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Chapter 18: Problem 129

Gold is produced electrochemically from an aqueous solution of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) containing an excess of \(\mathrm{CN}^{-}\). Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of \(\mathrm{O}_{2}\) will be produced during the production of \(1.00 \mathrm{~mol}\) gold?

Short Answer

Expert verified
During the electrochemical production of 1.00 mol of gold, 0.50 mol of oxygen gas will be produced.

Step by step solution

01

Half-reactions

Write down the half-reactions occurring at the electrodes for this electrochemical process. At the cathode (reduction): Au(CN)₂⁻ + e⁻ → Au + 2CN⁻ At the anode (oxidation): H₂O → 2H⁺ + 1/2O₂ + 2e⁻
02

Balance the half-reactions

Multiply the half-reactions by the appropriate factors so that the number of electrons lost and gained are the same (the electron transfer is conserved). For the reduction half-reaction: 2(Au(CN)₂⁻ + e⁻ → Au + 2CN⁻) For the oxidation half-reaction: 4(H₂O → 2H⁺ + 1/2O₂ + 2e⁻) Now, the number of electrons transferred is balanced.
03

Write the balanced overall reaction

Combine the balanced half-reactions to write the balanced overall reaction. 2Au(CN)₂⁻ + 4H₂O → 2Au + 4H⁺ + 2O₂ + 4CN⁻
04

Calculate the amount of O₂ produced per mol of gold

Utilize stoichiometry to determine the amount of O₂ that will be produced along with 1.00 mol of gold. According to the balanced overall reaction, 2 moles of gold are produced along with 2 moles of O₂. Thus, for the production of 1 mole of gold, we can determine the moles of O₂ formed: \(1.00 \thinspace \text{mol Au} \times \frac{1 \thinspace\text{mol O}_2}{2 \thinspace\text{mol Au}} = 0.50 \thinspace\text{mol O}_2\)
05

Conclusion

During the electrochemical production of 1.00 mol of gold, 0.50 mol of oxygen gas will be produced.

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Most popular questions from this chapter

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{~V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{cell}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{B}_{\text {coll }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=\) \(1.0 \times 10^{-5} M\)

It takes \(15 \mathrm{kWh}\) (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the HallHeroult process. Compare this to the amount of energy necessary to melt \(1.0 \mathrm{~kg}\) aluminum metal. Why is it economically feasible to recycle aluminum cans?

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

Consider the following half-reactions: $$\begin{aligned}\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & \mathscr{E}^{\circ}=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{C}^{\circ}=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{C}^{\circ}=0.96 \mathrm{~V}\end{aligned}$$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.
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