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Chapter 11: Problem 22

Consider the reaction \(\operatorname{Sn}(s)+\operatorname{Sn}^{4+}(a q) \rightleftharpoons\) \(2 \operatorname{Sn}^{2+}(a q) .\) If metallic tin is in equilibrium with a solution of \(\mathrm{Sn}^{2+}(a q)\) in which \(a_{S n^{2+}}=0.250,\) what is the activity of \(\mathrm{Sn}^{4+}(a q)\) at equilibrium at \(298.15 \mathrm{K} ?\)

Short Answer

Expert verified
The activity of $\mathrm{Sn}^{4+}(aq)$ at equilibrium at 298.15 K is approximately \(1.19 \times 10^{-6}\).

Step by step solution

01

Write down the balanced chemical equation

The balanced chemical equation for the given reaction is: \[ \operatorname{Sn}(s) + \operatorname{Sn}^{4+}(aq) \rightleftharpoons 2\operatorname{Sn}^{2+}(aq) \]
02

Write down the expression for the equilibrium constant (K) for the reaction

As this is a reduction-oxidation (redox) reaction, we'll use the reduction potentials to find the equilibrium constant. The relevant reduction half-reactions are: For Sn鈦粹伜: \[ \operatorname{Sn}^{4+}(aq) + 2e^- \rightarrow \operatorname{Sn}^{2+}(aq); \hspace{5mm} E^0_{Sn^{4+}/Sn^{2+}} = 0.15 \, \text{V} \] For Sn虏鈦�: \[ \operatorname{Sn}^{2+}(aq) + 2e^- \rightarrow \operatorname{Sn}(s); \hspace{5mm} E^0_{Sn^{2+}/Sn} = -0.14 \, \text{V} \] The overall reaction is achieved by adding these two half-reactions, so the overall standard reduction potential is: \[ E^0 = E^0_{Sn^{4+}/Sn^{2+}} - E^0_{Sn^{2+}/Sn} = 0.15 - (-0.14) = 0.29 \, \text{V} \] We can find the equilibrium constant, K, using the Nernst equation: \[ E = E^0 - \frac{RT}{nF} \ln{K} \] At equilibrium, E = 0. So, we can solve for K: \[ 0 = 0.29 - \frac{8.314 \times 298.15}{2 \times 96485} \ln{K} \]
03

Calculate the equilibrium constant (K)

Now, we can solve the equation from step 2 to find the equilibrium constant K: \[ \ln{K} = \frac{2 \times 96485 \times 0.29}{8.314 \times 298.15} \] \[ K = e^{\ln{K}} = e^{\frac{2 \times 96485 \times 0.29}{8.314 \times 298.15}} \approx 5.28 \times 10^4 \]
04

Use the equilibrium constant to find the activity of Sn鈦粹伜 ions

Now that we have the equilibrium constant, we can relate it to the activities of the ions in the reaction: \[ K = \frac{a_{Sn^{2+}}^2}{a_{Sn^{4+}}} \] We know the activity of Sn虏鈦� ions, a_{Sn^{2+}} = 0.250. So, we can solve for a_{Sn^{4+}}: \[ a_{Sn^{4+}} = \frac{a_{Sn^{2+}}^2}{K} = \frac{(0.250)^2}{5.28 \times 10^4} \] \[ a_{Sn^{4+}} \approx 1.19 \times 10^{-6} \] So, the activity of Sn鈦粹伜 ions at equilibrium at 298.15 K is approximately 1.19 脳 10鈦烩伓.

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Key Concepts

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

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between chemical species. In such reactions, one species gains electrons (is reduced) while another loses electrons (is oxidized). Understanding these processes is crucial because they govern numerous chemical reactions and are foundational in electrochemistry.
In the given exercise, we have a redox reaction where tin (Sn) changes its oxidation state. This involves two half-reactions: one for the reduction of Sn鈦粹伜 to Sn虏鈦�, and another for the reduction of Sn虏鈦� to solid Sn. These transformations showcase the electron transfer characteristic of redox processes.
  • Reduction: Sn鈦粹伜 gains electrons to become Sn虏鈦�.
  • Oxidation: In a reverse mentality, observing Sn虏鈦� reducing to Sn shows potential electron contributions if we revert Sn to solid form.
By understanding these half-reactions, it becomes easier to visualize how electrons flow in the system, guiding us through calculating things like the equilibrium constant.
Nernst Equation
The Nernst equation connects the dots between electrochemistry and reaction equilibria. It provides a way to calculate the reaction potential at any given concentrations of reactants and products. Mathematically, the Nernst equation is expressed as: \[E = E^0 - \frac{RT}{nF} \ln Q\] where:
  • \(E\) is the cell potential at non-standard conditions.
  • \(E^0\) is the standard cell potential.
  • \(R\) is the gas constant (8.314 J/mol路K).
  • \(T\) is the temperature in Kelvin.
  • \(n\) is the number of moles of electrons exchanged.
  • \(F\) is the Faraday constant (96485 C/mol).
  • \(Q\) is the reaction quotient.
In the exercise, the Nernst equation was crucial for determining the equilibrium constant \(K\) when \(E\) equals zero at equilibrium. This demonstrates how powerful the equation is in linking electronegativity with chemical equilibria.
Activity of Ions
"Activity" in chemistry refers to the "effective concentration" of ions in a solution, accounting for interactions that might affect them. Unlike concentration, which only considers the amount of solute in a solution, activity considers interactions that can affect ions' behavior. In diluted solutions, the activity is typically close to concentration, but in more complex or concentrated solutions, they can differ significantly.
In the given reaction, the activity coefficient adjusts the Sn虏鈦� ions' active behavior in the solution, making it a more measurable property. The exercise focuses on determining the activity of Sn鈦粹伜 ions at equilibrium. This activity is calculated using the equilibrium constant \(K\), illustrating how activity connects to equilibrium and redox principles.
Understanding activities helps chemists predict how ions behave in different environments, a valuable tool in various scientific and industrial applications.
Reduction Potentials
Reduction potentials, denoted by \(E^0\), let us assess how much a substance prefers to gain electrons, indicating its ability to act as an oxidizing agent. Every half-reaction in an electrochemical cell comes with a standard reduction potential which is universal and listed in electrochemical series tables.
In the exercise, the standard reduction potentials for the involved redox couples (Sn鈦粹伜/Sn虏鈦� and Sn虏鈦�/Sn) are used to calculate the overall potential \(E^0\) for the reaction and set the groundwork for determining the equilibrium constant. By knowing \(E^0\), we can predict the spontaneity of the reactions.
  • Higher reduction potential implies a stronger attraction for electrons.
  • Reduction potentials are essential in calculating the electromotive force (EMF) of cells, driving electrochemical reactions.
Understanding these potentials is vital for harnessing chemical energy, designing batteries, and advancing electrochemical technologies.

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Most popular questions from this chapter

The data in the following table have been obtained for the potential of the cell \(\operatorname{Pt}(s)\left|\mathrm{H}_{2}(g, f=1.00 \mathrm{atm})\right|\) \(\mathrm{HCl}(a q, m)|\operatorname{AgCl}(s)| \operatorname{Ag}(s)\) as a function of \(m\) at \(25^{\circ} \mathrm{C}\) $$\begin{array}{cccccc}m / \mathrm{mol} & & m / \mathrm{mol} & & m / \mathrm{mol} & \\ \mathrm{kg}^{-1} & E / V & \mathrm{kg}^{-1} & E / V & \mathrm{kg}^{-1} & E / V \\\ \hline 0.00100 & 0.57915 & 0.0200 & 0.43024 & 0.500 & 0.27231 \\ 0.00200 & 0.54425 & 0.0500 & 0.38588 & 1.000 & 0.23328 \\ 0.00500 & 0.49846 & 0.100 & 0.35241 & 1.500 & 0.20719 \\ 0.0100 & 0.46417 & 0.200 & 0.31874 & 2.000 & 0.18631\end{array}$$ a. Determine \(E^{\circ}\) using a graphical method. b. Calculate \(\gamma_{\pm}\) for \(\mathrm{HCl}\) at \(m=0.00100,0.0100,\) and \(0.100 \mathrm{mol} \mathrm{kg}^{-1}\).

The standard half-cell potential for the reaction \\[\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \text { is }+1.229 \mathrm{V} \text { at }\\] \(298.15 \mathrm{K} .\) Calculate \(E\) for a 0.300 -molal solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) for \(a_{\mathrm{O}_{2}}=1.00\) (a) assuming that the \(a_{H^{+}}\) is equal to the molality and (b) using the measured mean ionic activity coefficient for this concentration from the data tables. How large is the relative error if the concentrations, rather than the activities, are used?

Determine \(E^{\circ}\) for the reaction \(\mathrm{Cr}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(s)\) from the one-electron reduction potential for \(\mathrm{Cr}^{3+}(a q)\) and the three-electron reduction potential for \(\mathrm{Cr}^{3+}(a q)\) given in Table 11.1 (see Appendix B).

The half-cell potential for the reaction \\[\begin{array}{l}\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \text { is }+1.03 \mathrm{V} \text { at } \\ 298.15 \mathrm{K} \text { when } a_{\mathrm{O}_{2}}=1.00 . \text { Determine } a_{\mathrm{H}^{+}} \end{array}\\]

Determine \(K_{s p}\) for AgBr at \(298.15 \mathrm{K}\) using the electrochemical cell described by $$\operatorname{Ag}(s)\left|\operatorname{Ag}^{+}\left(a q, a_{A g^{+}}\right)\right| \operatorname{Br}^{-}\left(a q, a_{B r}-\right)|\operatorname{AgBr}(s)| \operatorname{Ag}(s)$$
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