A reaction cycle for an element is a series of reactions beginning and ending with that element. In the following copper reaction cycle, copper has either a 0 or a +2 oxidation state. Write balanced molecular and net ionic equations for each step. (1) Copper metal reacts with aqueous bromine to produce a greenblue solution. (2) Adding aqueous sodium hydroxide forms a blue precipitate. (3) The precipitate is heated and turns black (water is released). (4) The black solid dissolves in nitric acid to give a blue solution. (5) Adding aqueous sodium phosphate forms a green precipitate. (6) The precipitate forms a blue solution in sulfuric acid. (7) Copper metal is recovered from the blue solution when zinc metal is added.

Step by step solution

01

Copper Metal Reacts with Aqueous Bromine

First, write the molecular and net ionic equation for the reaction between copper metal and aqueous bromine. Molecular Equation: \[ Cu(s) + Br_2(aq) \rightarrow CuBr_2(aq) \] Net Ionic Equation: \[ Cu(s) + Br_2(aq) \rightarrow Cu^{2+}(aq) + 2Br^-(aq) \]

02

Adding Aqueous Sodium Hydroxide

Next, add sodium hydroxide to form a blue precipitate. Molecular Equation: \[ CuBr_2(aq) + 2NaOH(aq) \rightarrow Cu(OH)_2(s) + 2NaBr(aq) \] Net Ionic Equation: \[ Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s) \]

03

Heating the Precipitate

Then, heat the blue precipitate to form a black solid, releasing water. Molecular Equation: \[ Cu(OH)_2(s) \rightarrow CuO(s) + H_2O(l) \] Net Ionic Equation: \[ Cu(OH)_2(s) \rightarrow CuO(s) + H_2O(l) \]

04

Dissolving the Black Solid in Nitric Acid

Dissolve the black solid in nitric acid to yield a blue solution. Molecular Equation: \[ CuO(s) + 2HNO_3(aq) \rightarrow Cu(NO_3)_2(aq) + H_2O(l) \] Net Ionic Equation: \[ CuO(s) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + H_2O(l) \]

05

Adding Aqueous Sodium Phosphate

Now, add sodium phosphate to form a green precipitate. Molecular Equation: \[ 3Cu(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Cu_3(PO_4)_2(s) + 6NaNO_3(aq) \] Net Ionic Equation: \[ 3Cu^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Cu_3(PO_4)_2(s) \]

06

Dissolving the Precipitate in Sulfuric Acid

Dissolve the green precipitate in sulfuric acid to form a blue solution. Molecular Equation: \[ Cu_3(PO_4)_2(s) + 3H_2SO_4(aq) \rightarrow 3CuSO_4(aq) + 2H_3PO_4(aq) \] Net Ionic Equation: \[ Cu_3(PO_4)_2(s) + 6H^+(aq) \rightarrow 3Cu^{2+}(aq) + 2H_3PO_4(aq) \]

07

Recovering Copper Metal using Zinc

Finally, add zinc metal to recover copper from the blue solution. Molecular Equation: \[ CuSO_4(aq) + Zn(s) \rightarrow ZnSO_4(aq) + Cu(s) \] Net Ionic Equation: \[ Cu^{2+}(aq) + Zn(s) \rightarrow Zn^{2+}(aq) + Cu(s) \]

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States

Every element in a chemical reaction has a specific oxidation state, which helps in tracking electrons. Copper (Cu) within the given reaction cycle changes its oxidation state between 0 (in metallic form) and +2 (in ionic form). This means copper atoms either lose or gain electrons, depending on the reaction. An increase in the oxidation state implies oxidation (loss of electrons), while a decrease indicates reduction (gain of electrons). This back and forth is crucial in understanding redox reactions. For instance, in the first step, copper (Cu) with an oxidation state of 0 reacts with bromine (Br\(_2\)), resulting in copper(II) bromide (CuBr\(_2\)) where copper has an oxidation state of +2.

Net Ionic Equations

Net ionic equations simplify chemical reactions by showing only the species that actually participate in the chemical change. They exclude spectator ions that do not participate. For example, in the reaction where copper reacts with bromine: the molecular equation is: Cu(s) + Br\(_2\)(aq) \(\rightarrow\) CuBr\(_2\)(aq)
The net ionic equation simplifies this to focus on the actual ionic changes: Cu(s) + Br\(_2\)(aq) \(\rightarrow\) Cu\(^{2+}\)(aq) + 2Br\(^-\)(aq)
This focuses solely on copper and bromine's transformation, cutting out aqueous bromide ions that do not change during the reaction. Simplifying reactions in this way makes it easier to spot redox changes and understand the core chemistry.

Molecular Equations

Molecular equations represent all reactants and products in their undissociated forms. These equations are beneficial for recognizing the complete formula of compounds involved in the reactions. Throughout our copper cycle, each step's molecular equation presents a clear view of the initial compounds and the resulting products. For example, when copper is recovered using zinc metal, the molecular equation is: CuSO\(_4\)(aq) + Zn(s) \(\rightarrow\) ZnSO\(_4\)(aq) + Cu(s)
This helps to understand the whole process as it includes every single part of the reaction, even when some components do not change.

Precipitate Formation

A precipitate forms when a solid product emerges from a solution during a chemical reaction. These solids are usually insoluble in the given solvent and fall out of the solution as a solid mass. One key example in our copper reaction cycle is the formation of the blue precipitate, copper(II) hydroxide, from reacting sodium hydroxide with copper(II) bromide. The reaction is as follows: CuBr\(_2\)(aq) + 2NaOH(aq) \(\rightarrow\) Cu(OH)\(_2\)(s) + 2NaBr(aq)
Here, copper(II) ions from copper(II) bromide and hydroxide ions from sodium hydroxide come together to form insoluble copper(II) hydroxide, resulting in the blue solid. This illustrates how insolubility affects the reaction paths and results in precipitate formation.