[PDF] CHEM 1105 REDOX REACTIONS 1. Definition of Oxidation and





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CHEM 1105 REDOX REACTIONS 1. Definition of Oxidation and

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(H+ + Cl ) Solution d'hydroxyde de sodium (soude) (Na+ + HO ) Nitrate d'argent ( + ) I Test d'identification des ions chlorure 1 Expérience :

  • Comment s'appelle la solution qui contient des ions Cl et Na+?

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  • Comment identifier Na+ ?

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  • Comment calculer le coefficient d'ionisation de NaCl ?

    Les molécules sont chargées et la moitié d'entre elles sont dissociées, c'est à dire qu'elles sont constituées de molécules donnant chacune 2 ions selon la réaction : NaCl ‹ Na++Cl– Donc le coefficient d'activité est yO= 2. Le coefficient d'ionisation : i0 =1+ ?0 (?0?1)= 1+1/2(2?1)=3/2.
  • On constate qu'une mole du solide ionique dissout produit m moles du cation Xn+ et n moles de l'anion Ym?. On en déduit les relations entre la concentration en quantité de matière C du soluté apporté et les concentrations des ions en solution [Xn+] et [Ym?].
CHEM 1105 REDOX REACTIONS 1. Definition of Oxidation and

CHEM 1105 REDOX REACTIONS

1. Definition of Oxidation and Reduction

The old definition of oxidation was "addition of oxygen".

Consider the following reactions:

(a) 2Na + 4O 2 H Na 2

O (2Na

+ O 2- (b) Na + 4Cl 2

H NaCl (Na

+ Cl- By the old definition, reaction (a) is an oxidation. But in both cases Na is converted into Na . Hence, in reaction (b) Na is also oxidized. In both cases, Na lost electrons in forming Na ions. Oxidation and reduction occur simultaneously; when one species is oxidized, the other is reduced. Hence we call these redox reactions. In reaction (a), Na is oxidized by losing electrons and O is reduced by gaining electrons to form O 2- ions. Similarly, in reaction (b) Cl is reduced. New definition: Oxidation is the loss of electrons and reduction is the gain of electrons. LEO GER is a convenient way of remembering this (Loss of Electrons is Oxidation; Gain of Electrons is Reduction). [Another memory aid is OIL RIG]. In reaction (b), we can say that Na was oxidized by Cl and Cl was reduced by Na. Hence, we call Cl the oxidizing agent and Na the reducing agent. In general, the species oxidized is the reducing agent and the species reduced is the oxidizing agent.

2. Oxidation Numbers

In reaction (b), Na and Cl were changed to NaCl in a redox reaction. In NaCl, we have Na+ ions and Cl ions and we say that Na has an oxidation number of +1 and Cl has an oxidation number of -1. We can consider the free elements (in this case, Na and Cl) to have an oxidation number of 0. We can hence define oxidation as an increase in oxidation numner (Na: from 0 to +1) and reduction as a decrease in oxidation number (Cl: from 0 to -1) In more complicated species (compounds and polyatomic ions), we can make rules for calculating an oxidation number of an element in the ion or compound, but we must remember that in those cases the use of oxidation numbers is only a convenient bookkeeping device. Oxidation Number Rules1.The sum of all oxidation numbers of elements in a compound is zero.

2.The oxidation number of any free element is zero.

3.The oxidation number of a monatomic ion is the charge on the ion.

4.The sum of all oxidation numbers of elements in a polyatomic ion is the charge on the ion.

5.The oxidation number of any Group 1 metal in combination is +1.

6.The oxidation number of any Group 2 metal in combination is +2.7.The oxidation number of any halogen (Group 7) in combination with one other element other

than oxygen is -1.

8.The oxidation number of hydrogen in combination is +1 unless the H is present as H

-1 (hydride, as in NaH) in which case the oxidation number is -1.

9.The oxidation number of oxygen in combination is -2 unless the O is present as peroxide inwhich case the oxidation number is -1.

10. The oxidation number of fluorine is always -1.

Exercises

Calculate the oxidation number of the underlined element in each case. HCl H 2 SO 4 CuCl 2 S 2 O 32-
O 3 NH 4 +HNO 3 Cr 2 O 72-
K 2 XeF 6 N 2 O 3

Mg(ClO

4 2 -2-

3. Balancing Redox Reaction Equations in Aqueous Acid

The Half-Reaction (Ion-Electron) Method

This method divides the reaction into two half-reactions; one involving the species being oxidized and the other involving the species being reduced.

1) Use oxidation numbers to identify which species is oxidized and which is reduced.

2) Write the half-reactions. For each half-reaction, balance the element undergoing oxidation

number change, then the O (by adding H 2

O) and finally the H (by adding H

). Then balance the charge by adding electrons.

3) Multiply one or both half-reactions by the smallest integer(s) so that they have the same

number of electrons on opposite sides.

4) Add the multiplied half-reactions and cancel the charges and any other species that occur on

both sides.

5) Check that the resulting equation is balanced for mass (atoms) and charge.

Examples

Simple; no other atoms needed

Fe 3+ + Cl

H Fe + Cl

2

Solution

1)Fe 3+ is being reduced to Fe (ox. no. decreases from 3 to 0) Cl is being oxidized to Cl 2 (ox. no. increases from -1 to 0) 2)Fe 3+

H Fe

Fe 3+ + 3e

H Fe (balanced) (i)

Cl

H Cl

2 2Cl

H Cl

2 2Cl

H Cl

2 + 2e (balanced) (ii)

3) (i) x 2 gives: 2Fe

3+ + 6e

H 2Fe (iii)

(ii) x 3 gives: 6Cl

H 3Cl

2 + 6e (iv)

4) add (iii) + (iv): 2Fe

3+ + 6Cl

H 2Fe + 3Cl

2 (5) Check: LHS 2Fe; RHS 2Fe; LHS 6Cl; RHS 6Cl; LHS charge = 0; RHS charge = 0

EQUATION IS BALANCED

More complex; oxygen (from H

2

O) and H (from H

) needed MnO 4- + Cl

H Mn

2+ + Cl 2

Solution

1) Mn in MnO

4- is being reduced Mn 2+ (ox. no. decreases from 7 to 2) Cl is being oxidized to Cl 2 (ox. no. increases from -1 to 0) -3-

2) MnO

4-

H Mn

2+ MnO 4-

H Mn

2+ + 4H 2 O 8H + MnO 4-

H Mn

2+ + 4H 2 O 8H + MnO 4- + 5e

H Mn

2+ + 4H 2

O (balanced) (i)

Cl

H Cl

2 2Cl

H Cl

2 2Cl

H Cl

2 + 2e (balanced) (ii)

3) (i) x 2 gives: 16H

+ 2MnO 4- + 10e

H 2Mn

2+ + 8H 2

O (iii)

(ii) x 5 gives: 10Cl

H 5Cl

2 + 10e (iv)

4) add (iii) + (iv): 16H

quotesdbs_dbs29.pdfusesText_35
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