Section 10 Electrochemical Cells and Electrode Potentials.

14
Section 10 Electrochemical Cells and Electrode Potentials

Transcript of Section 10 Electrochemical Cells and Electrode Potentials.

Page 1: Section 10 Electrochemical Cells and Electrode Potentials.

Section 10

Electrochemical Cells and Electrode Potentials

Page 2: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryOxidation/Reduction Reactions

• “Redox” reactions involve electron transfer from one species to another

• Ox1 + Red2 Red1 + Ox2

• Ox1 + ne- Red1 (Reduction ½ reaction)

• Red2 Ox2 + ne- (Oxidation ½ reaction)

• “Reducing agent” donates electrons (is oxidezed)• “Oxidizing agent” accepts electrons (is reduced)

Page 3: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryOxidation/Reduction Reactions

• Typical oxidizing agents: Standard Potentials,V

– O2 + 4H+ + 4e- 2H2O +1.229

– Ce4+ + e- Ce3+ +1.6 (acid)

– MnO4- + 8H+ + 5e- Mn2+ + 4H2O +1.51

• Typical reducing agents:– Zn2+ + 2e- Zno -0.763– Cr3+ + e- Cr2+ -0.408– Na+ + e- Nao -2.714

Page 4: Section 10 Electrochemical Cells and Electrode Potentials.

Fig. 12.1. Voltaic cell.

The salt bridge allows charge transfer through the solution and prevents mixing.

The spontaneous cell reaction (Fe2+ + Ce4+ = Fe3+ + Ce4+) generates the cell potential.

The cell potential depends on the half-reaction potentials at each electrode.

The Nernst equation describes the concentration dependence.

A battery is a voltaic cell. It goes dead when the reaction is complete (Ecell = 0).

The salt bridge allows charge transfer through the solution and prevents mixing.

The spontaneous cell reaction (Fe2+ + Ce4+ = Fe3+ + Ce4+) generates the cell potential.

The cell potential depends on the half-reaction potentials at each electrode.

The Nernst equation describes the concentration dependence.

A battery is a voltaic cell. It goes dead when the reaction is complete (Ecell = 0).

©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

Page 5: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryStandard Reduction Potentials

• Half-Reaction Potentials:

• They are measured relative to each other

• Reference reduction half-reaction:

• standard hydrogen electrode (SHE)

• normal hydrogen electrode (NHE)

• 2H+(1.0) + 2e- H2(g 1atm) 0.0000 volts

Page 6: Section 10 Electrochemical Cells and Electrode Potentials.

The more positive the Eo, the better oxidizing agent is the oxidized form (e.g., MnO4-).

The more negative the Eo, the better reducing agent is the reduced form (e.g., Zn).

The more positive the Eo, the better oxidizing agent is the oxidized form (e.g., MnO4-).

The more negative the Eo, the better reducing agent is the reduced form (e.g., Zn).

©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

Page 7: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryReduction Potentials

• General Conclusions:

• 1. The more positive the electrode potential, the stronger an oxidizing agent the oxidized form is and the weaker a reducing agent the reduced form is

• 2. The more negative the reduction potential, the weaker the oxidizing agent is the oxidized formis and the stronger the reducing agent the reduced form is.

Page 8: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryOxidation/Reduction Reactions

• Typical oxidizing agents: Standard Potentials,V

– O2 + 4H+ + 4e- 2H2O +1.229

– Ce4+ + e- Ce3+ +1.6 (acid)

– MnO4- + 8H+ + 5e- Mn2+ + 4H2O +1.51

• Typical reducing agents:– Zn2+ + 2e- Zno -0.763– Cr3+ + e- Cr2+ -0.408– Na+ + e- Nao -2.714

Page 9: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryOxidation/Reduction Reactions

• Net Redox Reactions: Standard Potentials,V

• MnO4- Mn2+

• MnO4- + 8H+ + 5e- Mn2+ + 4H2O +1.51

• Sn4+ + 2e- Sn2+ +0.154

• Balanced Net Ionic Reaction:• 2MnO4

- + 16H+ + 5Sn2+ 2Mn2+ + 5Sn4+ + 8H2O

Page 10: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryVoltaic Cell

• The spontaneous (Voltaic) cell reaction is the one that gives a positive cell voltage when subtracting one half-reaction from the other.

• Eocell = Eo

right – Eo

left = Eocathode – Eo

anode =Eo+ - Eo

-

• Which is the Anode? The Cathode?• Convention:• The anode is the electrode where oxidation occurs

the more negative half-reaction potential• The cathode is the electrode where reduction occurs

the more positive half-reaction potential• anode solution cathode

Page 11: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryOxidation/Reduction Reactions

• Net Redox Reactions: Standard Potentials,V

• MnO4- Mn2+

• MnO4- + 8H+ + 5e- Mn2+ + 4H2O +1.51

• Sn4+ + 2e- Sn2+ +0.154

• Balanced Net Ionic Reaction:• 2MnO4

- + 16H+ + 5Sn2+ 2Mn2+ + 5Sn4+ + 8H2O

• Eocell = Eo

cat – Eoan = (+1.51 – (+0.154)) = +1.36 V

Page 12: Section 10 Electrochemical Cells and Electrode Potentials.

ElectrochemistryNernst Equation

• Effects of Concentrations on Potentials:• aOx + ne- bRed• E = Eo – (2.3026RT/nF) log([Red]b/[Ox]a

– Where E is the reduction at specific conc., – Eo is standard reduction potential, n is number of electrons

involved in the half reaction, – R is the gas constant (8.3143 V coul deg-1mol-1), – T is absolute temperature, – and F is the Faraday constant (96487 coul eq-1).

• At 25oC(298.16K) the value of 2.3026RT/F is 0.05916• Note: Concentrations should be activities

Page 13: Section 10 Electrochemical Cells and Electrode Potentials.

Electrochemistry

• Calculations:• MnO4

- + 8H+ + 5e- Mn2+ + 4H2O Eo = +1.51 V

• For [H+] = 1.0M, [MnO4-] = 0.10M, [Mn2+] = 0.010M

• E = Eo – 0.05916/5 (log ([Mn2+]/[MnO4-][H+]8)

• E = +1.51 – 0.1183(-1) = +1.63 V vs NHE• Note: This is more positive than Eo

• Greater tendency to be reduced compared to standard conditions.

Page 14: Section 10 Electrochemical Cells and Electrode Potentials.

Electrochemistry• Calculations:• Silver electrode/silver chloride deposit/0.010M NaCl• AgCl + 1e- Ago + Cl- E = ?• Ag+ + 1e- Ago Eo = +0.799 V

• AgCl Ag+ + Cl- Ksp= 1.8 x 10-10

• AgCl + e- Ago + Cl-

• E = Eo - (0.05916/1) Log (1/[Ag+])

• [Ag+] = Ksp/[Cl-] = 1.8 x 10-10/(0.010) = 1.8 x 10-8

• E = +0.799 – (0.05916)(7.74) = +0.341 V