Chem 127

November 29, 2004

Wed evening, 7:30, Optional Exam

ACS Standardized (Multiple Choice)

Comprehensive (all of General Chemistry)

No Risk; grade can replace an earlier exam

Electrochemistry

We can build a galvanic electrical cell

Observe the result of a spontaneous reaction

as a source of electrical power

consume reactants, use up the "battery"

voltage fixed by electrode E^o's (and conc)

Faraday's law determines current x time

as a laboratory probe

measure E, determine DG

measure E, determine a concentration

in lab system, negligible use of materials

Nernst's Law relates E and conc.

We can also operate an electrolytic cell

We force a non-spontaneous process to occur

We convert electrical energy into chemical products

electrolysis (e.g., H₂ from water, Cl₂ from NaCl)

reduction to metal (all Al, most Mg, Na-K-Li)

electroplating and refining (most pure Copper)

Not all reactions can be driven this way

(some cells can be recharged, others cannot)

Good example is the purification of raw copper

mine copper(II) carbonates, hydroxides

heat and form copper oxide (CuO)

heat with carbon and form Cu metal

also reduce other metals (Ag, Zn, Pb, Sn)

form alloys; but not very pure copper

bronze: Cu/Sn-Sb (the bronze age)

brass: Cu/Zn

set up an electrical cell

raw copper anode (will be oxidized)

cathode (will become pure copper)

solution (Cu²⁺ ions)

Anode reactions (reduction potential)

Zn --> Zn²⁺ + 2e- E^o = -0.76 easiest to oxidize

Cu --> Cu²⁺ + 2e- E^o= 0.34

Ag --> Ag⁺ + e- E^o= 0.80 hardest to oxidize

As a rule, the oxidation is of the most easily oxidized species

From the Table bottom (most negative E^o's)

Initially, we first oxidize Zn from anode surface

When gone, we oxidize Copper of the anode

The smaller amounts of Silver, Lead (-0.13) never oxidize

fall off as sludge as surrounding copper dissolves

(that valuable sludge is the largest source of Ag )

At the cathode, we reduce the species highest on the table

The largest Eo Reduction Potentials

if Eo of H2 is higher, we form H2 and not Metal

In the Electro- purification

Ag would plate out first-- but it's not in solution

Cu plates out before Zn (or before H2)

So we form nearly pure copper cathodes

solution gradually gets richer in Zn²⁺

may eventually need to replace solution

Notice that you cannot generally electroplate metals

like Zn (-0.76 V) since you form H2 instead

But we do electroplate Ni (-0.26V )

2H⁺ + 2e --> H₂ (g) should occur first (E^o = 0.00 V)

but this is only valid at 1.0 M H+

in pH 7 plating bath E = 0 - 0.059 log10( 1 /10^-7) =-0.4V

so 0.1M Ni²⁺ is reduced before pH 7 H⁺ is reduced

not much hope of doing this effectively for Zn (-0.76 V)

can be done with alkaline plating baths

hopeless for Mg (-2.37 V) or Na (-2.71) or Al (-1.66 V)

can reduce these from molten salts (no water)

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This is about where we stopped in class

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Decomposing water for form H2 and O2 gases

2 H+ + 2e --> H₂ (g) +0.00 V

O2(g) + 2 H2O +4e --> 4 OH- +0.40 V

Net reaction

O₂ (cathode) + 2H₂(anode) --> 4H+ (anode) + 4OH- (cathode)

Standard Cell Potential +0.40 - (0.0) = +0.40

can generate electrical power

gas bubbling over Pt electrodes in 1M HCl, 1 M NaOH

O₂ (cathode) + 2H₂(anode) --> 4H+ (anode) + 4OH- (cathode)

Standard Cell Potential +0.40 - (0.0) = +0.40

We can set up with separate electrolytes (solutions)

Reverse reaction--

decompose water to form H₂ and O₂ gases

probably not a standard cell

not both 1.0M NaOH and 1.0M HCl

in lab, one solution with both electrodes in it

(in lab we used 0.1 M H+ as H₂SO₄)

E = +0.40 -(0.059/4) log10 ( [OH-]⁴ [H+]⁴ /P_O2 P²_H2 )

seems a bit odd, but E is independent of [H+]

=0.40 - (0.015) log10 ( K⁴_w /P_O2 P²_H2)

=0.40 - (0.015) x 4 x (-14) if Pgases = 1 atm

=-0.44 V

DG = - nFE = - 4 * 96487 * (-0.44) = + 170,000 Joules

we need to supply (at least) 170 kJ to make O₂ + 2 H₂

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Using a Spontaneous Reaction as electrical power

ideally want a anode with large negative Eo

anode will be oxidized

value is standard REDUCTION potential

so large negative value is preferred

Zn is the dominant species (-0.76 V)

Al (-1.66 would be better)

not used, mostly due to stable oxide film

Li (-3.04 V) is also used

despite problems of packaging; no water

limited by safety rules to small batteries

(a gram of lithium could be quite hazardous)

(recent news story-- 80+ cell phones are known

to have exploded, some while in use,

mostly due to battery failures)

cathode is typically MnO₂ (s) < Mn(IV) >

reacts with Li+ to form LiMnO₂ (s) <Mn(III)>

one feature E = Eo - 0.059 log10 (1/1)

even as material is consumed, voltage stays constant

eventually battery dies, no "weak" stage

How much Li is in a battery?

Electronic Catalog lists a "coin" type with 100 mAh

milliAmp Hour

Faraday -- # mole electron = # mol Li

= 0.10 A x 3600 sec / 96485 = 3.7 x 10-3 moles

= 3.7 x 10-3 x 6.98 g/mol = 25.5 mg of Li

(may be more; often last 10-20% is unused if E drops)

An ordinary AA battery is rated 110 hr at 15 mA

# moles of Zn =(2)* (0.015)*(110*3600) / 96485 =0.12

or 64 g/mol x 0.12 mol = 7.9 g of Zn

more moles, denser metal

lower voltage cell

Zn(s) --> Zn2+

2MnO₂(s) + 2NH₄⁺(aq) + 2e --> Mn₂O₃ + 2NH₃ (aq) +H₂O(liq)

E = Eo - (0.059/2) log10 ( [NH₃]²/[NH₄]² )

as NH₄⁺ is used and NH₃ increases

E of cell decreases

1:100 going to 100:1 ratio is 10⁴

E = (1.5V) - 0.24 V = 1.2 V (a dying battery)

The Alkaline Cell (with NaOH or KOH)

for same amount of zinc, same current x time

nearly the same voltage

slightly higher current available if needed

less loss of Zn due to non-electrical reactions

storage life

batteries do go dead without being used

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