Balancing Redox Reactions: Examples 1 & 2

Podcast

Play an AI-generated podcast conversation about this lesson

Questions and Answers

The final equation shows that 2MnO₄^- reacts with 3CN^- to produce 2MnO₂ and 3OCN^-.

True

The charge balance for the final equation is -4 on each side.

False

H₂O is added to both sides to balance the number of hydrogen atoms in the equation.

True

For every H⁺ added, an equal number of OH^- ions must be added to the same side of the equation.

True Signup and view all the answers

In the reduction half-reaction, 4H⁺ and 4OH^- are added to maintain the balance.

False Signup and view all the answers

The oxidation half-reaction involves the conversion of CN^- to OCN^- with the loss of electrons.

True Signup and view all the answers

After the reactions, 8OH^- ions are produced in the final equation.

True Signup and view all the answers

The balancing process shows that there are 5 oxygen atoms on each side of the final equation.

False Signup and view all the answers

Removing electrons from the equation helps in achieving charge balance in redox reactions.

True Signup and view all the answers

The final balanced equation includes two different species of manganese.

False Signup and view all the answers

The oxidation state of sulfur in SO₃^2- is higher than in SO₄^2- after balancing the redox reaction.

False Signup and view all the answers

In the redox reaction, the reduction of MnO₄^- requires the addition of 5 electrons.

True Signup and view all the answers

During the balancing of redox reactions, oxygen is balanced using H₂O molecules.

True Signup and view all the answers

The final balanced equation for the reaction involving SO₃^2- and MnO₄^- produces 5 moles of OCN^-.

False Signup and view all the answers

In the oxidation half-reaction, SO₃^2- is converted to SO₄^2- and loses 2 electrons.

True Signup and view all the answers

The charge balance for the final equation is equal to -8 on both sides of the equation.

False Signup and view all the answers

In the example involving cyanate, MnO₄^- is reduced to MnO₂.

True Signup and view all the answers

The balancing of hydrogen atoms in the reaction requires the addition of H⁺ ions.

True Signup and view all the answers

The total number of moles of SO₄^2- produced in the final reaction is 6.

False Signup and view all the answers

In a redox reaction, balancing the number of electrons lost and gained is not necessary.

False Signup and view all the answers

Study Notes

Balancing Redox Reactions: Example 1

Sulfite (SO₃^2-) is oxidized to sulfate (SO₄^2-)
Permanganate (MnO₄^-) is reduced to manganese (II) ion (Mn²⁺)
The reaction takes place in an acidic solution
The half-reactions are balanced as follows:
- Oxidation: SO₃^2- + H₂O -> SO₄^2- + 2H⁺ + 2e^-
- Reduction: MnO₄^- + 8H⁺ + 5e^- -> Mn²⁺ + 4H₂O
The electrons are canceled out by multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2
The final balanced equation is: 5SO₃^2- + 2MnO₄^- + 6H⁺ -> 5SO₄^2- + 2Mn²⁺ + 3H₂O

Balancing Redox Reactions: Example 2

Cyanide (CN^-) is oxidized to cyanate (OCN^-)
Permanganate (MnO₄^-) is reduced to manganese dioxide (MnO₂)
The reaction is carried out in an aqueous basic solution
The half-reactions are balanced as follows:
- Oxidation: CN^- + 2OH^- -> OCN^- + H₂O + 2e^-
- Reduction: MnO₄^- + 2H₂O +3e^- -> MnO₂ + 4OH^-
The electrons are canceled out by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2
The final balanced equation is: 2MnO₄^- + 3CN^- + H₂O -> 2MnO₂ + 3OCN^- + 2OH^-

Studying That Suits You

Use AI to generate personalized quizzes and flashcards to suit your learning preferences.

Description

Test your knowledge on balancing redox reactions with these examples. This quiz covers the oxidation of sulfite to sulfate and cyanide to cyanate, along with the reduction of permanganate. Perfect for chemistry students aiming to master redox equations in acidic solutions.

Balancing Redox Reactions: Examples 1 & 2

Choose a study mode

Podcast

Questions and Answers

The final equation shows that 2MnO₄^- reacts with 3CN^- to produce 2MnO₂ and 3OCN^-.

The charge balance for the final equation is -4 on each side.

H₂O is added to both sides to balance the number of hydrogen atoms in the equation.

For every H⁺ added, an equal number of OH^- ions must be added to the same side of the equation.

In the reduction half-reaction, 4H⁺ and 4OH^- are added to maintain the balance.

The oxidation half-reaction involves the conversion of CN^- to OCN^- with the loss of electrons.

After the reactions, 8OH^- ions are produced in the final equation.

The balancing process shows that there are 5 oxygen atoms on each side of the final equation.

Removing electrons from the equation helps in achieving charge balance in redox reactions.

The final balanced equation includes two different species of manganese.

The oxidation state of sulfur in SO₃^2- is higher than in SO₄^2- after balancing the redox reaction.

In the redox reaction, the reduction of MnO₄^- requires the addition of 5 electrons.

During the balancing of redox reactions, oxygen is balanced using H₂O molecules.

The final balanced equation for the reaction involving SO₃^2- and MnO₄^- produces 5 moles of OCN^-.

In the oxidation half-reaction, SO₃^2- is converted to SO₄^2- and loses 2 electrons.

The charge balance for the final equation is equal to -8 on both sides of the equation.

In the example involving cyanate, MnO₄^- is reduced to MnO₂.

The balancing of hydrogen atoms in the reaction requires the addition of H⁺ ions.

The total number of moles of SO₄^2- produced in the final reaction is 6.

In a redox reaction, balancing the number of electrons lost and gained is not necessary.

Study Notes

Balancing Redox Reactions: Example 1

Balancing Redox Reactions: Example 2

Studying That Suits You

Related Documents

Description

More Like This

Balancing Redox Reactions

Balancing Redox Reactions - Examples 1 & 2

Identifying Oxidation and Reduction Species

Redox Reactions in Basic Conditions

Upgrade to continue

Balancing Redox Reactions: Examples 1 & 2

Choose a study mode

Podcast

Questions and Answers

The final equation shows that 2MnO4- reacts with 3CN- to produce 2MnO2 and 3OCN-.

The charge balance for the final equation is -4 on each side.

H2O is added to both sides to balance the number of hydrogen atoms in the equation.

For every H+ added, an equal number of OH- ions must be added to the same side of the equation.

In the reduction half-reaction, 4H+ and 4OH- are added to maintain the balance.

The oxidation half-reaction involves the conversion of CN- to OCN- with the loss of electrons.

After the reactions, 8OH- ions are produced in the final equation.

The balancing process shows that there are 5 oxygen atoms on each side of the final equation.

Removing electrons from the equation helps in achieving charge balance in redox reactions.

The final balanced equation includes two different species of manganese.

The oxidation state of sulfur in SO32- is higher than in SO42- after balancing the redox reaction.

In the redox reaction, the reduction of MnO4- requires the addition of 5 electrons.

During the balancing of redox reactions, oxygen is balanced using H2O molecules.

The final balanced equation for the reaction involving SO32- and MnO4- produces 5 moles of OCN-.

In the oxidation half-reaction, SO32- is converted to SO42- and loses 2 electrons.

The charge balance for the final equation is equal to -8 on both sides of the equation.

In the example involving cyanate, MnO4- is reduced to MnO2.

The balancing of hydrogen atoms in the reaction requires the addition of H+ ions.

The total number of moles of SO42- produced in the final reaction is 6.

In a redox reaction, balancing the number of electrons lost and gained is not necessary.

Study Notes

Balancing Redox Reactions: Example 1

Balancing Redox Reactions: Example 2

Studying That Suits You

Related Documents

Description

More Like This

Balancing Redox Reactions

Balancing Redox Reactions - Examples 1 & 2

Identifying Oxidation and Reduction Species

Redox Reactions in Basic Conditions

The final equation shows that 2MnO₄^- reacts with 3CN^- to produce 2MnO₂ and 3OCN^-.

H₂O is added to both sides to balance the number of hydrogen atoms in the equation.

For every H⁺ added, an equal number of OH^- ions must be added to the same side of the equation.

In the reduction half-reaction, 4H⁺ and 4OH^- are added to maintain the balance.

The oxidation half-reaction involves the conversion of CN^- to OCN^- with the loss of electrons.

After the reactions, 8OH^- ions are produced in the final equation.

The oxidation state of sulfur in SO₃^2- is higher than in SO₄^2- after balancing the redox reaction.

In the redox reaction, the reduction of MnO₄^- requires the addition of 5 electrons.

During the balancing of redox reactions, oxygen is balanced using H₂O molecules.

The final balanced equation for the reaction involving SO₃^2- and MnO₄^- produces 5 moles of OCN^-.

In the oxidation half-reaction, SO₃^2- is converted to SO₄^2- and loses 2 electrons.

In the example involving cyanate, MnO₄^- is reduced to MnO₂.

The balancing of hydrogen atoms in the reaction requires the addition of H⁺ ions.

The total number of moles of SO₄^2- produced in the final reaction is 6.