CHEM 180/181Chapter 20 Dana Roberts ([email protected]) Chapters covered: 18 and 20 Notes available online or in Resource Room (1 st floor). To print

CHEM 180/181 Chapter 20

Dana Roberts ([email protected])

Chapters covered: 18 and 20

Notes available online or in Resource Room (1st floor).

To print notes: From print window, select “print handouts” and 4 per page in black and white. Should turn out fine.

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Email me to set up an appointment. I’m almost always available!


Chapter 20 Chapter 20 ElectrochemistryElectrochemistry


Topics in ElectrochemistryTopics in Electrochemistry

• Oxidation-Reduction (Redox) Numbers• Balancing Redox Reactions • Voltaic Cells• Cell EMF• Spontaneity of Redox Reactions• Batteries• Corrosion• Electrolysis


Oxidation and ReductionOxidation and Reduction

Oxidation (loss of e-) Na Na+ + e-

Reduction (gain of e-) Cl + e- Cl-

Oxidation-reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced.

Electron transfer can produce electrical energy spontaneously, but sometimes electrical energy is used to make them occur (nonspontaneous).



Oxidation-Reduction (Redox) ReactionsOxidation-Reduction (Redox) Reactions

BOTH reduction and oxidation must occur.

A substance that gives up electrons is oxidized and is called a reducing agent or reductant (causes another substance to be reduced).

A substance that accepts electrons is reduced and is therefore called an oxidizing agent or oxidant (causes another substance to be oxidized).


Determine if it’s a redox reaction by keeping track of the oxidation states of all elements involved.

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

(0) (+1) (+2) (0)

Quick hint: If a reaction includes an ELEMENT that turns into an ion, it’s a redox reaction.

Oxidation-reduction (Redox) ReactionsOxidation-reduction (Redox) Reactions


1. Atoms in elemental form, oxidation number is zero. (Cl2, H2, P4, Ne are all zero)

2. Monoatomic ion, the oxidation number is the charge on the ion. (Na+: +1; Al3+: +3; Cl-: -1)

3. O is usually -2. But in peroxides (like H2O2 and Na2O2) it has an oxidation number of -1.

4. H is +1 when bonded to nonmetals and -1 when bonded to metals.

(+1 in H2O, NH3 and CH4; -1 in NaH, CaH2 and AlH3)5. The oxidation number of F is -1.6. The sum of the oxidation numbers for the molecule is the

charge on the molecule (zero for a neutral molecule).

Oxidation Number Guidelines


Example

Determine the oxidation state of all elements in ammonium thiosulfate (NH4)2(S2O3).

(NH4)2(S2O3)

NH4+ S2O3

2-

-3

+1 -2

-6total

+2

NH4+ S2O3

2-

+4sum = overallcharge on ion

-3

+4


Determining Oxidation StatesDetermining Oxidation States

What is the oxidation state of Mn in MnO4-?

Answer: +7


Balancing oxidation-reduction equationsBalancing oxidation-reduction equations

Balancing chemical equations follows law of conservation of mass.

AND, gains and losses of electrons must also be balanced.


Half-ReactionsHalf-Reactions

Separate oxidation and reduction processes in equation,

Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)

Oxidation: Sn2+(aq) Sn4+(aq) + 2e-

Reduction: 2Fe3+(aq) + 2e- 2Fe2+(aq)

Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half reaction.


Balancing Equations by the Method of Balancing Equations by the Method of Half-Reactions: AcidicHalf-Reactions: Acidic

Consider :

MnO4-(aq) + C2O4

2-(aq) Mn2+(aq) + CO2(g)

Unbalanced half-reactions: MnO4-(aq) Mn2+(aq)

C2O42-(aq) CO2(g)

First, balance everything EXCEPT hydrogen and oxygen.

Deal with half-reactions SEPARATELY.

(acidic)



To balance O:

Add 4H2O to products to balance oxygen in reactants.

MnO4-(aq) Mn2+(aq) + 4H2O(l)

To balance H:

Add 8H+ to reactant side to balance the 8H in water.8H+(aq) + MnO4

-(aq) Mn2+(aq) + 4H2O(l)



Balance charge: Add up charges on both sides. Add 5 electrons to reactant side.

5e- + 8H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l)

Mass balance of C in oxalate half-reaction.C2O4

2-(aq) 2CO2(g)

Balance charge by adding two electrons to the products.C2O4

2-(aq) 2CO2(g) + 2e-

Last step: Cancel electrons and add reactions together.



5e- + 8H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l)

C2O42-(aq) 2CO2(g) + 2e-

Top reaction times 2. Bottom reaction times 5. ALL PARTS!!

10e- + 16H+(aq) + 2MnO4-(aq) 2Mn2+(aq) + 8H2O(l)

5C2O42-(aq) 10CO2(g) + 10e- .

16H+(aq) + 2MnO4-(aq) + 5C2O4

2-(aq)

2Mn2+(aq) + 8H2O(l) + 10CO2(g)

**Electrons have cancelled out**


Balancing Equations by the Method of Balancing Equations by the Method of Half-Reactions: Acidic SummaryHalf-Reactions: Acidic Summary

1. Divide equation into two incomplete half-reactions.

2. Balance each half-reaction(a) balance elements other than H and O.(b) balance O atoms by adding H2O.

(c) balance H atoms by adding H+ (basic conditions will require further work at this step).(d) balance charge by adding e- to the side with greater overall positive charge.


Balancing Equations by the Method Balancing Equations by the Method of Half-Reactions: Summaryof Half-Reactions: Summary

3. Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other half-reaction.

4. Add the two half-reactions and cancel out all species appearing on both sides of the equation.

5. Check equation to make sure there are same number of atoms of each kind and the same total charge on both sides. Errors can be caught!!


Balancing Equations by the Method of Balancing Equations by the Method of Half-Reactions: BasicHalf-Reactions: Basic

Balancing process is started using H+ and H2O, then adjusting with OH- to uphold reaction conditions

(H+ does not exist in basic solutions).

Balance the following reaction:H2O2(aq) + ClO2(aq) ClO2

-(aq) + O2(g)

(basic)


Basic Redox ReactionsBasic Redox Reactions

Split into two half-reactions.H2O2(aq) O2(g)

ClO2(aq) ClO2-(aq)

Balance elements, then oxygen by adding H2O. Then, add H+ to balance H, just like an acidic redox reaction.

H2O2(aq) O2(g) + 2H+

ClO2(aq) ClO2-(aq)




Add OH- to both sides, enough to neutralize all H+ (basic reactions cannot support H+).

2OH- + H2O2(aq) O2(g) + 2H+ + 2OH-

ClO2(aq) ClO2-(aq)

Combine H+ and OH- to form H2O.

2OH- + H2O2(aq) O2(g) + 2H2O




Balance charge by adding e-.2OH- + H2O2(aq) O2(g) + 2H2O + 2e-

1e- + ClO2(aq) ClO2-(aq)

Multiply each reaction so both have same e-. Then add them together, cancelling where possible.

2OH- + H2O2(aq) + 2ClO2(aq)

O2(g) + 2H2O + 2ClO2

-(aq)

Double-check your answer!



Balance this redox equationBalance this redox equationCu(s) + NO3

-(aq) Cu2+(aq) + NO2(g) (acidic)

Ans: Cu(s) + 2NO3-(aq) + 4H+(aq) Cu2+(aq) + 2NO2(aq) +

2H2O(l)

Solution posted on notice board.


Balance this redox equationBalance this redox equation

NO2-(aq) + Al(s) NH3(aq) + Al(OH)4

-(aq) (basic)

2Al(s) + NO2-(aq) +OH-(aq) + 5H2O(aq) 2Al(OH)4

-(aq) + NH3(aq)

Solution posted on notice board.

Documents

CHEM 180/181Chapter 20 Dana Roberts ([email protected]) Chapters covered: 18 and 20 Notes available online or in Resource Room (1 st floor). To print