🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →

HomeStudy Guides420,000US230,000

CHE 1302 Study Guide - Final Guide: Electrochemistry, Ion, Faraday Constant

59 views7 pages

8 Mar 2018

School

Baylor University

Department

Course

Professor

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Oxidation reduction reactions cation // positive charge anion // negative charge. Once the electrons are donated, they are not shared; they are gone for good. Oxidation and reduction must both take place you cannot have one without the other. There are two half-reactions: oxidation: one species loses electrons to another, reduction: one species gains electrons from another (cid:862)oil rig(cid:863) Oxidation numbers (oxidation states) and rules for assigning them. Oxidizing agents like to gain electrons/become reduced. Reducing agents like to give away electrons/become oxidized. Adjust coefficients: we need the same # electrons in each half-reaction; then add together and simplify. Electrons must feature in half-reactions, but must not appear in the balanced redox reaction. Reaction may differ in acidic vs. basic solutions. In basic solutions: oh- and/or h2o may be needed. 18. 3 voltaic (galvanic) cells: generating electricity from spontaneous chemical reactions. In an electrochemical cell, two half-cells are connected to form an electrical circuit.

Related Documents

CHE 1302 Lecture Notes - Lecture 20: Redox, Sodium Chloride, Voltage Source

CHEM 107 Lecture 33: 3-6-17-Half-Reactions

CHM 1046 Lecture Notes - Lecture 43: Redox, Electrochemistry, Carbon Tetrachloride

Related Questions

Having trouble with my Pre-Lab questions I don't understand how to get them.

Introduction An important type of reaction mechanism is the transfer of electrons from one species to another effecting a change in oxidation state. This gain or loss of electrons is referred to as reduction and oxidation, respectively. For example, consider the two balanced, chemical equations (1 & 2) that follow Cu (s) D Cu (aq) 2e (1) Cl2(aq) 2e D 2Cl (aq) (2) The first equation (1) shows elemental copper metal with a oxidation state of zero forming aqueous copper(Il) ions having an oxidation state of +2. This gain in oxidation number, 0 to +2, is called oxidation. The second equation (2) show elemental chlorine having an oxidation state of zero forming chloride ions in aqueous solution with an oxidation state of This decrease in oxidation number, 0 to -1, is called reduction. In any chemical reaction, the electrons gained or lost are conserved, i.e.. there is no net gain or loss of electrons. In fact, the two reactions illustrated above could occur simultaneously with the electrons lost by copper, gained by the chlorine in aqueous solution. Since the copper provides the electrons that subsequently reduce the chlorine copper is referred to as the reducing agent. Likewise, since the chlorine is accepting the electrons from copper, which subsequently oxidizes the copper, chlorine is referred to as he oxidizing agent. The two reactions above (1 & 2) are called half-reactions, as much as they each constitute one-half of the total reaction Equation 1 is the oxidation half reaction showing Cu(s) losing two electrons. In Equation 2, Cla(aq) gains two electrons and is the reduction half-reaction. If these two half-reaction are added together the result is the total oxidation-reduction reaction or "redox" reaction. Adding the two equations yields (3). Cu(s) Cl2(aq) D Cu (aq) 2CI (aq) (3) Whether or not the above reaction is spontaneous proceeds from reactants to products without input of energy into the system, may be determined by observing the reaction mixture if the reactants and products differ greatly in their appearance. Copper metal is a bright copper color and chlorine in aqueous solution will appear yellow. The Cu ion may turn the liquid part of the mixture blue, however, the Cris colorless and cannot be observed directly. Copper, like many of the transition metals, exhibits a variety of oxidation states depending upon the chemical environment in which it is found. In a strong reducing environment copper may exist as the free metal, Cu(s). An example of such a situation is illustrated in the following chemical equation (4). 4Cuo(s) CH4(g) D 4Cu(s) CO2(g) 2H2O(l) (4)

I need help with questions 2-4 I can't figure them out.

Introduction An important type of reaction mechanism is the transfer of electrons from one species to another effecting a change in oxidation state. This gain or loss of electrons is referred to as reduction and oxidation, respectively. For example, consider the two balanced, chemical equations (1 & 2) that follow Cu (s) D Cu (aq) 2e (1) Cl2(aq) 2e D 2Cl (aq) (2) The first equation (1) shows elemental copper metal with a oxidation state of zero forming aqueous copper(Il) ions having an oxidation state of +2. This gain in oxidation number, 0 to +2, is called oxidation. The second equation (2) show elemental chlorine having an oxidation state of zero forming chloride ions in aqueous solution with an oxidation state of This decrease in oxidation number, 0 to -1, is called reduction. In any chemical reaction, the electrons gained or lost are conserved, i.e.. there is no net gain or loss of electrons. In fact, the two reactions illustrated above could occur simultaneously with the electrons lost by copper, gained by the chlorine in aqueous solution. Since the copper provides the electrons that subsequently reduce the chlorine copper is referred to as the reducing agent. Likewise, since the chlorine is accepting the electrons from copper, which subsequently oxidizes the copper, chlorine is referred to as he oxidizing agent. The two reactions above (1 & 2) are called half-reactions, as much as they each constitute one-half of the total reaction Equation 1 is the oxidation half reaction showing Cu(s) losing two electrons. In Equation 2, Cla(aq) gains two electrons and is the reduction half-reaction. If these two half-reaction are added together the result is the total oxidation-reduction reaction or "redox" reaction. Adding the two equations yields (3). Cu(s) Cl2(aq) D Cu (aq) 2CI (aq) (3) Whether or not the above reaction is spontaneous proceeds from reactants to products without input of energy into the system, may be determined by observing the reaction mixture if the reactants and products differ greatly in their appearance. Copper metal is a bright copper color and chlorine in aqueous solution will appear yellow. The Cu ion may turn the liquid part of the mixture blue, however, the Cris colorless and cannot be observed directly. Copper, like many of the transition metals, exhibits a variety of oxidation states depending upon the chemical environment in which it is found. In a strong reducing environment copper may exist as the free metal, Cu(s). An example of such a situation is illustrated in the following chemical equation (4). 4Cuo(s) CH4(g) D 4Cu(s) CO2(g) 2H2O(l) (4)