Redox Titration Notes

Redox Titration Introduction

This module reviews redox reactions, central to electrochemistry. Redox reactions either produce electric current to do work or require electric current to occur. The module covers calculating cell potential and determining spontaneity. It also studies the practical applications of redox titrations.

Learning Outcomes

Upon completion of this module, you should be able to:

  1. Define electrochemistry.

  2. Explain how chemical reactions produce electricity and how electricity can drive chemical reactions.

  3. Balance redox reactions in acidic or basic solutions using the half-reaction method.

  4. Determine whether a reaction is spontaneous or not.

  5. Solve redox titration problems.

Balancing Redox Reactions

Follow these steps to balance redox reactions:

  1. Write skeletal equations for the oxidation and reduction half-reactions.

  2. Balance each half-reaction for all elements except H and O.

  3. Balance each half-reaction for O by adding H2O.

  4. Balance each half-reaction for H by adding H+.

  5. Balance each half-reaction for charge by adding electrons.

  6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.

  7. Add the two half-reactions and simplify.

  8. If the reaction takes place in a basic medium, add OH− ions to the equation obtained in step 7 to neutralize the H+ ions (add in equal numbers to both sides of the equation) and simplify.

Example 16.1. Balancing Equations for Redox Reactions in Acidic Solutions

Write the balanced equation representing the reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.

Cu(s) + HNO3 (aq) \rightarrow NO(g) + Cu^{2+}(aq)

Example 16.2. Balancing Equations for Redox Reactions in Basic Solutions

Write the balanced equation representing the reaction between aqueous permanganate ion, MnO4^{-}, and solid chromium(III) hydroxide, Cr(OH)3, to yield solid manganese(IV) oxide, MnO2, and aqueous chromate ion, CrO4^{2-}. The reaction takes place in a basic solution.

Cr(OH)3 (s) + MnO4^- (aq) \rightarrow CrO4^{2-} (aq) + MnO2(s)

Redox Equilibria and the Nernst Equation

Unlike precipitation, acid-base, and complexation reactions, we rarely express the equilibrium position of a redox reaction with an equilibrium constant. Because a redox reaction involves a transfer of electrons, it is convenient to consider the reaction’s thermodynamics in terms of the electron.

For a reaction in which one mole of a reactant undergoes oxidation or reduction, the net transfer of charge, Q, in coulombs is

Q = nF

where n is the moles of electrons per mole of reactant, and F is Faraday’s constant (96,485 C/mol).

The free energy, \Delta G, to move this charge, Q, over a change in potential, E, is

\Delta G = EQ

The change in free energy (in kJ/mole) for a redox reaction, therefore, is

\Delta G = -nFE

where \Delta G has units of kJ/mol. The minus sign in the equation results from a different convention for assigning a reaction’s favorable direction. In thermodynamics, a reaction is favored when \Delta G is negative, but an oxidation‐reduction reaction is favored when E is positive.

The Nernst equation relates the potential, E, to the standard electrode potential, E^o, by the following equation where Q_r is the reaction quotient.

Note that at equilibrium, E = 0 (\Delta G = 0), and the logarithmic term is the equilibrium constant, K.

The standard potential for a redox reaction, E^o, is

E^o = E{red}^o - E{ox}^o

where E{red}^o and E{ox}^o are the standard reduction potentials for the reduction half-reaction and the oxidation half-reaction.

Example 6.4.4

Calculate (a) the standard potential, (b) the equilibrium constant, and (c) the potential when [Ag^+] = 0.020 M and [Cd^{2+}] = 0.050 M, for the following reaction at 25°C.

Cd(s) + 2Ag^+(aq) \rightleftharpoons 2Ag(s) + Cd^{2+}(aq)

Practice Exercises

  1. Balancing Redox Reactions

    a. MnO4^−(aq) + S2O3^{2−}(aq) \rightarrow Mn^{2+}(aq) + SO4^{2−}(aq); acidic solution
    b. Fe^{2+}(aq) + Cr2O7^{2−}(aq) \rightarrow Fe^{3+}(aq) + Cr^{3+}(aq); acidic solution
    c. Fe(s) + CrO4^{2−}(aq) \rightarrow Fe2O3(s) + Cr2O3(s); basic solution d. CO3^{2−}(aq) + N2H4(aq) \rightarrow CO(g) + N_2(g); basic solution

  2. The amount of Fe in a 0.4891-g sample of an ore is determined by titrating with K2Cr2O7. After dissolving the sample in HCl, the iron is brought into a +2 oxidation state using a Jones reductor. Titration to the diphenylamine sulfonic acid end point requires 36.92 mL of 0.02153 M K2Cr2O7. Report the ore’s iron content as %w/w Fe2O3 (Answer: 77.86%)

    The balanced reaction is
    K2Cr2O7(aq) + 6Fe^{2+}(aq) + 14H+(aq)\rightarrow2Cr^{3+}(aq) + 2K+(aq) + 6Fe^{3+}(aq) + 7H2O(l)

  3. The purity of a sample of sodium oxalate, Na2C2O4, is determined by titrating with a standard solution of KMnO4. If a 0.5116-g sample requires 35.62 mL of 0.0400 M KMnO4 to reach the titration’s end point, what is the %w/w Na2C2O4 in the sample.