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2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+
(aq)
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Reduction Potential and Cells
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• Explore the Standard Reduction Potential Chart
• Predict spontaneous using Reduction Potential.
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2 Ag+(aq) + Cu(s) →
Ag(s) + Cu2+(aq) →
Ag+ ions can oxidize Cu metal.Cu2+ cannot oxidize Ag metal.
2 Ag(s) + Cu2+(aq)
no reaction
Spontaneous rxns occurs without added energy.
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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)
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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
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(+) 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
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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.
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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
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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?
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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)
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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
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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
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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