Electrochemistry Notes

Electrochemical Reactions

• Chemical energy is converted into electrical energy.

• Electrons are transferred from one species to another.

Oxidation/Reduction

• Oxidation-reduction (redox) reactions are crucial in various aspects of life.

• Originally, oxidation meant combining with oxygen, and reduction meant the loss of oxygen.

• Today, these terms have a broader interpretation.

Oxidation and Reduction

• A redox reaction occurs when electrons are transferred.

  • Oxidation: Loss of electrons by an atom or ion.

  • Reduction: Gain of electrons by an atom or ion.

• Mnemonic: "L.E.O. the Lion says G.E.R."

  • L.E.O.: Lose Electrons Oxidation

  • G.E.R.: Gain Electrons Reduction

• Oxidation and reduction always occur together; hence, it's called a redox reaction.

Oxidation Numbers

• Oxidation numbers are assigned to keep track of electron loss or gain.

Oxidation and Reduction (Detailed)

• Oxidation:

  • A species is oxidized when it loses electrons.

  • Example: Zinc (Zn) loses two electrons to become $Zn^{2+}$.

• Reduction:

  • A species is reduced when it gains electrons.

  • Example: Hydrogen ions ($H^+$) gain electrons to form hydrogen gas ($H_2$).

• Oxidizing Agent:

  • The substance that is reduced.

  • Example: $H^+$ oxidizes Zn by taking electrons from it.

• Reducing Agent:

  • The substance that is oxidized.

  • Example: Zn reduces $H^+$ by giving it electrons.

Assigning Oxidation Numbers

1. Elements in their elemental form have an oxidation number of 0.

2. The oxidation number of a monatomic ion is the same as its charge.

3. Group 1 metals have an oxidation number of +1.

4. Group 2 metals have an oxidation number of +2.

5. Fluorine always has an oxidation number of -1.

6. Hydrogen is +1, except in metal hydrides where it is -1.

7. Oxygen is usually -2, except in the peroxide ion (-1) or in $OF_2$ (+2).

8. Most electronegative nonmetals have negative oxidation numbers satisfying the octet rule.

9. The sum of oxidation numbers in a neutral compound is 0.

10. The sum of oxidation numbers in a polyatomic ion equals the charge on the ion.

Recognizing Redox Reactions

• To identify a redox reaction, assign oxidation numbers to each element.

• If an element's oxidation number changes from reactants to products, electrons have been transferred.

• Synthesis, decomposition, single replacement, and combustion reactions are usually redox reactions.

  • The substance that loses electrons is oxidized and is the reducing agent.

  • The substance that gains electrons is reduced and is the oxidizing agent.

Balancing Oxidation-Reduction Equations

• The half-reaction method is an easy way to balance redox equations.

• This method treats oxidation and reduction as separate processes, balancing them individually and then combining them.

Half Reactions

• Chemical equations do not show the exchange of electrons.

• Half reactions show either the oxidation or reduction portion of the reaction.

• Half reactions must obey the law of conservation of matter, energy, and charge.

• Example: $Cu + 2AgNO3 \rightarrow Cu(NO3)_2 + 2Ag$

  • Oxidation: $Cu \rightarrow Cu^{2+} + 2e^-$ (Net charge = 0)

  • Reduction: $2Ag^+ + 2e^- \rightarrow 2Ag$ (Net charge = 0)

• The number of electrons lost must always equal the number of electrons gained.

Half-Reaction Method Steps

1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions.

3. Balance each half-reaction:

   • a. Balance elements other than H and O.

   • b. Balance O by adding $H_2O$.

   • c. Balance H by adding $H^+$.

   • d. Balance charge by adding electrons.

4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.

5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass.

7. Make sure the equation is balanced according to charge.

Half-Reaction Method Example

• Consider the reaction between $MnO<em>4^-$ and $C</em>2O<em>4^{2-}$. $MnO</em>4^-(aq) + C<em>2O</em>4^{2-}(aq) \rightarrow Mn^{2+}(aq) + CO_2(aq)$

Assign Oxidation Numbers (Example)

$MnO<em>4^- + C</em>2O<em>4^{2-} \rightarrow Mn^{2+} + CO</em>2$

$+7 +3 +2 +4$

Manganese is reduced (goes from +7 to +2), and carbon is oxidized (goes from +3 to +4).

Oxidation Half-Reaction (Example)

$C<em>2O</em>4^{2-} \rightarrow CO<em>2$ Balance carbon: $C</em>2O<em>4^{2-} \rightarrow 2CO</em>2$

Oxidation Half-Reaction: Balancing Charge (Example)

$C<em>2O</em>4^{2-} \rightarrow 2CO<em>2$ Balance the charge by adding 2 electrons to the right side: $C</em>2O<em>4^{2-} \rightarrow 2CO</em>2 + 2e^−$

Reduction Half-Reaction (Example)

$MnO<em>4^- \rightarrow Mn^{2+}$ Balance oxygen by adding 4 water molecules to the right side: $MnO</em>4^- \rightarrow Mn^{2+} + 4H_2O$

Reduction Half-Reaction: Balancing Hydrogen (Example)

$MnO<em>4^- \rightarrow Mn^{2+} + 4H</em>2O$

Add 8 $H^+$ to the left side to balance hydrogen:

$8H^+ + MnO<em>4^- \rightarrow Mn^{2+} + 4H</em>2O$

Reduction Half-Reaction: Balancing Charge (Example)

$8H^+ + MnO<em>4^- \rightarrow Mn^{2+} + 4H</em>2O$

Add 5 electrons to the left side to balance the charge:

$5e^- + 8H^+ + MnO<em>4^- \rightarrow Mn^{2+} + 4H</em>2O$

Combining the Half-Reactions

Oxidation: $C<em>2O</em>4^{2-} \rightarrow 2CO<em>2 + 2e^−$ Reduction: $5e^- + 8H^+ + MnO</em>4^- \rightarrow Mn^{2+} + 4H_2O$

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

Combining the Half-Reactions (Multiplied)

Oxidation: $5C<em>2O</em>4^{2-} \rightarrow 10CO<em>2 + 10e^−$ Reduction: $10e^- + 16H^+ + 2MnO</em>4^- \rightarrow 2Mn^{2+} + 8H<em>2O$ Combined: $10e^- + 16H^+ + 2MnO</em>4^- + 5C<em>2O</em>4^{2-} \rightarrow 2Mn^{2+} + 8H<em>2O + 10CO</em>2 + 10e^−$

Final Balanced Equation

$10e^- + 16H^+ + 2MnO<em>4^- + 5C</em>2O<em>4^{2-} \rightarrow 2Mn^{2+} + 8H</em>2O + 10CO<em>2 + 10e^−$ Subtract the electrons: $16H^+ + 2MnO</em>4^- + 5C<em>2O</em>4^{2-} \rightarrow 2Mn^{2+} + 8H<em>2O + 10CO</em>2$

Electrochemical Cells

• Electrochemical cells are a practical application of redox reactions.

• Two types of cells:

  • Voltaic (Galvanic) cell: Converts chemical energy into electrical energy (spontaneous).

  • Electrolytic cell: Requires an electric current to produce a chemical reaction (non-spontaneous).

• Electrochemical cells have two electrodes (surfaces that conduct electricity):

  • Anode: Site of oxidation.

  • Cathode: Site of reduction.

  • Mnemonic: "RED CAT and AN OX" (Reduction at Cathode, Anode Oxidation)

Voltaic Cells (Daniell cell)

• In spontaneous redox reactions, electrons are transferred, and energy is released.

• Example reaction:

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

Voltaic Cells: Harnessing Energy

• The energy from electron flow can be used to do work by making the electrons flow through an external device.

• Such a setup is called a voltaic cell.

Voltaic Cells Setup

• Oxidation occurs at the anode.

• Reduction occurs at the cathode.

Voltaic Cells: Charge Balance

• If electrons flow from the anode to the cathode, the charges become unbalanced, stopping the electron flow.

Voltaic Cells: Salt Bridge

• A salt bridge (usually a U-shaped tube containing a salt solution) is used to maintain charge balance.

  • Cations move toward the cathode.

  • Anions move toward the anode.

Voltaic Cells: Electron Flow

• Electrons leave the anode and flow through the wire to the cathode.

• As electrons leave the anode, cations formed dissolve into the solution in the anode compartment.

• A porous barrier or salt bridge allows ions to follow.

Voltaic Cells: Cathode Activity

• As electrons reach the cathode, cations in the cathode are attracted to the now-negative cathode.

• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

Voltaic Cells Summary

• Spontaneous reactions.

• Consist of two metallic electrodes in separate electrolytic solutions.

• Anode:

  • Where oxidation occurs.

  • Negative (metal is more active on Table J).

• Cathode:

  • Where reduction occurs.

  • Positive (the actual cathode is not reduced).

• Example: $Zn^{+2} + Mg \rightarrow Zn + Mg^{+2}$

  • Oxidation: $Mg \rightarrow Mg^{+2}$

  • Reduction: $Zn^{+2} \rightarrow Zn$

Electromotive Force (emf)

• Electrons spontaneously flow one way in a redox reaction, from higher to lower potential energy.

• More active to less active (Table J).

Electromotive Force (emf) Definition

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential, designated $E_{cell}$.

Cell Potential Measurement

• Cell potential is measured in volts (V).

• An electron has a charge of $1.6 × 10^{-19} C$

  • 1 C = amp/second

  • 1 V = 1 J/C

Standard Reduction Potentials

• Reduction potentials for many electrodes have been measured and tabulated.

• Examples:

  • $F_2(g) + 2e^- \rightarrow 2F^-(aq)$, Potential (V) = +2.87

  • $MnO<em>4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H</em>2O(l)$, Potential (V) = +1.51

  • $Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$, Potential (V) = +1.36

• The table shows a wide range of reduction half-reactions and their corresponding standard reduction potentials.

Standard Hydrogen Electrode (SHE)

• Values are referenced to a standard hydrogen electrode (SHE).

• By definition, the reduction potential for hydrogen is 0 V:

$2 H^+ (aq, 1M) + 2 e^− \rightarrow H_2 (g, 1 atm)$

Standard Cell Potentials Calculation

• The cell potential at standard conditions can be found through this equation:

$E<em>{cell}^° = E</em>{red}^° (cathode) − E_{red}^° (anode)$

• Cell potential is based on potential energy per unit of charge and is an intensive property (does not depend on amount).

$1 V = 1 J/C$

• Coefficients from balanced equation do not change voltage. Greater coefficients mean more joules, which would require more coulombs.

Cell Potentials Example

• For the oxidation in this cell, $E_{red}^° = −0.76 V$

• For the reduction, $E_{red}^° = +0.34 V$

Cell Potentials Calculation Example

$E<em>{cell}^° = E</em>{red}^° (cathode) − E_{red}^° (anode)$

$= +0.34 V − (−0.76 V) = +1.10 V$

Spontaneity and Cell Potential

• A positive cell potential ($E^o$) indicates a spontaneous reaction.

• A negative cell potential ($E^o$) indicates a non-spontaneous reaction.

Electrolytic Cell

• Non-spontaneous = requires a power source

• Electroplating: Metal we are plating is lower on Table J (forcing anode metal to be oxidized)

• Anode = oxidation but is positive

• Cathode = reduction but is negative

• Mnemonic: Red Cat gets Fat

Electrolytic Cell Diagram

• Diagram showing electron flow and ion movement in an electrolytic cell

Electrolysis of a Fused Salt

*Decompose Salt -> elements

• When the battery is switched on,

  • the + IONS move to the - CATHODE

  • the - IONS move to the + ANODE

  *This gives a way to SPLIT UP IONIC COMPOUNDS: "ELECTROLYSIS"

  Example: $NaCl \rightarrow Na + Cl<em>2$ Reduction (Cathode): $Na^+ + e^- \rightarrow Na$ Oxidation (Anode): $2Cl^- \rightarrow Cl</em>2 + 2e^-$

Electrolysis and Stoichiometry

• $I = Q/t$

  • 1 amp = 1 coulomb / 1 second

  • I = current (A)

  • Q = charge (c)

  • t = time (s)

• F = Faraday’s constant = 96485 Coulombs / mole of electrons

• Moles of electrons that travel through the wire in the given time.

Electrolysis and Stoichiometry Examples

• Example 1: A current of 0.511 amp for 672 s is used to electroplate nickel at the cathode of an electrochemical cell containing $Ni^{+2}(aq)$. Calculate the mass of nickel metal produced.

• Example 2: How long must a 20.0 amp current flow through a solution of $Zn^{+2}$ in order to produce 25.00 g of Zn metal.

Electrochemical Cells Summary

• Anode: site of oxidation

• Cathode: site of reduction

• Electrons flow from Anode to cathode

Voltaic (Electrochemical) Cells

• Chemical Energy converted to Electrical Energy

• Exothermic

• Cell potential > 0

Electrolytic Cells