Ch.4 - Chemical Quantities & Aqueous ReactionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: In the following reaction, the oxidation state (number) of chlorine changes from:2 Cr(OH)4-(aq) + ClO-(aq) + 2 OH-(aq) → 2 CrO42-(aq) + 3 Cl-(aq) + 5 H2O(I)A. -1 to -2B. +3 to +6C. +4 to +8D. +1 to -1E. -4 to +3

Problem

In the following reaction, the oxidation state (number) of chlorine changes from:

2 Cr(OH)4-(aq) + ClO-(aq) + 2 OH-(aq) → 2 CrO42-(aq) + 3 Cl-(aq) + 5 H2O(I)

A. -1 to -2

B. +3 to +6

C. +4 to +8

D. +1 to -1

E. -4 to +3

Solution

Determine the oxidation state/number of Cl in ClO- in the reactant side and Cl- in the product side

The rules for oxidation states are as follows:


A. General Rules:

1. For an atom in its elemental form (Zn, Cl2, C(graphite), etc.)    O.S. = 0

2. For an ion (Li+, Al3+, etc.)                                                       O.S. = charge


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