(a)
Interpretation:
The balanced equation for the
Concept introduction:
The oxidizer is the species whose oxidation state falls during the course of reaction and reducer is the species whose oxidation number increases. Oxidized product is the oxidation product of the reducer and reduced product is the reduction product of the oxidizer.
Answer to Problem 26E
The balanced equation for the redox reaction,
Explanation of Solution
The given redox reaction is shown below.
The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.
The oxidation state of manganese in
Step-1: Write down the oxidation number of every element and for unknown take “n”.
Step-2: Multiply the oxidation state with their number of atoms of an element.
Step-3: Add the oxidation numbers and set them equal to the charge of the species.
Calculate the value of n by simplifying the equation.
The oxidation state of manganese in
The oxidation state of manganese in
Step-1: Write down the oxidation number of every element and for unknown take “n”.
Step-2: Multiply the oxidation state with their number of atoms of an element.
Step-3: Add the oxidation numbers and set them equal to the charge of the species.
Calculate the value of n by simplifying the equation.
The oxidation state of manganese in
The oxidation number manganese decreases from
The reduction half-reaction for the above reaction is shown below.
The balancing of the half-reactions is done by the following the steps shown below.
Step-1: Identify and balanced getting oxidized or reduced.
The manganese is getting reduced and its number of atoms is balanced on both sides.
Step-2: Balance elements other than oxygen and hydrogen if any.
Step-3: Balance oxygen atoms by adding water on the appropriate side.
The number of oxygen atoms is balanced by adding two water molecules on the right-hand side of the equation.
Step-4: Balance the hydrogen atoms by adding
The number of hydrogen atoms is balanced by adding four
Step-5: Balance the charge by adding electrons to the appropriate side.
The charge is balanced by adding three electrons on the left-hand side
Step-6: Neutralize the all
Four hydroxide ions are added to both sides of the equation.
Simplify the above equation by making the water of neutralized protons and balance out water molecules.
Step-7: Recheck the equation to be sure that it is perfectly balanced.
The equation is completely balanced and is shown below.
The oxidation state of sulfur in
The oxidation number of sulfur increases from
The oxidation half-reaction for the above reaction is shown below.
The balancing of the half-reactions is done by the following the steps shown below.
Step-1: Identify and balance the element getting oxidized or reduced.
The sulfur is getting oxidized and their numbers of atoms are balanced on both sides.
Step-2: Balance elements other than oxygen and hydrogen if any.
Step-3: Balance oxygen atoms by adding water on the appropriate side.
Step-4: Balance the hydrogen atoms by adding
Step-5: Balance the charge by adding electrons to the appropriate side.
The charge is balanced by adding two electrons on the left-hand side of the equation.
Step-6: Recheck the equation to be sure that it is perfectly balanced.
The equation is completely balanced and is shown below.
The balanced redox equation is obtained by adding equation (1) and (2) in such a way that electrons are canceled out.
Multiply equation (1) by two and equation (2) by three and then add them.
The balance redox equation after adding these equations is shown below.
The
(b)
Interpretation:
The balanced equation for the redox reaction,
Concept introduction:
The oxidizer is the species whose oxidation state falls during the course of reaction and reducer is the species whose oxidation number increases. Oxidized product is the oxidation product of the reducer and reduced product is the reduction product of the oxidizer.
Answer to Problem 26E
The balanced equation for the redox reaction,
Explanation of Solution
The given redox reaction is shown below.
The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.
The oxidation state of the central metal atom is calculated by knowing the standard oxidation states of few elements.
The oxidation state of the chlorine in
Step-1: Write down the oxidation number of every element and for unknown take “n”.
Step-2: Multiply the oxidation state with their number of atoms of an element.
Step-3: Add the oxidation numbers and set them equal to the charge of the species.
Calculate the value of n by simplifying the equation as shown below.
The oxidation state of chlorine is
The oxidation number of chlorine is
The oxidation number of chlorine decreases from
The reduction half-reaction for the above reaction is shown below.
The balancing of the half-reactions is done by the following the steps shown below.
Step-1: Identify and balanced getting oxidized or reduced.
The chlorine is getting reduced and its number of atoms is balanced on both sides.
Step-2: Balance elements other than oxygen and hydrogen if any.
Step-3: Balance oxygen atoms by adding water on the appropriate side.
The number of oxygen atoms is balanced by adding one water molecule on the right-hand side of the equation.
Step-4: Balance the hydrogen atoms by adding
The number of hydrogen atoms is balanced by adding two
Step-5: Balance the charge by adding electrons to the appropriate side.
The charge is balanced by adding three electrons on the left-hand side
Step-6: Neutralize the all
Two hydroxide ions are added to both sides of the equation.
Simplify the above equation by making water of neutralized protons and balance out water molecules.
Step-7: Recheck the equation to be sure that it is perfectly balanced.
The equation is completely balanced and is shown below.
The oxidation number of copper is zero in
The oxidation number of copper is
The oxidation of copper increases from
The reduction half-reaction for the above reaction is shown below.
The balancing of the half-reactions is done by the following the steps shown below.
Step-1: Identify and balance the element getting oxidized or reduced.
The copper is getting oxidized and its number of atoms is balanced on both sides.
Step-2: Balance elements other than oxygen and hydrogen if any.
Step-3: Balance oxygen atoms by adding water on the appropriate side.
Step-4: Balance the hydrogen atoms by adding
Step-5: Balance the charge by adding electrons to the appropriate side.
The charge is balanced by adding two electrons on the right-hand side of the equation.
Step-6: Recheck the equation to be sure that it is perfectly balanced.
The equation is completely balanced and is shown below.
The balanced redox equation is obtained by adding equation (1) and (2) in such a way that electrons are canceled out.
Add equation (1) and equation (2).
The balance redox equation after adding these equations is shown below.
The balanced equation of redox reaction is shown below.
Want to see more full solutions like this?
Chapter 17 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
- Write balanced net ionic equations for the following reactions in acid solution. (a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed. (b) Solid phosphorus (P4) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate (H2PO4-) ions. (c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.arrow_forwardFour metals, A, B, C, and D, exhibit the following properties: (a) Only A and C react with 1.0 M hydrochloric acid to give H2(g). (b) When C is added to solutions of the ions of the other metals, metallic B, D, and A are formed. (c) Metal D reduces Bn+ to give metallic B and Dn+. Based on this information, arrange the four metals in order of increasing ability to act as reducing agents.arrow_forwardThe iron content of hemoglobin is determined by destroying the hemoglobin molecule and producing small water-soluble ions and molecules. The iron in the aqueous solution is reduced to iron(II) ion and then titrated against potassium permanganate. In the titration, iron(ll) is oxidized to iron(III) and permanganate is reduced to manganese(II) ion. A 5.00-g sample of hemoglobin requires 32.3 mL of a 0.002100 M solution of potassium permanganate. The reaction with permanganate ion is MnO4(aq)+8H+(aq)+5Fe2+(aq)Mn2+(aq)+5Fe3+(aq)+4H2O What is the mass percent of iron in hemoglobin?arrow_forward
- Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: IO3(aq) + I(aq) I3(aq) Triiodide ion concentration is determined by titration with a sodium thiosulfate (Na2S2O3) solution. The products are iodide ion and tetrathionate ion (S4O6). a. Balance the equation for the reaction of IO3 with I ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HQ required to convert all of the IO3 ions to I ions? c. Write and balance the equation for the reaction of S2O32 with I3 in acidic solution. d. A 25.00-mL sample of a 0.0100 M solution of KIO. is reacted with an excess of KI. It requires 32.04 mL of Na2S2O3 solution to titrate the I3 ions present. What is the molarity of the Na2S2O3 solution? e. How would you prepare 500.0 mL of the KIO3 solution in part d using solid KIO3?arrow_forwardThe exposed electrodes of a light bulb are placed in a solution of H2SO4 in an electrical circuit such that the light bulb is glowing. You add a dilute salt solution, and the bulb dims. Which of the following could be the salt in the solution? a. Ba(NO3)2 b. NaNO3 c. K2SO4 d. Ca(NO3)2 Justify your choices. For those you did not choose, explain why they are incorrect.arrow_forwardWrite the net ionic equation for the reaction, if any, that occurs on mixing (a) solutions of sodium hydroxide and magnesium chloride. (b) solutions of sodium nitrate and magnesium bromide. (c) magnesium metal and a solution of hydrochloric acid to produce magnesium chloride and hydrogen. Magnesium metal reacting with HCl.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning