ElectrochemistryApplications of Redox

Oxidation reduction reactions involve a transfer of electrons.

OIL- RIG Oxidation Involves Loss Reduction Involves Gain LEO-GER Lose Electrons Oxidation Gain Electrons Reduction

Review

Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. Which substance is a reductant (reducing agent) and which is an oxidant (oxidizing agent)?  

A. PbS, reductant; O2, oxidant  B. PbS, reductant; SO2, oxidant  C. Pb2+, reductant; S2- oxidant  D. PbS, reductant; no oxidant  E. PbS, oxidant; SO2, reductant

Moving electrons is electric current. 8H++MnO4

-+ 5Fe+2 +5e- ® Mn+2 + 5Fe+3

+4H2O Helps to break the reactions into half reactions. 8H++MnO4

-+5e- ® Mn+2 +4H2O 5(Fe+2 ® Fe+3 + e- ) In the same mixture it happens without doing useful

work, but if separate

Applications

H+

MnO4-

Fe+2

Connected this way the reaction starts Stops immediately because charge builds up.

e-e- e-

e-e-

H+

MnO4-

Fe+2

Galvanic CellSalt Bridge allows current to flow

H+

MnO4-

Fe+2e-

Electricity travels in a complete circuit

H+

MnO4-

Fe+2

Porous Disk

Instead of a salt bridge

Reducing Agent

Oxidizing Agent

e-

e-

e- e-

e-

e-

Anode Cathode

Oxidizing agent pulls the electron. Reducing agent pushes the electron. The push or pull (“driving force”) is called

the cell potential Ecell Also called the electromotive force (emf) Unit is the volt(V) = 1 joule of work/coulomb of charge Measured with a voltmeter

Cell Potential

Zn+2 SO4-

2

1 M HCl

Anode

0.76

1 M ZnSO4

H+ Cl-

H2 in

Cathode

1 M HCl

H+ Cl-

H2 in

Standard Hydrogen Electrode This is the reference all

other oxidations are compared to

Eº = 0 º indicates standard

states of 25ºC, 1 atm, 1 M solutions.

Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s) The total cell potential is the sum of the

potential at each electrode.

Eºcell = EºZn® Zn+2 + EºCu+2 ® Cu We can look up reduction potentials in a

table. One of the reactions must be reversed, so

change it sign.

Cell Potential

Determine the cell potential for a galvanic cell based on the redox reaction.

Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq)

Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V 2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V

Cell Potential

More negative Eº more easily electron is added More easily reduced Better oxidizing agent

More positive Eº more easily electron is lost More easily oxidized Better reducing agent

Reduction potential

solid½Aqueous½½Aqueous½solid Anode on the left½½Cathode on the right Single line different phases. Double line porous disk or salt bridge. If all the substances on one side are

aqueous, a platinum electrode is indicated.

Line Notation

Cu2+ Fe+2

For the last reactionCu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

In a galvanic cell, the electrode that acts as a source of electrons to the solution is called the __________; the chemical change that occurs at this electrode is called________.  

a.  cathode, oxidation  b.  anode, reduction  c.  anode, oxidation  d.  cathode, reduction

Under standard conditions, which of the following is the net reaction that occurs in the cell?

Cd|Cd2+ || Cu2+|Cu  a.  Cu2+ + Cd → Cu + Cd2+  b.  Cu + Cd → Cu2+ + Cd2+  c.  Cu2+ + Cd2+ → Cu + Cd  d.  Cu + Cd 2+ → Cd + Cu2+ 

The reaction always runs spontaneously in the direction that produced a positive cell potential.

Four things for a complete description.1) Cell Potential2) Direction of flow3) Designation of anode and cathode4) Nature of all the components- electrodes

and ions

Galvanic Cell

Completely describe the galvanic cell based on the following half-reactions under standard conditions.

MnO4- + 8 H+ +5e- ® Mn+2 + 4H2O

Eº=1.51 V

Fe+3 +3e- ® Fe(s) Eº=0.036V

Practice

emf = potential (V) = work (J) / Charge(C)E = work done by system / chargeE = -w/q Charge is measured in coulombs. -w = q E Faraday = 96,485 C/mol e-

q = nF = moles of e- x charge/mole e-

w = -qE = -nFE = DG

Potential, Work and DG

DGº = -nFEº if Eº > 0, then DGº < 0 spontaneous if Eº< 0, then DGº > 0 nonspontaneous In fact, reverse is spontaneous. Calculate DGº for the following reaction: Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)

Fe+2(aq) + e-® Fe(s) Eº = 0.44 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

Potential, Work and DG

Qualitatively - Can predict direction of change in E from LeChâtelier.

2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M

if [Al+3] = 1.0 M and [Mn+2] = 1.5M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

Cell Potential and Concentration

DG = DGº +RTln(Q) -nFE = -nFEº + RTln(Q)

E = Eº - RTln(Q)

nF 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)

Eº = 0.48 V Always have to figure out n by balancing. If concentration can gives voltage, then from

voltage we can tell concentration.

The Nernst Equation

As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and

Ecell = 0 0 = Eº - RTln(K)

nFEº = RTln(K)

nF nF Eº = ln(K)

RT

The Nernst Equation

Car batteries are lead storage batteries. Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O

Batteries are Galvanic Cells

Dry Cell Zn + NH4

+ +MnO2 ®

Zn+2 + NH3 + H2O + Mn2O3

Batteries are Galvanic Cells

Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base)

Batteries are Galvanic Cells

NiCad NiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

Batteries are Galvanic Cells

Rusting - spontaneous oxidation. Most structural metals have reduction

potentials that are less positive than O2 .

Fe ® Fe+2 +2e- Eº= 0.44 V O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V

Fe+2 + O2 + H2O ® Fe2O3 + H+ Reactions happens in two places.

Corrosion

WaterRust

Iron Dissolves- Fe ® Fe+2

e-

Salt speeds up process by increasing conductivity

O2 + 2H2O +4e- ® 4OH-

Fe2+ + O2 + 2H2O ® Fe2O3 + 8 H+

Fe2+

Coating to keep out air and water. Galvanizing - Putting on a zinc coat Has a lower reduction potential, so it is more

easily oxidized. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces

of an active metal like magnesium that get oxidized instead.

Preventing Corrosion

Running a galvanic cell backwards. Put a voltage bigger than the potential and

reverse the direction of the redox reaction. Used for electroplating.

Electrolysis

1.0 M Zn+2

e- e-

Anode Cathode

1.10

Zn Cu1.0 M Cu+2

1.0 M Zn+2

e- e-

AnodeCathode

A battery >1.10V

Zn Cu1.0 M Cu+2

Have to count charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second q = I x t q/nF = moles of metal Mass of plated metal How long must 5.00 amp current be applied

to produce 15.5 g of Ag from Ag+

Calculating plating

1. Current x time = charge2. Charge ∕Faraday = mole of e-

3. Mol of e- to mole of element or compound4. Mole to grams of compoundOr the reverse if you want time to plate

Calculating plating

Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.

How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO3)3 at a current of 2.00 A?

Electrolysis of water. Separating mixtures of ions. More positive reduction potential means the

reaction proceeds forward. We want the reverse. Most negative reduction potential is easiest

to plate out of solution.

Other uses

Know the table2. Recognized by change in oxidation state.3. “Added acid”4. Use the reduction potential table on the

front cover.5. Redox can replace. (single replacement)

Redox

6. Combination Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element.I- + IO3

- + H+ ® I2 + H2O7. Decomposition.

a) peroxides to oxidesb) Chlorates to chloridesc) Electrolysis into elements.d) carbonates to oxides

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1. A piece of solid bismuth is heated strongly in oxygen.

2. A strip or copper metal is added to a concentrated solution of sulfuric acid.

3. Dilute hydrochloric acid is added to a solution of potassium carbonate.

Examples

45

23. Hydrogen peroxide solution is added to a solution of iron (II) sulfate.

24. Propanol is burned completely in air.25. A piece of lithium metal is dropped into a

container of nitrogen gas.26. Chlorine gas is bubbled into a solution of

potassium iodide.

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5. A stream of chlorine gas is passed through a solution of cold, dilute sodium hydroxide.

6. A solution of tin ( II ) chloride is added to an acidified solution of potassium permanganate

7. A solution of potassium iodide is added to an acidified solution of potassium dichromate.

Examples

47

70. Magnesium metal is burned in nitrogen gas.

71. Lead foil is immersed in silver nitrate solution.

72. Magnesium turnings are added to a solution of iron (III) chloride.

73. Pellets of lead are dropped into hot sulfuric acid

74. Powdered Iron is added to a solution of iron(III) sulfate.

An Ox – anode is where oxidation occurs Red Cat – Reduction occurs at cathode Galvanic cell- spontaneous- anode is

negative Electrolytic cell- voltage applied to make

anode positive

A way to remember

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(a) Draw a diagram of this cell. (b) Describe what is happening at the cathode

(Include any equations that may be useful.)

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(c) Describe what is happening at the anode. (Include any equations that may be useful.)

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(d) Write the balanced overall cell equation. (e) Write the standard cell notation.

A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.(f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH3)+ (aq). The student remeasures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu2+ (aq) concentration in the cell after the ammonia has been added?