Oxidation-Reduction Reactions - AP Chemistry

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Question

What is the oxidation number of nitrogen in ?

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Answer

First, note that the molecule does not have a charge, meaning that the oxidation numbers of each atom must add up to zero. Hydrogen has an oxidation number of and oxygen has an oxidation number of . Thus, if we call the oxidation number of nitrogen , we can get the equation $1+x+3\times(-2)=0$ . Solving this gives , so the oxidation number of nitrogen is in this molecule.

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Question

The following is a modified true/false question. In it, you must decide if each individual statement is true (T) or false (F). If both are true, then you must also decide if the second statement is a correct explanation (CE) of the first statement.

I: An $\text{Al}^{3+}$ ion would undergo reduction to form $\text{Al}$ metal

BECAUSE

II: reduction is a loss of electrons

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Answer

Statement II is false: reduction is a gain of electrons. Looking at statement I, the ion's charge would decrease by 3 in becoming a metal, which suggests that the ion gains 3 electrons. Thus, it is undergoing reduction.

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Question

What is the coefficient on sulfur dioxide if the following redox reaction is balanced in an acidic solution?

$SO_{2} + Cr_{2}O_{7}^{2-} \rightarrow SO_{4}^{2-} + Cr^{3+}$

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Answer

Balancing redox reactions involves the following steps:

1. Divide the reaction into oxidation and reduction half reactions and balance all atoms that are not oxygen and hydrogen:

$SO_{2} \rightarrow SO_{4}^{2-}$

$Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}$

2. Balance the oxygens by adding water molecules on the opposite side of the reactions:

$2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-}$

$Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

3. Balance the hydrogens by adding protons to the opposite side of the equation:

$2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-} + 4H^{+}$

$14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

4. Add electrons in order to equal the charges on both sides of the equation:

$2H_{2}O + SO_{2} \rightarrow SO_{4}^{2-} + 4H^{+} + 2e^{-}$

$6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

5. In order to make the electron exchange equal in each half step, we must multiply the top half reaction by 3:

$6H_{2}O + 3SO_{2} \rightarrow 3SO_{4}^{2-} + 12H^{+} + 6e^{-}$

$6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O$

6. Add up the reactants and products while cancelling out substances on opposite sides of the reactions. For example: we will cancel the 6 water molecules as reactants and be left with only one water molecule as a product. In addition, only 2 protons will be left on the reactant's side after canceling the 12 from the product's side.

$Cr_{2}O_{7}^{2-} + 3SO_{2} + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_{4}^{2-} + H_{2}O$

In the balanced redox reaction, the coefficient on sulfur dioxide is 3.

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Question

For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?

$PF_2I_{(l)}+Hg_{(l)}\rightarrow P_2F_4_{(g)}+Hg_2I_2_{(s)}$

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Answer

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.

Now we can begin to look at the half-reactions.

$PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

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Question

Oxidation is the of electrons, reduction is the of electrons.

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Answer

An oxidation-reduction (redox) reaction is a reaction where electrons are transferred between two substances. When an atom is oxidized, it is called the reducing agent, and it loses a number of electrons. Similarly, a reduced atom is called the oxidizing agent, and gains the same number of electrons. A popular mnemonic to help remember is OIL RIG, or Oxidation is Loss of electrons, Reduction is Gain of electrons.

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Question

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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Answer

When you balance the redox reaction in acidic conditons, there are 6Fe2+, 1 Cr2O72–, 14 H+, 6 Fe3+, 2 Cr3+, and 7 H2O. Don't forget to add the 1 in front of the Cr2O72–

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Question

For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?

$PF_2I_{(l)}+Hg_{(l)}\rightarrow P_2F_4_{(g)}+Hg_2I_2_{(s)}$

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Answer

To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.

Now we can begin to look at the half-reactions.

$PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Balance the atoms.

$2PF_2I_{(l)}\rightarrow P_2F_4_{(g)}$

$2Hg_{(l)}\rightarrow Hg_2I_2_{(s)}$

Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.

$2PF_2I_{(l)}+2e^-\rightarrow P_2F_4_{(g)}$

$2Hg \rightarrow Hg_2I_2_{(s)}+2e^{-}$

We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.

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Question

How many electrons are involved in the following reaction?

$MnO_4^-+ I^-\rightarrow I_2+Mn^{2+}$

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Answer

The common factor between 2 e- and 5 e- is 10. Therefore the number of electrons involved is 10 e-.

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Question

How many electrons are involved in the following reaction?

$MnO_4^-+ Fe^{2+} \rightarrow Fe^{3+}+Mn^{2+}$

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Answer

The common factor between 1 e- and 5 e- is 5. Therefore the number of electrons involved is 5 e-.

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Question

What is the balanced coefficient on OH- for the following reaction:

$Br^- + MnO_4^- \rightarrow MnO_2 + BrO_3^-$ (under basic conditions)

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Answer

Add them together:

$3 H_2 O + Br^- + 2 MnO_4^- + 8 H^+ + 6 e^- \rightarrow 2 MnO_2 + BrO_3^- + 6 H^+ + 4 H_2 O + 6 e^-$

Simplify:

$Br^- + 2 MnO_4^- + 2 H^+ \rightarrow 2 MnO_2 + BrO_3^- + H_2 O$

Add Hydroxides to each side to counter H+.

$Br^- + 2 MnO_4^- + 2 H_2 O \rightarrow 2 MnO_2 + BrO_3^- + H_2 O+2 OH^-$

Simplify:

$Br^- + 2 MnO_4^- + H_2 O \rightarrow 2 MnO_2 + BrO_3^- +2 OH^-$

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Question

What is the sum of all the balanced coefficients in the following reaction:

$O_2 + Sb \rightarrow H_2 O_2 + SbO_2^-$ (basic conditions)

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Answer

Add the equations together

$2Sb+4H_2 O + 3 O_2 + 6 H^+ + 6e^- \rightarrow 2SbO_2^-+8H^++3 H_2 O_2 + 6e^-$

Simplify

$2Sb+4H_2 O+3 O_2 \rightarrow 2SbO_2^-+2H^++3H_2 O_2$

Add 2 OH- to each side to cancel out the H+.

$2Sb+4H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+2H_2 O+3H_2 O_2$

Simplify:

$2Sb+2 H_2 O+3 O_2 + 2 OH^- \rightarrow 2SbO_2^-+3H_2 O_2$

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Question

What is the balanced equation when heptane is combusted?

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Answer

Heptane: C7H16. Combustion is when a molecule reacts with O2 and the products are CO2 and H2O. Balancing gives 7 CO2, 8 H2O, 1 heptane, and 11 O2

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Question

Which of the following reactions has the most exothermic heat of reaction?

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Answer

The longer the hydrocarbon chain, the greater the amount of combustion products (CO2 and H20) generated. Branched molecules such as isopropane and isobutane are more difficult to combust than their straight-chain counterparts.

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Question

Which of the following conditions would describe a combustion reaction?

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Answer

An exothermic reaction will have a negative $\Delta H$ value, indicating that it releases heat. Conversely, an endothermic reaction will have a positive $\Delta H$ value, indicating a consumption of heat.

A combustion reaction releases heat; thus it must have a negative $\Delta H$ value and be exothermic.

Exergonic reactions have a negative $\Delta G$ value, indicating spontaneity, while endergonic reactions are non-spontaneous. While most combustion reactions will be non-spontaneous, it is impossible to draw this conclusion for certain without knowing more about the reaction. The only thing we know for certain is that heat is released, and the reaction is exothermic.

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Question

What is the oxidation number of chlorine in perchlorate?

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Answer

Perchlorate is a complex ion with the formula .

The most important rule for oxidation number is that the sum of the oxidation states of the atoms must equal the overall molecular charge.

This compound has four oxygens, which always have a oxidation state.

Chlorine usually has an oxidation state of , but in this case it must balance out the oxygens.

Chlorine must be .

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Question

$Fe_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_{2(g)}$

For the redox reaction shown, which of the following half reactions occurs in the anode?

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Answer

Recall that oxidation always occurs at the anode (in both the electrochemical and galvanic cells). loses two electrons in this case to become $Fe^{2+}$ . The presence of $Fe^{2+}$ is hinted by the ionic compound .

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Question

When is the oxidation number of H (-1)?

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Answer

We typically think of Hydrogen as having an oxidation number of +1. However when it is bonded to a less electronegative element such as Na it is actually assigned an oxidation number of -1.

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Question

What is the oxidation number of Cr, S, and Fe in the following substances: (a) K2Cr2O7 (b) H2SO4 (c) Fe2O3.

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Answer

(a) Since O has a –2 oxidation number and K has a +1 oxidation number (1 valence
electron it gives up), that means that Cr must have an oxidation number of +6. (b) Since H
has a +1 oxidation number and O has a –2 oxidation number, S has a +6 oxidation number.
(c) Fe has an oxidation number of +3 in order for it to have a net 0 oxidation state.

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Question

Consider the following balanced equation:

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

What is the difference between the oxidation state of aluminum on the right side of the equation versus the left?

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Answer

$2Al_{(s)}+3H_2SO_{4\hspace{1 mm}(aq)}\rightarrow Al_2(SO_4){3\hspace{1 mm}(aq)}+\hspace{1 mm}H{2\hspace{1 mm}(g)}$

On the left side of the equation, $Al_{(s)}$ is a solid, so its oxidation state is zero, but on the right side it is in a salt, so it is not in its zero state.

Sulfate, $SO_4^{-2}$ , is an anionic salt, and there are three sulfate ions in each complex, yielding a net charge of -6. The two aluminum ions must have a net charge of +6, which, divided over two aluminum ions, yields an oxidation state of +3 for each aluminum ion.

The difference comes from simple subtraction: $3-0=3$ .

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Question

What is the oxidation number of manganese in ?

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Answer

Potassium always has an oxidation state of , while oxygen always has an oxidation state of . Since we have four oxygens, there is a total charge of from them.

The most important rule of oxidation numbers is that their sum must equal the charge on the molecule. In this compound, we have .

Manganese needs to cancel out the charge from potassium and oxygen in order to give us a neutral compound.

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