Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 114

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 \(1.00 \mathrm{~mole}\) of gold?

Short Answer

Expert verified
In the balanced equation, 2 moles of \(\mathrm{Au}\) are produced for every 1 mole of \(\mathrm{O}_{2}\). When 1.00 mole of gold is produced, \(\frac{1}{2}\) mole of \(\mathrm{O}_{2}\) will be produced. So, \(0.50\ \text{mole}\) of oxygen gas will be produced during the production of 1.00 mole of gold.

Step by step solution

01

Determine the half-reactions and overall reaction

First, we need to write down the half-reactions taking place at the electrodes. We know that gold is produced from \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) and oxygen gas is produced. At the anode (oxidation): \[\mathrm{CN}^{-}(aq) \rightarrow \frac{1}{2}\mathrm{O}_{2}(g) + \mathrm{CNO}^{-}(aq) + e^{-}\] At the cathode (reduction): \[\mathrm{Au}(\mathrm{CN})_{2}^{-}(aq) + e^{-} \rightarrow \mathrm{Au}(s) + 2\mathrm{CN}^{-}(aq)\] Now, we can write down the overall reaction by combining the half-reactions: \[\mathrm{CN}^{-}(aq) + \mathrm{Au}(\mathrm{CN})_{2}^{-}(aq) \rightarrow \frac{1}{2}\mathrm{O}_{2}(g) + \mathrm{CNO}^{-}(aq) + \mathrm{Au}(s) + 2\mathrm{CN}^{-}(aq)\]
02

Balance the half-reactions and determine stoichiometric coefficients

We now balance the half-reactions by adjusting the number of electrons transferred and equalizing the charges. We want to eliminate the electrons so that there is no charge in the balanced reaction: \[\mathrm{CN}^{-}(aq) + 2\mathrm{Au}(\mathrm{CN})_{2}^{-}(aq) \rightarrow \mathrm{O}_{2}(g) + 2\mathrm{CNO}^{-}(aq) + 2\mathrm{Au}(s) + 4\mathrm{CN}^{-}(aq)\] Given this balanced equation, we see that 2 moles of \(\mathrm{Au}\) are produced for every 1 mole of \(\mathrm{O}_{2}\).
03

Use stoichiometry to find the amount of \(\mathrm{O}_{2}\) produced

Now, we can use the stoichiometry from the balanced equation to find the amount of oxygen gas produced when 1 mole of gold is produced: \[\frac{1\ \text{mole of}\ \mathrm{O}_{2}}{2\ \text{moles of}\ \mathrm{Au}} \times 1.00\ \text{mole of}\ \mathrm{Au} = \frac{1}{2} \text{mole of}\ \mathrm{O}_{2}\] Therefore, \(0.50 \ \text{mole}\) of \(\mathrm{O}_{2}\) will be produced during the production of 1.00 mole of gold.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume \(\mathrm{Al}\) is oxidized.)

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. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

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 \(\mathscr{E}^{\circ}\) 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.

It took \(2.30\) min using a current of \(2.00 \mathrm{~A}\) to plate out all the silver from \(0.250 \mathrm{~L}\) of a solution containing \(\mathrm{Ag}^{+}\). What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free