2O + 2 e− ® H2 + 2 OH− E°= −0.41 V oxidation occurs at the anode and reduction at the cathode anode: 2 H2O(l) ® O2(g) + 4 e− + 4H+(aq) cathode: 2 H2O(l) + 2 e− ® H2(g) + 2 OH−(aq) Tro: Chemistry: A Molecular Approach, 2/e

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 Tro: Chemistry: A Molecular Approach, 2/e

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 Plan: Relationships: t(s), amp charge (C) mol e− mol Au g Au Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e− Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the amperage required to plate 2
Practice – Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the amperage required to plate 2
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: 3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s current, A Conceptual Plan: Relationships: g Au mol Au mol e− charge amps Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e− Tro: Chemistry: A Molecular Approach, 2/e

Corrosion Corrosion is the spontaneous oxidation of a metal by chemicals in the environment mainly O2 Because many materials we use are active metals, corrosion can be a very big problem metals are often used for their strength and malleability, but these properties are lost when the metal corrodes for many metals, the product of corrosion also does not adhere to the metal, and as it flakes off more metal can corrode Tro: Chemistry: A Molecular Approach, 2/e

Reduction of O2 O2 is very easy to reduce in moist conditions
O2(g) + 2 H2O(l) + 4 e−  2 OH−(aq) Eº = 0.40 V O2 is even easier to reduce under acidic conditions O2(g) + 4 H+ + 4 e−  2 H2O(l) Eº = 1.23 V Because the reduction of most metal ions lies below O2 on the Table of Standard Reduction Potentials, the oxidation of those metals by O2 is spontaneous Tro: Chemistry: A Molecular Approach, 2/e

Rusting At the anodic regions, Fe(s) is oxidized to Fe2+
The electrons travel through the metal to a cathodic region where O2 is reduced in acidic solution from gases dissolved in the moisture The Fe2+ ions migrate through the moisture to the cathodic region where they are further oxidized to Fe3+ which combines with the oxygen and water to form rust rust is hydrated iron(III) oxide, Fe2O3•nH2O the exact composition depends on the conditions moisture must be present water is a reactant required for ion flow between cathodic and anodic regions Electrolytes promote rusting enhances current flow Acids promote rusting lowering pH will lower E°red of O2 Tro: Chemistry: A Molecular Approach, 2/e

Preventing Corrosion One way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment paint some metals, such as Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding Another method to protect one metal is to attach it to a more reactive metal that is cheap sacrificial electrode galvanized nails Tro: Chemistry: A Molecular Approach, 2/e

Sacrificial Anode Tro: Chemistry: A Molecular Approach, 2/e

Chapter 18 Electrochemistry

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Chapter 18 Electrochemistry

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Presentation on theme: "Chapter 18 Electrochemistry"— Presentation transcript:

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About project

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