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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 \(\mathrm{PbSO}_{4} ?\)

Short Answer

Expert verified
(a) The concentration of Pb虏鈦 in the anode half-cell is approximately 0.0717 M. (b) The Ksp of PbSO鈧 is approximately 0.0717.

Step by step solution

01

(Step 1: Identify the half-reactions for the voltaic cell)

In order to use the Nernst equation, we must first identify the half reactions for the voltaic cell. For this cell, the half-reactions can be determined as: Anode: Pb(s) -> Pb2+(aq) + 2e- Cathode: Sn2+(aq) + 2e- -> Sn(s) Now that we have the half-reactions, we can proceed to the next step.
02

(Step 2: Determine the standard reduction potentials)

By looking up the values in a table of reduction potentials, we can find: E掳(Pb2+/Pb) = -0.126 V E掳(Sn2+/Sn) = -0.136 V
03

(Step 3: Calculate the standard cell potential)

We can find the standard cell potential by adding the reduction potentials for the anode and cathode: E掳(cell) = E掳(cathode) - E掳(anode) = -0.136 - (-0.126) V = -0.010 V
04

(Step 4: Apply the Nernst Equation to find the Pb虏鈦 concentration)

The Nernst equation relates the cell potential, standard cell potential, concentrations, and temperature: E(cell) = E掳(cell) - \(\frac{RT}{nF}\) lnQ Where R is the gas constant, T is the temperature, n is the number of electrons transferred, F is the Faraday's constant, and Q is the reaction quotient. In this case, E(cell) = 0.22 V, E掳(cell) = -0.010 V, n = 2 (as per the half-reactions), and Q = \(\frac{[Pb^{2+}]}{[Sn^{2+}]}\). Assuming the temperature is 298 K, we can plug the values into the equation: 0.22 V = -0.010 V - \(\frac{8.314\ \text{J}\ (\text{mol} \ \text{K})^{-1}\times 298\ \text{K}}{2\times 96485\ \text{C}\ \text{mol}^{-1}} \) ln\(\frac{[Pb^{2+}]}{1.00}\) Solve for [Pb虏鈦篯: [Pb虏鈦篯 鈮 0.0717 M
05

(Step 5: Identify the Ksp expression for PbSO鈧)

For PbSO鈧 dissolved in water, the equilibrium reaction can be written as: PbSO4(s) -鈭掆垝> Pb虏鈦 (aq) + SO4虏鈦 (aq) The Ksp expression for PbSO鈧 is: Ksp = [Pb虏鈦篯 [SO鈧劼测伝]
06

(Step 6: Calculate the Ksp for PbSO4)

Using the determined concentration of Pb虏鈦 (0.0717 M) and the provided concentration of SO鈧劼测伝 (1.00 M), we can calculate the Ksp value for PbSO鈧: Ksp = (0.0717)(1.00) = 0.0717 Therefore, the Ksp of PbSO鈧 is approximately 0.0717.

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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 an essential tool used in electrochemistry to determine the cell potential of a voltaic cell under non-standard conditions. It accounts for changes in concentration or temperature, giving us the actual cell potential rather than just the standard potential.

The equation is expressed as follows: \[ E_{cell} = E^{掳}_{cell} - \frac{RT}{nF} \ln Q \]Here, \( E_{cell} \) is the cell potential at specific conditions, \( E^{掳}_{cell} \) is the standard cell potential, \( R \) is the universal gas constant \((8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1})\), \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is Faraday鈥檚 constant \((96485 \, \text{C} \, \text{mol}^{-1})\), and \( Q \) is the reaction quotient.

This equation helps chemists understand how changes in conditions like concentration can influence the electromotive force (emf) generated by the cell. Tuning these conditions can optimize or adjust the performance of electrochemical processes or devices.
Standard Reduction Potential
Standard reduction potentials are key in evaluating the tendency of a chemical species to acquire electrons and undergo reduction. They are measured under standard conditions (1 M concentration, 1 atm pressure, and 25掳C). These potentials are often tabulated and allow for comparison of different reactions.

Each half-equation in a redox reaction has a standard potential, denoted as \( E^{掳} \). For example, the half-reaction for lead \( \text{Pb}^{2+}/\text{Pb} \) has a standard reduction potential \( E^{掳} = -0.126 \text{ V} \). Comparing these values for two full reactions in a voltaic cell allows us to calculate the standard cell potential \( E^{掳}_{cell} \).

To find \( E^{掳}_{cell} \), we subtract the anode's potential from the cathode's potential. If the value is positive, the reaction is spontaneous, making it suitable for harnessing electricity, as seen in voltaic cells. This is central to determining the feasibility and direction of redox reactions.
Reaction Quotient
In electrochemistry, the reaction quotient \( Q \) is a dimensionless number that helps quantify the ratio of concentrations or pressures of products to reactants at any point other than equilibrium. Its expression resembles that of the equilibrium constant \( K \) but applies to a system that has not reached equilibrium.

For the studied voltaic cell system, \( Q \) is expressed as:\[ Q = \frac{[\text{Pb}^{2+}]}{[\text{Sn}^{2+}]} \]where square brackets denote concentration.

By plugging \( Q \) into the Nernst equation, one can evaluate how far a reaction is from equilibrium and quantify its effect on cell potential.
  • If \( Q < K \), the forward reaction is favored, and the cell potential is higher than \( E^{掳}_{cell} \).
  • If \( Q = K \), the reaction is at equilibrium, and the cell potential equals zero (\( E_{cell} = 0 \)).
  • If \( Q > K \), the reverse reaction is favored, dropping the cell potential below \( E^{掳}_{cell} \).
Understanding \( Q \) is crucial for predicting how changes in conditions affect reaction spontaneity and cell potential.
Solubility Product (Ksp)
The solubility product constant, \( K_{sp} \), describes the solubility of sparingly soluble ionic compounds in solution. It characterizes a specific type of equilibrium constant that pertains to the dissolution of solids.

It is derived from the equilibrium reaction for a soluble salt dissociating in water. For lead sulfate \( \text{PbSO}_{4} \), the dissociation is:\[ \text{PbSO}_{4} (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{SO}_{4}^{2-} (aq) \]\( K_{sp} \) is then given by:\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_{4}^{2-}] \]This expression reflects the concentration of the ions that result when the compound dissolves.

Calculating \( K_{sp} \) enables chemists to predict how much of a solute can dissolve in a solvent before reaching saturation. This parameter is often vital in contexts such as pharmaceuticals, environmental science, and materials engineering where solubility influences effectiveness, contamination levels, or material properties.

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Most popular questions from this chapter

A plumber's handbook states that you should not connect a brass pipe directly to a galvanized steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use instead an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for \(2.00 \mathrm{~min}\), \(12.3 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(25.5^{\circ} \mathrm{C},\) and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

A cell has a standard cell potential of \(+0.177 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction (a) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. $$ \begin{array}{l} \text { (a) } \mathrm{PBr}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HBr}(a q) \\ \text { (b) } \mathrm{NaI}(a q)+3 \mathrm{HOCl}(a q) \longrightarrow \mathrm{NaIO}_{3}(a q)+3 \mathrm{HCl}(a q) \\ \text { (c) } 3 \mathrm{SO}_{2}(g)+2 \mathrm{HNO}_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ 3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NO}(g) \\ \text { (d) } 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \\ \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper buses that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 hr?

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