Principles of Oxidation-Reduction Reactions - AP Chemistry

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Question

Oomycetes of the genus Phytophthora are capable of causing economic damage by destroying crops. The name Phytopthora from the Greek phyton = plant, and phthora = destruction literally means plant-destroyer. However, chemists may battle their destructive effects.

The compound named copper(II) carbonate can be used to destroy the plant-destroyer, and can be produced as shown in the following equation:

$CuSO_4_{(aq)} + Na_2CO_3_{(aq)} \rightarrow CuCO_3_{(s)} + Na_2SO_4_{(aq)}$

What is the net ionic equation for the given reaction that produces copper(II) carbonate?

Answer

An ionic equation is a chemical equation in which ionic compounds in an aqueous solution are written as dissociated ions.

A complete ionic equation shows all the ions present in the solution (which includes both the reactive and spectator ions), as the dissociated ions. The complete ionic equation for this problem's reaction is:

$2Na^+_{(aq)}+CO_3^{2-}_{(aq)}+Cu^{2+}_{(aq)}+SO_4^{2-}_{(aq)}\rightarrow2Na^+_{(aq)}+SO_4^{2-}_{(aq)}+CuCO_3_{(s)}$

It is worth noting that in a complete ionic equation, it becomes readily apparent which ions are involved in the oxidation-reduction reaction, and which ions are the spectator ions (those that do not change during the reaction). For example since $2Na^+_{(aq)}$ and $SO_4^{2-}_{(aq)}$ are present on both sides of the reaction, they are spectator ions that go through the reaction unchanged. Also because the $Cu^{2+}_{(aq)}$ and $CO_3^{2-}_{(aq)}$ are only present as dissolved ions on the reactants side of the equation and form the solid precipitate $CuCO_3_{(s)}$ on the product side, they are the reactive species in this chemical reaction.

The net ionic equation shows only the reactive ions in solution and the product(s) formed by them. Therefore a net ionic equation can most simply be made by looking at the complete ionic equation and omitting the spectator ions as shown below:

Net ionic equation:

$CO_3^{^{^{2-}}}_{(aq)} + Cu^{2+}_{(aq)}\rightarrow CuCO_{3}_{(s)}$

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Question

In the following reaction, which compound is being oxidized?

$3CuS+8HNO_3\rightarrow3CuSO_4+8NO_{(g)}+4H_2O$

Answer

Hydrogen doesn't change. Cu2+ doesn't change (partnered with S2- then with SO42-). Sulfur goes from S2- and S6+(paired with 6 O2- with a 2– charge), showing an oxidation. Nitrogen goes from N5+ to N2+ meaning it was reduced.

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Question

Which of the following reactions involves the transfer of electrons?

Answer

Redox reactions occur when one species is oxidized—loses electrons—and another is reduced—gains electrons. Thus, this is the only one of the given options that involves the transfer of electrons.

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Question

Consider the following reaction:

$Cu_{(s)}+2Ag^{+}{(aq)}\rightarrow Cu^{2+}{(aq)}+2Ag_{(s)}$

What is the oxidizing agent, and what is the reducing agent?

Answer

Let's break down the reaction into two separate reactions:

$Cu_{(s)}\rightarrow Cu^{2+}_{(aq)}+2e^-$ and

$2Ag^+{(aq)}+2e^- \rightarrow 2Ag{(s)}$

We can see that copper loses electrons, while silver gains electrons. Recall that oxidation is loss and reduction is gain, with regard to electrons. Copper is oxidized and silver is reduced.

However, this question asks for the oxidizing agent and reducing agent. Recall that the oxidizing agent is reduced, while the reducing agent is oxidized. Since copper is oxidized, it is the reducing agent. Similarly, since silver is reduced, it is the oxidizing agent.

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Question

Which statement about this reaction is FALSE?

$Cr_2O_3(s)+2Al(s)\rightarrow 2Cr(s)+Al_2O_3$

Answer

In Cr2O3, Cr has a charge of +3. The chromium metal product is neutral, and therefore has a charge of 0. So, Cr is reduced from +3 to 0.

Remember, OIL RIG. Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons).

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Question

Methane combusts in the presence of oxygen according to the following reaction:

$CH_4 + 2O_2 \to CO_2 + 2H_2O$

Which of the following statements is true concerning the reaction?

Answer

By comparing the oxidation number of an atom as a reactant and its oxidation number as a product, we can determine if the atom has been oxidized or reduced. In increase in oxidation number indicates a loss of electrons, or oxidation. A decrease in oxidation number signals a gain of electrons, or reduction.

For electrochemistry, you should familiarize yourself with the traditional oxidation states of hydrogen $\small (+1)$ , halogens $\small (-1)$ , oxygen $\small (-2)$ , and elemental atoms $\small (0)$ .

Carbon is initially in the form of methane, meaning that it is attached to four hydrogen atoms. The molecule is neutral, and each hydrogen has an oxidation number of $\small +1$ . Carbon must have an initial oxidation state of $\small -4$ in order to balance the molecular charge.

In the a product, carbon is attached to two oxygens, each with a charge of $\small -2$ . Again, the molecule is neutral, so carbon must balance these charges. This means that carbon's final oxidation state is $\small +4$ .

Since carbon went from an oxidation state of $\small -4$ to $\small +4$ , we can conclude that carbon has been oxidized in the reaction.

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Question

Consider the following reaction:

$Zn_(_s_) + CuSO_4_(_a_q_) \to Cu_(_s_) + ZnSO_4_(_a_q_)$

Which of the following statements is true?

Answer

For any oxidation-reduction reaction, the oxidized atom will lose electrons and the reduced atom will gain electrons. Zinc is originally in elemental form, so it has an initial oxidation state of zero. As a product, it has a charge of $\small +2$ . This means that zinc has been oxidized in the reaction; its oxidation number increased due to a loss of electrons.

Copper originally has an oxidation state of $\small +2$ , but changes to its elemental form as a product. This gives copper a final oxidation state of zero. Since it has gained electrons, copper has been reduced.

Sulfur remains in the sulfate complex ion, which retains its overall $\small -2$ charge throughout the reaction. It is neither oxidized, nor reduced.

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Question

Consider the following reaction:

$Zn_(_s_) + CuSO_4_(_a_q_) \to Cu_(_s_) + ZnSO_4_(_a_q_)$

Which of the following substances is the oxidizing agent?

Answer

In an oxidation-reduction reaction, the oxidizing agent is the reactant that accepts electrons, becoming reduced. The oxidizing or reducing agent is always the whole reagent, and is never just the atom in the reagent that is being oxidized or reduced.

is the reagent that accepts two electrons from elemental zinc. This allows zinc to become oxidized, changing oxidation state from zero to $\small +2$ . Copper accepts these electrons, changing from an oxidation state of $\small +2$ to zero. As a result, we say that is the oxidizing agent for the reaction.

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Question

$Ca(s) + 2HCl (aq) \rightarrow CaCl_{2} (aq) + H_{2} (g)$

In this reaction, electrons are transferred from _ to _.

Answer

A transfer of electrons indicates an oxidation-reduction reaction. The element losing electrons is oxidized, while the element gaining electrons is reduced.

Calcium is initially a neutral solid, but as a product has a charge of $\small +2$ . Calcium's charge increased, meaning it must have lost electrons.

Hydrogen initially has a charge of $\small +1$ , but as a product is a neutral diatomic gas. Hydrogen's charge decreased, meaning it must have gained electrons.

We can thus conclude that calcium was oxidized and hydrogen was reduced, indicating that electrons were transferred from calcium to hydrogen.

The charge on chlorine remains $\small -1$ throughout the entire reaction.

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Question

What does oxidation mean?

Answer

Think "OIL RIG" Oxidation Is Loss of electrons, Reduction Is Gain of electrons. The loss and gain of electrons is what changes the charge of the atom or molecule. In redox reactions, electrons are transferred. Changing the number of protons changes the identity of the element ,which is not the case. Since chemical reactions involve valence electrons, the nucleus is unchanged.

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Question

$2MnO_{4}^{-} + 5SO_{3}^{2-} + 6H^{+}\rightarrow 2Mn^{2+} + 5SO_{4}^{2-} + 3H_{2}O$

Which statement about the reaction equation is correct?

Answer

$SO_{3}^{2-}$ is converted to $SO_4^{-2}$ . Thus, an additional oxygen has been added to the original molecule, thus $SO_{3}^{2-}$ was oxidized. The substance that is oxidized is also called the reducing agent because it causes the reduction of the other molecule to which the oxygen originally was bound , which is the oxidizing agent.

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Question

Oomycetes of the genus Phytophthora are capable of causing economic damage by destroying crops. The name Phytopthora from the Greek phyton = plant, and phthora = destruction literally means plant-destroyer. However, chemists may battle their destructive effects.

The compound named copper(II) carbonate can be used to destroy the plant-destroyer, and can be produced as shown in the following equation:

$CuSO_4_{(aq)} + Na_2CO_3_{(aq)} \rightarrow CuCO_3_{(s)} + Na_2SO_4_{(aq)}$

What is the net ionic equation for the given reaction that produces copper(II) carbonate?

Answer

An ionic equation is a chemical equation in which ionic compounds in an aqueous solution are written as dissociated ions.

A complete ionic equation shows all the ions present in the solution (which includes both the reactive and spectator ions), as the dissociated ions. The complete ionic equation for this problem's reaction is:

$2Na^+_{(aq)}+CO_3^{2-}_{(aq)}+Cu^{2+}_{(aq)}+SO_4^{2-}_{(aq)}\rightarrow2Na^+_{(aq)}+SO_4^{2-}_{(aq)}+CuCO_3_{(s)}$

It is worth noting that in a complete ionic equation, it becomes readily apparent which ions are involved in the oxidation-reduction reaction, and which ions are the spectator ions (those that do not change during the reaction). For example since $2Na^+_{(aq)}$ and $SO_4^{2-}_{(aq)}$ are present on both sides of the reaction, they are spectator ions that go through the reaction unchanged. Also because the $Cu^{2+}_{(aq)}$ and $CO_3^{2-}_{(aq)}$ are only present as dissolved ions on the reactants side of the equation and form the solid precipitate $CuCO_3_{(s)}$ on the product side, they are the reactive species in this chemical reaction.

The net ionic equation shows only the reactive ions in solution and the product(s) formed by them. Therefore a net ionic equation can most simply be made by looking at the complete ionic equation and omitting the spectator ions as shown below:

Net ionic equation:

$CO_3^{^{^{2-}}}_{(aq)} + Cu^{2+}_{(aq)}\rightarrow CuCO_{3}_{(s)}$

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Question

In the following reaction, which compound is being oxidized?

$3CuS+8HNO_3\rightarrow3CuSO_4+8NO_{(g)}+4H_2O$

Answer

Hydrogen doesn't change. Cu2+ doesn't change (partnered with S2- then with SO42-). Sulfur goes from S2- and S6+(paired with 6 O2- with a 2– charge), showing an oxidation. Nitrogen goes from N5+ to N2+ meaning it was reduced.

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Question

Which of the following reactions involves the transfer of electrons?

Answer

Redox reactions occur when one species is oxidized—loses electrons—and another is reduced—gains electrons. Thus, this is the only one of the given options that involves the transfer of electrons.

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Question

Consider the following reaction:

$Cu_{(s)}+2Ag^{+}{(aq)}\rightarrow Cu^{2+}{(aq)}+2Ag_{(s)}$

What is the oxidizing agent, and what is the reducing agent?

Answer

Let's break down the reaction into two separate reactions:

$Cu_{(s)}\rightarrow Cu^{2+}_{(aq)}+2e^-$ and

$2Ag^+{(aq)}+2e^- \rightarrow 2Ag{(s)}$

We can see that copper loses electrons, while silver gains electrons. Recall that oxidation is loss and reduction is gain, with regard to electrons. Copper is oxidized and silver is reduced.

However, this question asks for the oxidizing agent and reducing agent. Recall that the oxidizing agent is reduced, while the reducing agent is oxidized. Since copper is oxidized, it is the reducing agent. Similarly, since silver is reduced, it is the oxidizing agent.

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Question

Which statement about this reaction is FALSE?

$Cr_2O_3(s)+2Al(s)\rightarrow 2Cr(s)+Al_2O_3$

Answer

In Cr2O3, Cr has a charge of +3. The chromium metal product is neutral, and therefore has a charge of 0. So, Cr is reduced from +3 to 0.

Remember, OIL RIG. Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons).

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Question

Methane combusts in the presence of oxygen according to the following reaction:

$CH_4 + 2O_2 \to CO_2 + 2H_2O$

Which of the following statements is true concerning the reaction?

Answer

By comparing the oxidation number of an atom as a reactant and its oxidation number as a product, we can determine if the atom has been oxidized or reduced. In increase in oxidation number indicates a loss of electrons, or oxidation. A decrease in oxidation number signals a gain of electrons, or reduction.

For electrochemistry, you should familiarize yourself with the traditional oxidation states of hydrogen $\small (+1)$ , halogens $\small (-1)$ , oxygen $\small (-2)$ , and elemental atoms $\small (0)$ .

Carbon is initially in the form of methane, meaning that it is attached to four hydrogen atoms. The molecule is neutral, and each hydrogen has an oxidation number of $\small +1$ . Carbon must have an initial oxidation state of $\small -4$ in order to balance the molecular charge.

In the a product, carbon is attached to two oxygens, each with a charge of $\small -2$ . Again, the molecule is neutral, so carbon must balance these charges. This means that carbon's final oxidation state is $\small +4$ .

Since carbon went from an oxidation state of $\small -4$ to $\small +4$ , we can conclude that carbon has been oxidized in the reaction.

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Question

Consider the following reaction:

$Zn_(_s_) + CuSO_4_(_a_q_) \to Cu_(_s_) + ZnSO_4_(_a_q_)$

Which of the following statements is true?

Answer

For any oxidation-reduction reaction, the oxidized atom will lose electrons and the reduced atom will gain electrons. Zinc is originally in elemental form, so it has an initial oxidation state of zero. As a product, it has a charge of $\small +2$ . This means that zinc has been oxidized in the reaction; its oxidation number increased due to a loss of electrons.

Copper originally has an oxidation state of $\small +2$ , but changes to its elemental form as a product. This gives copper a final oxidation state of zero. Since it has gained electrons, copper has been reduced.

Sulfur remains in the sulfate complex ion, which retains its overall $\small -2$ charge throughout the reaction. It is neither oxidized, nor reduced.

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Question

Consider the following reaction:

$Zn_(_s_) + CuSO_4_(_a_q_) \to Cu_(_s_) + ZnSO_4_(_a_q_)$

Which of the following substances is the oxidizing agent?

Answer

In an oxidation-reduction reaction, the oxidizing agent is the reactant that accepts electrons, becoming reduced. The oxidizing or reducing agent is always the whole reagent, and is never just the atom in the reagent that is being oxidized or reduced.

is the reagent that accepts two electrons from elemental zinc. This allows zinc to become oxidized, changing oxidation state from zero to $\small +2$ . Copper accepts these electrons, changing from an oxidation state of $\small +2$ to zero. As a result, we say that is the oxidizing agent for the reaction.

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Question

$Ca(s) + 2HCl (aq) \rightarrow CaCl_{2} (aq) + H_{2} (g)$

In this reaction, electrons are transferred from _ to _.

Answer

A transfer of electrons indicates an oxidation-reduction reaction. The element losing electrons is oxidized, while the element gaining electrons is reduced.

Calcium is initially a neutral solid, but as a product has a charge of $\small +2$ . Calcium's charge increased, meaning it must have lost electrons.

Hydrogen initially has a charge of $\small +1$ , but as a product is a neutral diatomic gas. Hydrogen's charge decreased, meaning it must have gained electrons.

We can thus conclude that calcium was oxidized and hydrogen was reduced, indicating that electrons were transferred from calcium to hydrogen.

The charge on chlorine remains $\small -1$ throughout the entire reaction.

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