Chapter 18

Chapter 18: Electrochemistry

Redox Reactions

• Redox reactions involve changes in oxidation numbers of one or more elements.
  • Includes single displacement and combustion reactions, and some synthesis and decomposition reactions.
  • Always involve both oxidation and reduction.
  • Can be split into oxidation and reduction half-reactions.
  • Also known as electron transfer reactions.
  • Half-reactions include electrons.
• 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 and Reduction

• Oxidation:
  • Oxidation number of an element increases.
  • Element loses electrons.
  • Compound adds oxygen.
  • Compound loses hydrogen.
  • Half-reaction has electrons as products.
• Reduction:
  • Oxidation number of an element decreases.
  • Element gains electrons.
  • Compound loses oxygen.
  • Compound gains hydrogen.
  • Half-reaction has electrons as reactants.

Rules for Assigning Oxidation States

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.

• 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
  • C: -4 \rightarrow +4 (oxidation)
  • O: 0 \rightarrow -2 (reduction)

Oxidizing and Reducing Agents (Revisited)

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

Identifying Oxidizing and Reducing Agents

• Example 1: 3H2S + 2NO3^- + 2H^+ \rightarrow 3S + 2NO + 4H_2O
  • N is reduced (oxidizing agent).
  • S is oxidized (reducing agent).
• Example 2: MnO2 + 4HBr \rightarrow MnBr2 + Br2 + 2H2O
  • Mn is reduced (oxidizing agent).
  • Br is oxidized (reducing agent).

Balancing Redox Reactions

1. Assign oxidation numbers.
   • Determine the element oxidized and the element reduced.
2. Write oxidation and reduction half-reactions, including electrons.
   • Oxidation: electrons on the right.
   • Reduction: electrons on the left.
3. Balance half-reactions by mass.
   • First, balance elements other than H and O.
   • Add H_2O where needed to balance O.
   • Add H^+ where needed to balance H.
   • Neutralize H^+ with OH^- in base (if in basic solution).
4. Balance half-reactions by charge.
   • Balance charge by adjusting electrons.
5. Balance electrons between half-reactions.
6. Add half-reactions.
7. Check.

Examples of Balancing Redox Reactions

• Example 1: I^-(aq) + MnO4^-(aq) \rightarrow I2(aq) + MnO_2(s) in basic solution.
• Example 2: Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq) in acidic solution.

Electrical Current

• Analogy: Current of a liquid in a stream is the amount of water that passes by in a given period of time.
• Electric current is the amount of electric charge that passes a point in a given period of time.
  • Whether as electrons flowing through a wire or ions flowing through a solution.

Redox Reactions and Current

• Redox reactions involve the transfer of electrons from one substance to another.
• Therefore, redox reactions have the potential to generate an electric current.
• To use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring.

Electric Current Flowing Directly Between Atoms

• Spontaneous Redox Reaction: Zn + Cu^{2+}
• Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

Electrochemical Cells

• Electrochemistry is the study of redox reactions that produce or require an electric current.
• The conversion between chemical energy and electrical energy is carried out in an electrochemical cell.
• Spontaneous redox reactions take place in a voltaic cell (aka galvanic cells).
• Nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy.

Electrochemical Cells (cont.)

• Oxidation and reduction reactions are kept separate in half-cells.
• Electron flow through a wire, along with ion flow through a solution, constitutes an electric circuit.
• Requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons through an external circuit.
• Ion exchange occurs between the two halves of the system via an electrolyte.

Electrodes

• Anode:
  • Electrode 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:
  • Electrode 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

• A salt bridge is required to complete the circuit and maintain charge balance.
• Example: Zinc/Copper Voltaic Cell
  • Anode: Zn(s) \rightarrow Zn^{2+} + 2e^- (oxidation)
  • Cathode: Cu^{2+} + 2e^- \rightarrow Cu(s) (reduction)

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
  • The voltage needed to drive electrons through the external circuit.
  • The amount of force pushing the electrons through the wire is called the electromotive force (emf).

Cell Potential

• The difference in potential energy between the anode and the cathode in a voltaic cell is called the cell potential.
• The cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode.
• The cell potential under standard conditions is called the standard emf, E°_{cell}.
  • Standard conditions: 25°C, 1 atm for gases, 1 M concentration for solutions.
  • E°_{cell} is the sum of the cell potentials for the half-reactions.

Cell Notation

• Shorthand description of a voltaic cell.
• electrode | electrolyte || electrolyte | electrode
• Oxidation half-cell on the left, reduction half-cell on the right.
• Single | = phase barrier
  • If multiple electrolytes are in the same phase, a comma is used rather than |.
  • Often, an inert electrode is used.
• Double line || = salt bridge

Inert Platinum Electrode

• Example: Fe(s) | Fe^{2+}(aq) || MnO_4^-(aq), Mn^{2+}(aq), H^+(aq) | Pt(s)
  • Anode: Fe(s) \rightarrow Fe^{2+}(aq) + 2e^- (oxidation)
  • Cathode: MnO4^-(aq) + 5e^- + 8H^+(aq) \rightarrow Mn^{2+}(aq) + 4H2O(l)

Standard Reduction Potential

• A half-reaction with a strong tendency to occur has a large positive half-cell potential.
• When two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur.
• We cannot measure the absolute tendency of a half-reaction; we can only measure it relative to another half-reaction.
• We select as a standard half-reaction the reduction of H^+ to H_2 under standard conditions, which we assign a potential difference = 0 V.
  • Standard Hydrogen Electrode (SHE).

Measuring Half-Cell Potential with SHE

• Example: Zinc half-cell connected to SHE.
  • Anode: Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-
  • Cathode: 2H^+(aq) + 2e^- \rightarrow H_2(g)

Half-Cell Potentials

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

Standard Reduction Potentials Table

• Table of Standard Reduction Potentials at 25°C lists various half-reactions and their corresponding E° (V) values.
• The table shows the relative strengths of oxidizing and reducing agents.
  • Stronger oxidizing agents are at the top left.
  • Stronger reducing agents are at the bottom right.

Practice Problems

• Calculate the E°_{cell} for a given reaction at 25°C.
• Predict if the following reaction is spontaneous under standard conditions:
  • Fe(s) + Mg^{2+}(aq) \rightarrow Fe^{2+}(aq) + Mg(s)
• Sketch and label a voltaic cell, writing the half-reactions and overall reaction, and determine the cell potential under standard conditions.

Predicting Whether a Metal Will Dissolve 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 table will dissolve in acid.
  • Example: Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)

E°_{cell}, \Delta G° and K

• For a spontaneous reaction (proceeds in the forward direction with chemicals in their standard states):
  • \Delta G° < 0 (negative)
  • E°_{cell} > 0 (positive)
  • K > 1
• Relationships:
  • \Delta G° = -RT \ln K = -nFE°_{cell}
  • E°_{cell} = (0.0592 V / n) \log K
    • n = number of electrons transferred.
    • F = Faraday's Constant = 96,485 C/mol e-

Calculating \Delta G° and K

• Calculate \Delta G° for the reaction: I2(s) + 2Br^-(aq) \rightarrow 2I^-(aq) + Br2(l)
• 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° + RT \ln Q
• E = E° - (0.0592/n) \log Q at 25°C
• When Q = K, E = 0
• Used to calculate E when concentrations are not 1 M.

E° at Nonstandard Conditions (Example)

• Standard conditions: [Zn^{2+}] = 1 M, [Cu^{2+}] = 1 M, E = 1.10 V
• Nonstandard conditions: [Zn^{2+}] = 0.010 M, [Cu^{2+}] = 2 M, E = 1.17 V
  • Anode: Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-
  • Cathode: Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)

Nonstandard Conditions and the Nernst Equation (cont.)

• \Delta G = \Delta G° + RT \ln Q
• E = E° - (0.0592/n) \log Q at 25°C
• When Q = K, E = 0
• Use to calculate E when concentrations are not 1 M
• Calculate E_{cell} at 25°C for the reaction:
  • 3 Cu(s) + 2 MnO4^-(aq) + 8 H^+(aq) \rightarrow 2 MnO2(s) + 3Cu^{2+}(aq) + 4 H_2O(l)
  • [Cu^{2+}] = 0.010 M, [MnO_4^-] = 2.0 M, [H^+] = 1.0 M