   # The overall reaction for the production of Cu(OH) 2 from Cu in oxygenated water can be broken into three steps: an oxidation half-reaction, a reduction half-reaction, and a precipitation reaction. a. Complete and balance the two missing half-reactions to give the overall equation for the oxidation of cooper in seawater. Oxidation half-reaction: ? Reduction half-reaction: ? Precipitation: Cu 2+ ( a q ) + 2OH − (aq) → Cu(OH) 2 ( s ) Overall: Cu(s ) + 1 2 O 2 (g) + H 2 O( l ) → Cu(OH) 2 ( s ) b. Determine the equilibrium constant for the overall reaction at 25 °C using standard reduction potentials and the solubility product constant ( K sp ) of Cu(OH) 2 (s). ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19.8, Problem 4Q
Textbook Problem
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## The overall reaction for the production of Cu(OH)2 from Cu in oxygenated water can be broken into three steps: an oxidation half-reaction, a reduction half-reaction, and a precipitation reaction. a. Complete and balance the two missing half-reactions to give the overall equation for the oxidation of cooper in seawater. Oxidation half-reaction: ?Reduction half-reaction: ? Precipitation: Cu 2+ ( a q )  + 2OH − (aq)  → Cu(OH) 2 ( s ) Overall: Cu(s )  + 1 2  O 2 (g)  + H 2 O( l ) → Cu(OH) 2 ( s ) b. Determine the equilibrium constant for the overall reaction at 25 °C using standard reduction potentials and the solubility product constant (Ksp) of Cu(OH)2(s).

a.

Interpretation Introduction

Interpretation:

The overall reaction for the production of Cu(OH)2 from Cu in oxygenated water can be broken into three steps. An oxidation half reaction, a reduction half reaction and precipitation reaction.

The two missing half –reactions to give the overall equation for the oxidation of copper in seawater has to be completed and balanced.

The given precipitation and overall reaction is as follows.

Precipitation: Cu2+(aq) + 2OH-(aq)  Cu(OH)2(s)Overall : Cu(s) + 12O2(g) + H2O(l)   Cu(OH)2(s)

Concept introduction:

Electrolysis:

It is a decomposition of ionic compounds by passing electricity through molten compounds or aqueous solutions of compounds.

Electricity used to produce chemical changes. The apparatus used for electrolysis is called an electrolytic cell.

Generally, oxidation occurs at anode and reduction occurs a cathode.

### Explanation of Solution

The given precipitation and redution half reactions are as follows.

Precipitation: Cu2+(aq) + 2OH-(aq)  Cu(OH)2(s)Overall : Cu(s) + 12O2(g) + H2O(l)   Cu(OH)2(s)

Let’s write the half reactions are as follows.

Oxidation half reaction:Cu(s)  Cu2+(aq) + 2e-

Reduction half reaction:12O2(g) + H2O(l) + 2e-  2OH-

b.

Interpretation Introduction

Interpretation:

The overall reaction for the production of Cu(OH)2 from Cu in oxygenated water can be broken into three steps. An oxidation half reaction, a reduction half reaction and precipitation reaction.

The equilibrium constant for the overall reaction at 250 C using standard reduction potentials and the solubility product constant (Ksp) of Cu(OH)2(s) has to be determined.

Concept introduction:

Electrolysis:

It is a decomposition of ionic compounds by passing electricity through molten compounds or aqueous solutions of compounds.

Electricity used to produce chemical changes. The apparatus used for electrolysis is called an electrolytic cell.

Generally, oxidation occurs at anode and reduction occurs a cathode.

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