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Ch.19 - Electrochemistry
Chapter 19, Problem 164

Consider the redox titration of 100.0 mL of a solution of 0.010 M Fe2+ in 1.50 M H2SO4 with a 0.010 M solution of KMnO4, yielding Fe3+ and Mn2+. The titration is carried out in an electrochemical cell equipped with a platinum electrode and a calomel reference electrode consisting of an Hg2Cl2/Hg electrode in contract with a saturated KCl solution having [Cl-] = 2.9M. Using any data in Appendixes C and D, calculate the cell potential after addition of (a) 5.0 mL, (b) 10.0mL, (c) 19.0 mL, and (d) 21.0 mL of the KMnO4 solution.

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1
Identify the half-reactions involved in the redox titration: Fe^{2+} \rightarrow Fe^{3+} + e^- and MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O.
Calculate the initial moles of Fe^{2+} in the solution using the concentration and volume: moles = 0.010 \text{ M} \times 0.100 \text{ L}.
Determine the moles of KMnO_4 added at each volume increment (5.0 mL, 10.0 mL, 19.0 mL, 21.0 mL) using its concentration: moles = 0.010 \text{ M} \times \text{volume in L}.
Use the stoichiometry of the balanced redox reaction to find the moles of Fe^{2+} and MnO_4^- remaining or reacted at each stage.
Apply the Nernst equation to calculate the cell potential at each stage, considering the concentrations of Fe^{2+}, Fe^{3+}, MnO_4^-, and Mn^{2+} and the standard electrode potentials from Appendixes C and D.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Redox Reactions

Redox reactions involve the transfer of electrons between two species, where one species is oxidized (loses electrons) and the other is reduced (gains electrons). In this titration, Fe2+ is oxidized to Fe3+, while MnO4- is reduced to Mn2+. Understanding the oxidation states and half-reactions is crucial for calculating the cell potential at various stages of the titration.
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Electrochemical Cell and Cell Potential

An electrochemical cell consists of two half-cells connected by a salt bridge or a porous membrane, allowing the flow of ions. The cell potential (E) is determined by the difference in reduction potentials of the two half-reactions. In this titration, the potential changes as KMnO4 is added, affecting the concentrations of Fe2+ and Fe3+, which must be accounted for in the Nernst equation to calculate the cell potential.
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Nernst Equation

The Nernst equation relates the cell potential to the concentrations of the reactants and products in a redox reaction. It is expressed as E = E° - (RT/nF) ln(Q), where E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient. This equation is essential for calculating the cell potential at different volumes of KMnO4 added during the titration.
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Related Practice
Textbook Question

The half-reactions that occur in ordinary alkaline batteries can be written as In 1999, researchers in Israel reported a new type of alkaline battery, called a 'super-iron' battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO42- ion (from K2FeO4) to solid Fe(OH)3 at the cathode. (b) Write a balanced equation for the cathode half-reaction in a super-iron battery. The half-reaction occurs in a basic environment.

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Textbook Question

The half-reactions that occur in ordinary alkaline batteries can be written as In 1999, researchers in Israel reported a new type of alkaline battery, called a 'super-iron' battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO42- ion (from K2FeO4) to solid Fe(OH)3 at the cathode. (c) A super-iron battery should last longer than an ordinary alkaline battery of the same size and weight because its cathode can provide more charge per unit mass. Quan-titatively compare the number of coulombs of charge released by the reduction of 10.0 g K2FeO4 to Fe(OH)3 with the number of coulombs of charge released by the reduction 10.0 g of MnO2 to MnO(OH).

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Textbook Question
Gold metal is extracted from its ore by treating the crushed rock with an aerated cyanide solution. The unbalanced equation for the reaction is (b) Use any of the following data at 25 °C to calculate ∆G° for this reaction at 25 °C: Kf for Au(CN)2- = 6.2 x 10^38, Ka for HCN = 4.9 x 10^-10, and standard reduction potentials are
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Textbook Question

We've said that the +1 oxidation state is uncommon for indium but is the most stable state for thallium. Verify this statement by calculating E ° and ΔG ° (in kilojoules) for the disproportionation reaction

3 M+1aq2S M3+1aq2 + 2 M1s2 M = In or Tl

Is disproportionation a spontaneous reaction for In+ and/orTl+? Standard reduction potentials for the relevant halfreactions are

In3+1aq2 + 2 e- S In+1aq2 E° = -0.44 V

In+1aq2 + e- S In1s2 E° = -0.14 V

Tl3+1aq2 + 2 e- S Tl+1aq2 E° = +1.25 V

Tl+1aq2 + e- S Tl1s2 E° = -0.34 V

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