1chemistry 2c lecture 2: march 31 st, 2010 i.acidic redox balancing ii.net ionic equations iii....

1Chemistry 2C Lecture 2: March 31st, 2010

I. Acidic Redox BalancingII. Net Ionic Equations

III. Balancing Redox rxs. in basic solutionsIV. Electrochemistry Intro

Lecture 2: Electrochemistry

Should be able to balance the Thermite reactionFe2O3 (s) + Al (s) -> Fe (l) + Al2O3 (s)

http://en.wikipedia.org/wiki/Thermite

OIL RIG


From last lecture: Balancing Redox Rxs.

General method (for acidic aqueous rxs):

1. Write half reactions separately by first identifying the oxidized and reduced species

2. Balance element other than H or O, if any

3. Balance O by adding H2O molecules

4. Balance H by adding H+

5. Balance charge by adding electrons to the more positive side of the half-reaction

6. Multiple the half reaction by the whole number factors so that the number of electrons lost equal the number of electrons gained.

7. Add the two half reactions together and cancel common terms occurring on both sides of the arrow.


Balancing Redox Rxs.Fe2+ (aq) + Cr2O7

2- (aq) Fe3+ (aq) + Cr3+ (aq)Unbalanced Eq:

Step 1. Write half reactions separately by first identifying the oxidized and reduced species

Fe is being oxidized (loss of electrons)

Cr is being reduced (gain of electrons)

Oxidation state Change: +2 +3

Oxidation state Change: +6 +3

Fe2+ (aq) Fe3+ (aq)

Cr2O72- (aq) Cr3+ (aq)

Eq. 1

Eq. 2




Step 2. Balance element other than H or O, if any

Fe2+ (aq) Fe3+ (aq)

Cr2O72- (aq) Cr3+ (aq)

Eq. 1

Eq. 2 (old)

Cr2O72- (aq) 2Cr3+ (aq)Eq. 2 (new)




Step 3. Balance O by adding H2O molecules

Fe2+ (aq) Fe3+ (aq)Eq. 1

Cr2O72- (aq) 2Cr3+ (aq)Eq. 2 (old)

Cr2O72- (aq) 2Cr3+ (aq) + 7H20Eq. 2 (new)




Step 4. Balance H by adding H+

Fe2+ (aq) Fe3+ (aq)Eq. 1

Cr2O72- (aq) 2Cr3+ (aq) + 7H20Eq. 2 (old)

Eq. 2 (new) 14H+ (aq) + Cr2O72- (aq) 2Cr3+ (aq) + 7H20




Step 5. Balance charge by adding electrons to the more positive side of the half-reaction

Fe2+ (aq) Fe3+ (aq)Eq. 1 (old)

Fe2+ (aq) Fe3+ (aq) + e-Eq. 1 (new)

Eq. 2 (old)

6e- + 14H+ (aq) + Cr2O72- (aq) 2Cr3+ (aq) + 7H20Eq. 2 (new)

14H+ (aq) + Cr2O72- (aq) 2Cr3+ (aq) + 7H20




Step. 6 Multiple the half reaction by the whole number factors so that the number of electrons lost equal the

number of electrons gained

Fe2+ (aq) Fe3+ (aq) + e-Eq. 1 (old)

6e- + 14H+ (aq) + Cr2O72- (aq) 2Cr3+ (aq) + 7H20Eq. 2

6Fe2+ (aq) 6Fe3+ (aq) + 6e-Eq. 1 (new)

The number of electrons in oxidized half reaction is equal to the number of electrons in the reduced half

reaction (6).




Step 7: Add the two half reactions together and cancel common terms occurring on both sides of the arrow

6e- + 14H+ (aq) + Cr2O72- (aq) 2Cr3+ (aq) + 7H20Eq. 2

6Fe2+ (aq) 6Fe3+ (aq) + 6e-Eq. 1

Electrons are never a species in a balanced redox equation!!!!

Sum 6e- + 14H+ (aq) + Cr2O72- (aq) + 6Fe2+ (aq)

2Cr3+ (aq) + 7H20 + 6Fe3+ (aq) + 6e-


Net Ionic Equations• What is a net ionic equation? An equations that

concentrates on the true reacting species and ignores non-interacting species that are just “hanging out”

• Suppose Mg metal is treated with hydrochloric acid, H2 gas is evolved and MgCl2 can be isolated from the evaporated solution (not precipitated).

• The balanced reaction may look like this:

Mg (s) + 2HCl (aq) -> MgCl2 (aq) + H2(g)0 +1 -1 +2 -1 0

Oxidized (loss of electrons)

Reduced (gain of electrons)


Net Ionic Equations

• But HCl is a strong acid, (hence a strong electrolyte) so it really should be written as H+ (aq) + Cl- (aq)

• Similarly MgCl2 is soluble in H20, so this should be written as Mg+2 (aq) and Cl- (aq)

• So overall the balance reaction shouldn’t beMg (s) + 2HCl (aq) -> MgCl2 (aq) + H2(g)

But really Mg (s) + 2H+ (aq) + 2Cl- (aq) -> Mg+2 (aq) + Cl- (aq)

+ H2(g)


Net Ionic EquationsIn this equation

Mg (s) + 2H+ (aq) + 2Cl- (aq) -> Mg+2 (aq) + Cl- (aq) + H2(g)

The chloride ion is not participating with the redox reaction (it is neither being reduced nor being oxidized). This is commonly referred to as a

“spectator ion” and the “net ionic equation” is really

Mg (s) + 2H+ (aq) -> Mg+2 (aq) + H2(g)


ElectrochemistryPremise: We know that a large number of reactions fall

into the redox class, which involve the transfer of electrons from a species (atom) that is being oxidized

to a species (atom) that is being reduced.

Such charge transfer reactions are ubiquitous in nature!- Photosynthesis (light to chemical energy)

- Photovoltiacs (light to electric energy)- Muscles (energy released by redox rx. causes motion)

- Dynamite (redox rx. that generate a lot of energy)- Combustion of fuels (oxidation)

- Batteries (redox rx. To make electricity)- Corrosion, chrome plating, “tin” cans (sacrificial metals with

boats)


ElectrochemistryPremise: We know that a large number of reactions fall

into the redox class, which involve the transfer of electrons from a species (atom) that is being

oxidized to a species (atom) that is being reduced.

Terms: Oxidizing agents: Species that take electrons and are

themselves reduced (e.g. they oxidize species)

Reducing agents: Species that give electrons and are themselves oxidized (e.g. they reduce species)


ElectrochemistryFrom the list above, it is clear that many redox

reactions don’t occur in the simplified acidic aqueous environment that was discussed last

lecture. For example, we may require the balancing of a redox reaction in a basic solution.

How? The easiest way is to follow the protocol for balancing redox rxs. for acidic aqueous solutions, but you must use OH- to neutralize the H+ species

(after step 4).


Balancing basic Aquaeous redox rxs.

Following the protocol with acidic aqueous rxs.

S2O32- (aq) + Cl2 (g) -> HSO4

- (aq) + Cl- (aq)+2

The sulfur in S2O32- is being oxidized (loses e-)

The chlorine in Cl2 is being reduced (gains e-)

0 +6 -1

The strength of a oxidizing agent (or reducing agent) varies depending on the nature (and thermodynamics) of the

species. Cl2 is a stronger oxidizing agent than I2 which was used in your lab.

Step 1. Write half reactions separately by identifying the oxidized and reduced species



S2O32- (aq) + Cl2 (g) -> HSO4

- (aq) + Cl- (aq)+2 0 +6 -1

Cl2 (g) -> Cl- (aq)

S2O32- (aq) -> HSO4

- (aq)

Eq. 1 (reduction):Eq. 2 (oxidation):

Step 1. Write half reactions separately by identifying the oxidized and reduced species

Step 2. Balance element other than H or O, if any

Cl2 (g) -> 2Cl- (aq)

S2O32- (aq) -> 2HSO4

- (aq)




Cl2 (g) -> 2Cl- (aq)

5H20 (l) + S2O32- (aq) -> 2HSO4

- (aq)



Step 3. Balance O by adding H2O molecules

Step 4. Balance H by adding H+

Cl2 (g) -> 2Cl- (aq)

5H20 (l) + S2O32- (aq) -> 2HSO4

- (aq) + 8H+ (aq)



Step 4New. Add OH- to both sides to neutralize any H+


Cl2 (g) -> 2Cl- (aq)

8OH- (aq) + 5H20 (l) + S2O32- (aq)

-> 2HSO4- (aq) + 8H+ (aq) + 8OH- (aq)

Eq. 2 (oxidation): 8OH- (aq) + 5H20 (l) + S2O32- (aq)

-> 2HSO4- (aq) + 8H2O (l)

The new step of adding OH- to each side of the equation to neutralize H+ species. This is the primary difference between balancing redox

equations in acidic or basic aqueous environments!



Eq. 1 (reduction):

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

Eq. 2 (oxidation):8OH- (aq) + S2O32- (aq) -> 2HSO4

- (aq) + 3H20 (l) + 8e-

Step 5. Balance charge by adding electrons to the more positive side of the half-reaction

Step. 6 Multiple the half reaction by the whole number factors so that the number of electrons lost equal the

number of electrons gained

Multiple Eq. 1 by 4 to get equal number of electrons gained and lost.

Eq. 1 (reduction):

8e- + 4Cl2 (g) -> 8Cl- (aq)



Eq. 1 (reduction):

8e- + 4Cl2 (g) -> 8Cl- (aq)

Eq. 2 (oxidation):8OH- (aq) + S2O32- (aq) -> 2HSO4

- (aq) + 3H20 (l) + 8e-

Step 7: Add the two half reactions together and cancel common terms occurring on both sides of the arrow

Total rx: 8OH- (aq) + S2O32- (aq) + 8e- + 4Cl2 (g) -> 2HSO4

- (aq)

+ 8Cl- (aq) + 3H20 (l) + 8e-


Disproportionation RxssSpecial reactions in which the same substance is both oxidized and reduced (obviously not the same specific

atom)2H2O2 (aq) -> 2H2O (l) + O2

(g)

2Cu+ (aq) -> Cu (s) + Cu+2 (aq)

Exercise: Balance in a basic solution

ClO2 -> Cl03- (aq) +Cl- (aq)

H2O2 (aq) -> H2O (l)

H2O2 (aq) -> O2 (g)

Eq. 1 (reduction):

Eq. 2 (oxidation):

Cu+ (aq) -> Cu (s)

Cu+ (aq) -> Cu+2 (aq)

Eq. 1 (reduction):

Eq. 2 (oxidation):


ElectrochemistryIs an old, but still important science!!!

1786: Many say that eletrochemistry had its start

with Luigi Galvani who plugged in poor froggies with different metals. The froggies muscles moved and thus he came to the conclusion that electricity was a “life force.”

Mid-18th century: Ben Franklin had many theories about positive and negative electricity. His famous kite expt. was in 1752. He pushed the approach that electricity was like a

fluid that passed from one body to another

http://www.corrosion-doctors.org/Biographies/VoltaBio.htm


Electrochemistry

Late 1700’s: Luigi Galvani friend Alessandro Volta realized that the animal electricity was

really a reaction from different metals. He made the first battery by stacking Ag and Zn disks separated by a wet cloth containing a

salt or weak acid solution.

These people gave us their names: “Galvanic cell”, “Voltaic cell”,

Volts, galvanization.

Question for next lecture: Why was is important to use wet, salty

clothes in Volta’s experiment?

1chemistry 2c lecture 2: march 31 st, 2010 i.acidic redox balancing ii.net ionic equations iii....

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