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Chapter 20: Problem 72

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\(M\) 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 \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

Short Answer

Expert verified
The concentration of Pb²⁺ in the anode half-cell is approximately 0.015 M, and the Ksp of PbSO₄ is 0.015.

Step by step solution

01

Identify the two half-reactions

First, let's identify the two half-reactions occurring in the voltaic cell: 1. Cathode (reduction): Sn2+(aq) + 2e- → Sn(s) 2. Anode (oxidation): Pb(s) → Pb2+(aq) + 2e-
02

Look up the standard reduction potentials

Now, we need to look up the standard reduction potentials for Sn2+(aq) and Pb(s). From standard reduction potential tables, we find: Eº(Sn2+/Sn) = -0.14 V Eº(Pb2+/Pb) = -0.13 V
03

Calculate the overall cell potential

Using the Nernst equation, we can calculate the cell potential: Ecell = Eº(cell) + \(\frac{RT}{nF}\) * ln(Q) We know that Ecell is given as +0.22 V. The number of electrons transferred in the reaction, n, is 2. The reaction quotient Q is \(\frac{[Pb^{2+}]}{[Sn^{2+}]}\). We are also given that the concentration of Sn2+ is 1.00 M. Eº(cell) = Eº(cathode) - Eº(anode) = (-0.14 V) - (-0.13 V) = -0.01 V Now, we can plug in the values and solve for [Pb2+].
04

Solve for the concentration of Pb2+

Plugging in the values in the Nernst equation: 0.22 V = -0.01 V + \(\frac{(RT)(2.303)}{2F}\) * log\(\frac{[Pb^{2+}]}{1}\) 0.23 V = \(\frac{(RT)(2.303)}{2F}\) * log([Pb2+]) Now, we can solve for [Pb2+]: log([Pb2+]) = \(\frac{0.23(2F)}{(2.303)(RT)}\) [Pb2+] = 10^(\(\frac{0.23(2F)}{(2.303)(RT)}\)) ≈ 0.015 M
05

Find the Ksp value of PbSO4

Using the information provided in part (b), we know that the concentration of SO42- is 1.00 M in equilibrium with PbSO4(s). The Ksp expression for PbSO4 is: Ksp = [Pb2+][SO42-] Using the concentration of [Pb2+] we found in step 4 and the given [SO42-]: Ksp = (0.015)(1.00) = 0.015 Therefore, the Ksp of PbSO4 is 0.015.

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Most popular questions from this chapter

During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from Zn to \(\mathrm{MnO}_{2} ?\)

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 \(\left[\mathrm{Ni}^{2+}\right]=3.00 M\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?(\mathbf{c})\) 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) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metal hydride batteries over nickel-cadmium batteries?

A common shorthand way to represent a voltaic cell is $$ \text {anode} | \text {anode solution} | | \text {cathode solution} | \text {cathode} $$ A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fel Fe \(^{2+} \| \operatorname{Ag}^{+} | A g;\) calculate the standard cell emf using data in Appendix E. (b) Write the half-reactions and overall cell reaction represented by Zn \(\left|Z \mathrm{n}^{2+}\right| \mathrm{H}^{+} | \mathrm{H}_{2} ;\) calculate the standard cell emf using data in Appendix E and use Pt for the hydrogen electrode. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \\ \longrightarrow & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the ClO \(_{3}^{-}\) and \(\mathrm{Cl}^{-} .\) Calculate the standard cell emf given: \(\mathrm{ClO}_{3}^{-}(a q)+\) \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l); E^{\circ}=1.45 \mathrm{V}\).

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent?
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