Lecture 41 - Electrochemistry V. Review Galvanic Cells: Reaction is spontaneous E o cell > 0 The...

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Lecture 41 - Electrochemistry V

Review

Galvanic Cells:

Reaction is spontaneous

Eocell > 0

The “product” is an electrical current

Some can be reversed (recharged)

Electrolytic Cells

Reaction is not spontaneous Eo

cell < 0 Used for producing a chemical product

Electrolysis of Water

We want:

2 H2O(l) 2 H2(g) + O2(g)

2 H2O(l) O2(g) + 4 H+(aq) + 4 e- Eo = -1.23 V

2 H2O(l) + 2 e- H2(g) + 2 OH-(aq) Eo = -0.83 V

Eocell = -2.06 V

i.e. non-spontaneous!

Electrolysis of Water

H2O(l) +salt

Electrolysis of Water

-+

2 H2O + 2 e-H2(g) + 2 OH-

H2O(l) +salt

cathodeanode

Note reversal of polarity! battery

2 H2O

O2(g) + 4 H+ + 4 e-

-+

anode cathode

CuCl2(aq)

Electrowinning of Metals

at the cathode:

Cu+2 Cu(s) ??

or

H2O H2(g) ??

Rule of thumb…

The cathode reaction with the most positive reduction potential will occur

preferentially

Cu+2(aq) + 2 e- Ž Cu(s) Eo = +0.15 V

2 H2O(l) + 2 e- Ž H2(g) + 2 OH- Eo = - 0.83 V

at the anode:

Cl- Cl2(g) ??

or

H2O O2(g) ??

Same rule of thumb…

The anode reaction with the most positive oxidation potential will occur

preferentially

Same rule of thumb…

2 Cl-(aq) Ž Cl2(g) + 2 e- Eo = -1.36 V

2 H2O(l) Ž O2(g) + 4 H+ + 4 e- Eo = - 1.23 V

Thus, we predict water gets oxidized at the anode.

Overpotential

Erequired > Eocell for electrolysis

In this case, Cl- is oxidized

-+

anode cathode

CuSO4(aq)

2 Cl- Cl2(g) + 2 e-

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

Electrowinning of Metals

Overall Reaction:

Cu+2(aq) + 2 Cl-(aq) Ž Cu(s) + Cl2(g)

Eocell = -1.21 V

Electrorefining of Metals

-+Impure Cu

Pure Cu

Cu

Making Fluorine, F2(g)

Electrolysis of HF 2 KF(l)

oxidation: 2 F- F2(g) + 2 e-

reduction: 2 H+ + 2 e- H2(g)

Making Chlorine, Cl2(g)

Electrolysis of brine (NaCl(aq))

oxidation:

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

reduction:

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

Overall,

2 NaCl(aq) + 2 H2O(l) Cl2(g) + H2(g)

+ 2 NaOH(aq)

disinfection, plastics (PVC)

cleaners,paper, etc

Manufacture of Aluminum

Why not just electrowin Al+3(aq)?

Al+3(aq) + 3 e- Al(s) Eo = -1.66 V

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

Eo = -0.83 V

(reduction of water is favored)

Manufacture of Aluminum

Bauxite = impure Al2O3(s)

m.p. = 2045oC

Solution?1. Lower the melting point2. Perform electrolysis

Manufacture of Aluminum

Al2O3(s) + AlF3(s) + 3 NaF(s)

Na3AlF6, cryolite

m.p. 900oC

Manufacture of Aluminum

reduction: AlFxOy3-x-2y + n e- Al(l) + ?

oxidation: AlFxOy3-x-2y O2(g) + n e- + ?

sinks!oxidizes Celectrodes!