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The selectivity coefficient, \(K_{\mathrm{Li}^{+}, \mathrm{H}^{+}}^{\mathrm{Pot}}\), for a \(\mathrm{Li}^{+}\)ion-selective electrode is \(4 \times 10^{-4}\). When this electrode is placed in \(3.44 \times 10^{-4} \mathrm{M}\) \(\mathrm{Li}^{+}\)solution at \(\mathrm{pH} 7.2\), the potential is \(-0.333 \mathrm{~V}\) versus S.C.E. What would be the potential if the \(\mathrm{pH}\) were lowered to \(1.1\) and the ionic strength were kept constant?

Short Answer

Expert verified
The potential would be approximately -0.397 V.

Step by step solution

01

Understand the Problem

You need to determine the potential of a Li鈦 ion-selective electrode when the pH changes from 7.2 to 1.1 while maintaining constant ionic strength. The selectivity coefficient is given, which indicates the preference of the electrode for Li鈦 over H鈦 ions.
02

Apply the Nernst Equation for the Electrode Potential

The electrode potential can be calculated using the Nernst equation, which for a monovalent cation like Li鈦 is:\[ E = E^0 + \frac{RT}{F} \ln \left( a_{\mathrm{Li}^+} + K_{\mathrm{Li}^+, \mathrm{H}^+}^{\mathrm{Pot}} \cdot a_{\mathrm{H}^+} \right) \]where \( E^0 \) is the standard electrode potential, \( a_{\mathrm{Li}^+} \) and \( a_{\mathrm{H}^+} \) are the activities of Li鈦 and H鈦 ions, respectively.
03

Calculate the Activity of H鈦 Ions

At \( \text{pH} = 1.1 \), the concentration of H鈦 ions is \( 10^{-1.1} \) M, which is approximately \( 7.94 \times 10^{-2} \) M. Assuming the activity coefficient is 1, the activity of H鈦 ions is approximately equal to its concentration.
04

Determine the Li鈦 Ion Activity

Given that the Li鈦 concentration is \( 3.44 \times 10^{-4} \) M, its activity is approximately this concentration, assuming the ionic strength remains constant.
05

Substitute Values into the Nernst Equation

Substitute the values into the Nernst equation:\[ E = -0.333 + \frac{RT}{F} \ln \left( 3.44 \times 10^{-4} + 4 \times 10^{-4} \times 7.94 \times 10^{-2} \right) \]Given that \( R = 8.314 \) J/(mol K), \( T = 298 \) K, and \( F = 96485 \) C/mol, calculate the potential.
06

Simplify and Solve for the Potential

Simplify the expression inside the logarithm:\( 3.44 \times 10^{-4} + 4 \times 10^{-4} \times 7.94 \times 10^{-2} \approx 6.5824 \times 10^{-4} \).Substitute in the constants:\[ E = -0.333 + \frac{8.314 \, \times \, 298}{96485} \ln (6.5824 \times 10^{-4}) \]Solving this gives \( E \approx -0.397 \) V.

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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 a fundamental tool in electrochemistry that allows you to calculate the electrode potential of a cell. It relates the concentration (or activity) of ions in the solution to the potential across the electrode. For a monovalent cation like lithium ion \[E = E^0 + \frac{RT}{F} \ln \left( a_{\mathrm{Li}^+} + K_{\mathrm{Li}^+, \mathrm{H}^+}^{\mathrm{Pot}} \cdot a_{\mathrm{H}^+} \right)\]
  • \(E^0\): Standard electrode potential.
  • \(R\): Universal gas constant, 8.314 J/(mol路K).
  • \(T\): Temperature in Kelvin.
  • \(F\): Faraday's constant, 96485 C/mol.
  • \(a_{\mathrm{Li}^+}\): Activity of lithium ions in solution.
  • \(a_{\mathrm{H}^+}\): Activity of hydrogen ions, related to pH.
The Nernst equation highlights how even small changes in ion concentrations can substantially alter the electrode potential, which is vital for understanding how sensors and electrodes function in various chemical environments.
pH Effect
pH is a measure of the acidity or basicity of a solution; it directly influences the activity of hydrogen ions \((a_{\mathrm{H}^+})\) in the Nernst equation. When the pH changes, it affects the ion activity and thus, the electrode potential. The formula \(a_{\mathrm{H}^+} = 10^{-\text{pH}}\) shows us the relationship between pH and hydrogen ion activity.
In the context of ion-selective electrodes, a shift in pH can change the potential reading significantly, especially if the electrode has some sensitivity to hydrogen ions. As seen in the problem, decreasing the pH from 7.2 to 1.1 markedly increases the hydrogen ion activity, which affects the measured potential. Thus, pH control or compensation may be necessary for accurate measurements in analytical chemistry.
Selectivity Coefficient
The selectivity coefficient \((K_{\mathrm{Li}^+, \mathrm{H}^+}^{\mathrm{Pot}})\) is a parameter that quantifies how much more an ion-selective electrode responds to its primary ion over interfering ions, such as hydrogen ions. For the lithium ion (\(Li^+\)) electrode, a low selectivity coefficient means that it has a high preference for lithium ions and minimal interference from hydrogen ions.
This is crucial in contexts where the solution includes different ions, as it affects the accuracy of measurements. A lower selectivity coefficient indicates that the electrode is more selective to \(Li^+\) and less affected by the presence of \(H^+\). In our exercise, the selectivity coefficient of \(4 \times 10^{-4}\) illustrates some degree of interference by hydrogen ions, influencing how the potential changes as pH varies.
Electrode Potential Calculation
Electrode potential calculation involves substituting known values into the Nernst equation to determine the overall potential of the electrode under specified conditions. For the problem at hand, the potential was calculated for a lithium-ion electrode when the pH was altered.
This process starts by determining the activities of the ions involved, utilizing both given concentrations and mathematical expressions for the activities, such as \(a_{\mathrm{H}^+} = 10^{-\text{pH}}\). Then, these activities and constants (\(R, T, F\)) are plugged into the Nernst equation to solve for the new potential \(E\).
The step-by-step substitution of these values allowed the calculation to yield a potential of approximately \(-0.397 \mathrm{~V}\), illustrating the impact of changing pH on electrode readings.

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Most popular questions from this chapter

The \(\mathrm{Pb}^{2+}\) ion buffer used inside the electrode for the colored curve in Figure 14-27 was prepared by mixing \(1.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(100.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Na}_{2}\) EDTA. At the measured \(\mathrm{pH}\) of \(4.34, \alpha_{\mathrm{Y}^{4}}=1.46 \times 10^{-8}\) (Equation 11-4). Show that \(\left[\mathrm{Pb}^{2+}\right]=\) \(1.4 \times 10^{-12} \mathrm{M}\).

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into \(0.10 \mathrm{M} \mathrm{CuSO}_{4}\) solution. The Cu wire was attached to the positive terminal of a potentiometer and the calomel electrode was attached to the negative terminal. (a) Write a half-reaction for the Cu electrode. (b) Write the Nernst equation for the Cu electrode. (c) Calculate the cell voltage.

Activity problem. Citric acid is a triprotic acid \(\left(\mathrm{H}_{3} \mathrm{~A}\right)\) whose anion \(\left(\mathrm{A}^{3-}\right)\) forms stable complexes with many metal ions. $$ \mathrm{Ca}^{2+}+\mathrm{A}^{3-} \stackrel{K_{i}}{\rightleftharpoons} \mathrm{CaA}^{-} $$ When a \(\mathrm{Ca}^{2+}\) ion-selective electrode with a slope of \(29.58 \mathrm{mV}\) was immersed in a solution having \(\mathcal{A}_{\mathrm{Ca}^{2+}}=1.00 \times 10^{-3}\), the reading was \(+2.06 \mathrm{mV}\). Calcium citrate solution was prepared by mixing equal volumes of solutions 1 and 2 . Solution 1: $$ \left[\mathrm{Ca}^{2+}\right]=1.00 \times 10^{-3} \mathrm{M}, \mathrm{pH}=8.00, \mu=0.10 \mathrm{M} $$ Solution 2: \([\text { Citrate }]_{\text {total }}=1.00 \times 10^{-3} \mathrm{M}, \mathrm{pH}=8.00, \mu=0.10 \mathrm{M}\) The calcium citrate solution gave an electrode reading of \(-25.90 \mathrm{mV}\). (a) Refer to the discussion with Figure B-2 in Appendix B. Calculate the activity of \(\mathrm{Ca}^{2+}\) in the calcium citrate solution. (b) Calculate the formation constant, \(K_{\mathrm{f}}\), for \(\mathrm{CaA}^{-}\). Assume that the size of \(\mathrm{CaA}^{-}\)is \(500 \mathrm{pm}\). At pH \(8.00\) and \(\mu=0.10 \mathrm{M}\), the fraction of free citrate in the form \(\mathrm{A}^{3-}\) is \(0.998\).

Explain the principle of operation of ion-selective electrodes. How does a compound electrode differ from a simple ion-selective electrode?

What makes a liquid-based ion-selective electrode specific for one analyte?

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