(a)
Interpretation:
The balanced equations for the following reaction in basic solution needs to be determined.
Concept introduction:
Here are the rules to
- Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
- Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
- Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.
Answer to Problem 7QAP
Explanation of Solution
Determining the oxidation numbers:
The oxidation number of I in I2 and I- is 0 and -1 respectively. As the oxidation number of I is decreasing this is the reduction half reaction.
Reduction half reaction:
Oxidation half reaction:
To balance the excess O atoms on the product side we add one H2 O on the reactant side, and then we balance the excess H atom on the reactant side by adding two H+ on the product side. Finally, we balance the charge by adding two electrons to the product side of the half reaction.
Thus, the balanced oxidation half reaction:
Now, balance the charge by adding two electrons to the reactant side of the half reaction.
Thus, the balanced reduction half reaction:
Net reaction:
By adding both the half reaction, the net reaction is obtained as follows:
Basic medium:
Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. 2 OH- ions are added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction.
Thus, the balanced net reaction in the basic medium will be:
Or,
(b)
Interpretation:
The balanced equations for the following reaction in basic solution needs to be determined.
Concept introduction:
Here are the rules to balance
- Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
- Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
- Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.
Answer to Problem 7QAP
Explanation of Solution
Determining the oxidation numbers:
The oxidation number of Zn in Zn and Zn2+ is 0 and +2 respectively. As the oxidation number of Zn is increasing this is the oxidation half of the reaction.
Oxidation half reaction:
Reduction half reaction:
Balance the charge by adding two electrons to the product side of the half reaction.
Thus, the balanced oxidation half reaction will be:
To balance the excess O atoms on the product side we add three H2 O on the product side, and then we balance the excess H atom on the product side by adding nine H+ on the reactant side. Finally, we balance the charge by adding eight electrons to the reactant side of the half reaction.
Thus, the balanced reduction half reaction will be:
Net reaction:
Multiply the oxidation half reaction by 4, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.
Basic medium:
Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction. Thus, the balanced net reaction will be:
Or,
(c)
Interpretation:
The balanced equations for the following reaction in basic solution needs to be determined.
Concept introduction:
Here are the rules to balance redox reactions in basic medium:
- Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
- Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
- Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.
Answer to Problem 7QAP
Explanation of Solution
Determining the oxidation numbers:
The oxidation number of Cl in ClO- and Cl- is +1 and -1 respectively. As the oxidation number of Cl is decreasing this is the reduction half of the reaction.
The reduction half reaction:
Oxidation half reaction:
To balance the excess O atoms on the product side we add two H2 O on the reactant side, and then we balance the excess H atom on the reactant side by adding four H+ on the product side. Finally, we balance the charge by adding three electrons to the reactant side of the half reaction.
Thus, the balanced oxidation half reaction will be:
Similarly, to balance the excess O atoms on the reactant side we add one H2 O on the product side, and then we balance the excess H atom on the product side by adding two H+ on the reactant side. Finally, we balance the charge by adding two electrons to the reactant side of the half reaction.
Thus, the balanced reduction half reaction will be:
Net reaction:
Multiply the oxidation half reaction by 2 and reduction half reaction by 3, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.
Basic medium:
Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction.
Or,
(d)
Interpretation:
The balanced equations for the following reaction in basic solution needs to be determined.
Concept introduction:
Here are the rules to balance redox reactions in basic medium:
- Determine the oxidation numbers of the elements and write the oxidation and reduction half equation.
- Balance the atoms of elements other than O and H. Use H2 O to balance O, H+ to balance H, electrons to balance the charges. Equalize the number of electrons on both the half reactions and then add the half reactions.
- Add OH- to neutralize the excess protons since this is a basic medium. Add OH- to both side of the equation to balance it.
Answer to Problem 7QAP
Explanation of Solution
Determining the oxidation numbers:
The oxidation number of K in K and K+ is 0 and +1 respectively. As the oxidation number of K is increasing this is the oxidation half of the reaction.
Oxidation half reaction:
Reduction half reaction:
Balance the charge by adding one electron to the product side of the half reaction.
Thus, the balanced oxidation half reaction will be:
To balance the excess O atoms on the reactant side we add one H2 O on the product side, and then we balance the excess H atom on the product side by adding two H+ on the reactant side. Finally, we balance the charge by adding two electrons to the reactant side of the half reaction.
Thus, the balanced reduction half reaction will be:
Net reaction:
Multiply the oxidation half reaction by 2, to cancel out the electrons in the net reaction. We add both the half reaction to get the net reaction.
Basic medium:
Since the reaction takes place in basic medium the excess proton on the reactant side of the net reaction is neutralized with OH-. OH- is added on both side of the equation to even out the charges. H+ and OH- combines to form H2 O, which is then cancelled out on both side of the net reaction.
Or,
Want to see more full solutions like this?
Chapter 17 Solutions
OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:
- Calculate the equilibrium constant at 25 C for the reaction 2 Ag+(aq) + Hg() 2 Ag(s) + Hg2+(aq)arrow_forwardOrder the following oxidizing agents by increasing strength under standard-state conditions: Mg2+(aq), Hg2+(aq), Pb2+(aq).arrow_forwardAn electrode is prepared from liquid mercury in contact with a saturated solution of mercury(I) chloride, Hg2Cl, containing 1.00 M Cl . The cell potential of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half-cell as the anode is 0.268 V. What is the solubility product of mercury(I) chloride?arrow_forward
- A 1.0-L sample of 1.0 M HCl solution has a 10.0 A current applied for 45 minutes. What is the pH of the solution after the electricity has been turned off?arrow_forwardFollow the directions of Question 23 for the following species: Cu+ Zn Ni2+ Fe2+ H+ (acidic)arrow_forwardConsider the following cell running under standard conditions: Fe(s)Fe2+(aq)Al3+(aq)Al(s) a Is this a voltaic cell? b Which species is being reduced during the chemical reaction? c Which species is the oxidizing agent? d What happens to the concentration of Fe3+(aq) as the reaction proceeds? e How does the mass of Al(s) change as the reaction proceeds?arrow_forward
- What is the standard cell potential you would obtain from a cell at 25C using an electrode in which Hg22+(aq) is in contact with mercury metal and an electrode in which an aluminum strip dips into a solution of Al3+(aq)?arrow_forwardOrder the following oxidizing agents by increasing strength under standard-state conditions: O2(g); MnO4(aq); NO3 (aq) (in acidic solution ).arrow_forwardCalcium metal can be obtained by the direct electrolysis of molten CaCl2, at a voltage of 3.2 V. (a) How many joules of electrical energy are required to obtain 12.0 1b of calcium? (b) What is the cost of the electrical energy obtained in (a) if electrical energy is sold at the rate of nine cents per kilowatt hour?arrow_forward
- Assign an oxidation number to the underlined atom in each ion or molecule. (a) Fe2O3, (b) H2SO4, (C) CO32- (C) NO2+arrow_forwardAnswer the following questions by referring to standard electrode potentials at 25C. a Will oxygen, O2, oxidize iron(II) ion in solution under standard conditions? b Will copper metal reduce 1.0 M Ni2(aq) to metallic nickel?arrow_forwardCalculate the standard cell potential of the following cell at 25C. Cr(s)Cr3(aq)Hg22(aq)Hg(l)arrow_forward
- Chemistry & 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 LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning