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Chapter 20: Problem 119

The \(K_{s p}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-28} .\) By using this value together with an electrode potential from Appendix \(\mathrm{E}\), determine the value of the standard reduction potential for the reaction $$\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)$$

Short Answer

Expert verified
The standard reduction potential for the half-cell reaction \(\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\) is approximately -0.414 V, as calculated using the Nernst equation and the given solubility product constant, \(K_{sp}\), value of \(8.0 \times 10^{-28}\).

Step by step solution

01

Write the balanced equation

First, we need to write the balanced equation for the dissolution of PbS(s) into its ions: \(\mathrm{PbS}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{S}^{2-}(a q)\)
02

Define the half-cell reaction

For the given half-cell reaction, we have: \(\mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q)\)
03

Use the Nernst equation

The Nernst equation relates the standard reduction potential (Eº) of a half-cell reaction to the reduction potential (E) under non-standard conditions, the number of electrons involved in the reaction (n), the reaction quotient (Q) and the solubility product constant (Ksp). The Nernst equation is: \[E = Eº -\dfrac{RT}{nF} * \ln Q\] Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin (assumed to be 298 K), n is the number of electrons involved in the reaction (in this case, 2), and F is the Faraday constant (96485 C/mol).
04

Determine the reaction quotient (Q)

For our half-cell reaction, the reaction quotient (Q) can be written as: \[Q = \dfrac{[\mathrm{S^{2-}}]}{[\mathrm{Pb^{2+}}]}\] Since the reaction involves the dissolution of PbS, the concentrations of each ion are equal at equilibrium. Therefore, we can express the reaction quotient as: \[Q = \dfrac{[\mathrm{S^{2-}}]}{[\mathrm{S^{2-}}]} = 1\]
05

Substitute Q and Ksp into the Nernst equation

Now we can substitute the value of Q into the Nernst equation: \[E = Eº - \dfrac{RT}{2F} * \ln 1\] Since \(\ln 1 = 0\), it simplifies to: \[E = Eº\] Now, we can use the solubility product constant (Ksp) to find the value of Eº: \[K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{S^{2-}}] = [\mathrm{Pb^{2+}}]^{1}[\mathrm{S^{2-}}]^{1} \] Since Ksp is equal to \(8.0 \times 10^{-28}\), we can write: \([S^{2-}] = \sqrt{K_{sp}} = \sqrt{8.0 \times 10^{-28}}\)
06

Find the standard reduction potential (Eº)

Now we can find the standard reduction potential (Eº) using the value of E: \[E = Eº = \dfrac{RT}{2F} * \ln \sqrt{K_{sp}}\] Plug in the value of Ksp and the constants to find the standard reduction potential: \[Eº = \dfrac{(8.314)(298)}{2(96485)} * \ln \sqrt{8.0 \times 10^{-28}}\] Calculate the value of Eº: \[Eº \approx -0.414 V\] So, the value of the standard reduction potential for the half-cell reaction is -0.414 V.

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Most popular questions from this chapter

A voltaic cell is constructed that is based on the following reaction: $$\mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q)$$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is \(1.00 \mathrm{M}\) and the cell generates an emf of \(+0.22 \mathrm{~V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? (b) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 \mathrm{M}\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{~K}\) : $$\mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s)$$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)

(a) A voltaic cell is constructed with all reactants and products in their standard states. Will this condition hold as the cell operates? Explain. (b) Can the Nernst equation be used at temperatures other than room temperature? Explain. (c) What happens to the emf of a cell if the concentrations of the products are increased?

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) have an atypical oxidation state.) (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow\) $$ \mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q) \text { (acidic solution) } $$ (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) $$ \mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q) \text { (acidic solution) } $$ (d) \(\mathrm{BrO}_{3}^{-}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{N}_{2}(g)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?
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