(a)
Interpretation:
The given half reaction should be balanced
Concept Introduction:
The
The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of
The following rules must be followed in balancing redox reaction by half equation method:
- Initially, redox reaction is separated into two half equations; oxidation and reduction.
- Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
- Oxygen atoms are balanced by addition of water on either side of the reaction.
- Hydrogen ion/s is added to balance the hydrogen atom.
- Electrons are added to balance the charge.
- Half reactions are added to get the net total equation.
- The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 100AP
Explanation of Solution
The given half reaction is as follows:
In the above reaction, add 2 water molecules to the right to balance the number of oxygen atoms:
Now, add 4 hydrogen ions to the left to balance the hydrogen atoms,
To balance the charge add 4 electrons to the left side of the reaction arrow thus,
The above reaction is balanced half reaction.
(b)
Interpretation:
The given half reaction should be balanced
Concept Introduction:
The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.
The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.
The following rules must be followed in balancing redox reaction by half equation method:
- Initially, redox reaction is separated into two half equations; oxidation and reduction.
- Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
- Oxygen atoms are balanced by addition of water on either side of the reaction.
- Hydrogen ion/s is added to balance the hydrogen atom.
- Electrons are added to balance the charge.
- Half reactions are added to get the net total equation.
- The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 100AP
Explanation of Solution
The given half reaction is as follows:
In the above reaction, add 2 hydrogen ions to the left to balance the hydrogen atoms.
To balance the charge add two electrons to the left:
The above half reaction is balanced.
(c)
Interpretation:
The given half reaction should be balanced
Concept Introduction:
The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.
The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.
The following rules must be followed in balancing redox reaction by half equation method:
- Initially, redox reaction is separated into two half equations; oxidation and reduction.
- Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
- Oxygen atoms are balanced by addition of water on either side of the reaction.
- Hydrogen ion/s is added to balance the hydrogen atom.
- Electrons are added to balance the charge.
- Half reactions are added to get the net total equation.
- The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 100AP
Explanation of Solution
The given half reaction is as follows:
In the above reaction, add 1 water molecule to the right to balance the number of oxygen atoms first.
Add 3 hydrogen ions to the left to balance the hydrogen atoms,
Last step is to balance the charge, thus, 2 electrons are added to left side of the reaction arrow:
The above half reaction is balanced.
(d)
Interpretation:
The given half reaction should be balanced
Concept Introduction:
The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.
The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half-reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.
The following rules must be followed in balancing redox reaction by half equation method:
- Initially, redox reaction is separated into two half equations; oxidation and reduction.
- Atoms other than hydrogen and oxygen are balanced first in the unbalanced half equations.
- Oxygen atoms are balanced by addition of water on either side of the reaction.
- Hydrogen ion/s is added to balance the hydrogen atom.
- Electrons are added to balance the charge.
- Half reactions are added to get the net total equation.
- The further addition of hydroxide ion takes place on both side of the reaction, if the solution is basic in nature to neutralise the hydrogen ion present in the solution.

Answer to Problem 100AP
Explanation of Solution
The given half reaction is as follows:
To balance the above reaction, add 2 water molecules to right side of the reaction arrow.
Now, hydrogen atoms can be balanced by adding 4 hydrogen ions to the left side of the reaction arrow:
Last step is to balance the charge,
Thus, 3 electrons are added to left side of the reaction arrow and balanced half reaction will be:
Want to see more full solutions like this?
Chapter 18 Solutions
INTRODUCTORY CHEMISTRY
- -AG|F=2E|V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 3-1.12 -1.57 -2.05 -0.89 PO HPO →→H2PO2 P PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forward
- Predict the products of this organic reaction: Explanation Check IN NaBH3CN H+ ? Click and drag to start drawing a structure. D 5 C +arrow_forwardPredict the products of this organic reaction: H3O+ + ? • Draw all the reasonable products in the drawing area below. If there are no products, because no reaction will occur, check the box under the drawing area. • Include both major and minor products, if some of the products will be more common than others. • Be sure to use wedge and dash bonds if you need to distinguish between enantiomers. No reaction. Click and drag to start drawing a structure. dmarrow_forwardIarrow_forward
- Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)arrow_forward
- 2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation constant for [Fe(CN)6] 4 is ~1032, and that: Fe3+ (aq) + e = Fe²+ (aq) E° = +0.77 V [Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V (4 points)arrow_forward5. Consider the compounds shown below as ligands in coordination chemistry and identify their denticity; comment on their ability to form chelate complexes. (6 points) N N A B N N N IN N Carrow_forward1. Use standard reduction potentials to rationalize quantitatively why: (6 points) (a) Al liberates H2 from dilute HCl, but Ag does not; (b) Cl2 liberates Br2 from aqueous KBr solution, but does not liberate C12 from aqueous KCl solution; c) a method of growing Ag crystals is to immerse a zinc foil in an aqueous solution of AgNO3.arrow_forward
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