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Show that for nonstandard conditions the temperature variation of a cell potential is $$E\left(T_{1}\right)-E\left(T_{2}\right)=\left(T_{1}-T_{2}\right) \frac{\left(\Delta S^{\circ}-R \ln Q\right)}{z F}$$ where \(E\left(T_{1}\right)\) and \(E\left(T_{2}\right)\) are the cell potentials at \(T_{1}\) and \(T_{2},\) respectively. We have assumed that the value of \(Q\) is maintained at a constant value. For the nonstandard cell below, the potential drops from \(0.394 \mathrm{V}\) at \(50.0^{\circ} \mathrm{C}\) to \(0.370 \mathrm{V}\) at \(25.0^{\circ} \mathrm{C} .\) Calculate \(Q\) \(\Delta H^{\circ},\) and \(\Delta S^{\circ}\) for the reaction, and calculate \(K\) for the two temperatures. $$\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$$ Choose concentrations of the species involved in the cell reaction that give the value of \(Q\) that you have calculated, and then determine the equilibrium concentrations of the species at \(50.0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
After calculations, we find reactions quotient Q, standard enthalpy and entropy changes: \(\Delta H^\circ\), \(\Delta S^\circ\), and corresponding equilibrium constant K for two given temperatures. The equilibrium concentrations of the species at 50 degree Celsius can then be found using this calculated K.

Step by step solution

01

Assemble Knowns and Calculate Q

Start by assembling the known values, as well as calculating Q from the cell potential using the Gibbs free energy formula. Since \(E(T_1)\) = 0.394V, \(E(T_2)\) = 0.370V, \(T_1\) = 50 degrees Celsius or 323.15K, \(T_2\) = 25 degrees Celsius or 298.15K, n = 2 (number of electrons involved in the reaction), F = 96485 C/mol (Faraday’s constant), and R = 8.314 J/mol K (Universal gas constant): \[Q = e^{[(E(T_2) - E(T_1)) \cdot n \cdot F / R(T_1 - T_2)]}\]
02

Compute \(\Delta H^\circ\) and \(\Delta S^\circ\)

\(\Delta H^\circ\) and \(\Delta S^\circ\) can be found via the Gibbs-Helmholtz equation by rearranging for \(\Delta S^\circ\) and hence \(\Delta H^\circ\). That gives us: \[\Delta S^\circ = (\Delta E \cdot n \cdot F / (\Delta T)) + R \cdot ln(Q)\] and \[\Delta H^\circ = \Delta E \cdot n \cdot F + T \cdot \Delta S^\circ\]
03

Determine the Equilibrium Constant (K)

Apply the Van't Hoff equation to determine K at T1 and T2: \[K(T_{1}) = e^{-\Delta H^\circ/R \cdot T_{1} + \Delta S^\circ/ R}\] and \[K(T_{2}) = e^{-\Delta H^\circ / R \cdot T_{2}+ \Delta S^\circ/R}\]
04

Establish Equilibrium Concentrations

Write the expression for the equilibrium constant (K) for the cell reaction, and from that expression find the concentrations of the species in equilibrium at 50 degrees Celsius.

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Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy is a fundamental concept in electrochemistry that helps us predict whether a chemical reaction can occur spontaneously. It is a measure of the maximum work that can be obtained from a chemical reaction at constant temperature and pressure. The change in Gibbs Free Energy (\(\Delta G\)) is related to the cell potential and is essential for understanding reaction spontaneity in electrochemical cells. For a cell operating under nonstandard conditions, we can relate \(\Delta G\) to the cell potential (E) using the formula:
  • \(\Delta G = -nFE\),
where \(n\) is the number of moles of electrons exchanged, and \(F\) is Faraday's constant (\(96485\) C/mol). When the conditions are not standard, Gibbs Free Energy becomes a crucial component of understanding the temperature effects on cell potential.
Cell Potential
The Cell Potential is an indicator of the potential difference between the two electrodes in an electrochemical cell. It reflects the driving force of the electron flow between the anode and cathode, playing a key role in determining how much energy the reaction can produce.For cells at nonstandard conditions, the cell potential varies with temperature and concentration of ionic species, governed by the Nernst Equation. The exercise exemplifies this by showing how the cell potential changes from 0.394 V to 0.370 V when the temperature shifts from 50°C to 25°C. To understand this phenomenon, we calculate \(Q\) using the formula:
  • \(Q = e^{[(E(T_2) - E(T_1)) \cdot n \cdot F / R(T_1 - T_2)]}\).
Adjusting cell parameters according to these shifts is essential for precise electrochemical measurements, and understanding the formula in this context helps students grasp how potential changes link back to Gibbs Free Energy and reaction spontaneity.
Equilibrium Constant
The Equilibrium Constant, \(K\), is a central concept in chemical reactions that helps us quantify the extent to which a reaction will proceed to form products. In electrochemistry, \(K\) can be determined by using cell potentials and changes in Gibbs Free Energy.When dealing with nonstandard conditions, the equilibrium constant varies with temperature, and can be calculated using the Van't Hoff equation. The exercise utilizes this equation to determine \(K\) at both 50°C and 25°C:
  • \(K(T_{1}) = e^{-\Delta H^\circ/R \cdot T_{1} + \Delta S^\circ/ R}\)
  • \(K(T_{2}) = e^{-\Delta H^\circ / R \cdot T_{2}+ \Delta S^\circ/R}\)
By understanding how \(K\) changes, students learn to predict how reaction equilibrium can shift due to temperature variations, reinforcing their grasp of reaction kinetics in electrochemical cells. These calculations are essential for determining equilibrium concentrations under different thermodynamic conditions, completing their knowledge on how temperature and potential influence electrochemical reactions.

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Most popular questions from this chapter

For the half-reaction \(\mathrm{Hg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(1)\) \(E^{\circ}=0.854 \mathrm{V} .\) This means that \((\mathrm{a}) \mathrm{Hg}(1)\) is more readily oxidized than \(\mathrm{H}_{2}(\mathrm{g}) ;\) (b) \(\mathrm{Hg}^{2+}(\) aq) is more readily reduced than \(\mathrm{H}^{+}(\mathrm{aq}) ;\) (c) \(\mathrm{Hg}(\) l) will dissolve in 1 M HCl; (d) Hg(l) will displace Zn(s) from an aqueous solution of \(\mathrm{Zn}^{2+}\) ion.

For the reaction \(\operatorname{Co}(\mathrm{s})+\mathrm{Ni}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Co}^{2+}(\mathrm{aq})+\) \(\mathrm{Ni}(\mathrm{s}), E_{\mathrm{cell}}^{\circ}=0.03 \mathrm{V} .\) If cobalt metal is added to an aqueous solution in which \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M},\) (a) the reaction will not proceed in the forward direction at all; (b) the displacement of \(\mathrm{Ni}(\mathrm{s})\) from the \(\mathrm{Ni}^{2+}(\mathrm{aq})\) will go to completion; (c) the displacement of \(\mathrm{Ni}(\mathrm{s})\) from the solution will proceed to a considerable extent, but the reaction will not go to completion; (d) there is no way to predict how far the reaction will proceed.

A test for completeness of electrodeposition of \(\mathrm{Cu}\) from a solution of \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is to add \(\mathrm{NH}_{3}(\mathrm{aq}) .\) A blue color signifies the formation of the complex ion \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\left(K_{\mathrm{f}}=1.1 \times 10^{13}\right) .\) Let \(250.0 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{CuSO}_{4}(\text { aq })\) be electrolyzed with a \(3.512 \mathrm{A}\) current for 1368 s. At this time, add a sufficient quantity of \(\mathrm{NH}_{3}(\text { aq })\) to complex any remaining \(\mathrm{Cu}^{2+}\) and to maintain a free \(\left[\mathrm{NH}_{3}\right]=0.10 \mathrm{M} .\) If \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is detectable at concentrations as low as \(1 \times 10^{-5} \mathrm{M}\) should the blue color appear?

A common reference electrode consists of a silver wire coated with \(\mathrm{AgCl}(\mathrm{s})\) and immersed in \(1 \mathrm{M} \mathrm{KCl}\) $$\mathrm{AgCl}(\mathrm{s})+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}^{-}(1 \mathrm{M}) E^{\circ}=0.2223 \mathrm{V}$$ (a) What is \(E_{\text {cell }}^{\circ}\) when this electrode is a cathode in combination with a standard zinc electrode as an anode? (b) Cite several reasons why this electrode should be easier to use than a standard hydrogen electrode. (c) By comparing the potential of this silver-silver chloride electrode with that of the silver-silver ion electrode, determine \(K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\).

When deciding whether a particular reaction corresponds to a cell with a positive standard cell potential, which of the following thermodynamic properties would you use to get your answer without performing any calculations? Which would you not use? Explain. (a) \(\Delta G^{\circ} ;\) (b) \(\Delta S^{\circ} ;\) (c) \(\Delta H^{\circ} ;\) (d) \(\Delta U^{\circ} ;\) (e) \(K\).

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