Lecture 12 Notes: Ksp and Electrolysis

Lecture 12: Hydrogen, Membrane Power Supply, Ksp and Electrolysis

Learning Objectives

  • Calculate the equilibrium constant for a redox reaction given the cell potential.
  • Apply the Galvanic cell to determine solubility products for insoluble salts.
  • Identify the 1/2 reactions involved in an electrolytic cell.
  • Calculate quantities of products formed in electrolysis using the equation for electron stoichiometry.
  • Recognize the importance of electrolysis reactions in our everyday lives.

Review with ALEKS Example

  • A certain half-reaction has a standard reduction potential E_{red}^0 = +0.74V.
  • An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 0.60 V of electrical power.
  • The cell will operate under standard conditions.
  • The task is to determine the minimum and maximum standard reduction potential that the half-reaction used at the anode of this cell can have, if any.
  • A balanced equation describing a half-reaction that could be used at the anode of this cell will be provided as: H_2(g) \rightarrow 2H^+(aq) + 2e^-.

Equilibrium Constants of Redox Reactions

  • From the Nernst Equation, for the reaction: aA + bB \rightleftharpoons cC + dD
  • K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}
  • \Delta G^0 = -RT \ln K_{eq}
  • E = E^0 - \frac{RT}{nF} \ln Q
  • At equilibrium, E = 0, Q = K{eq}, so E^0 = \frac{RT}{nF} \ln K{eq}
  • At 298K: E^0 = \frac{0.0591V}{n} \log K_{eq}
  • Note: Using log makes it easier to find the pH because pH = -\log[H^+].

Discharging a Battery

  • Discharging a battery results in equilibrium state between reactants and products: LiC6 + CoO2 \rightleftharpoons C6 + LiCoO2

Calculate the Equilibrium Constant

  • For the reaction: NO3^- + 4H^+ + 3Ag(s) \rightarrow NO + 2H2O + 3Ag^+
  • Half-reactions:
    • NO3^- + 4H^+ + 3e^- \rightarrow NO + 2H2O E^0 = 0.96V
    • Ag^+ + e^- \rightarrow Ag(s) E^0 = 0.80V
  • E^0 = 0.96V - 0.80V = 0.16V
  • \log K_{eq} = \frac{nE^0}{0.0591V} = \frac{3(0.16V)}{0.0591V} = 8.122
  • K_{eq} = 1.32 \times 10^8

The Spontaneity Triangle

  • \Delta G^0: Available total energy of the forward reaction to do work.
  • K_{eq}: Composition of the mixture of reactants and products at equilibrium.
  • E^0: Available electrical energy of the forward reaction to do work.
  • \Delta G^0 = -nFE^0
  • \Delta G^0 = -RT \ln K_{eq}
  • \log K_{eq} = \frac{nE^0}{0.0591V}

Ksp: Solubility Product Constant

  • K_{sp} is the equilibrium constant for salts that are “insoluble”.
  • Non-spontaneous process.
  • K should be less than 1.

Electrochemical Determination of Ksp

  • Ag(s) + Cl^- \rightarrow AgCl(s) + e^-
  • Ag^+ + e^- \rightarrow Ag(s)
  • Net reaction: Ag^+ + Cl^- \rightarrow AgCl(s) E^0 = 0.58V
  • Reverse reaction: AgCl(s) \rightarrow Ag^+ + Cl^- E^0 = -0.58V
  • This is a K_{sp} reaction!
  • Therefore, K{sp} can be calculated from E^0 by: \log K{eq} = \frac{nE^0}{0.0591V} = \frac{1(-0.58V)}{0.0591V} = -9.814
  • K_{sp} = 1.54 \times 10^{-10}

Using Table 11.1 to Determine Ksp

  • Pb^{2+}(aq) + 2e^- \rightarrow Pb(s) E^0 = -0.13V
  • PbSO4(s) + 2e^- \rightarrow Pb(s) + SO4^{2-}(aq) E^0 = -0.35V
  • Find the K_{sp} of lead(II) sulfate from these two ½ rxns.

Electrolytic Cells

  • Zn^{2+} + 2e^- \rightarrow Zn
  • Cu \rightarrow Cu^{2+} + 2e^-
  • Outside power source drives a non-spontaneous reaction.
  • Voltage must be > E^0 of the spontaneous reaction.
  • The moles of electrons used to power the process can be determined from the current and the time.
  • Quantities of products formed can be predicted by knowing the moles of the electrons transferred.

Stoichiometric Relationship

  • Between moles of electrons supplied and quantities of products formed:
  • Anode reaction: 2H2O \rightarrow O2 + 4H^+ + 4e^- E^0 = -1.23V
  • Cathode reaction: 4H2O + 4e^- \rightarrow 2H2 + 4OH^- E^0 = -0.83V
  • Total reaction: 6H2O \rightarrow 2H2 + O_2 + 4H^+ + 4OH^- E^0 = -0.83 + (-1.23) = -2.06V
  • 4 moles of electrons transferred produces 2 moles of H2 and 1 mole of O2.
  • Note the 2:1 volume relationship between gases…

Stoichiometry of Electrolysis

  • n_e = \frac{I \times t}{F}
  • What is the mass of Cl_2 produced during a 6 hours electrolysis of molten NaCl using a current of 3 amp.?
    • n_e = \frac{I \times t}{F} = \frac{(3 amps)(6 hr)(60 min/hr)(60 sec/min)}{96485 coul/mol} = 0.672 mol
    • moles Cl_2 = \frac{1}{2} \times 0.672 = 0.336 mol
    • mass Cl_2 = (0.336 mol)(70.9 g/mol) = 23.8 g

ALEKS Example 1

  • An electric current of 0.930 A flows for 33.0 minutes. Calculate the amount of electric charge transported.
  • 1.84 kC

ALEKS Example 2

  • Suppose 750. mmol of electrons must be transported from one side of an electrochemical cell to another in 39.0 minutes. Calculate the size of electric current that must flow.
  • 3.09 \times 10^{-1} A

ALEKS Example 3

  • Suppose a current of 810. mA is passed through an electroplating cell with an aqueous solution of Ag2SO4 in the cathode compartment for 89.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.
  • 8.06 \times 10^{-2} g

Another Tougher Example

  • A 40.0 amp current flowed through molten iron(III) chloride for 10.0 hours. Determine the mass of iron and the volume of chlorine gas (measured at 25oC and 1 atm) that is produced during this time.
  • Calculate the number of moles of electrons.
  • Write the half-reactions that take place at the anode and at the cathode.
  • Calculate the moles of iron and of chlorine produced using the number of moles of electrons calculated and the stoichiometries from the balanced half-reactions.
  • Calculate the mass of iron using the molar mass and calculate the volume of chlorine gas using the ideal gas law (PV = nRT).

Electrolysis and the Refining of Aluminum

  • Copper, lead and tin have been used by humans for thousands of years.
  • Aluminum in its metallic form does not occur naturally and was not produced commercially through electrolysis until 1854.
  • Large sources of electricity were developed, like hydropower, which allowed aluminum refining to occur on a large scale.

Important Points to Remember

  • \log_{10} and 10^x are inverses of each other.
  • A K_{sp} reaction is written as a metal salt turning into its ions.
  • The E^0 for a K{sp} reaction will be negative since a K{sp} reaction is non-spontaneous.
  • In electrolysis, the ½ rxn that includes the substance being formed by electrolysis, will indicate the stoichiometric ratios.
  • To solve complicated electrolysis problems, set up with units and let unit conversion lead to the correct answer.