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Chapter 18: Problem 72

Calculate the \(\mathrm{pH}\) of the cathode compartment for the following reaction given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\) when \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), and \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow\) \(2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
Given the provided information and by following the steps, we can calculate the pH of the cathode compartment using the Nernst equation, half-reactions, and cell potentials. First, find the standard reduction and oxidation potentials (\(E_{\text{red}}^{\circ}\) and \(E_{\text{ox}}^{\circ}\)) from a reference table. Then, substitute the given values for the concentrations of Chromium (III), Aluminum (III), and Dichromate ions in the equation from Step 4. Solve for the H⁺ concentration, \([\text{H}^{+}]\). Finally, use the pH formula (\(pH = -\log_{10} [\text{H}^{+}]\)) to calculate the pH of the cathode compartment.

Step by step solution

01

Identify the Half-Reactions

The overall redox reaction is given as: \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\) The half-reactions are: Reduction: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2}\mathrm{O}(l)\) Oxidation: \(2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+6 \mathrm{e}^{-}\)
02

Write the Nernst Equation for the Reduction Half-Reaction

The Nernst equation relates the half-cell potentials to the concentrations of the species involved in the reaction. For the reduction half-reaction, the Nernst equation is: \(E_{\text{red}}=E_{\text{red}}^{\circ}-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{[\text{Cr}^{3+}]^{2}}{[\text{Cr}_{2}\text{O}_{7}^{2-}][\text{H}^{+}]^{14}}\right)\)
03

Write the Nernst Equation for the Oxidation Half-Reaction

For the oxidation half-reaction, the Nernst equation is given by: \(E_{\text{ox}}=E_{\text{ox}}^{\circ}-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{1}{[\text{Al}^{3+}]^{2}}\right)\)
04

Use the Given Overall Cell Potential to Relate the Half-Cell Potentials

The overall cell potential can be found by subtracting the oxidation half-cell potential from the reduction half-cell potential: \(\mathscr{E}_{\text {cell }} = E_{\text{red}} - E_{\text{ox}}\) Substitute the Nernst equations for \(E_{\text{red}}\) and \(E_{\text{ox}}\): \(\mathscr{E}_{\text {cell}} =[E_{\text{red}}^\circ-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{[\text{Cr}^{3+}]^{2}}{[\text{Cr}_{2}\text{O}_{7}^{2-}][\text{H}^{+}]^{14}}\right)] - [E_{\text{ox}}^\circ-\dfrac{0.05916}{6} \log_{10}\left(\dfrac{1}{[\text{Al}^{3+}]^{2}}\right)]\)
05

Solve for the H⁺ Concentration

Given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\), \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\), we will need to find the standard reduction and oxidation potentials (\(E_{\text{red}}^{\circ}\) and \(E_{\text{ox}}^{\circ}\)) from a reference table. Then, substitute the given values in the equation from Step 4 and solve for the H⁺ concentration, \([\text{H}^{+}]\).
06

Calculate the pH

Once the H⁺ concentration is found, use the pH formula to calculate the pH: \(pH = -\log_{10} [\text{H}^{+}]\) Substitute the H⁺ concentration into the formula and find the pH of the cathode compartment.

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Most popular questions from this chapter

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 \mathrm{M}\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)

The equation \(\Delta G^{\circ}=-n F \mathscr{E}^{\circ}\) also can be applied to halfreactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ}\right.\) for \(\left.\mathrm{e}^{-}=0 .\right)\)

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu} $$ The mass of each electrode is \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?

Given the following two standard reduction potentials, $$ \begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{~V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{~V} \end{array} $$ solve for the standard reduction potential of the half-reaction $$ \mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)
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