CHEM1101 Study Guide - Midterm Guide: Gibbs Free Energy, Roman Numerals, Manganese

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Electro-Chemistry

Electrochemistry is the study of the relationships between electricity and chemical reactions. It

includes the study of both spontaneous and nonspontaneous processes.

Synopsis of Assigning Oxidation Numbers

1. Elements = 0

2. Monatomic ion = charge

3. F: –1

4. O: –2 (unless peroxide = –1)

5. H: +1 (unless a metal hydride = –1)

6. The sum of the oxidation numbers equals the overall charge (0 in a compound).

Oxidation Numbers

• To keep trak of hat loses eletros ad hat gains them, we assign oxidation numbers.

• If the oidatio uer ireases for a eleet, that eleet is oidised.

• If the oidatio uer dereases for a eleet, that eleet is redued

Oxidation and Reduction

• A speies is oidised he it loses electrons. – Zinc loses two

electrons, forming the Zn2+ ion.

• A speies is redued he it gais eletros. – H+ gains an

electron, forming H2.

• A oidisig aget auses soethig else to e oidized H+; a

reducing agent causes something else to be reduced (Zn).

Half-Reactions

• The oidatio ad redutio are ritte ad alaed separatel.

• We ill use the to alae a redo reatio.

• For eaple, he S+ ad Fe+ reat,

Balancing Redox Equations: The Half-Reactions Method (a Synopsis)

1) Make two half-reactions

(oxidation and reduction).

2) Balance atoms other than

O and H. Then, balance

O and H using H2O/H+.

3) Add electrons to balance charges.

4) Multiply by common factor

to make electrons in half-

reactions equal.

5) Add the half-reactions.

6) Simplify by dividing by common factor or converting H+ to

OH– if basic.

7) Double-check atoms and charges balance!

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Document Summary

Electrochemistry is the study of the relationships between electricity and chemical reactions. It includes the study of both spontaneous and nonspontaneous processes. Oxidation and reduction: a spe(cid:272)ies is o(cid:454)idised (cid:449)he(cid:374) it loses electrons. Half-reactions: the o(cid:454)idatio(cid:374) a(cid:374)d redu(cid:272)tio(cid:374) are (cid:449)ritte(cid:374) a(cid:374)d (cid:271)ala(cid:374)(cid:272)ed separatel(cid:455), we (cid:449)ill use the(cid:373) to (cid:271)ala(cid:374)(cid:272)e a redo(cid:454) rea(cid:272)tio(cid:374), for e(cid:454)a(cid:373)ple, (cid:449)he(cid:374) s(cid:374)(cid:1006)+ a(cid:374)d fe(cid:1007)+ rea(cid:272)t, Balancing redox equations: the half-reactions method (a synopsis: make two half-reactions (oxidation and reduction), balance atoms other than. O and h using h2o/h+: add electrons to balance charges, multiply by common factor to make electrons in half- reactions equal, add the half-reactions, simplify by dividing by common factor or converting h+ to. Oh if basic: double-check atoms and charges balance! O= -2 ( x 4) + mg = -(cid:1005), thus mg is + 7 . 2 x 4 atoms + 2c = -2, -8 + 2c = -2 thus c = 4 as the whole thing has to equal -2.

Related Questions

Answer true or false.

(a) When a substance is oxidized, it loses electrons.

(b) When a substance gains electrons, it is reduced.

(d) In a redox reaction. the reducing agent becomes oxidized.

(e) When Zn is converted to Zn²⁺ ion, zinc is oxidized.

(f) Oxidation can also be defined as the loss of oxygen atoms and/or gain of hydrogen atoms.

(g) Reduction can also be defined as the gain of oxygen atoms and/or loss of hydrogen atoms.

(h) When oxygen, O₂, is converted to hydrogen peroxide, H₂O₂, we say that O₂ is reduced.

(i) Hydrogen peroxide, H₂O₂, is an oxidizing agent.

(j) All combustion reactions are redox reactions.

(k) The products of complete combustion (oxidation) of hydrocarbon fuels are carbon dioxide, water, and heat.

(l) In the combustion of hydrocarbon fuels, oxygen is the oxidizing agent and the hydrocarbon fuel is the reducing agent.

(m) Incomplete combustion of hydrocarbon fuels can produce significant amounts of carbon monoxide.

(n) Most common bleaches are oxidizing agents.

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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)

CHEM1101 Study Guide - Midterm Guide: Gibbs Free Energy, Roman Numerals, Manganese

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