Electrochemistry Lecture Notes

Introduction to Electrochemical Processes

  • Electrochemistry is the study of the branch of chemistry that deals with the relationship between electrical energy and chemical changes.

  • Chemical processes in electrochemistry are oxidation-reduction (redox) reactions where:

    • The energy released by a spontaneous chemical reaction is converted into electricity.

    • Electrical energy is used to drive a nonspontaneous chemical reaction to occur.

  • Fundamental definitions of redox reactions:

    • Oxidation: The process of losing electrons ($e^-$). The substance that loses electrons also serves as the reducing agent.

    • Reduction: The process of gaining electrons ($e^-$). The substance that gains electrons also serves as the oxidizing agent.

  • Example chemical equation and its half-reactions:

    • Overall: 2Mg(s)+O2(g)→2MgO(s)2Mg(s) + O_2(g) \rightarrow 2MgO(s)

    • Atomic breakdown: $2Mg^0$ and $O_2^0$ react to form $2Mg^{2+}$ and $2O^{2-}$.

    • Oxidation half-reaction: 2Mg→2Mg2++4e−2Mg \rightarrow 2Mg^{2+} + 4e^-

    • Reduction half-reaction: O2+4e−→2O2−O_2 + 4e^- \rightarrow 2O^{2-}

Rules for Determining Oxidation Numbers

  • The oxidation number (oxidation state) represents the charge an atom would have in a molecule (or an ionic compound) if electrons were completely transferred.

  • Rule 1: Free elements in their uncombined state always have an oxidation number of zero. Examples include $Na$, $Be$, $K$, $Pb$, $H_2$, $O_2$, and $P_4$.

  • Rule 2: In monatomic ions, the oxidation number is exactly equal to the charge of the ion. For example, in $Li^+$, the number is $+1$; in $Fe^{3+}$, it is $+3$; and in $O^{2-}$, it is $-2$.

  • Rule 3: The oxidation number of oxygen is usually $-2$. Significant exceptions occur in hydrogen peroxide ($H_2O_2$) and the peroxide ion ($O_2^{2-}$), where it is $-1$.

  • Rule 4: The oxidation number of hydrogen is $+1$, except when it is bonded to metals in binary compounds (e.g., $LiH$), in which case the oxidation number is $-1$.

  • Rule 5: Group IA metals (Alkali metals) are always $+1$ in compounds. Group IIA metals (Alkaline earth metals) are always $+2$ in compounds. Fluorine ($F$) is always $-1$ in all its compounds.

  • Rule 6: The sum of the oxidation numbers of all atoms in a neutral molecule is zero. For a polyatomic ion, the sum must equal the net charge of the ion.

  • Example Calculation ($HCO_3^-$):

    • Oxygen: 3×(−2)=−63 \times (-2) = -6

    • Hydrogen: +1+1

    • Sum equation: (+1)+C+(−6)=−1(+1) + C + (-6) = -1

    • Result for Carbon: C=+4C = +4

Balancing Redox Equations using the Ion-Electron Method

  • Step 1: Write the unbalanced equation for the reaction in ionic form.

    • Example: Oxidation of $Fe^{2+}$ to $Fe^{3+}$ by dichromate ($Cr_2O_7^{2-}$) in acidic solution: Fe2++Cr2O72−→Fe3++Cr3+Fe^{2+} + Cr_2O_7^{2-} \rightarrow Fe^{3+} + Cr^{3+}

  • Step 2: Separate the equation into two half-reactions based on oxidation states.

    • Reduction (Cr changes from $+6$ to $+3$): Cr2O72−→Cr3+Cr_2O_7^{2-} \rightarrow Cr^{3+}

    • Oxidation (Fe changes from $+2$ to $+3$): Fe2+→Fe3+Fe^{2+} \rightarrow Fe^{3+}

  • Step 3: Balance atoms other than Oxygen and Hydrogen in each half-reaction.

    • Cr2O72−→2Cr3+Cr_2O_7^{2-} \rightarrow 2Cr^{3+}

  • Step 4: For reactions in acid, add H2OH_2O to balance Oxygen atoms and H+H^+ to balance Hydrogen atoms.

    • Oxygen balance: Cr2O72−→2Cr3++7H2OCr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O

    • Hydrogen balance: 14H++Cr2O72−→2Cr3++7H2O14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O

  • Step 5: Add electrons ($e^-$) to one side of each half-reaction to balance the total charges.

    • Oxidation half: Fe2+→Fe3++1e−Fe^{2+} \rightarrow Fe^{3+} + 1e^-

    • Reduction half: 6e−+14H++Cr2O72−→2Cr3++7H2O6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O

  • Step 6: Equalize the number of electrons in the two half-reactions by multiplying the reactions by appropriate coefficients.

    • Multiply oxidation by 6: 6Fe2+→6Fe3++6e−6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-

  • Step 7: Add the half-reactions together. The electrons on both sides must cancel out.

    • Final combined equation: 14H++Cr2O72−+6Fe2+→6Fe3++2Cr3++7H2O14H^+ + Cr_2O_7^{2-} + 6Fe^{2+} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O

  • Step 8: Verify charge and atom balance.

    • Net charge on left: (14×1)−2+(6×2)=24(14 \times 1) - 2 + (6 \times 2) = 24

    • Net charge on right: (6×3)+(2×3)=24(6 \times 3) + (2 \times 3) = 24

  • Step 9 (Special Case for Basic Solutions): Perform the balancing as if in acid, then add OH−OH^- to both sides for every H+H^+ present to form water ($H_2O$), canceling where possible.

Galvanic Cells and Cell Potential

  • A Galvanic (or Voltaic) cell is a device that generates electricity from a spontaneous redox reaction.

  • Key Components:

    • Anode: The electrode where oxidation occurs (negative sign in a galvanic cell). Example: Zn(s)→Zn2+(aq)+2e−Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-.

    • Cathode: The electrode where reduction occurs (positive sign). Example: Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s).

    • Salt Bridge: A tube filled with electrolyte solution (e.g., $KCl$) that allows ions to flow to maintain electrical neutrality.

    • Voltmeter: Measures the electrical potential difference between the electrodes.

  • Net Reaction: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

  • Cell Potential Definitions:

    • Cell Voltage: The difference in electrical potential between the anode and cathode.

    • Electromotive Force (emf): Calculated as Ecell0=Ecathode0−Eanode0E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0.

  • Cell Diagram Notation:

    • Convention: Anode on the left, cathode on the right. A single line | represents a phase boundary, and a double line || represents the salt bridge.

    • Example: Zn(s)∣Zn2+(1M)∣∣Cu2+(1M)∣Cu(s)Zn(s) | Zn^{2+}(1 M) || Cu^{2+}(1 M) | Cu(s)

Standard Reduction Potentials ($E^0$)

  • Standard reduction potential is the voltage associated with a reduction reaction at an electrode when all solutes are at 1M1 M concentration and all gases are at 1 atm1 \text{ atm}.

  • Standard Hydrogen Electrode (SHE): Used as a universal reference. The potential for the reduction of hydrogen is defined as exactly zero.

    • Half-reaction: 2e−+2H+(1M)→H2(1 atm)2e^- + 2H^+(1 M) \rightarrow H_2(1 \text{ atm})

    • E0=0 VE^0 = 0 \text{ V}

  • Calculations using SHE:

    • For the cell Zn(s)∣Zn2+(1M)∣∣H+(1M)∣H2(1 atm)∣Pt(s)Zn(s) | Zn^{2+}(1 M) || H^+(1 M) | H_2(1 \text{ atm}) | Pt(s), the measured Ecell0=0.76 VE_{\text{cell}}^0 = 0.76 \text{ V}.

    • Since Ecell0=EH+/H20−EZn2+/Zn0E_{\text{cell}}^0 = E_{H^+/H_2}^0 - E_{Zn^{2+}/Zn}^0, then 0.76 V=0−EZn2+/Zn00.76 \text{ V} = 0 - E_{Zn^{2+}/Zn}^0.

    • Therefore, the standard reduction potential for zinc is EZn2+/Zn0=−0.76 VE_{Zn^{2+}/Zn}^0 = -0.76 \text{ V}.

  • Properties of Standard Reduction Potentials:

    • The more positive the E0E^0, the greater the tendency for the substance to be reduced (acting as a stronger oxidizing agent).

    • Half-cell reactions are reversible.

    • Reversing a reaction changes the sign of E0E^0.

    • E0E^0 is an intensive property; changing the stoichiometric coefficients of a half-reaction does not change the numerical value of E0E^0.

Thermodynamics and Spontaneity of Redox Reactions

  • Relationship between Free Energy and Cell potential:

    • Standard condition: ΔG0=−nFEcell0\Delta G^0 = -nFE_{\text{cell}}^0

    • Non-standard: ΔG=−nFEcell\Delta G = -nFE_{\text{cell}}

    • Variables:

      • nn = number of moles of electrons transferred.

      • FF = Faraday constant = 96,500 J(V×mol)−1=96,500 C/mol96,500 \text{ J} (\text{V} \times \text{mol})^{-1} = 96,500 \text{ C/mol}.

  • Relationship with Equilibrium Constant ($K$):

    • ΔG0=−RT ln K\Delta G^0 = -RT \text{ ln } K

    • Ecell0=RTnF ln KE_{\text{cell}}^0 = \frac{RT}{nF} \text{ ln } K

    • At 298 K298 \text{ K}, the converted formulas are:

      • Ecell0=0.0257 Vn ln KE_{\text{cell}}^0 = \frac{0.0257 \text{ V}}{n} \text{ ln } K

      • Ecell0=0.0592 Vn log KE_{\text{cell}}^0 = \frac{0.0592 \text{ V}}{n} \text{ log } K

  • Spontaneity Correlation:

    • If \Delta G^0 < 0, then K > 1 and E_{\text{cell}}^0 > 0: Reaction is spontaneous (favors products).

    • If ΔG0=0\Delta G^0 = 0, then K=1K = 1 and Ecell0=0E_{\text{cell}}^0 = 0: At equilibrium.

    • If \Delta G^0 > 0, then K < 1 and E_{\text{cell}}^0 < 0: Reaction is nonspontaneous (favors reactants).

The Nernst Equation (Concentration Effects)

  • The Nernst equation allows the calculation of cell potential under non-standard conditions.

  • Formula: E=E0−RTnF ln QE = E^0 - \frac{RT}{nF} \text{ ln } Q

  • At standard temperature (298 K298 \text{ K}):

    • E=E0−0.0257 Vn ln QE = E^0 - \frac{0.0257 \text{ V}}{n} \text{ ln } Q

    • E=E0−0.0592 Vn log QE = E^0 - \frac{0.0592 \text{ V}}{n} \text{ log } Q

    • Where QQ is the reaction quotient.

  • Example: Spontaneity check for Co(s)+Fe2+(aq)→Co2+(aq)+Fe(s)Co(s) + Fe^{2+}(aq) \rightarrow Co^{2+}(aq) + Fe(s).

    • Conditions: [Co2+]=0.15M[Co^{2+}] = 0.15 M, [Fe2+]=0.68M[Fe^{2+}] = 0.68 M.

    • E0=EFe2+/Fe0−ECo2+/Co0=−0.44 V−(−0.28 V)=−0.16 VE^0 = E_{Fe^{2+}/Fe}^0 - E_{Co^{2+}/Co}^0 = -0.44 \text{ V} - (-0.28 \text{ V}) = -0.16 \text{ V}.

    • E=−0.16 V−0.0257 V2 ln (0.150.68)=−0.14 VE = -0.16 \text{ V} - \frac{0.0257 \text{ V}}{2} \text{ ln } (\frac{0.15}{0.68}) = -0.14 \text{ V}.

    • Since E < 0, the reaction is nonspontaneous.

Practical Applications: Batteries and Corrosion

  • Dry Cell (Leclanché cell):

    • Anode: Zn(s)→Zn2+(aq)+2e−Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-

    • Cathode: 2NH4+(aq)+2MnO2(s)+2e−→Mn2O3(s)+2NH3(aq)+H2O(l)2NH_4^+(aq) + 2MnO_2(s) + 2e^- \rightarrow Mn_2O_3(s) + 2NH_3(aq) + H_2O(l)

  • Mercury Battery:

    • Anode: Zn(Hg)+2OH−(aq)→ZnO(s)+H2O(l)+2e−Zn(Hg) + 2OH^-(aq) \rightarrow ZnO(s) + H_2O(l) + 2e^-

    • Cathode: HgO(s)+H2O(l)+2e−→Hg(l)+2OH−(aq)HgO(s) + H_2O(l) + 2e^- \rightarrow Hg(l) + 2OH^-(aq)

  • Lead Storage Battery:

    • Anode: Pb(s)+SO42−(aq)→PbSO4(s)+2e−Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-

    • Cathode: PbO2(s)+4H+(aq)+SO42−(aq)+2e−→PbSO4(s)+2H2O(l)PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)

  • Fuel Cell:

    • An electrochemical cell requiring a continuous supply of reactants.

    • Reaction: 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

  • Corrosion of Iron:

    • Oxidation occurs at the anode: Fe(s)→Fe2+(aq)+2e−Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-.

    • Further oxidation: Fe2+(aq)→Fe3+(aq)+e−Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-.

    • Reduction at the cathode (air/water interface): O2(g)+4H+(aq)+4e−→2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l).

    • Cathodic Protection: Preventing rust by connecting iron to a more active metal (sacrificial anode like Magnesium).

      • Mg reaction: Mg(s)→Mg2+(aq)+2e−Mg(s) \rightarrow Mg^{2+}(aq) + 2e^-.

Electrolysis and Quantitative Aspects

  • Electrolysis: The use of electrical energy to drive a nonspontaneous chemical reaction.

  • Electrolysis of Water:

    • Oxidation (Anode): 2H2O(l)→O2(g)+4H+(aq)+4e−2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-

    • Reduction (Cathode): 4H+(aq)+4e−→2H2(g)4H^+(aq) + 4e^- \rightarrow 2H_2(g)

  • Quantitative Relationship:

    • charge (C)=current (A)×time (s)\text{charge (C)} = \text{current (A)} \times \text{time (s)}

    • 1 mole e−=96,500 C1 \text{ mole } e^- = 96,500 \text{ C}

  • Example Calculation (Molten CaCl2CaCl_2):

    • Current = 0.452 A0.452 \text{ A}, Time = 1.5 hours1.5 \text{ hours}.

    • Moles of CaCa produced:

      • 0.452 C/s×(1.5×3600 s)=2440.8 C0.452 \text{ C/s} \times (1.5 \times 3600 \text{ s}) = 2440.8 \text{ C}

      • 2440.8 C×(1 mol e−/96,500 C)×(1 mol Ca/2 mol e−)=0.0126 mol Ca2440.8 \text{ C} \times (1 \text{ mol } e^- / 96,500 \text{ C}) \times (1 \text{ mol Ca} / 2 \text{ mol } e^-) = 0.0126 \text{ mol Ca}

      • Mass = 0.0126 mol×40.08 g/mol=0.50 g0.0126 \text{ mol} \times 40.08 \text{ g/mol} = 0.50 \text{ g}.

Questions & Discussion

  • Scenario: Predicting reactions with Bromine ($Br_2$)

    • Question: Predict what happens if Br2Br_2 is added to a solution of NaClNaCl and NaINaI at standard states.

    • Data: ECl2/Cl−0=1.36 VE^0_{Cl_2/Cl^-} = 1.36 \text{ V}, EBr2/Br−0=1.07 VE^0_{Br_2/Br^-} = 1.07 \text{ V}, EI2/I−0=0.53 VE^0_{I_2/I^-} = 0.53 \text{ V}.

    • Analysis (Diagonal Rule): Br2Br_2 is a stronger oxidizing agent than I2I_2 but weaker than Cl2Cl_2.

    • Result: Br2Br_2 will oxidize I−I^- but will not oxidize Cl−Cl^-.

    • Reaction: 2I−(1M)+Br2(l)→I2(s)+2Br−(1M)2I^-(1 M) + Br_2(l) \rightarrow I_2(s) + 2Br^-(1 M).

  • Topic: Dental Filling Discomfort

    • Discomfort is caused by electrochemical reactions between different metals in fillings like Amalgam ($Ag/Hg/Sn$).

    • Potentials involved: Hg22+/Ag2Hg3Hg_2^{2+}/Ag_2Hg_3 ($+0.85 ext{ V}$) and Sn2+/Ag3SnSn^{2+}/Ag_3Sn ($-0.05 ext{ V}$).

    • The resulting small electric currents in the mouth cause a metallic taste and pain (galvanic shock).