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Chapter 18: Problem 145

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}{\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)} & {\mathscr{E}^{\circ}=-0.440 \mathrm{V}} \\ {2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)} & {\mathscr{E}^{\circ}=0.000 \mathrm{V}}\end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, $P_{\mathrm{H}_{2}}=1.00 \mathrm{atm},$ and a weak acid, HA, at an initial concentration of 1.00 \(\mathrm{M}\) . If the observed cell potential is 0.333 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Short Answer

Expert verified
The equilibrium constant (\(K_a\)) for the weak acid HA, given the parameters of the galvanic cell and the observed cell potential, is approximately \(1.0 \times 10^{-8}\).

Step by step solution

01

Determine the redox reaction

The galvanic cell is based on the given half-reactions: $$\begin{array}{ll}{\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)} & {\mathscr{E}^{\circ}=-0.440 \mathrm{V}} \\\ {2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)} & {\mathscr{E}^{\circ}=0.000 \mathrm{V}}\end{array}$$ Combine the half-reactions to obtain the overall redox reaction. $$\mathrm{Fe}^{2+} +2 \mathrm{H}^{+} \longrightarrow \mathrm{Fe}(s) + \mathrm{H}_{2}(g)$$
02

Calculate the standard cell potential.

Add the standard potential (\(\mathscr{E}^{\circ}\)) of the two half-cell reactions to calculate the standard cell potential. $$\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ}$$ $$\mathscr{E}_{cell}^{\circ}= 0.000 -(- 0.440)$$ $$\mathscr{E}_{cell}^{\circ}= 0.440 V$$
03

Use the Nernst equation

Apply the Nernst equation to find the relationship between cell potential, standard cell potential, and concentrations of the species involved: $$\mathscr{E}_{cell} = \mathscr{E}_{cell}^{\circ} - \frac{RT}{nF} \ln Q$$ Where: - \(\mathscr{E}_{cell}\) is the cell potential at non-standard conditions (0.333 V) - \(\mathscr{E}_{cell}^{\circ}\) is the standard cell potential (0.440 V) - R is the gas constant: 8.314 J/(mol·K) - T is the temperature in Kelvin: 25°C + 273.15 = 298.15 K - n is the number of electrons transferred in the reaction (2 electrons) - F is Faraday's constant: 96485 C/mol - Q is the reaction quotient Given the concentrations of the species involved: $$Q = \frac{[\mathrm{Fe}^{2+}][\mathrm{H}^{+}]^2}{1}$$
04

Calculate the reaction quotient Q

The concentrations of Fe2+ and H+ are given as: $$[\mathrm{Fe}^{2+}] = 1.00 \times 10^{-3} M$$ $$[\mathrm{H}^{+}] = x$$ The initial concentration of the weak acid, HA, is given as 1.00 M, and its dissociation is: $$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$$ Assuming x moles of HA have dissociated: $$[\mathrm{HA}] = 1.00 - x$$ The concentration of H+ contributed from the weak acid and the acidic solution are considered, so: $$Q= [\mathrm{Fe}^{2+}][\mathrm{H}^{+}]^2 = (1.00 \times 10^{-3})(x)^2$$
05

Calculate the concentration of H+

Substitute the values of cell potential and standard cell potential into the Nernst equation, and solve for the concentration of H+: $$0.333 = 0.440 - \frac{8.314 \times 298.15}{2 \times 96485} \ln (1.00 \times 10^{-3} \times x^2)$$ Solve this equation to find x (the concentration of H+). The value of x = 1.00 × 10^{-4} M.
06

Calculate the Ka value for the weak acid HA

Given the dissociation reaction of the weak acid HA and the concentration of H+, find the Ka value: $$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$$ $$[\mathrm{HA}] = 1.00 - x$$ $$[\mathrm{H}^{+}] = x$$ $$[\mathrm{A}^{-}] = x$$ $$Ka = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]} = \frac{(1.00 \times 10^{-4})(1.00 \times 10^{-4})}{1.00 - (1.00 \times 10^{-4})}$$ Calculate Ka: $$Ka \approx 1.0 \times 10^{-8}$$ Thus, the equilibrium constant (\(K_a\)) for the weak acid HA is approximately \(1.0 \times 10^{-8}\).

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Most popular questions from this chapter

Which of the following statement(s) is(are) true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow$ Al and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}\) , aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

When magnesium metal is added to a beaker of \(\mathrm{HCl}(\mathrm{aq})\), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when \(\mathrm{Mg}\) is added directly to the beaker of \(\mathrm{HCl}\)? How can you harness this reaction to do useful work?

Which of the following statements concerning corrosion is(are) true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{E}_{\text { meas }}=\mathscr{E}_{\text { ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text { ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that $\mathscr{E}_{\text { meas }}=0.480 \mathrm{V}$ a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the nncertainty in the measured potential is \(+1 \mathrm{mV}\) \(( \pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 \(\mathrm{K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83}$$ a. Calculate \(8^{\circ}\) and \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.
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