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Ch.20 - Electrochemistry
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
All textbooksBrown 14th EditionCh.20 - ElectrochemistryProblem 97
Chapter 20, Problem 97

Complete and balance the following disproportionation reactions: (a) Ni⁺(aq) → Ni²⁺(aq) + Ni(s) (acidic solution) (c) H₂SO₃(aq) → S(s) + HSO₄⁻(aq) (acidic solution) (d) Cl₂(aq) → Cl⁻(aq) + ClO⁻(aq) (basic solution)

Verified step by step guidance
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Step 1: Identify the oxidation states of the elements involved in each reaction. For (a) Ni⁺(aq) → Ni²⁺(aq) + Ni(s), Ni⁺ has an oxidation state of +1, Ni²⁺ has an oxidation state of +2, and Ni(s) has an oxidation state of 0. For (c) H₂SO₃(aq) → S(s) + HSO₄⁻(aq), sulfur in H₂SO₃ is +4, in S(s) is 0, and in HSO₄⁻ is +6. For (d) Cl₂(aq) → Cl⁻(aq) + ClO⁻(aq), Cl₂ is 0, Cl⁻ is -1, and ClO⁻ is +1.
Step 2: Write the half-reactions for oxidation and reduction. For (a), the oxidation half-reaction is Ni⁺(aq) → Ni²⁺(aq) and the reduction half-reaction is Ni⁺(aq) → Ni(s). For (c), the oxidation half-reaction is H₂SO₃(aq) → HSO₄⁻(aq) and the reduction half-reaction is H₂SO₃(aq) → S(s). For (d), the oxidation half-reaction is Cl₂(aq) → ClO⁻(aq) and the reduction half-reaction is Cl₂(aq) → Cl⁻(aq).
Step 3: Balance the atoms other than oxygen and hydrogen in each half-reaction. For (a), both half-reactions are already balanced for Ni. For (c), both half-reactions are balanced for sulfur. For (d), both half-reactions are balanced for chlorine.
Step 4: Balance the oxygen atoms by adding H₂O molecules. For (a), no oxygen atoms are involved. For (c), add H₂O to balance the oxygen in the oxidation half-reaction: H₂SO₃(aq) → HSO₄⁻(aq) + H₂O(l). For (d), add H₂O to balance the oxygen in the oxidation half-reaction: Cl₂(aq) + H₂O(l) → ClO⁻(aq).
Step 5: Balance the hydrogen atoms by adding H⁺ ions in acidic solutions or OH⁻ ions in basic solutions. For (a), add H⁺ to balance the hydrogen in the reduction half-reaction: Ni⁺(aq) + H⁺(aq) → Ni(s). For (c), add H⁺ to balance the hydrogen in the oxidation half-reaction: H₂SO₃(aq) + H⁺(aq) → HSO₄⁻(aq) + H₂O(l). For (d), add OH⁻ to balance the hydrogen in the oxidation half-reaction: Cl₂(aq) + H₂O(l) + OH⁻(aq) → ClO⁻(aq).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Disproportionation Reactions

Disproportionation reactions are a type of redox reaction where a single substance is both oxidized and reduced, resulting in two different products. In these reactions, one species undergoes an increase in oxidation state while another undergoes a decrease. Understanding the oxidation states of the elements involved is crucial for identifying how to balance the reaction correctly.
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Balancing Redox Reactions

Balancing redox reactions involves ensuring that the number of atoms and the charge are equal on both sides of the equation. This can be achieved using the half-reaction method, where the oxidation and reduction half-reactions are balanced separately before combining them. It is essential to account for electrons transferred during the oxidation and reduction processes to maintain charge balance.
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Acidic and Basic Solutions

The pH of the solution affects the species present and the balancing of redox reactions. In acidic solutions, protons (H⁺) are abundant, which can influence the formation of products and the balancing of charges. Conversely, in basic solutions, hydroxide ions (OH⁻) are present, which can also affect the reaction dynamics. Recognizing the conditions of the solution is vital for correctly balancing the disproportionation reactions.
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Related Practice
Textbook Question

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of Sn to Sn2+ by I2 (to form I-), (b) reduction of Ni2+ to Ni by I- (to form I2), (c) reduction of Ce4+ to Ce3+ by H2O2, (d) reduction of Cu2+ to Cu by Sn2+ (to form Sn4+).

Textbook Question
A mixture of copper and gold metals that is subjected toelectrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowestcommon oxidation state, Te4+, isTe4+1aq2 + 4 e- ¡ Te1s2 E°red = 0.57 VGiven this information, describe the probable fate of tellurium impurities during electrorefining. Do the impuritiesfall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If theygo into solution, do they plate out on the cathode?
Textbook Question

A disproportionation reaction is an oxidation–reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions:

(b) MnO42-(aq) → MnO4-(aq) + MnO2(s) (acidic solution)