Gen Chem II Ch 18: Electrochemistry

Electrochemistry

Redox Reactions

• Definition: Reactions where one or more elements change oxidation number.
  • Examples include single displacement and combustion reactions.
  • Some synthesis and decomposition reactions can also be redox reactions.
• Key Feature: Always involve both oxidation and reduction.
• Analysis: Can be split into oxidation and reduction half-reactions.
• Mechanism: Also known as electron transfer reactions.
• Half-Reactions: Include electrons in the equation.
• Oxidizing Agent:
  • The reactant molecule that causes oxidation.
  • Contains the element that is reduced.
• Reducing Agent:
  • The reactant molecule that causes reduction.
  • Contains the element that is oxidized.

Oxidation

• Definition: Process that occurs when:
  • The oxidation number of an element increases.
  • An element loses electrons.
  • A compound adds oxygen.
  • A compound loses hydrogen.
• Half-Reaction Representation: Electrons appear as products.

Reduction

• Definition: Process that occurs when:
  • The oxidation number of an element decreases.
  • An element gains electrons.
  • A compound loses oxygen.
  • A compound gains hydrogen.
• Half-Reaction Representation: Electrons appear as reactants.

Rules for Assigning Oxidation States

• The rules are applied in order of priority.

1. Free Elements: Have an oxidation state of 0.

• Example: Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g).

• Example: Na = +1 and Cl = -1 in NaCl.

• Example: Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0.

• Example: N = +5 and O = -2 in NO_3, (+5) + 3(-2) = -1.

• Example: Na = +1 in NaCl.

• Example: Mg = +2 in MgCl_2.

• Note: Nonmetals higher on the table take priority.

Oxidation State Changes

• Oxidation: Occurs when an atom's oxidation state increases during a reaction.
• Reduction: Occurs when an atom's oxidation state decreases during a reaction.
• Example:
  CH4 + 2O2 \rightarrow CO2 + 2H2O
  Oxidation numbers:
  C: -4 to +4 (oxidation)
  O: 0 to -2 (reduction)

Oxidizing and Reducing Agents

• Oxidation and reduction must occur simultaneously.
  • If an atom loses electrons, another atom must gain them.
• Reducing Agent:
  • The reactant that reduces an element in another reactant.
  • Contains the element that is oxidized.
• Oxidizing Agent:
  • The reactant that oxidizes an element in another reactant.
  • Contains the element that is reduced.
• Example: 2 Na(s) + Cl_2(g) \rightarrow 2 NaCl(s)
  • Na is oxidized, so Na is the reducing agent.
  • Cl2 is reduced, so Cl2 is the oxidizing agent.

Identifying Oxidizing and Reducing Agents

• Example 1:

  3 H2S + 2 NO3^- + 2 H^+ \rightarrow 3 S + 2 NO + 4 H_2O

  • N in NO_3^- is reduced (oxidizing agent).
  • S in H_2S is oxidized (reducing agent).

• Example 2:

  MnO2 + 4 HBr \rightarrow MnBr2 + Br2 + 2 H2O

  • Mn in MnO_2 is reduced (oxidizing agent).
  • Br in HBr is oxidized (reducing agent).

Balancing Redox Reactions

1. Assign oxidation numbers: Determine the elements oxidized and reduced.
2. Write half-reactions: Include electrons.
   • Oxidation: Electrons on the right side of the arrow.
   • Reduction: Electrons on the left side of the arrow.
3. Balance half-reactions by mass:
   • Balance elements other than H and O first.
   • Add H_2O where O is needed.
   • Add H^+ where H is needed.
   • Neutralize H^+ with OH^- in basic solutions.
4. Balance half-reactions by charge: Adjust electrons to balance charge.
5. Balance electrons: Make sure the number of electrons lost in oxidation equals the number of electrons gained in reduction.
6. Add half-reactions: Combine the balanced half-reactions.
7. Check: Verify that the equation is balanced.

• Examples:
  Example 1:
  I^-(aq) + MnO4^-(aq) \rightarrow I2(aq) + MnO2(s) in basic solution Example 2: Fe^{2+}(aq) + MnO4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq) in acidic solution

Electrical Current

• Definition: Amount of electric charge that passes a point in a given period.
  • Analogous to the amount of water flowing in a stream.
  • Can be electrons flowing through a wire or ions flowing through a solution.

Redox Reactions and Current

• Redox reactions involve the transfer of electrons.
• This transfer has the potential to generate an electric current.
• To use the current, oxidation and reduction must occur in separate locations.

Electric Current Flow Between Atoms

• Spontaneous Redox Reaction Example:

  Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

• Voltaic Cell Setup:

  • Anode: Where oxidation occurs. Zinc strip in Zn(NO3)2(aq). Zn(s) \rightarrow Zn^{2+} + 2e^-
  • Cathode: Where reduction occurs. Copper strip in Cu(NO3)2(aq). Cu^{2+} + 2e^- \rightarrow Cu(s)
  • Salt Bridge: Containing KNO_3(aq), to maintain charge balance.
  • Electrons flow from the anode to the cathode.

Electrochemical Cells

• Electrochemistry: Study of redox reactions that produce or require an electric current.
• Electrochemical Cell: Device where conversion between chemical energy and electrical energy takes place.
• Voltaic Cell (Galvanic Cell): Spontaneous redox reactions occur.
• Electrolytic Cell: Nonspontaneous redox reactions are made to occur with the addition of electrical energy.

Electrochemical Cells Details

• Oxidation and reduction reactions are kept separate in half-cells.
• Electron flow through a wire and ion flow through a solution form an electric circuit.
• Requires a conductive solid electrode (metal or graphite) for electron transfer.
• Ion exchange occurs between the two halves of the system through an electrolyte.

Electrodes

• Anode:
  • Where oxidation occurs.
  • Anions are attracted to it.
  • Connected to the positive end of the battery in an electrolytic cell.
  • Loses weight in an electrolytic cell.
• Cathode:
  • Where reduction occurs.
  • Cations are attracted to it.
  • Connected to the negative end of the battery in an electrolytic cell.
  • Gains weight in an electrolytic cell.
  • Electrode where plating takes place in electroplating.

Voltaic Cells Components

• Salt Bridge: Required to complete the circuit and maintain charge balance.

Current and Voltage

• Current: The number of electrons that flow through the system per second.
  • Unit: Ampere (A)
  • 1 A = 1 Coulomb/second
  • 1 A = 6.242 x 10^{18} electrons/second
  • Electrode surface area dictates the number of electrons that can flow.
• Potential Difference: The difference in potential energy between the reactants and products.
  • Unit: Volt (V)
  • 1 V = 1 J/Coulomb
  • Voltage drives electrons through the external circuit.
• Electromotive Force (emf): The amount of force pushing the electrons through the wire.

Cell Potential

• Definition: Difference in potential energy between the anode and the cathode in a voltaic cell.
• Dependence: Depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode.
• Standard emf (E^{\circ}_{cell}): Cell potential under standard conditions (25°C, 1 atm for gases, 1 M concentration of solutions).
  • E^{\circ}_{cell} is the sum of the cell potentials for the half-reactions.

Cell Notation

• Description: Shorthand representation of a voltaic cell.
• Format: electrode | electrolyte || electrolyte | electrode
  • Oxidation half-cell on the left, reduction half-cell on the right.
  • Single | = phase boundary.
    • If multiple electrolytes are in the same phase, a comma is used instead of |.
  • Double line || = salt bridge.
  • Inert electrodes are often used.

Inert Platinum Electrode

• Platinum is often used as an inert electrode.
• Example:
  Fe(s) | Fe^{2+}(aq) || MnO_4^-(aq), Mn^{2+}(aq), H^+(aq) | Pt(s)

Standard Reduction Potential

• Definition: A half-reaction with a strong tendency to occur has a large positive half-cell potential.
• Electron Flow: When two half-cells are connected, electrons flow so that the half-reaction with the stronger tendency occurs.
• Measurement: Half-reaction tendencies are measured relative to a standard half-reaction.
• Standard Hydrogen Electrode (SHE): Reduction of H^+ to H_2 under standard conditions is assigned a potential difference = 0 V.

Measuring Half-Cell Potential with SHE

• Connected to a voltmeter to measure the potential.
• Oxidation: Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-
• Reduction: 2H^+(aq) + 2e^- \rightarrow H_2(g)

Half-Cell Potentials

• SHE reduction potential is defined as exactly 0 V.
• Half-reactions with a stronger tendency toward reduction than SHE have a positive value for E_{red}^\circ.
• Half-reactions with a stronger tendency toward oxidation than SHE have a negative value for E_{red}^\circ.
• E{cell}^\circ = E{oxidation}^\circ + E_{reduction}^\circ
• E{oxidation}^\circ = -E{reduction}^\circ
• When adding E^\circ values for the half-cells, do not multiply the half-cell E^\circ values, even if you need to multiply the half-reactions to balance the equation.

Predicting Metal Dissolution in Acid

• Acids dissolve metals if the reduction of the metal ion is easier than the reduction of H^+(aq).
• Metals whose ion reduction reaction lies below H^+ reduction on the standard reduction potential table will dissolve in acid.
• Example:
  Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)

E_{cell}^\circ, \Delta G^\circ and K

• Spontaneous Reaction: Proceeds in the forward direction with the chemicals in their standard states.
• \Delta G^\circ < 0 (negative)
• E_{cell}^\circ > 0 (positive)
• K > 1
• \Delta G^\circ = -RT \ln K = -nFE_{cell}^\circ
• E_{cell}^\circ = \frac{0.0592 V}{n} \log K
• Where:
  • n is the number of electrons transferred in the balanced redox reaction.
  • F = Faraday's Constant = 96,485 C/mol e-

Calculating \Delta G^\circ and K

• Example 1:
  Calculate \Delta G^\circ for the reaction:
  I2(s) + 2Br^-(aq) \rightarrow 2I^-(aq) + Br2(l)
• Example 2:
  Calculate K at 25°C for the reaction:
  Cu(s) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + H_2(g)

Nonstandard Conditions and the Nernst Equation

• \Delta G = \Delta G^\circ + RT \ln Q
• E = E^\circ - (\frac{0.0592}{n}) \log Q at 25°C
• When Q = K, E = 0
• Used to calculate E when concentrations are not 1 M.

E^\circ at Nonstandard Conditions

• Nernst Equation Example:
  3 Cu(s) + 2 MnO4^-(aq) +8 H^+(aq) \rightarrow 2 MnO2(s) + 3Cu^{2+}(aq) +4 H2O(l) [Cu^{2+}] = 0.010 M, [MnO4^-] = 2.0 M, [H^+] = 1.0 M