1 9 & 19. ELECTROCHEMISTRY 1. 2 Electron Transfer Reactions 1. Electron transfer reactions are redox...

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119 & 19. 9 & 19. ELECTROCHEMISTRY 1ELECTROCHEMISTRY 1

22Electron Transfer ReactionsElectron Transfer ReactionsElectron Transfer ReactionsElectron Transfer Reactions

1. Electron transfer reactions are 1. Electron transfer reactions are redoxredox

reactions.reactions.

2. Results in the generation of an electric 2. Results in the generation of an electric

current (electricity)current (electricity) or or be caused by be caused by

imposing an electric current. imposing an electric current.

3. Therefore, this field of chemistry is often 3. Therefore, this field of chemistry is often

called called ELECTROCHEMISTRY.ELECTROCHEMISTRY.

Two industrial applications:

• I. Voltaic Cells ____ Produces energy from a spontaneous chemical reaction.

• II. Electrolytic Cells____ Uses energy in order to promote a non spontaneous chemical reaction.

Voltaic Cells/ Electrochemical Voltaic Cells/ Electrochemical CellsCells

A device that obtains A device that obtains electrical energy from a electrical energy from a spontaneousspontaneous chemical chemical reactionreaction

Batteries are voltaic Batteries are voltaic cellscells

Terms Used for Voltaic Terms Used for Voltaic CellsCells

Zn2+ ions

Cu2+ ions

saltbridge

electrons

Zn2+ ions

Cu2+ ions

saltbridge

electrons

••Electrons travel thru external wire.Electrons travel thru external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.

Zn --> ZnZn --> Zn2+2+ + 2e- + 2e- CuCu2+2+ + 2e- --> Cu + 2e- --> Cu

<--Anions<--AnionsCations-->Cations-->

OxidationOxidationAnodeAnodeNegativeNegative

RedReductionuctionCatCathodehodePositivePositive

RED CATRED CAT

AN OX chases a RED CAT

Zn/Cu Electrochemical CellZn/Cu Electrochemical Cell

Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e- EEoo = +0.76 V = +0.76 VCuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s) EEoo = +0.34 V = +0.34 V------------------------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) ---> Zn(aq) + Zn(s) ---> Zn2+2+(aq) + Cu(s) (aq) + Cu(s)

EEoo = +1.10 V = +1.10 V

Cathode, Cathode, positive, positive, sink for sink for electronselectrons

Anode, Anode, negative, negative, source of source of electronselectrons

Zn2+ ions

Cu2+ ions

saltbridge

electrons

Zn2+ ions

Cu2+ ions

saltbridge

electrons ++

Cd Salt Bridge

Cd --> CdCd --> Cd2+2+ + 2e- + 2e-oror

CdCd2+2+ + 2e- --> Cd + 2e- --> Cd

Fe --> FeFe --> Fe2+2+ + 2e- + 2e-oror

FeFe2+2+ + 2e- --> Fe + 2e- --> Fe

EEoo for a Voltaic Cell for a Voltaic Cell

All ingredients are present. Which way does All ingredients are present. Which way does reaction proceed?reaction proceed?

CHEMICAL CHANGE --->CHEMICAL CHANGE --->ELECTRIC CURRENTELECTRIC CURRENT

Zn metal

Cu2+ ions

Zn metal

Cu2+ ions

With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.”

•Zn is oxidized Zn is oxidized and is the reducing agent and is the reducing agent Zn(s) ---> ZnZn(s) ---> Zn2+2+(aq) + 2e-(aq) + 2e-•CuCu2+2+ is reduced is reduced and is the oxidizing agentand is the oxidizing agentCuCu2+2+(aq) + 2e- ---> Cu(s)(aq) + 2e- ---> Cu(s)

1212More About More About Calculating Cell VoltageCalculating Cell Voltage

Assume IAssume I-- ion can reduce water. ion can reduce water.

2 H2O + 2e- ---> H2 + 2 OH- Cathode2 I- ---> I2 + 2e- Anode-------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2

Assuming reaction occurs as written, Assuming reaction occurs as written,

E˚ = E˚E˚ = E˚catcat+ E˚+ E˚anan= (-0.828 V) - (- +0.535 V) = = (-0.828 V) - (- +0.535 V) = -1.363 V-1.363 V

Minus E˚ means rxn. occurs in opposite Minus E˚ means rxn. occurs in opposite

directiondirection

(the connection is backwards or you are (the connection is backwards or you are

recharging the battery)recharging the battery)

1313http://ibchem.com/IB/ibnotes/full/red_htm/19.2.htm

• Write the equation for the spontaneous reaction that will occur when a magnesium half cell is connected to na aluminum half cell.

Charging a BatteryCharging a BatteryWhen you charge a battery, you are When you charge a battery, you are forcing the electrons backwards (from forcing the electrons backwards (from the + to the -). To do this, you will the + to the -). To do this, you will need a higher voltage backwards than need a higher voltage backwards than forwards. This is why the ammeter in forwards. This is why the ammeter in your car often goes slightly higher your car often goes slightly higher while your battery is charging, and then while your battery is charging, and then returns to normal.returns to normal.

In your car, the battery charger is In your car, the battery charger is called an alternator. If you have a called an alternator. If you have a dead battery, it could be the dead battery, it could be the battery needs to be replaced OR battery needs to be replaced OR the alternator is not charging the the alternator is not charging the battery properly.battery properly.

Dry Cell BatteryDry Cell Battery

Anode (-)Anode (-)

Zn ---> ZnZn ---> Zn2+2+ + 2e- + 2e-

Cathode (+)Cathode (+)

2 NH2 NH44++ + 2e- ---> + 2e- --->

2 2 NHNH33 + H + H22

Alkaline BatteryAlkaline Battery

Nearly same reactions as Nearly same reactions as in common dry cell, but in common dry cell, but under basic conditions.under basic conditions.

Anode (-): Anode (-): Zn + 2 OHZn + 2 OH-- ---> ZnO + H ---> ZnO + H22O + 2e-O + 2e-

Cathode (+): Cathode (+): 2 MnO2 MnO22 + H + H22O + 2e- ---> O + 2e- --->

MnMn22OO33 + 2 OH + 2 OH--

Mercury BatteryMercury Battery

Anode:Anode:

Zn is reducing agent under basic conditionsZn is reducing agent under basic conditions

Cathode:Cathode:

HgO + HHgO + H22O + 2e- ---> Hg + 2 OHO + 2e- ---> Hg + 2 OH--

Lead Storage BatteryLead Storage Battery

Anode (-) Anode (-) EEoo = +0.36 V = +0.36 V

Pb + HSOPb + HSO44-- ---> PbSO ---> PbSO44 + H + H++ + 2e- + 2e-

Cathode (+) Cathode (+) EEoo = +1.68 V = +1.68 V

PbOPbO22 + HSO + HSO44-- + 3 H + 3 H++ + 2e- + 2e-

---> PbSO---> PbSO44 + 2 H + 2 H22OO

Ni-Cad BatteryNi-Cad Battery

Anode (-)Anode (-)

Cd + 2 OHCd + 2 OH-- ---> Cd(OH) ---> Cd(OH)22 + 2e- + 2e-

Cathode (+) Cathode (+)

NiO(OH) + HNiO(OH) + H22O + e- ---> Ni(OH)O + e- ---> Ni(OH)22 + OH + OH--

HH22 as a Fuel as a Fuel

Cars can use electricity generated by HCars can use electricity generated by H22/O/O22

fuel cells.fuel cells.HH22 carried in tanks or generated from carried in tanks or generated from

hydrocarbonshydrocarbons

Balancing Equations Balancing Equations for Redox Reactionsfor Redox Reactions

Some redox reactions have equations that must be balanced by Some redox reactions have equations that must be balanced by special techniques.special techniques.

MnOMnO44-- + 5 Fe + 5 Fe2+2+ + 8 H + 8 H++

---> Mn---> Mn2+ 2+ + 5 Fe+ 5 Fe3+3+ + 4 H + 4 H22OO

Mn = +7Mn = +7 Fe = +2Fe = +2Fe = +3Fe = +3Mn = +2Mn = +2

Balancing Balancing EquationsEquations

Consider the Consider the reduction of Agreduction of Ag++ ions with copper ions with copper metal.metal.

Cu + AgCu + Ag++ --give--> Cu --give--> Cu2+2+ + Ag + Ag

2323Balancing Balancing EquationsEquations

Step 1:Step 1: Divide the reaction into half-reactions, one for Divide the reaction into half-reactions, one for oxidation and the other for reduction.oxidation and the other for reduction.

OxOx Cu ---> CuCu ---> Cu2+2+

RedRed Ag Ag++ ---> Ag ---> Ag

Step 2:Step 2: Balance each element for mass. Already done Balance each element for mass. Already done in this case.in this case.

Step 3:Step 3: Balance each half-reaction for charge by Balance each half-reaction for charge by adding electrons.adding electrons.

OxOx Cu ---> Cu Cu ---> Cu2+2+ + + 2e-2e-

RedRed Ag Ag++ + + e- e- ---> Ag---> Ag

Step 4:Step 4: Multiply each half-reaction by a factor so that Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the the reducing agent supplies as many electrons as the oxidizing agent requires.oxidizing agent requires.

Reducing agentReducing agent Cu ---> Cu Cu ---> Cu2+2+ + 2e- + 2e-Oxidizing agentOxidizing agent 22 Ag Ag++ + + 22 e- ---> e- ---> 22 Ag AgStep 5:Step 5: Add half-reactions to give the overall equation.Add half-reactions to give the overall equation.Cu + 2 AgCu + 2 Ag++ ---> Cu ---> Cu2+2+ + 2Ag + 2Ag

The equation is now balanced for both The equation is now balanced for both charge and mass.charge and mass.

Balance the following in acid solution—Balance the following in acid solution—

VOVO22++ + Zn ---> VO + Zn ---> VO2+ 2+ + Zn+ Zn2+2+

Step 1:Step 1:Write the half-reactionsWrite the half-reactions

OxOx Zn ---> ZnZn ---> Zn2+2+

RedRed VOVO22++ ---> VO ---> VO2+2+

Step 2:Step 2:Balance each half-reaction for mass.Balance each half-reaction for mass.

OxOx Zn ---> ZnZn ---> Zn2+2+

RedRedVOVO22

++ ---> VO ---> VO2+2+ + + HH22OO2 H2 H++ ++

Add HAdd H22O on O-deficient side and add HO on O-deficient side and add H++

on other side for H-balance.on other side for H-balance.

Step 3:Step 3: Balance half-reactions for charge.Balance half-reactions for charge.OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e- 2e-

RedRed e- e- + 2 H+ 2 H++ + VO + VO22++ ---> VO ---> VO2+2+ + H + H22OO

Step 4:Step 4: Multiply by an appropriate factor.Multiply by an appropriate factor.OxOx Zn ---> ZnZn ---> Zn2+2+ + + 2e-2e-

RedRed 22e-e- + + 44 H H++ + + 22 VO VO22++

---> ---> 22 VO VO2+2+ + + 22 H H22OO

Step 5:Step 5: Add Add balancedbalanced half-reactions half-reactions

Zn + 4 HZn + 4 H++ + 2 VO + 2 VO22++

---> Zn ---> Zn2+2+ + 2 VO + 2 VO2+2+ + 2 H + 2 H22OO

Tips on Balancing Tips on Balancing EquationsEquations

• Never add ONever add O22, O atoms, or , O atoms, or OO2-2- to balance oxygen. to balance oxygen.

• Never add HNever add H22 or H atoms to or H atoms to balance hydrogen.balance hydrogen.

• Be sure to write the correct Be sure to write the correct charges on all the ions.charges on all the ions.

• Check your work at the end Check your work at the end to make sure mass and to make sure mass and charge are balanced.charge are balanced.

• PRACTICE!PRACTICE!

• 1. Can an acidified aqueous solution of potassium dichromate spontaneously oxidize a solution of bromide ions to bromine?

• 2. Can a solution of tin II ions reduce a solution of iron III ions? If so, are the iron III ions reduced to iron II or to iron metal?

• 3. What will happen when copper I sulfate(s) dissolves in water?

3131II. Electrolysishttp://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/

faraday.php

Electrolysis is the situation when redox cells are forced to run in reverse by attaching an electricity source to overcome the potential difference.

Salgema-Maceio/Brasil

Some applications:

3434a)Electrolysis of molten NaCl: NaCl(l)

Species present: Na+ and Cl-

electrolyte

3535Electrolysis of molten NaCl: NaCl(l)Species present: Na+ and Cl-

Positive electrode=> oxidation

2 Cl- (l) => Cl2 (g) + 2e

Negative electrode=> reduction

Cathode

Na+ (l) + e => Na(l)

Overall:

2 Cl- (l)+ 2 Na+ (l) => Cl2 (g) + 2 Na(l) electrolyte

3636b)Electrolysis of aqueous NaCl, NaCl(aq)

Species: Na+1 , H+1 , Cl-1 , H2O

This time,H+ will be reduced instead of Na+

Cathode (-): Reduction

2 H2O(l) + 2 e- =>H2(g) + 2 OH-(aq)

Anode (+): Oxidation

2 Cl- => Cl2 (g)+ 2 e-

Overall:

2 NaCl(aq) + 2 H2O(l) => 2 Na+(aq) + 2 OH-(aq) + H2(g) + Cl2(g)

• Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially.

• Electrolysis of an aqueous NaCl solution has two other advantages: It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

Electrolysis Al

9.5.3. How electric current is conducted.

• http://www.youtube.com/watch?v=Y9qMR3GV7WA

• In an electrolytic cell, current is conducted by electrons in the wire and by ions in the electrolyte.

Hodder Q5

4343II. Electrolytic CellThe ions that are successfully released at the electrodes depend on three factors

1)The position of the ion in the electrochemical series.

As a rule of thumb, if the metal appears below hydrogen in the electrochemical series then it will be preferentially deposited.

2)The concentration of the ion in the solution.

3)The nature of the electrode:

Platinum, graphite

• http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php

9.5.4. Deduce the products of the electrolysis of any molten salt, PbBr2

• http://www.youtube.com/watch?v=kINjUBolU3M&feature=related

• Hodder Q1b and Q2

4545The ions that are successfully released at the electrodes depend on three factors

The ions that are successfully released at the electrodes depend on three factors

1)The position of the ion in the electrochemical series.

As a rule of thumb, if the metal appears below hydrogen in the electrochemical series then it will be preferentially deposited.

2)The concentration of the ion in the solution.

3)The nature of the electrode:

Platinum, graphite

• http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php

IB Questions

4848http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php

• Because the demand for chlorine is much larger than the demand for sodium, electrolysis of aqueous sodium chloride is a more important process commercially.

• Electrolysis of an aqueous NaCl solution has two other advantages. It produces H2 gas at the cathode, which can be collected and sold. It also produces NaOH, which can be drained from the bottom of the electrolytic cell and sold.

• Solid sodium chloride doesn't conduct electricity, because there are no electrons which are free to move. When it melts, sodium chloride undergoes electrolysis, which involves conduction of electricity because of the movement and discharge of the ions. In the process, sodium and chlorine are produced. This is a chemical change rather than a physical process.

5050Predict the products of electrolysis of strong calcium chloride solution

• At the cathode

• Species present Ca2+ and H+. Ca2+ is higher in the reactivity seriers than hydrogen and therefore cannot be released.

• The reaction is therefore: 2H+(aq) + 2e H2(g)

• At the anode

• Species present OH- and Cl- . The chloride concentration is strong and so it is preferentially oxidised and the reaction is: 2Cl-

(aq) Cl2(g) + 2e

• Species remaining in solution: Calcium ions and hydroxide ions

5151Quantities produced by electrolysis

• Faraday's law states that the mass of product produced will be proportional to the charge passed.

• The charge (Coulombs)= current (amps) x time (seconds)

Q=It F = Q/96500

• Faraday's law may also be restated as...the number of faradays required to discharge 1 mol of an ion at an electrode equals the number of charges on that ion.

1 Faraday = 96500 Coulombs # Faradays, F = Q / 96500

• A metallic object to be plated with copper is placed in a solution of CuSO4. a) To which electrode of a direct current power supply should the object be connected?

b) What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?

Solution:

• A metallic object to be plated with copper is placed in a solution of CuSO4. a) To which electrode of a direct current power supply should the object be connected?b) What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?

Solution:

a) Since Cu2+ ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!)

b) The amount of charge passing through the cell is

Q = I x t = (0.22 amp) × (5400 sec) = 1200 c

F = Q / 96500 = (1200 ) ÷ (96500 ) = 0.012 F

Since the reduction of one mole of Cu2+ ion requires the addition of two moles of electrons, the mass of Cu deposited will be

n =0.012/2 = 0.006 moles

m = n x MM = (63.54 g mol–1) (0.006) = 0.39 g of copper

• Example: Calculate the mass of copper released by a current of 10A passing for 200 seconds through a Copper II sulphate solution.

5555http://www.chem1.com/acad/webtext/elchem/ec8.html

• Example: Calculate the mass of copper released by a current of 10A passing for 200 seconds through a Copper II sulphate solution.

• Q = I t =Charge =10 x 200 = 2000 coulombs

• Number of Faradays = 2000/96500 = 0,0207 Faradays

• The reaction occurring at the cathode is:

• Cu2+(aq) + 2e- Cu(s)

• Therefore: 2 Faradays will produce 1 mole of copper

• 0,0207 Faradays will produce 0,0207/2 moles of copper = 0,0104 moles

• Therefore the mass of copper produced = 0,0104 x 63,5 = 0,658g

• Calculate the number of moles of hydrogen released when 5 amps of current passes for 3000 seconds through a solution of sulphuric acid

Q = It

Q = 5 x 3000 = 15000 C

Q = 15000/96500 Faradays

Q = 0.155 F

5757Factors affecting the relative amounts of products formed during electrolysis

1. Charge on the ionsNa+ (l) + e => Na(l) 1 mol of e required to produce 1 mol of Na atoms

Cu 2+ (aq) + 2e => Cu(s) 2 mol of e to produce 1 mol Cu atoms

2. Quantity of electrons

The quantity of electrons depends on the current and time.

If two electrolytic cells are connected in series, the same current will flow through both and for the same length of time.

In the electrolysis the hydrogen gas is released at the cathode as follows:

2H+ + 2e => H2

• 2 moles of electrons release 1 mole of gas

• 2 Faradays of charge are needed to release 1 mole of hydrogen

• 0.155 F releases 0.155/2 moles of hydrogen

• moles of hydrogen released = 0.0777 moles

5959Electrolysis of CuSO4 (aq)

With graphite electrodes With Cu electrodes

6161Electroplating

• Use of electrolysis in electroplating:

• The object to be electroplated is made the negative electrode, and it is placed in a solution of the ions of the metal used to plate it. Electroplating different from electrolysis in that the metal deposited from electrolysis plates out on the surface of another metal. The electrolyte contains the plating metal in the form of dissolved ions and the anode usually is made of the plating metal. The object to be plated is the cathode.

1 9 & 19. ELECTROCHEMISTRY 1. 2 Electron Transfer Reactions 1. Electron transfer reactions are redox...

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