“minimal” galvanic cells

only need reactants

Zn

H+

o = 0.34 V

Zn2+(aq) + 2e- Zn(s) o = -0.76 V

2H+(aq) + 2e- H2(g) o = 0.00 Vreduction

b) oxidationZn Zn2+ + 2e- = 0.76 V

2H+ (aq)

0 = .76 V = very largeQ =

a) oxidationCu Cu2+(aq) + 2e- Cu(s)

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

0

Electrolytic Cells

cell > 0

cell < 0

G < 0 spontaneous galvanic cellG > 0 non-spontaneous electrolytic cell

2 H2(g) G = -474 kJ

redox reaction:O

spontaneous

H 00

+ O2(g) 2 H2O(l)

1+ 2-

Electrolytic Cells

2 H2O(l) G = 474 kJ

oxygen half-cell: H2O

H2O O2

a) oxidationb) reduction

a) anodeb) cathode

2 + 4 H+ + 4 e-

O2

reaction

2 H2(g) + O2(g)

Electrolytic Cells

2 H2O(l) 2 H2(g) + O2(g) G = 474 kJ

hydrogen half-cell: H2O H2

H2O

reduction reaction cathode

+ 2e- H2+ OH-22

Electrolytic Cells

2 H2O

2 H2O

oxidation: anode

reduction: cathode2( )____________________________________

6 H2O

2 H2O

O2 + 2 H2+ 4H++ 4 OH-

O2+ 2 H2

O2 + 4 H+ + 4 e-

+ 2e- H2 + 2 OH-

Electrolysis of water

oxidation reduction2H2O O2+ 4H++ 4e- 4H2O + 4e- 2H2 + 4OH-

Pt electrodes

battery+ -

e-

e-

anode cathode

acid base1 mol gas 2 mol gas

Electrolysis of water

2.5 amp Power source current =

charge mol e- molproduct

gramproduct

current and time

A(C/s) x s x 1mol e-

96,500 Cx mol product mol e-

x g product mol product

3.2 g O2

amperes (A)= coulombs/sec (C/s)

Electrolysis of water

charge mol e- molproduct

gramproduct

current and time

2 H2O O2 + 4 H+ + 4 e- 2.5 A, 3.2 g O2

(C/s) x s

2.5 A 14

32.0 g/mol 3.2

x mol e-

Cx mol O2

mol e- mol O2

x g O2 = g O2

1 mol e-

96500 C

Electrolysis of water

2 H2O O2 + 4 H+ + 4 e- 2.5 A, 3.2 g O2

3.2 g O2 x

2.5 C x s

15440 s x

1 mol O2 x32 g O2

4 mol e- x1 mol O2

96500 C =1 mol e-

38600 C

= 38600 C

s = 15440 s

1 min x 60 s

1 hr60 min

= 4.3 hr

s

Electroplating

Cu2+ + 2e- Cu

anode cathode

oxidation reduction

Cu(s) Cu2+(aq) Cu2+(aq)+ 2e- + 2e- Cu(s)

o = 0.34 V

Electroplating

anode cathode

oxidationreduction

Cu(s) Cu2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)

0.75 Aatomic mass of Cu =

for 25 min. deposits 0.37 g Cu

Electroplating

Cu2+(aq) + 2e- Cu(s)

0.75 A for 25 min deposits 0.37 g Cu

A (C) s

x s x 1 mol e-

96500 Cx 1 mol Cu 2 mol e-

x g Cu mol Cu

Electroplating

0.75(C) s

x 25 min

0.37 g Cu

= 1.125 x 103 C

1.125 x 103 C x 1 mol e-

96500 C= 1.17 x 10-2 mol e-

1.17 x 10-2 mol e- x 1mol Cu 2mol e-

=5.83 x 10-3mol Cu

= 63.5 g mol5.8x10-3 mol Cu

atomic mass Cu

x 60s min