Reactions and Equilibrium - AP Chemistry

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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?

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?

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

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?

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

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?

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

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)

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)

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

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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?

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

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?

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

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?

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

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)

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)

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

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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–

Compare your answer with the correct one above

Question

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

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

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.

Compare your answer with the correct one above

Question

How many electrons are involved in the following reaction?

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

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?

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

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)

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)

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

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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–

Compare your answer with the correct one above

Question

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

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

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.

Compare your answer with the correct one above

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Reactions and Equilibrium - AP Chemistry

The following ReDox reaction takes place in acidic solution: Fe2+ + Cr2O72– → Fe3+ + Cr3+ What is the sum of coefficients in this redox reaction?

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–

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

How many electrons are involved in the following reaction?

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

How many electrons are involved in the following reaction?

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

What is the balanced coefficient on OH- for the following reaction: (under basic conditions)

Add them together: Simplify: Add Hydroxides to each side to counter H+. Simplify:

What is the sum of all the balanced coefficients in the following reaction: (basic conditions)

Add the equations together Simplify Add 2 OH- to each side to cancel out the H+. Simplify:

The following ReDox reaction takes place in acidic solution: Fe2+ + Cr2O72– → Fe3+ + Cr3+ What is the sum of coefficients in this redox reaction?

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–

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

How many electrons are involved in the following reaction?

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

How many electrons are involved in the following reaction?

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

What is the balanced coefficient on OH- for the following reaction: (under basic conditions)

Add them together: Simplify: Add Hydroxides to each side to counter H+. Simplify:

What is the sum of all the balanced coefficients in the following reaction: (basic conditions)

Add the equations together Simplify Add 2 OH- to each side to cancel out the H+. Simplify:

The following ReDox reaction takes place in acidic solution: Fe2+ + Cr2O72– → Fe3+ + Cr3+ What is the sum of coefficients in this redox reaction?

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–

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

How many electrons are involved in the following reaction?

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

How many electrons are involved in the following reaction?

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

What is the balanced coefficient on OH- for the following reaction: (under basic conditions)

Add them together: Simplify: Add Hydroxides to each side to counter H+. Simplify:

What is the sum of all the balanced coefficients in the following reaction: (basic conditions)

Add the equations together Simplify Add 2 OH- to each side to cancel out the H+. Simplify:

The following ReDox reaction takes place in acidic solution: Fe2+ + Cr2O72– → Fe3+ + Cr3+ What is the sum of coefficients in this redox reaction?

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–

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

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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

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

How many electrons are involved in the following reaction?

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

How many electrons are involved in the following reaction?

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

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

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

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)

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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

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

How many electrons are involved in the following reaction?

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

How many electrons are involved in the following reaction?

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

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

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

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)

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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

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

How many electrons are involved in the following reaction?

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

How many electrons are involved in the following reaction?

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

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

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

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)

The following ReDox reaction takes place in acidic solution:

Fe2+ + Cr2O72– → Fe3+ + Cr3+

What is the sum of coefficients in this redox reaction?

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

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