Notes on Electrochemistry
(Handout 18.2: 4/15/01)

• General reaction: AB + C ----> AC + B (single replacement)
• Oxidation: loss of electrons or hydrogen; addition of oxygen
• Reduction: addition of electrons or hydrogen; loss of oxygen
• Oxidation and reduction are coupled, thus the term redox.
• Redox reaction must be mass balanced and charge balanced.
• Oxidizing agents are reduced, Reducing agents are oxidized.
• Example: 2 HCl(aq) + Mg(s) ---> H₂(g) + MgCl₂(aq)
  • The net ionic equation is: 2 H⁺(aq) + Mg(s) ---> H₂(g) + Mg⁺²(aq)
  • H⁺(aq) is reduced in the “half-reaction”: 2 H⁺(aq) + 2 e^- ---> H₂(g)
  • Mg(s) is oxidized in the “half reaction”: Mg(s) ---> Mg⁺²(aq) + 2 e^-
    
  • H⁺(aq) is the oxidizing agent, Mg(s) is the reducing agent
• Example: Indicate the net ionic equation, the oxidized and reduced species and the oxidizing and reducing agent in:
  • CuSO₄(aq) + Zn(s) ---> ZnSO₄(aq) + Cu(s)?

• Why do we care about oxidation numbers?
  • Example: C₆H₁₂O₆ + 6 O₂ ---> 6 CO₂ + 6 H₂O
  • This is a redox reaction since oxygen is added. But there are no ion charges to indicate the electrons lost and gained.
  • Chemists use an arbitrary but logical set of rules to artificially assign charges to atoms in formulas.
• Oxidation Number Rules
  • Elements = 0; monoatomic ions = charge on ion.
  • H = +1 for covalent hydrides; H = -1 in ionic hydrides.
  • O = -2 for all compounds except peroxides (-1).
  • The sum of the oxidation numbers for all elements in a formula equals the total charge for the formula.
• Example: C in C₆H₁₂O₆ = 0 and in CO₂ = +4, so carbon is oxidized. O in O₂ = 0 and in H₂O = -2, so oxygen is reduced.

• Determine the oxidation numbers of the elements in the reactants and products, then use the change in oxidation numbers in going from reactant to product to determine which species is oxidized and which is reduced, then balance the reaction equation.
• If the redox reaction takes place in acid, balanace the hydrogen and oxygen with H⁺ and H₂O. If it takes place in base, balanace the hydrogen and oxygen with OH^- and H₂O.
• Example: balance KMnO₄(aq) + H₂C₂O₄(aq) ---> Mn⁺²(aq)+ K⁺(aq) + CO₂(g)
  • This reaction takes place in acid, so we will balance H and O with H⁺ and H₂O
  • The oxidation number of Mn in KMnO₄(aq) is +7 and in Mn⁺²(aq) it is +2.
  • The oxidation number of C in H₂C₂O₄(aq) is +3 and in CO₂(g) it is +4.
  • Thus the Mn is reduced and the C is oxidized.
  • Write half reaction and mass balance all atoms except H and O:
    • KMnO₄(aq) ---> Mn⁺²(aq)+ K⁺(aq)
    • H₂C₂O₄(aq) ---> 2 CO₂(g)
  • Balance H and O:
    • 8 H⁺(aq) + KMnO₄(aq) ---> Mn⁺²(aq)+ K⁺(aq) + 4 H₂O(l)
    • H₂C₂O₄(aq) ---> 2 CO₂(g) + 2 H⁺(aq)
  • Balance charge with electrons so that reduction occurs:
    • 5 e^- + 8 H⁺(aq) + KMnO₄(aq) ---> Mn⁺²(aq)+ K⁺(aq) + 4 H₂O(l)
    • H₂C₂O₄(aq) ---> 2 CO₂(g) + 2 H⁺(aq) + 2 e^-
  • Use the least common multiple to make the number of electrons equal in each
    • 2 x [5 e^- + 8 H⁺(aq) + KMnO₄(aq) ---> Mn⁺²(aq)+ K⁺(aq) + 4 H₂O(l)]
    • 5 x [H₂C₂O₄(aq) ---> 2 CO₂(g) + 2 H⁺(aq) + 2 e^-]
  • These reduce to:
    • 10 e^- + 16 H⁺(aq) + 2 KMnO₄(aq) ---> 2 Mn⁺²(aq)+ 2 K⁺(aq) + 8 H₂O(l)
    • 5 H₂C₂O₄(aq) ---> 10 CO₂(g) + 10 H⁺(aq) + 10 e^-
  • Add the half reactions:
    • 10 e^- + 16 H⁺(aq) + 2 KMnO₄(aq) + 5 H₂C₂O₄(aq) ---> 2 Mn⁺²(aq)+ 2 K⁺(aq) + 8 H₂O(l) + 10 CO₂(g) + 10 H⁺(aq) + 10 e^-
  • This reduces to:
    • 10 e^- + 6 16 H⁺(aq) + 2 KMnO₄(aq) + 5 H₂C₂O₄(aq) ---> 2 Mn⁺²(aq)+ 2 K⁺(aq) + 8 H₂O(l) + 10 CO₂(g) + 10 H⁺(aq) + 10 e^-

• Produces electricity from a spontaneous reaction: Zn(s) + Cu⁺²(aq) ---> Zn⁺²(aq) + Cu(s)
• The driving force for this reaction is called electromotive force or emf and it is measured in volts (= 1 J/coulomb).
• The emf is a measure of the potential energy or “electrical pressure” produced.
• The oxidation and reduction reactions are kept in separate containers or cells. Only ions can migrate between cells.
• The anode is the electrode where oxidation occurs, the cathode is where reduction occurs.
• Electrons flow through a wire from the anode to the cathode. The circuit is completed through a salt bridge.
• Cell Notation for a Voltaic Cell: Zn(s) / Zn⁺²(aq) // Cu⁺²(aq) / Cu(s)

• A Standard Reduction Potential is:
  • E° measured under standard state conditions : 1 M, 1 atm, 298 K
  • Relative to the SHE which has an E° = 0.0 v
  • Used to calculate E° of a cell (= E°_reduction + E°_oxidation)
  • ®G°= -nFE°; F=9.65x10⁴ J/v*mole (C/mol), n=no. of e^-.
• From a table of standard potentials note that:
  • All reactions are written as reduction reactions.
  • The more positive the E°, the more likely the reaction will be used as written (a reduction reaction).
  • All reactions could occur in either direction depending on the potential of the partner.
  • Any species on the left will oxidize any species on the right if it is below it.
  • Changing the stoichiometric factors doesn’t change E°.

• Batteries
  • voltaic cells used to do work
  • primary cells such as dry cells produce electricity from the chemicals sealed inside
  • secondary cells such as a car battery must be charged before use
  • fuel cells use the reaction between hydrogen and oxygen to produce electricity
• Corrosion
  • Any metal lower than -0.83 volts in the electrochemical series can be oxidized by water. This is the driving force for corrosion
  • Galvanization and using a sacrificial anode help prevent corrosion.

• We can use measurements of current to count the moles of electrons used in a chemical reaction.
• Current is measured in amps or coulombs/second (C/s)
• Since 1 electron has a charge of 1.6022 x 10^-19 then 1 mole of electrons (6.022 x 10²³) has a charge of 9.65x10⁴ C.
• Electrolytic cells contain a non-spontaneous reaction driven by an outside power source.
• In aqueous solution, a species can only be oxidized if its oxidation potential is more positive than -1.2 volts.
• In aqueous solution, a species can only be reduced if its reduction potential is more positive than -0.83 volts.

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