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Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{Ce}^{4+}\) in \(1 \mathrm{M}\) \(\mathrm{HClO}_{4}\) by \(0.0400 \mathrm{M} \mathrm{Cu}^{+}\)to give \(\mathrm{Ce}^{3+}\) and \(\mathrm{Cu}^{2+}\), using Pt and saturated \(\mathrm{Ag} \mid \mathrm{AgCl}\) electrodes to find the end point. (a) Write a balanced titration reaction. (b) Write two different half-reactions for the indicator electrode. (c) Write two different Nernst equations for the cell voltage. (d) Calculate \(E\) at the following volumes of \(\mathrm{Cu}^{+}: 1.00,12.5,24.5\), \(25.0,25.5,30.0\), and \(50.0 \mathrm{~mL}\). Sketch the titration curve.

Short Answer

Expert verified
Ce\(^{4+}\) + Cu\(^+\) \(\rightarrow\) Ce\(^{3+}\) + Cu\(^{2+}\); calculate E using Nernst equation at different volumes; plot potential versus volume.

Step by step solution

01

Write the balanced titration reaction

The titration process involves the reduction of \( \mathrm{Ce}^{4+} \) to \( \mathrm{Ce}^{3+} \) and the oxidation of \( \mathrm{Cu}^{+} \) to \( \mathrm{Cu}^{2+} \). The balanced reaction is:\[ \mathrm{Ce}^{4+} + \mathrm{Cu}^{+} \rightarrow \mathrm{Ce}^{3+} + \mathrm{Cu}^{2+} \]
02

Write half-reactions for the indicator electrode

Two half-reactions occur at the platinum electrode:1. Reduction half-reaction for cerium: \[ \mathrm{Ce}^{4+} + e^- \rightarrow \mathrm{Ce}^{3+} \]2. Oxidation half-reaction for copper: \[ \mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+} + e^- \]
03

Write the Nernst equations for the cell voltage

The standard Nernst equation for an electrochemical cell is:\[ E = E^\circ - \frac{0.0592}{n} \log Q \]For each half-reaction:1. Cerium half-cell potential: \[ E_{\mathrm{Ce}} = E^\circ_{\mathrm{Ce}} - \frac{0.0592}{1} \log\left( \frac{[\mathrm{Ce}^{3+}]}{[\mathrm{Ce}^{4+}]} \right) \]2. Copper half-cell potential: \[ E_{\mathrm{Cu}} = E^\circ_{\mathrm{Cu}} - \frac{0.0592}{1} \log\left( \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Cu}^{+}]} \right) \]
04

Determine initial concentrations before endpoint

Initially, you have 100 mL of 0.0100 M \( \mathrm{Ce}^{4+} \). Calculate the moles:\[ n_{\mathrm{Ce}^{4+}} = 0.100 \times 0.0100 = 0.00100 \text{ mol} \]
05

Calculate E before the endpoint (at 1.00 mL, 12.5 mL, and 24.5 mL)

At each stage, use the Nernst equation to evaluate the potential, adjusting for concentrations:- At 1.00 mL, 12.5 mL, and 24.5 mL of 0.0400 M \( \mathrm{Cu}^{+} \), update the moles: \[ n_{\mathrm{Cu}^{+}} = V \times 0.0400 \]\Find \([\mathrm{Ce}^{4+}]\), \([\mathrm{Ce}^{3+}]\), \([\mathrm{Cu}^{+}]\), and \([\mathrm{Cu}^{2+}]\) and substitute into the Nernst equation to get \( E_{\text{cell}} \).
06

Calculate E at equivalence point (25.0 mL)

At the equivalence point, all \( \mathrm{Ce}^{4+} \) is converted to \( \mathrm{Ce}^{3+} \);- Moles of \( \mathrm{Cu}^{+} = 0.025 \times 0.0400 = 0.00100 \text{ mol} \)Concentrations are equivalent; use Nernst equation to find \( E_{\text{cell}} \).
07

Calculate E after equivalence (25.5 mL, 30.0 mL, 50.0 mL)

After 25.0 mL, additional \( \mathrm{Cu}^{+} \) does not react with \( \mathrm{Ce}^{4+} \), use the excess:- Find moles of excess \( \mathrm{Cu}^{+} \)Calculate concentrations, and substitute into the Nernst equation to find \( E_{\text{cell}} \).
08

Sketch the titration curve

Plotting \( E \) against the volume of added \( \mathrm{Cu}^{+} \), the graph's key features include a sharp increase in potential at the equivalence point, providing visual validation of the calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst Equation
The Nernst Equation is crucial in electrochemistry. It helps us calculate the electrochemical cell potential under non-standard conditions. This equation relates the standard electrode potential and concentrations (or pressures) of the reacting species. It's expressed as:
\[ E = E^\circ - \frac{0.0592}{n} \log Q \]where:
  • \( E \) is the cell potential.
  • \( E^\circ \) is the standard electrode potential.
  • \( n \) is the number of moles of electrons transferred in the reaction.
  • \( Q \) is the reaction quotient, akin to the equilibrium constant but applies to any point in the reaction.In titration, we apply the Nernst Equation to calculate potentials at various stages. It's especially valuable for determining the potential at the equivalence point, where reactants are stoichiometrically balanced. The role of the Nernst Equation in interpreting titration curves allows for predicting how potential changes throughout the process.
Electrode Potentials
Electrode potentials are a measure of the tendency of chemical species to acquire electrons and therefore be reduced. When discussing redox titrations, like in our example with cerium and copper ions, electrode potentials determine how driving or spontaneous the reactions are. Each half-reaction in the titration has its own potential:- **Cerium half-cell (reduction):** \( \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+} \)- **Copper half-cell (oxidation):** \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \)The standard electrode potentials can be found in electrochemical series tables. By subtracting the cathode potential from the anode potential, the cell potential \( E \) can be calculated. This value indicates the voltage difference between the two electrodes, guiding the flow of electric current in the cell.
Electrode potentials influence the overall energy exchange in a redox reaction and are central to understanding and controlling titration processes.
Titration Curve
A titration curve graphically represents how cell potential changes with the addition of titrant, in this example, as \(\text{Cu}^+\) is gradually added to \(\text{Ce}^{4+}\). This curve offers a clear snapshot of the titration process. Key elements shown include:
  • **Initial Region:** Slow change in potential as the titration starts.
  • **Equivalence Point:** Rapid increase in potential, indicating the stoichiometric completion of reaction.
  • **Post-Equivalence Region:** Potential stabilizes as excess titrant is introduced.By visualizing these regions, we can pinpoint the equivalence point, crucial in determining the exact volume of titrant needed to completely react with the analyte. This point is represented by a sharp inflection, simplifying the estimation of endpoint potential and concentration changes.
Electrochemical Cells
Electrochemical cells are setups used to conduct redox reactions; they generate electrical energy or allow for the consumption of electricity in chemical processes. In our example, the cell is comprised of two electrodes—Pt for cerium and the saturated \( \text{Ag} | \text{AgCl} \) for copper. These form two half-cells connected by an ionic pathway.In such cells:
  • **Anode:** Site of oxidation, where electrons are released.
  • **Cathode:** Site of reduction, where electrons are gained.
  • The flow of electrons from anode to cathode generates a current detectable via the electrode potential.
    The design of electrochemical cells ensures that reactions progress in a controlled manner, measured by voltmeters. Reaction progress in titrations can be traced effectively, allowing for detailed potential calculations at each stage of analyte addition, transforming chemical energy into measurable electrical outputs.

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Most popular questions from this chapter

What is a Jones reductor and what is it used for?

Here is a description of an analytical procedure for superconductors containing unknown quantities of \(\mathrm{Cu}(\mathrm{I}), \mathrm{Cu}(\mathrm{II})\), \(\mathrm{Cu}(\mathrm{III})\), and peroxide \(\left(\mathrm{O}_{2}^{2-}\right):^{32}\) " The possible trivalent copper and/or peroxide-type oxygen are reduced by \(\mathrm{Cu}(\mathrm{I})\) when dissolving the sample (ca. \(50 \mathrm{mg}\) ) in deoxygenated \(\mathrm{HCl}\) solution (1 \(\mathrm{M}\) ) containing a known excess of monovalent copper ions ( \(\mathrm{ca} .25 \mathrm{mg} \mathrm{CuCl}\) ). On the other hand, if the sample itself contained monovalent copper, the amount of \(\mathrm{Cu}(\mathrm{I})\) in the solution would increase upon dissolving the sample. The excess \(\mathrm{Cu}(\mathrm{I})\) was then determined by coulometric back-titration ... in an argon atmosphere." The abbreviation "ca." means "approximately." Coulometry is an electrochemical method in which the electrons liberated in the reaction \(\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{e}^{-}\) are measured from the charge flowing through an electrode. Explain with your own words and equations how this analysis works.

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Calcium fluorapatite \(\left(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6} \mathrm{~F}_{2}, \mathrm{FM} 1008.6\right)\) laser crystals were doped with chromium to improve their efficiency. It was suspected that the chromium could be in the \(+4\) oxidation state. 1\. To measure the total oxidizing power of chromium in the material, a crystal was dissolved in \(2.9 \mathrm{M} \mathrm{HClO}_{4}\) at \(100^{\circ} \mathrm{C}\), cooled to \(20^{\circ} \mathrm{C}\), and titrated with standard \(\mathrm{Fe}^{2+}\), using \(\mathrm{Pt}\) and \(\mathrm{Ag} \mid \mathrm{AgCl}\) electrodes to find the end point. Chromium oxidized above the \(+3\) state should oxidize an equivalent amount of \(\mathrm{Fe}^{2+}\) in this step. That is, \(\mathrm{Cr}^{4+}\) would consume one \(\mathrm{Fe}^{2+}\), and each atom of \(\mathrm{Cr}^{6+}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) would consume three \(\mathrm{Fe}^{2+}\) : $$ \begin{aligned} \mathrm{Cr}^{4+}+\mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+} \\ \frac{1}{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{Fe}^{3+} \end{aligned} $$ 2\. In a second step, the total chromium content was measured by dissolving a crystal in \(2.9 \mathrm{M} \mathrm{HClO}_{4}\) at \(100^{\circ} \mathrm{C}\) and cooling to \(20^{\circ} \mathrm{C}\). Excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and \(\mathrm{Ag}^{+}\)were then added to oxidize all chromium to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). Unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) was destroyed by boiling, and the remaining solution was titrated with standard \(\mathrm{Fe}^{2+}\). In this step, each \(\mathrm{Cr}\) in the original unknown reacts with three \(\mathrm{Fe}^{2+}\). $$ \begin{aligned} \mathrm{Cr}^{x+} & \stackrel{\mathrm{S}_{2} \mathrm{O}_{8}^{2-}}{\longrightarrow} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \\ \frac{1}{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3 \mathrm{Fe}^{2+} & \rightarrow \mathrm{Cr}^{3+}+3 \mathrm{Fe}^{3+} \end{aligned} $$ In step \(1,0.4375 \mathrm{~g}\) of laser crystal required \(0.498 \mathrm{~mL}\) of \(2.786 \mathrm{mM}\) \(\mathrm{Fe}^{2+}\) (prepared by dissolving \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) in \(2 \mathrm{M} \mathrm{HClO}_{4}\) ). In step \(2,0.1566 \mathrm{~g}\) of crystal required \(0.703 \mathrm{~mL}\) of the same \(\mathrm{Fe}^{2+}\) solution. Find the average oxidation number of \(\mathrm{Cr}\) in the crystal and find the total micrograms of \(\mathrm{Cr}\) per gram of crystal.

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