Spontaneous Redox Reactions

• Describe single replacement reactions in terms of oxidation and reduction.

• Explore the Standard Reduction Potential Chart

• Predict spontaneous.

2 Ag+(aq) + Cu(s) →

Ag(s) + Cu2+(aq) →

Ag+ ions can oxidize Cu metal.Cu2+ cannot oxidize Ag metal.

Cu metal can reduce Ag+ ions.Ag metal cannot reduce Cu2+ ions.

2 Ag(s) + Cu2+(aq)

no reaction

Spontaneous rxns occurs without added energy.

Electrochemical cells are described in terms of their voltage.

(formerly called Emf – electromotive force).

Voltage (Eocell) found using Reduction Potentials.

Potentials (Eo) for each substance are calculated by competing with a hydrogen standard in a cell.

2H+(aq) + 2e– ↔ H2(g)

Reduction Potential ChartOrder of ability to react with other compounds.

NO3¯ + 4 H+ + 3e¯ NO(g) + 2 H2O +0.96Hg2+ + 2e¯ Hg(l) +0.85Ag+ + e¯ Ag(s) +0.80

1/2 Hg22+ + e¯ Hg(l) +0.80

NO3¯ + 2 H+ + e¯ NO2(g) + H2O +0.78Fe3+ + e¯ Fe2+ +0.77I2(s) + 2e¯ 2 I¯ +0.53Cu+ + e¯ Cu(s) +0.52

Cu2+ + 2e¯ Cu(s) +0.34SO4

2¯ + 4 H+ + 2e¯ SO2(g)+ 2 H2O +0.17Sn4+ + 2e¯ Sn2+ +0.15

S + 2 H+ + 2e¯ H2S(g) +0.142 H+ + 2e¯ H2(g) 0.00

Fe3+ + 3e¯ Fe(s) –0.04Pb2+ + 2e¯ Pb(s) –0.13Sn2+ + 2e¯ Sn(s) –0.14Ni2+ + 2e¯ Ni(s) –0.25Co2+ + 2e¯ Co(s) –0.28Cd2+ + 2e¯ Cd(s) –0.40

Se + 2 H+ + 2e¯ H2Se(g) –0.40Fe2+ + 2e¯ Fe(s) –0.44Cr2+ + 2e¯ Cr(s) –0.56

Ag2S + 2e¯ 2 Ag(s) + S2¯ –0.69Cr3+ + 3e¯ Cr(s) –0.74

(+) Eo - greater tendency to accept electrons vs. H+.X – reduced Hydrogen –

oxidized

(-) Eo - lesser tendency to accept electrons. X – oxidized Hydrogen –

reduced

Greater positive number is reduced.

H2(g) / H+(aq) // Cu2+

(aq) / Cu(s) Eo = +0.34 V

Zn(s) / Zn2+(aq) // H+

(aq) / H2(g) Eo = -0.76 V

E°cell - sum of potentials of each half-cell.

E°cell = E°ox + E°red

Note: table lists reduction potentials. Oxidation potentials are the reverse – switch sign.

(+) E°cell – spontaneous reaction.

(-) E°cell – non – spontaneous reaction.

What is the cell potential for a silver-copper cell?

→

Ag+(aq) + 1e– → Ag(s)

E°cell = E°ox + E°red =

E° = +0.80 V

E° = 0.34 VCu2+(aq) + 2e– Cu(s) –+ox

red

- 0.34 + 0.80 + 0.46

A cell of zinc and gold metal as electrodes: a) What is the cathode and what is the anode?b) What is the net reaction?c) What is the line notation for the cell?d) What is the cell potential?

We do NOT multiply the voltage.

[ ] ×2

[ ] ×3  →

Au3+(aq) + 3e– → Au(s)

E°cell = E°ox + E°red =

E° = +1.50 V

E° = 0.76 VZn2+(aq) + 2e– Zn(s) +-ox

red

+ 0.76 + 1.50+ 2.26

Zn(s) / Zn2+(aq) // Au3+

(aq) / Au(s)

2 Au3+(aq) + 3 Zn(s) → 2 Au(s) + 3 Zn2+(aq)

Hg2+ + 2e¯ Hg(l) +0.85

Ag+ + e¯ Ag(s) +0.801/2 Hg2

2+ + e¯ Hg(l) +0.80NO3¯ + 2 H+ + e¯ NO2(g) + H2O +0.78

Fe3+ + e¯ Fe2+ +0.77I2(s) + 2e¯ 2 I¯ +0.53Cu+ + e¯ Cu(s) +0.52

Cu2+ + 2e¯ Cu(s) +0.34SO4

2¯ + 4 H+ + 2e¯ SO2(g)+ 2 H2O +0.17Sn4+ + 2e¯ Sn2+ +0.15

S + 2 H+ + 2e¯ H2S(g) +0.142 H+ + 2e¯ H2(g) 0.00

Fe3+ + 3e¯ Fe(s) –0.04Pb2+ + 2e¯ Pb(s) –0.13Sn2+ + 2e¯ Sn(s) –0.14

2 Ag+(aq) + Cu(s) →

Ag(s) + Cu2+(aq) →

2 Ag(s) + Cu2+(aq)

no reaction

E°c = +0.46

E°c = - 0.46

Will tin strips in hydrochloric acid react?

Sn(s) + H+(aq) → ??

Sn(s) + H+(aq) → Sn2+(aq) + H2(g) 2

Hg2+ + 2e¯ Hg(l) +0.85

Ag+ + e¯ Ag(s) +0.801/2 Hg2

2+ + e¯ Hg(l) +0.80NO3¯ + 2 H+ + e¯ NO2(g) + H2O +0.78

Fe3+ + e¯ Fe2+ +0.77I2(s) + 2e¯ 2 I¯ +0.53Cu+ + e¯ Cu(s) +0.52

Cu2+ + 2e¯ Cu(s) +0.34SO4

2¯ + 4 H+ + 2e¯ SO2(g)+ 2 H2O +0.17Sn4+ + 2e¯ Sn2+ +0.15

S + 2 H+ + 2e¯ H2S(g) +0.142 H+ + 2e¯ H2(g) 0.00

Fe3+ + 3e¯ Fe(s) –0.04Pb2+ + 2e¯ Pb(s) –0.13Sn2+ + 2e¯ Sn(s) –0.14

Given the following experimental data, arrange the following in increasing order of oxidizing ability.

1. + In(s) → Co(s) + In2+

2. + Co(s) → Cu(s) + Co2+

3. Cu2+ + → no reaction

Weak Oxidizer In2+ + 2e– → In(s)

Co2+ + 2e– → Co(s)

Cu2+ + 2e– → Cu(s)

Strong Oxidizer Pd2+ + 2e– → Pd(s)

Co2+

Cu2+ Pd(s)

Remember the strong oxidizer is reduced