Skip to main content
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 97b
Chapter 20, Problem 97b

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)

Verified step by step guidance
1
Identify the oxidation states of manganese in each compound: MnO42- has Mn in +6 oxidation state, MnO4- has Mn in +7 oxidation state, and MnO2 has Mn in +4 oxidation state.
Write the half-reactions for the oxidation and reduction processes. For oxidation, Mn in MnO42- goes from +6 to +7. For reduction, Mn in MnO42- goes from +6 to +4.
Balance the manganese atoms in each half-reaction. Since there is one Mn atom on each side of the half-reactions, they are already balanced for Mn.
Balance the oxygen atoms by adding water (H2O) molecules. For the oxidation half-reaction, add one water molecule on the right side. For the reduction half-reaction, add two water molecules on the left side.
Balance the hydrogen and charge in each half-reaction by adding hydrogen ions (H+) and electrons (e-). For the oxidation half-reaction, add two H+ on the right and one e- on the left. For the reduction half-reaction, add four H+ on the right and two e- on the left.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
Was this helpful?

Key Concepts

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

Disproportionation Reaction

A disproportionation reaction is a specific type of redox reaction where a single substance undergoes both oxidation and reduction simultaneously. In this process, one part of the molecule loses electrons (is oxidized) while another part gains electrons (is reduced), resulting in the formation of two different products from the same reactant.
Recommended video:
Guided course
01:30
Alcohol Reactions: Dehydration Reactions

Oxidation States

Oxidation states (or oxidation numbers) are a way to keep track of electron transfer in redox reactions. They indicate the degree of oxidation of an atom in a compound, helping to identify which atoms are oxidized and which are reduced. Understanding oxidation states is crucial for balancing redox reactions, including disproportionation reactions.
Recommended video:
Guided course
02:42
Oxidation Numbers

Balancing Redox Reactions

Balancing redox reactions involves ensuring that the number of electrons lost in oxidation equals the number of electrons gained in reduction. This can be achieved using the half-reaction method, where the oxidation and reduction processes are balanced separately before combining them to form a complete balanced equation. This is essential for accurately representing the stoichiometry of the reaction.
Recommended video:
Guided course
01:04
Balancing Basic Redox Reactions
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

Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are Au+1aq2 + e- ¡ Au1s2 Ered ° = +1.69 V Au3+1aq2 + 3 e- ¡ Au1s2 Ered ° = +1.50 V (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction 4 Au1s2 + 8 NaCN1aq2 + 2 H2O1l2 + O21g2 ¡ 4 Na3Au1CN2241aq2 + 4 NaOH1aq2 What is being oxidized, and what is being reduced in this reaction?

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?