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Questions and Answers
The reaction between Zn and NO3- produces Zn(OH)42- and methane.
The reaction between Zn and NO3- produces Zn(OH)42- and methane.
False
The half-reaction for the reduction of NO3- involves a gain of 8 electrons.
The half-reaction for the reduction of NO3- involves a gain of 8 electrons.
True
The balancing of charges in the reaction requires adding protons (H+).
The balancing of charges in the reaction requires adding protons (H+).
True
In basic conditions, water is produced as a product of the reaction.
In basic conditions, water is produced as a product of the reaction.
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The final balanced equation includes 4Zn, 6H2O, and 8OH- on the reactants side.
The final balanced equation includes 4Zn, 6H2O, and 8OH- on the reactants side.
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To balance the oxygen atoms in the reaction, H2O is added.
To balance the oxygen atoms in the reaction, H2O is added.
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The total charge on the reactant side of the final balanced equation is neutral.
The total charge on the reactant side of the final balanced equation is neutral.
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The compound formed by zinc and hydroxide is Zn(OH)42-.
The compound formed by zinc and hydroxide is Zn(OH)42-.
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The reduction of NO3- to NH3 involves the loss of electrons.
The reduction of NO3- to NH3 involves the loss of electrons.
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The net reaction indicates that 4 moles of Zn and 1 mole of NO3- react together.
The net reaction indicates that 4 moles of Zn and 1 mole of NO3- react together.
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The balanced reaction shows that 4 moles of Zn and 6 moles of H2O react with 1 mole of NO3-.
The balanced reaction shows that 4 moles of Zn and 6 moles of H2O react with 1 mole of NO3-.
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In the half-reaction of the reduction process, NO3- loses 6 electrons.
In the half-reaction of the reduction process, NO3- loses 6 electrons.
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To balance the number of electrons in both half-reactions, the half-reaction involving Zn is multiplied by 2.
To balance the number of electrons in both half-reactions, the half-reaction involving Zn is multiplied by 2.
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The resulting reaction produces a total of 9 moles of hydroxide ions (OH-) on the product side.
The resulting reaction produces a total of 9 moles of hydroxide ions (OH-) on the product side.
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In basic conditions, the overall charge of the reactants in the final equation remains neutral.
In basic conditions, the overall charge of the reactants in the final equation remains neutral.
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The formation of Zn(OH)42- involves the loss of protons (H+).
The formation of Zn(OH)42- involves the loss of protons (H+).
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Six moles of water are required to balance the final equation on the reactant side.
Six moles of water are required to balance the final equation on the reactant side.
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The total number of hydrogen atoms equals 17 in the final balanced reaction.
The total number of hydrogen atoms equals 17 in the final balanced reaction.
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The compound formed from the reaction of Zn and hydroxide is simply Zn(OH)2.
The compound formed from the reaction of Zn and hydroxide is simply Zn(OH)2.
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The oxidation half-reaction of Zn requires a total of 8 electrons to balance the charge.
The oxidation half-reaction of Zn requires a total of 8 electrons to balance the charge.
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Study Notes
Identifying Oxidation and Reduction Species
- The provided reaction involves the oxidation of zinc (Zn) and the reduction of nitrate (NO3-) ions.
- Zinc loses electrons and forms zinc hydroxide ions (Zn(OH)42-), indicating oxidation.
- Nitrate ions gain electrons and form ammonia (NH3), indicating reduction.
Balancing Redox Reactions in Basic Solution
- The steps to balance the redox reaction in a basic solution are outlined below:
- Skeleton Equations: Separate the overall reaction into two half-reactions, focusing on Zn and NO3-.
- Balance Elements: Balance all elements in the half-reactions except hydrogen (H) and charge.
- Balance Oxygen: Add water molecules (H2O) to balance oxygen atoms.
- Balance Hydrogen: Add hydrogen ions (H+) to balance hydrogen atoms.
- Convert to Basic Conditions: To convert acidic conditions (presence of H+) to basic, add hydroxide ions (OH-) on both sides of the half-reactions.
- Combine H+ and OH-: Combine H+ and OH- to form water molecules (H2O) and shift them to either side of the reaction.
- Balance Charge: Balance the charge in each half-reaction by adding electrons (e-).
- Equalize Electrons: Multiply the half-reactions by appropriate factors to make the number of electrons in each equal.
- Combine Half-Reactions: Add the multiplied half-reactions together, canceling out common species.
Final Balanced Redox Reaction
- The final balanced reaction is: 4Zn + 7OH- + NO3- + 6H2O → 4Zn(OH)42- + NH3
- This reaction demonstrates the oxidation of zinc to zinc hydroxide ions and the reduction of nitrate ions to ammonia.
- The equation is balanced by ensuring the same number of atoms of each element and the same charge on both sides.
Balancing Redox Reactions
- Zn + NO3- → Zn(OH)42- + NH3 This is a redox reaction occurring in a basic solution.
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Balancing the half-reactions involves multiple steps:
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Identify the oxidizing and reducing species:
- Zinc (Zn) is oxidized, its oxidation state changes from 0 to +2.
- Nitrate (NO3-) is reduced, its oxidation state changes from +5 to -3.
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Write skeletal half-reactions:
- Zn → Zn(OH)42-
- NO3- → NH3
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Balance all elements except H and charge by inspection:
- Zn + 4H2O → Zn(OH)42-
- NO3- + 4H2O → NH3 + 3H2O
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Balance oxygen by adding water:
- Zn + 4H2O → Zn(OH)42-
- NO3- + 9H2O → NH3 + 3H2O
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Balance hydrogen by adding protons (H+)
- Zn + 4H2O → Zn(OH)42- + 4H+
- NO3- + 9H+ + 9H2O → NH3 + 3H2O
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Since the reaction is basic, convert to basic conditions by adding hydroxide ions (4OH- and 9OH- respectively) to both sides:
- Zn + 4H2O + 4OH- → Zn(OH)42- + 4H+ + 4OH-
- NO3- + 9H+ + 9OH- + 9H2O → NH3 + 3H2O + 9OH-
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Combine H+ and OH- to form water and place H2O to one side:
- Zn + 4H2O + 4OH- → Zn(OH)42- + 4H2O
- NO3- + 9H2O → NH3 + 3H2O + 9OH-
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Simplify by canceling water on both sides:
- Zn + 4OH- → Zn(OH)42-
- NO3- + 6H2O → NH3 + 9OH-
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Balance charge by adding electrons:
- Zn + 4OH- → Zn(OH)42- + 2e-
- NO3- + 6H2O + 8e- → NH3 + 9OH-
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Multiply the zinc half-reaction by 4 to have an equal number of electrons in both half-reactions:
- 4Zn + 16OH- → 4Zn(OH)42- + 8e-
- NO3- + 6H2O + 8e- → NH3 + 9OH-
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Combine the half-reactions as simultaneous equations and eliminate common species:
- 4Zn + 16OH- + NO3- + 6H2O → 4Zn(OH)42- + NH3 + 9OH-
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Simplify the final equation by cancelling common species:
- 4Zn + 7OH- + NO3- + 6H2O → 4Zn(OH)42- + NH3
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Identify the oxidizing and reducing species:
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Check the final balanced equation:
- All atoms are balanced: 4xZn, 1xNO3-, 16xO, 19xH
- The charge is balanced: 8x negative charge on both sides
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The final balanced equation is:
- 4Zn + 7OH- + NO3- + 6H2O → 4Zn(OH)42- + NH3
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Description
This quiz focuses on identifying oxidation and reduction species in redox reactions, specifically involving zinc and nitrate ions. It also covers the steps necessary to balance redox reactions in basic solutions, guiding you through the process of balancing half-reactions and adjusting for different conditions.
