(a)
Interpretation:
The following skeletal equation has to be balanced using
Concept Introduction:
Net ionic equation:
Net ionic equation is defined as the specific species that only involves to a particular reaction. This type of equations is generally used in acid-base neutralization reactions and redox reactions.
Oxidizing agent:
The material which gains electron in a
Reducing agent:
The material, which loses electrons in a chemical reaction, is called reducing agent. In this reaction, the oxidation number will be increased.
(a)

Answer to Problem 6K.6E
The balanced reaction for the production of chlorite ions from dichlorine heptoxide by reaction with hydrogen peroxide solution is given below,
Here, the oxidizing agent is
Explanation of Solution
The unbalanced skeletal equation for the reaction is
Oxidation half-reaction:
The oxidation number of
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Therefore, the balanced oxidation half-reaction is
Here,
Reduction half-reaction:
The oxidation number of
Balance the equation except
Balance the
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Therefore, the balanced reduction half-reaction is
Here, the
Now add the two half reactions together. Match the number of electrons in each side. Because in oxidation half reaction
Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.
Divide by 2 each side of the arrow.
Therefore, the balanced net ionic equation the above reaction is
(b)
Interpretation:
The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.
Concept introduction:
Refer to part (a).
(b)

Answer to Problem 6K.6E
The balanced reaction of permanganate ions with sulfide ions is given below,
Here, the oxidizing agent is
Explanation of Solution
The unbalanced skeletal equation for the reaction is
Oxidation half-reaction:
The oxidation number of
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Therefore, the balanced oxidation half-reaction is
Here,
Reduction half-reaction:
The oxidation number of
Balance the equation except
Balance the
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Here, the
Now add the two half reactions together. Match the number of electrons in each side. Because in oxidation half reaction
Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.
Therefore, the balanced net ionic equation the above reaction is
(c)
Interpretation:
The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.
Concept introduction:
Refer to part (a).
(c)

Answer to Problem 6K.6E
The balanced reaction of hydrazine with chlorate ions is given below,
Here, the oxidizing agent is
Explanation of Solution
The unbalanced skeletal equation for the reaction is
Oxidation half-reaction:
The oxidation number of
Balance the equation except
Balance the
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Therefore, the balanced oxidation half-reaction is
Here,
Reduction half-reaction:
The oxidation number of
Balance the
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Here, the
Now add the two half reactions together. Match the number of electrons in each side. Because in oxidation half reaction
Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.
Divide by
Therefore, the balanced net ionic equation the above reaction is
(d)
Interpretation:
The following skeletal equation has to be balanced using oxidation and reduction half reactions and also the oxidizing agent, reducing agent has to be identified.
Concept introduction:
Refer to part (a).
(d)

Answer to Problem 6K.6E
The balanced reaction of plumblate ions and hypochlorite ions is given below,
Here, the oxidizing agent is
Explanation of Solution
The unbalanced skeletal equation for the reaction is
Oxidation half-reaction:
The oxidation number of
Balance the equation except
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Therefore, the balanced oxidation half-reaction is
Here,
Reduction half-reaction:
The oxidation number of
Balance the equation except
Balance the
Balance the
Balance the net charges by adding the electrons.
Here, in left side, the net charge is
Here, the
Now add the two half reactions together. Match the number of electrons in each side. Because in oxidation half reaction
Add the equation and cancel the common ions, electrons and water molecules in each side of the arrow.
Divide by
Therefore, the balanced net ionic equation the above reaction is
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Chapter 6 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
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- Use the reaction coordinate diagram to answer the below questions. Type your answers into the answer box for each question. (Watch your spelling) Energy A B C D Reaction coordinate E A) Is the reaction step going from D to F endothermic or exothermic? A F G B) Does point D represent a reactant, product, intermediate or transition state? A/ C) Which step (step 1 or step 2) is the rate determining step? Aarrow_forward1. Using radii from Resource section 1 (p.901) and Born-Lande equation, calculate the lattice energy for PbS, which crystallizes in the NaCl structure. Then, use the Born-Haber cycle to obtain the value of lattice energy for PbS. You will need the following data following data: AH Pb(g) = 196 kJ/mol; AHƒ PbS = −98 kJ/mol; electron affinities for S(g)→S¯(g) is -201 kJ/mol; S¯(g) (g) is 640kJ/mol. Ionization energies for Pb are listed in Resource section 2, p.903. Remember that enthalpies of formation are calculated beginning with the elements in their standard states (S8 for sulfur). The formation of S2, AHF: S2 (g) = 535 kJ/mol. Compare the two values, and explain the difference. (8 points)arrow_forwardIn the answer box, type the number of maximum stereoisomers possible for the following compound. A H H COH OH = H C Br H.C OH CHarrow_forward
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