Chapter 19:Electrochemistry

Overview of the Chapter◆ review oxidation-reduction chemistry basics

◆ galvanic cells spontaneous chemical reaction generates a voltage

set-up of galvanic cell & identification of:anode (and half-reaction)cathode (and half-reaction)net cell reactioncell potential (E or E°)

relationship between cell potential and work, free energy, Q and K

◆ Electrolytic cells electric current drives a non-spontaneous chemical rxn

stoichiometric calculations based on cell rxn, current, time

Oxidation-Reduction Reactionsaka Redox Reactions

oxidation: loss of electrons

reduction: gain of electrons

Consider the reaction of calcium metal with oxygen:

2 Ca (s) + O2 (g) ! 2 CaO (s)

What happens to each element during the course of this reaction?

Redox Review Oxidation and Reduction Half Reactions

for the reaction of calcium with oxygen:

oxidation ! reaction: Ca ! Ca2+ + 2 e–

note: in oxidation ! rxn e–’s are products

reduction ! reaction: O2 + 4 e– ! 2 O2–

note: in reduciton ! rxn e–’s are reactants

net redox reaction: 2 Ca + O2 ! 2 Ca2+ + 2 O2–

note: add together the !-reactions; multiply as necessary to have same number of electrons in each half reaction before adding

Oxidation Numbers (Nox)

An oxidation number indicates the amount of electropositive or electronegative character of an atom - particularly as part of a polyatomic species.

Determining Oxidation Numbers:

ex. Determine Nox of sulfur in SO3.Nox S + 3(Nox O) = 0

Nox S + 3(–2) = 0∴ Nox S = +6

ex. Determine Nox of sulfur in SO32–.Nox S + 3(Nox O) = –2

Nox S + 3(–2) = –2∴ Nox S = +4

ex. Determine Nox N, H, P, and O in (NH4)3PO4. consider NH4+ and PO43– separately:

Nox N + 4(Nox H) = +1 Nox P + 4(Nox O) = –3 Nox N + 4(+1) = +1 Nox P + 4(–2) = –3 ∴ Nox N = –3 ∴ Nox P = +5

Recognizing Oxidation and Reduction

Compare oxidation numbers of elements in reactants and products:

◆ if Nox increases - the element is oxidized

◆ if Nox decreases - the element is reduced

ex: 2 Al (s) + Cr2O3 (s) ! Al2O3 (s) + 2 Cr (s)

Identify the oxidizing and reducing agents:

Recognizing Oxidation and Reduction

oxidizing agent

◆ reactant that facilitates oxidation by taking on (i.e. gaining) electrons lost during oxidation process

∴ oxidizing agent is the reactant that contains the element that is reduced

reducing agent

◆ reactant that facilitates reduction by providing (i.e. losing) electrons that are gained during reduction process

∴ reducing agent is the reactant that contains the element that is oxidized

Recognizing Oxidation and Reduction

Consider the following reaction:

2 Ca3(PO4)2 + 6 SiO2 + 10 C ! P4 + 6 CaSiO3 + 10 CO

Identify the following:

element oxidized

element reduced

oxidizing agent

reducing agent

Balancing Redox Equations: the Half-Reaction Methodin acidic solution

1. Identify the element oxidized and the element reduced.2. Write the skeletal oxidation & reduction !-reactions.3. Balance elements other than H and O.4. Balance O’s by adding H2O.5. Balance H’s by adding H+.6. Balance charge by adding e–’s.*at this point the individual !-reactions are balanced*7. Prepare to add the !-reactions together;

multiply as necessary so that the # of e–’s in oxidation !-reaction equals # of e–’s in the reduction !-reaction.

8. Add !-reactions together; clean up.

Balance the following redox reaction that occurs in acidic solution using the !-reaction method:

I2 + NO3– ! IO3– + NO2step 2:

ox ! rxn: I2 ! IO3–

red ! rxn: NO3– ! NO2

step 3:ox ! rxn: I2 ! 2 IO3–

red ! rxn: NO3– ! NO2

step 4:ox ! rxn: 6 H2O + I2 ! 2 IO3–

red ! rxn: NO3– ! NO2 + H2O

step 5:ox ! rxn: 6 H2O + I2 ! 2 IO3– + 12 H+ red ! rxn: 2 H+ + NO3– ! NO2 + H2O

Balance the following redox reaction that occurs in acidic solution using the !-reaction method (continued):

I2 + NO3– ! IO3– + NO2

step 6:ox ! rxn: 6 H2O + I2 ! 2 IO3– + 12 H+ + 10 e–

red ! rxn: 1 e– + 2 H+ + NO3– ! NO2 + H2O

step 7:ox ! rxn: 6 H2O + I2 ! 2 IO3– + 12 H+ + 10 e–

red ! rxn: 10"(1 e– + 2 H+ + NO3– ! NO2 + H2O)

step 8:6 H2O + I2 + 20 H+ + 10 NO3– ! 2 IO3– + 12 H+ + 10 NO2 + 10 H2O

net rxn: I2 + 8 H+ + 10 NO3– ! 2 IO3– + 10 NO2 + 4 H2O

◆ net ionic equation:Cu + 2 Ag+ → Cu2+ + 2 Ag

Cu (s) + 2 AgNO3 (aq) → Cu(NO3)2 (aq) + 2 Ag (s)

◆ copper metal in solution of silver nitrate

◆ With this set-up, any energy released by the reaction goes into sol’n causing ∆T.

◆ If we want the energy produced to do something, we need to change the experimental set-up.

Redox Reaction in an Electrochemical Cell

◆ galvanic (or voltaic) cell: a spontaneous chemical reaction generates an electric current

◆ separate the oxidation and reduction processes

◆ each !-reaction occurs in its own compartment oxidation occurs at the anodereduction occurs at the cathode

◆ The electrons produced in the oxidation process are transferred across an electrically conducting wire to compartment where reduction will occur.

◆ consider the rxn: Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)

anode compartment◆ oxidation occurs here◆ electrons produced◆ anode ! reaction:

Zn → Zn2+ + 2 e–

◆ designated – on battery

salt bridgecathode compartment

◆ reduction occurs here◆ electrons consumed◆ cathode ! reaction:

Cu2+ + 2 e– → Cu◆ designated + on battery

◆ consider the rxn: Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)

anode compartmentsalt bridge or porous barrier

◆ allows migration of spectator ions between compartments

◆ maintain charge neutrality◆ anions migrate toward the anode◆ cations migrate toward the cathode

cathode compartment

◆ the electrodes must be connected by an electrically conducting wire to complete the circuit

◆ need a complete circuit for:e– movementchemical reaction

◆ the actual, physical electrodes in this cell are a strip of Zn (the anode) and a strip of Cu (the cathode)

◆ in a galvanic cell, the direction of e– flow is always:

anode → cathode

◆ the e– transfer generates a potential difference (measured in volts, V)

◆ the specific voltage generated is dependent on the anode & cathode used, and [ ]’s or P’s

example:

A galvanic cell is constructed based on the following redox reaction:

Fe (s) + 2 Fe3+ (aq) → 3 Fe2+ (aq)

◆ Write the balanced !-reactions that occur at the anode and the cathode.

◆ Sketch this galvanic cell indicating the following:the anode and the cathode

the species in the anode & cathode compartments

the direction of electron flow

the direction of cation and anion migration

Line Notation for Galvanic Cells

◆ short-hand representation of galvanic cells

◆ use single vertical lines to indicate/separate species of different phases within same compartment

◆ use double vertical lines to indicate separation of anode and cathode compartments

anode | anode compartment || cathode compartment | cathode

◆ examples:

Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)

Fe (s) | Fe2+ (aq) || Fe2+ (aq), Fe3+ (aq) | Pt (s)

Cell Potential, E and E°◆ oxidation-reduction reaction results in e– transfer

as e–’s move from anode to cathode, a voltage is generated

◆ cell potential, E is measured in volts

1 volt = 1 joule of work done as 1 Coulomb of charge moves from a point of higher potential to a point of lower potential

1 V = 1 J/C

◆ standard cell potential, E° – voltage generated when:solids and liquids are in their pure form

solutions at 1 M concentration

gases at 1 atm pressure

Cell Potential, E and E°

E°cell = E°cathode – E°anode

◆ standard half-cell designated as reference electrode

E°’s of other half-cells determined and tabulated relative to this reference electrode

◆ standard hydrogen electrode (SHE):

2 H+ (aq) + 2 e– → H2 (g)[H+] = 1 MPH2 = 1 atmT = 298 K

E° = 0.000 V

Standard Hydrogen Electrode Table of Standard Reduction Potentials

◆ !–reactions written as reductions

oxidizing agent + e– → reducing agent

◆ top to bottom, increasing E°

increasing tendency for reductiondecreasing tendency for oxidation

◆ relative strengths of oxidizing and reducing agentsstrongest reducing agents are higher in the tablestronger oxidizing agents are lower in the table

◆ correlation between sign of E° and spontaneity+ E°, reduction is spontaneous– E°, oxidation is spontaneous

Spontaneous Reactions and Cell Potential

◆ in a galvanic cell, a spontaneous chemical reaction generates a + potential

E°cell must be positive

E°cathode > E°anode

◆ comparing any two !–reactions in the Table of Standard Reduction Potentials:

the !–reaction higher in the table (smaller E°) will be the anode, oxidation !–reaction

the !–reaction lower in the table (larger E°) will be the cathode, reduction !–reaction

example:

Consider a galvanic cell based on the following half-cells:

Ni2+ (aq) + 2 e– → Ni (s); E° = –0.23 VCd2+ (aq) + 2 e– → Cd (s); E° = –0.40 V

Write the anode !–reaction, the cathode !–reaction, and the overall cell reaction. Then determine E°cell.

example:

Consider a galvanic cell based on the following half-cells:

Pb2+ + 2 e– → Pb; E° = –0.13 VPbO2 + SO42– + 4 H+ + 2 e– → PbSO4 + 2 H2O; E° = 1.69 V

Write the anode !–reaction, the cathode !–reaction, and the overall cell reaction. Then determine E°cell.

example:

Can nitrate ion in an acidic solution oxidize iron (II) ions?

example:

Can MnO4– in an acidic solution oxidize Ni? Ag?

example:

Is Cr or Al a stronger reducing agent?

Complete Description of a Galvanic Cell

At this point for a galvanic cell, you should be able to:

◆ write the balanced anode and cathode !–reactions

◆ write the balanced cell reaction

◆ determine the standard cell potential (E°) in volts

◆ line notation

◆ sketch or draw the set-up of the cell showing:anode and cathodespecies in anode and cathode compartmentssalt bridgedirection of e– flowdirection of ion migration

Work, Free Energy, and Electrochemistry◆ work is done in a galvanic cell as e–’s are transferred

from anode to cathode

work is done by the cell ∴ exothermic ; w is –

◆ maximum work that can be done by a galvanic cell is equal to the free energy change for the cell reaction

wmax = ∆G°

wmax = (quantity of charge transferred)(operating voltage of cell)

wmax = ∆G° = –nFE°n = mol e–’s transferred

F = Faraday constant = 96485 C/mol e–

◆ note: When E° is +, ∆G° is – . This confirms that the chemical reaction in a galvanic cell is spontaneous.

Work, Cell Potential, and Efficiency

◆ the maximum work that can be done by a cell is given by:

wmax = –nFE°

BUT . . . this maximum work is never achieved

◆ to calculate the actual work done by a cell:wactual = –nFE

◆ can also determine the cell’s efficiency:

efficiency = ––––– x 100

efficiency = –––– x 100

wactualwmax

EE°

example:

Consider a galvanic cell based on these !-reactions:

Cu2+ + 2 e– → Cu; E°=0.34VAg+ + e– → Ag; E°=0.80V

Calculate ∆G° (in kJ) for this galvanic cell’s reaction.

If the cell operates at a potential of 0.35 V, determine its efficiency.

The Nernst Equation:Dependence of E on Composition

◆ we have discussed the relationship between ∆G° and E°:

∆G° = –nFE° at standard conditions: s and l in pure form

sol’ns at 1M concentration gases at 1 atm pressure

◆ galvanic cells can operate at conditions other than standard:

∆G = –nFE

◆ relationship between ∆G, ∆G°, E, E°, composition?

◆ Nernst equation: E = E° – –––– ln QRTnF

Cell Potential and the Equilibrium Constant◆ starting with the Nernst equation:

E = E° – –––– ln Q

◆ for a system at equilibrium, E = 0 and Q = K:

E° = –––– ln K

◆ rework the equations a bit: at T = 25°C combine R, T, and F convert ln to log

E = E° – ––––––– log Q; E° = ––––––– log K

RTnF

nFRT

.0592 V .0592 Vn n

example:

Consider a galvanic cell based on the following cell rxn:

Cu (s) + 2 Fe3+ (aq) → Cu2+ (aq) + 2 Fe2+ (aq); E°cell = 0.43 V

Determine ∆G and E at 25°C if the cell is run with the following ion concentrations:

[Fe2+] = 0.20 M [Fe3+] = 1.0 x 10–4 M [Cu2+] = 0.25 M

Determine the equilibrium constant, K, for this reaction at 25°C.

example:

Consider the following galvanic cell:

anode: Tl (s) → Tl+ (aq) + e–; E°anode = –.34Vcathode: Ni2+ (aq) + 2 e– → Ni (s); E°cathode = –.24V

Determine the concentration of Tl+ (aq) if the cell operates at E = 0.15V at 25°C when [Ni2+] = 3.3 M.

Electrolysis◆ an electrical current is used to drive a

nonspontaneous chemical reaction

◆ in an electrolytic cell:w is +∆G is +Ecell is –

◆ we will use the relationships between:mol e– transferred and mol product formed

stoichiometry of balanced !–reactionsmol e– transferred and quantity of charge

Faraday constant, Fquantity of charge and time

current, amperes (1A = 1C/s)

example:

Determine the time required to plate 85.5 g Zn onto a frame if 23.0 A current is passed through a solution of ZnSO4 (aq).

Zn2+ (aq) + 2 e– → Zn (s)

strategy?

g Zn → mol Zn → mol e– → C → time

example:

Determine the current required to plate 2.86 g Cr in 2.5 min onto a bathroom fixture from Cr2(SO4)3 (aq)

Cr3+ (aq) + 3 e– → Cr (s)

strategy?

g Cr → mol Cr → mol e– → C → current, A

example:

Determine the mass of nickel that can be produced in an experiment when 13.7 A passed through a solution of Ni(NO3)2 (aq) for 5.0 min.

Ni2+ (aq) + 2 e– → Ni (s)

strategy?

time → C → mol e– → mol Ni → g Ni